THC Science

; Archives
	Archive 1: July 2004; Archive 2: July 2004 – July 2005; Archive 3: August 2005 – February 2006; Archive 4: March 2006 – November 2006; Archive 5: November 2006 – April 2007; Archive 6: March 2007 – February 2008; Archive 7: February 2008 – August 2008;

Statistics Unassessed

	This article is within the scope of WikiProject Statistics, a collaborative effort to improve the coverage of statistics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.StatisticsWikipedia:WikiProject StatisticsTemplate:WikiProject StatisticsStatistics articles
???	This article has not yet received a rating on Wikipedia's content assessment scale.
???	This article has not yet received a rating on the importance scale.

Monty Hall problem is a featured article; it (or a previous version of it) has been identified as one of the best articles produced by the Wikipedia community. Even so, if you can update or improve it, please do so.

This article appeared on Wikipedia's Main Page as Today's featured article on July 23, 2005.

Article milestones
Date	Process	Result
May 3, 2005	Peer review	Reviewed
June 25, 2005	Featured article candidate	Promoted
January 29, 2007	Featured article review	Kept
May 18, 2008	Featured article review	Kept
Current status: Featured article

Mathematics FA‑class Low‑priority

	Mathematics portal This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.MathematicsWikipedia:WikiProject MathematicsTemplate:WikiProject Mathematicsmathematics articles
FA	This article has been rated as FA-class on Wikipedia's content assessment scale.
Low	This article has been rated as Low-priority on the project's priority scale.

Template:Game Show Project

Very important topic

The below suggested simple "frequency" solution should replace the current confusing conditional probability solution
See discussion further down for reasons why it should replace the current solution.

Since these have had plenty of expert attention, I am going to disable the templates. — Carl (CBM · talk) 12:16, 16 September 2008 (UTC)[reply]

Summary and Solution

There are three doors. Two doors have a goat behind them and one door has a car behind.
(Remember goat is bad and car is good)

1) Choose one door
2) Game host opens one door containing a goat
3) Do you change door?

For example:

Door 1	Door 2	Door 3
Goat	Goat	Car

Now you have two opions: Not switch or Switch

Not switch

pick	show	outcome
1	2	lose
2	1	lose
3	2 or 1	win

If you do not switch the probability of winning is 1/3

Switch

pick	show	outcome
1	2	win
2	1	win
3	2 or 1	lose

If you switch the probability of winning is 2/3

You should therefore choose to switch doors

Note that there also exist an simpler and more elegant solution to the above problem.

From above we know that the probability of winning if we do not switch doors is 1/3.

We also know that we only have two options: Not switch or Switch

This means that:

P(switch) + P(not switch) = 1

which means that

P(switch) = 1 - P(not switch)= 1 - 1/3 = 2/3

Which is the same solution we had previously

tildes ( 82.39.51.194 (talk) 10:40, 17 August 2008 (UTC) )[reply]

I propose a very compact, straightforward solution with no mention of probability per se. The problem is, in fact, simple so we should treat it that way. After reading most of this discussion and some of the references about how to explain it in an understandable way, I've yet to see the following type of explanation.

The problem is easy to solve - every time the game is played, one of two sequences occur:

A)Contestant picks a door that has the car behind it; host opens one of the other doors revealing a goat; behind the remaining door is the remaining goat.

B)Contestant picks a door that has a goat behind it; host opens the other door with a goat behind it; behind the remaining door is the car.

Sequence A happens 1 time in 3, so, doing the math, sequence B happens 2 times in 3.

Stickers win in sequence A and switchers win in sequence B.

I told you it was easy.

Millbast5 (talk) 22:03, 17 September 2008 (UTC)[reply]

Hang on a second.

Not switch

pick	show	outcome
1	2	lose
2	1	lose
3	2 or 1	win

3 win? 3 is the goat, which is a known lose... Did I miss something? And this seems to presume knowing the outcome. If I pick 2, & switch, & 2 is the winning door, how, exactly, is this a win? TREKphiler ^{hit me ♠} 05:36, 22 October 2008 (UTC)[reply]

Yes you are missing something and that is your brain! :-) Door 1 & 2 are goats —Preceding unsigned comment added by 92.41.207.32 (talk) 10:42, 6 February 2009 (UTC)[reply]

Repetitive new solution section

I've undone this change twice. It renames the existing "Solution" section as "Discussion" and adds an additional "Solution" section that basically repeats what's in the existing Solution section - without references, and with a variety of style issues per WP:MOS (such as directly addressing the reader). If there's some shortcoming in the current Solution section please say what it is and we can work on addressing the concern. -- Rick Block (talk) 18:57, 14 August 2008 (UTC)[reply]

I've undone it myself too, and it's back. Can we perhaps lock the article until this user responds in the talk page?The Glopk (talk) 14:13, 15 August 2008 (UTC)[reply]

--

The problem I have with your so called solution section is that it is general confussing! You are confussed, your glossy matrix and tree are confussed. It is not about where the car is !! The solution will look the same irregardles. It is a question whether you choose to switch or not and which door the contestant choose.
The whole point of the Montey hall excercice is to answer the question wheter you switch or not and still you refuse to take that into account !

Therefor you should divide your matrix it into

Switch
Not switch

and then review

1)all the possible doors the contestant can choose

2)the response of the host

3)the outcome of such a choice

Your original matrix might be glossy and fancy but regret to inform you that it is inaccurate

It is better if you put your glossy matrix under the section "Sources of confusion"

I can also inform you that I am going to change back my solution until you either, stop removing my solution or redesign your matrix.

Ps I found it quite strange that I have to fight for such an obvious point ! It just reaffirm the notion that wikipedia is good in theory but does not work in practice. Now you have to fight a war with every moron with a computer ! Where are the credentials, expertise and self critic ? Just because you have a computer doesn mean that you are an expert !
I have attached the original post
tildes ( 82.39.51.194 (talk) 10:40, 17 August 2008 (UTC) )[reply]

Discussion

—Preceding unsigned comment added by 82.39.51.194 (talk) 08:35, 16 August 2008 (UTC)[reply]

The existing "solution" section is clearer than the above explanation. Cretog8 (talk) 15:04, 16 August 2008 (UTC)[reply]

The solution proposed above keeps the arrangement of goats and car the same and varies the player's initial pick rather than keeping the player's initial pick the same and varying the location of the car. Both of these approaches simplify the actual 9 cases (or 18 assuming the goats can be distinguished) down to three to better show the 2/3 probability of winning. In the rest of the article (in particular, in the Bayesian analysis section) the player's pick is kept as door 1 rather than the alternative approach of examining all picks given a single arrangement of goats and cars - and this approach is the most commonly presented in the cited references (and matches the presentation of the problem published in the vos Savant Parade column).

Another reason this approach is used is that the problem as generally presented can be considered a conditional probability problem given both a specific initial pick and a specific door the host has opened. This interpretation of the problem (see the Morgan et al reference) is discussed in the last two paragraphs of the solution section and involves examining the situation after the player's initial pick and given which door the host opens. Initially using the approach where the goat/car configuration is constant but the player's pick varies makes the connection to this conditional analysis extremely difficult to see.

The solution section was actually changed fairly recently to use the "fixed pick, varying car location" approach (rather than the "fixed configuration, varying pick" approach) to be consistent with the majority of the references and with the rest of the article. We can certainly discuss whether including the alternative solution is worthwhile, however given that they are effectively equivalent (both relying on a "without loss of generality" assumption), and that most sources use the "fixed pick, varying car location" approach, and that the approach currently used easily extends to the conditional analysis, and that the article is quite long already, I don't think you'll find much support for including this alternate solution. -- Rick Block (talk) 19:14, 16 August 2008 (UTC)[reply]

I dont agree with what you are saying ! It is about presenting an easy to understand and approachable article. as of now none of this is taking place. Who gives a crap about "keeping the player's initial pick the same and varying the location of the car". That is not what the Monty Hall problem is about. The sooner you will realize this the better of this article will be! You need to keep the most critical stuff and just remove everything ells ! Howevere I am starting to lose interest in all this bull crap ! This is all about you ! You have written something that you protect like hamster. This is exacly the reason why wikipedia will never gain any credential in academic circles because it is not about the optimal soluton but rather individual ego! —Preceding unsigned comment added by 82.39.51.194 (talk) 19:53, 16 August 2008 (UTC)[reply]

Please 82.39.51.194, no personal attacks. Feel free to discuss your displeasure and/or disagreement with the article but refrain from attacking the editors. -hydnjo talk 20:41, 16 August 2008 (UTC)[reply]

yes, I agree no personal attacks! It is just that I get so frustrated when the solution is staring them in the face and they still refuse to accept it because they are biased towards their own writing (which apparently is a lot). Just because someone wrote something first doesn't mean that such a person has the right to "validate" (oohh daddy please can I) or delete everything that comes after. You seam to have a lot of rules! Dont you have a rule for that? —Preceding unsigned comment added by 82.39.51.194 (talk) 21:26, 16 August 2008 (UTC)[reply]

There is indeed a rule for that: WP:OWN says that nobody should behave as if they "own" a particular article. I've put a welcome message at your talk page which you can look over to find out more stuff. All the same, you have to have some humility, too, and accept that maybe the reason your suggestions aren't being implemented is because several people don't think they're an improvement. Cretog8 (talk) 22:47, 16 August 2008 (UTC)[reply]

As a Featured (July 23, 2005) and high-traffic (392 thousand views, Jan-Jun '08) and heavily edited (500 edits to date in '08) article, there are lots of inputs for further improvement. The article has been scrutinized by the community at large in a process called Featured article review and the criticisms from that review have been successfully addressed so that the article meets today's FA standards. Given that bit of perspective, the article is absolutely not frozen and the editors that who are watching continue to to demonstrate (IMO) an open mind when it comes to addressing suggestions for improvement. It may be helpful to scan the archives of this talk page to gain further perspective about the responses to your suggestions. -hydnjo talk 23:26, 16 August 2008 (UTC)[reply]

Moreover, the current version attempts to stay as close as possible to the academic references. Have you read the references that are in the article and do you have other references that present the solution using your preferred approach? Even if you're not going to respond to the explanation above about why the current approach is used, a suggestion that you find "so and so's explanation from this paper or book" easier to understand would be received quite differently from "I like my explanation better". -- Rick Block (talk) 23:33, 16 August 2008 (UTC)[reply]

With all due respect I am not convinced that you (any of you) fully understand the rules of the Monty Hall Game.

1) First the allocation of goats and the car is taking place (remains fixed for the rest of the game).

2) Then the contestant choose one door.

Your "fixed pick, varying car location" approach totaly contradict this line of reasoning. The location of the car IS NOT changing. The allocation is stationary, it does NOT change over the period of the game. Again the first thing that happens is the allocation of the car. Again that location remaine FIXED for the ENTIRE game. But still you in insist like some stubborn child that the location is varying. It is NOT ! and it never will be ! The Monty Hall game is a SEQUENTIAL GAME which means that you CAN NOT just switch around the location of the car based upon your preferences in a later stage of the game. This means that the ONLY valid way of approaching this is the way I have put forward which means that you keep the allocation fixed and you evaluate each pick individually (the pick, the game show respons, and the outcome).

Also the argument that "majority of the references" do it one way isn't valid either. This is not an excercise in burping up some solution from some jerk off text book. Fine! you should have references to other articles but in the end it is about evaluating which approach is the most correct and not the least which approach is the easiest to understand. I hate to burst you bubble but the current approach dosent fulfill neither of these criterias irregardles if you have won a nobel price for the article.

Further, it is not about Bayesian statistics ! To allow Bayesian statistics to dictate how the original problem and its solution is presented is WRONG. Especially when such a solution contradict(is incorrect)the original game and the sequential steps such a game is based upon. If you want to have a section of Bayesian statistics that is fine! But you need to very carefully point out the different assumption such an analysis is based on! This is not taking place at the moment! The way I see it is that you have all been so blinded by the complexity of the Bayesian analysis that you have completely surrendered to its preachings! --82.39.51.194 (talk) 10:37, 17 August 2008 (UTC)[reply]

Perhaps it is not clear to you what is the meaning of a clause like "Assume, without loss of generality, X.". To put it simply, it means: "I could repeat the same argument for cases W,Y,Z,etc., But since these are obviously equivalent to X, as they differ only by a change of names, I'll just write down the case for X and be done with it". For example, consider the proof of the Pythagorean theorem: the logic of the proof would be unchanged if you rotated the names of the triangle's vertexes so that A becomes B, B becomes C, and C becomes A. That is, the particular labeling of the vertexes is irrelevant to the proof. Similarly, in the analysis of the Monty Hall problem, the naming of the doors is unimportant, so long as it is kept consistent within the reasoning. Because of this, for example, the Bayesian analysis of the problem is written concisely using the "Assume, without loss of generality" clause.The Glopk (talk) 16:48, 17 August 2008 (UTC)[reply]

(to 82.39.51.194) First, please read Wikipedia:No original research. Taken to its extreme what this policy says is that even if the preponderance of references are provably incorrect (which, just to be clear, is not at all the case here), the Wikipedia article must summarize what they say rather than presenting something we make up ourselves. Second, although you're perfectly correct about the sequencing of the problem we're talking about the probability of winning by switching, either over all iterations of the game or (in the Morgan et al interpretation) given a specific player's initial pick and the host's specific response. Also note that the Parade description of the problem specifically includes the words:

You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?"

If the goal is to determine the overall chance of winning by switching, over all possible scenarios, in a strict sense we should enumerate all possible car locations, player picks, and host responses (per the fully expanded decision tree as presented in Grinstead and Snell). On the other hand, from the player's point of view, the player (who doesn't know where the car is) picks a door and then the host opens a door, so following through what happens when the player picks a specific door (say #1) over all possible car locations (the approach currently in the article and in most references) is entirely sufficient. This is not varying the location of the car after the player has picked, but examining all possibilities of where the car might be to determine the probability of winning. Indeed, given the Parade version of the problem your preferred approach makes almost no sense at all (why are we considering what happens if the player picks door 2 when it's stated the player has picked door 1?). Morgan et al take this one step further, and consider only the scenario where the player has picked door 1 and the host has opened door 3 (eliminating the possibility that the car is behind door 3! - and, even in only this one case, the probability of winning by switching is still 2/3 [if the host chooses which of two goat doors to open with equal probability]).

I suspect you're thinking the problem is asking about the probability of winning over all possible car locations, player picks, and host responses. What would your analysis be if the problem is about the chances of winning for a player who's literally initially picked door #1 followed by the host opening door #3? -- Rick Block (talk) 17:38, 17 August 2008 (UTC)[reply]

You dont seem to understand! The most critical question we should answer is NOT what happens if the contestant chooses one specific door and the the host opens another door! The problem is more general than that ! The whole point with the Monty Hall exercise is to answer the question whether or not the contestant should switch doors in GENERAL. This is also indicated in your quote

"You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?"

The word "say" that appear before the door numbers indicate to me that this is just an example. The individual case of door 1 and door 3 is not that important! It could have been any configuration! Again the important question is whether or not the contestant should switch doors in GENERAL. Also note the sequential nature of your quote. 1)first the allocation has taken place (which remain fixed for the rest of the game 2)Then the contestant choose one door. When you start to shuffle around the location of the car it CONTRADICT the sequential nature of the game. It also obscure the whole purpose of the Monty Hall exercise which is to prove that the contestant will benefit from switching doors in ANY situation not just "for a player who's literally initially picked door #1 followed by the host opening door #3? "--82.39.51.194 (talk) 08:39, 18 August 2008 (UTC)[reply]

My points are:

If the goal is to answer the general question then all goat/car configurations, all initial picks, and all host responses should be considered. Your approach is considering only one configuration using an assumption that the same logic pertains to all other configurations. The approach currently in the article considers only one initial pick and uses an assumption that the same logic applies to all other picks. Your assertion seems to be that the latter approach is wrong. Either one of these is valid (they are like looking at two sides of the same coin), although most references use the approach currently in the article. If you're not seeing that these approaches are fundamentally equivalent, I think it is you who is not understanding the problem. In addition the current approach is (IMO) more consistent with the form of the Parade problem statement which uses the "say No. 1" terminology (which encourages thinking through scenarios involving a given initial pick).
Some references (e.g. Morgan et al) explicitly consider the situation at the point the player has picked a door and after the host has opened a door. Your approach makes this analysis difficult, however the current approach makes this a relatively simple extension of the analysis.

The approach currently in the article is used because most references use it and this approach easily extends to cover the "conditional" (Morgan et al) interpretation of the problem. Even though you think the critical question is the general question, the conditional interpretation exists in the literature (the Morgan et al and Gillman references) so it is considered in the article. I'll ask again - how would you address this interpretation? -- Rick Block (talk) 19:15, 18 August 2008 (UTC)[reply]

First of all the current solution is NOT more consistent with the form of the Parade problem statement. Actually my proposed solution is MORE consistent with the Parade problem! The reason for that is that it is more logically consistent (sequential game nature which means that the location of the car remain fixed), more general (consider a larger amount of cases) and easier to understand (iteration nerver suns out of styl).

Secondly, How I would deal with the conditional interpretation? For me that is an interpretation that is not critical for the Monty Hall problem nor its solution. You can and should understand the Monty Hall problem and its solution only with the basic frequency statistics approach (without Bayesian statistics). Note that my proposed solution uses frequency which is the most consistent with the original formulation of the Mony Hall game since again

1) It DOES NOT contradict the logic and sequential nature of the original Monty Hall game. The location of the car (and the goats for that matter) remain fixed for the entire game.
The logic is intact
2) Is not based upon any prior assumtions about the door selection. what you see it what you get.
3) Does not start in the middle of the game (door selecton), again sequential nature
3) The formulation is much more general (consider a larger amount of cases)
4) Much easier to extend to alternative configurations for example D1=C,D2=G,D3=G or D1=G,D2=C,D3=C or D1=G,D2=G,D3=C

As I said previously if you want to include a section on conditional probabilities as an extra bonus you have to very carefully point out the different assumption such an analysis is based on. If you dont correctly point out such differences the Bayesian section will actualy contradict the original sequential reasong of the Monty Hall game.

1) Firstly you you need to explain that in order for a conditional probability approach to work we have to take the steep from frequency statistics to Bayesian statistics. This means that we have to depart form the sequential nature of the problem where we first have the allocation of the car and goats and then we we have the door selection. Now instead we are going to assume that the player already has done his door selection. In order to evaluate such a decision in the middle of the game we have to vary the location of the car. So basically we are inverting the origial order of the game. 1) door selection 2) allocation

2)Secondly you need to explain the reason for that). You need point out that Bayesian statistics in the form of conditional probabilities are based upon the assumtion of serial correlation (normal distribution with fat tails) which means that the observations are not independent. For example

WHOA THERE!!! The above paragraph is pure nonsense. The Bayesian formulation of probability theory, based on Cox's axioms, is completely general and has been shown to be equivalent to any other standard formulation (e.g. Kolmogorov's). See the references ited in that section, e.g., E.T. Jaynes's "Probability Theory as Logic". Please do not make absurd statements to (try to) make a point. Also, please stop talking about "Bayesian statistics". There is not statistics involved in the Monty Hall problem: rather, it is purely a problem of probability theory.The Glopk (talk) 01:25, 21 August 2008 (UTC)[reply]

P(A∩B) is the probability of A and B happening
P(B) is the probability of B happening
P(A|B) is the probability of A happening given that event B has happened

Bayesian expression: P(A∩B)=P(B)*P(A|B)

When we have serial independence ( A and B are independent) then the P(A|B) expression is reduced to P(A)

Which means that our Bayesian expression is reduced to the frequency expression

P(A∩B)=P(B)*P(A)

These two explanations are important because it helps the reader to understand the some what contradicting and confussing set up! Here I am open for suggestions! Any solid and easy to understand explanations that can simplify the transition is most welcomed !

I now feel that I have put forward a solid argument why the current solution is not optimal. If you still fail to take my points into consideration I suggest that we seek outside help on the matter since otherwise we will continue this discussion for all eternity. To tell you the truth I have more productive things to do that arguing with you about these things!
--82.39.51.194 (talk) 10:50, 19 August 2008 (UTC)[reply]

I will not go into the details of what you wrote here; others can do that much better than I can.

First, I want to applaud your change of attitude instead of changing the article time and time again, you are now discussing the matter.

However, you should note that you seem to be the only one arguing this point. That doesn't in itself mean you are wrong, of course; but you should appreciate the fact that many editors, some of them very well versed in this theory, have already been over this article in great detail. That in turn means you should consider the possibility that you are in fact mistaken. In your first posts here you seemed to take the stance that you were inquestionably correct, but as I said before, you have changed your attitude a bit, instead putting forward arguments to support your opinion. Kudos.

One point that you haven't responded to yet is the tenet of no original research. The references cited in the article support the problem statement we currently use, and so far you have proposed no references to support your version. Oliphaunt (talk) 11:57, 19 August 2008 (UTC)[reply]

You are right I have to considered the possibility that I might be wrong! I have done that and the answer was that I am not! It is not that much to wrong about! Further, the best references is the original Monty Hall game and its corresponding rules! I can probably dig up a reference on the frequency statistics approach (which I assume is my suggested approach) but I am not sure it is necessary. It is like asking for a reference for why 2+2=4. It is a simple exercise on calculating probabilities! If you find that difficult then you should probably not take on the conditional probability approach which dominates the current article ! --82.39.51.194 (talk) 13:31, 19 August 2008 (UTC)[reply]

What you're not right about is your assertion that the current approach contradicts the sequence of the game. It absolutely follows the sequence, from the player's perspective. First two goats and a car are placed behind 3 closed doors, but the player doesn't know what is behind each door. In your solution, the configuration is given. Why doesn't the player just pick door 3 and stick with their initial choice? The reason of course is that the player does not know the configuration. In the version currently in the article, the configuration is left unknown (matching the situation from the player's point of view). The configuration is certainly fixed before the player picks a door, but the player doesn't know the configuration. What the player does know is the door he or she initially picks. The current solution, most sources, and the Parade problem description implicitly say (and the Bayesian analysis section explicitly says) "let's call the door the player picks door 1 (renumbering the doors if necessary)". The analysis proceeds given the unknown, but fixed, configuration of goats and car with the (now fixed) initial player choice. We examine all possible configurations of the goats and car not because we're moving them around after the player has picked (which would indeed violate the sequence of the game) but to enumerate all possible scenarios using the same frequency based approach your solution uses. This approach

1) exactly matches the player's view of the game. The location of the car is fixed, but unknown (in your solution, the car location is known but the player's pick is treated as variable - this is distinctly not the player's view and, literally shows only that switching wins with 2/3 probability if the car is behind door 3)

2) is not based on a specific arrangement of goats and car, which (again) matches the player's view of the game.

3) does not start in the middle of the game (if the current section is at all confusing in this regard, we could certainly work on clarifying it)

4) easily extends to the conditional analysis, without needing to switch to Bayesian logic (indeed, the current solution section includes the conditional analysis, without mentioning anything about Bayesian analysis)

5) considers all initial configurations (and, through renumbering of the doors, all initial picks and host responses)

6) is consistent with the view of the problem throughout the article (all the images and all the text consistently call the player's initial choice "door 1" and the door the host opens "door 3")

You're quite welcome to seek outside help, although I would suggest the best help would be sources. The existing references are (as far as I know) the best, most authoritative sources on the topic of the Monty Hall problem. The existing approach follows the solution typically presented in these sources. I've said this about 3 times already, but your solution and the solution currently in the article are essentially equivalent (both rely on a "without loss of generality" assumption). Given two essentially equivalent approaches, using the one that more closely matches the references, more closely matches the player's view of the game, and more easily extends to cover the conditional analysis seems like the obvious choice. -- Rick Block (talk) 16:20, 19 August 2008 (UTC)[reply]

yeahh yeahh what ever !--82.39.51.194 (talk) 17:29, 19 August 2008 (UTC)[reply]

I've clarified the sequencing in the Solution section. I hope this helps to address your concerns. -- Rick Block (talk) 02:54, 21 August 2008 (UTC)[reply]

I actually agree with 82.39.51.194 I also find the current article difficult to understand. I think a frequency statistic approach is easier to understand and more consistent with the original game.--92.41.172.75 (talk) 08:50, 21 August 2008 (UTC)[reply]

Suggestion to replace existing solution section

It appears 92.41.17.172 is suggesting that the current solution section be replaced with the contents from #Summary and Solution (above). My understanding was that per #Discussion (also above) we basically came to an agreement, albeit not very enthusiastic on the part of 82.39.51.194, about this. I won't repeat the discussion from above, but does anyone have anything more to say about this? -- Rick Block (talk) 18:30, 28 August 2008 (UTC)[reply]

The proposed change (diff) replaces the existing solution with text which is less detailed, has no supporting references, and which includes less appealing figures. The existing section should remain. TenOfAllTrades(talk) 18:56, 28 August 2008 (UTC)[reply]

I tend to agree with Rick, too, that the existing solution should remain. In my 35 years of teaching mathematics at a major university, including many, many probability courses, I've dealt with the Monty Hall problem many times in classes. What intrigues me most, though, is not so much the fact that the direct solution given seems to bother so many (perhaps because of the supposed veridical nature of the statement of the problem), but the fact that if one merely steps back and looks at the entire problem, there is a remarkably simple solution. The key to this solution is not to focus on switching but instead to focus on not switching. Choosing the strategy of not switching is tantamount to ignoring any and all extra information with the problem, and it is thus manifestly clear that:

P(not switching) = 1/3

and thus we have

P(switching) = 1 - P(not switching) = 2/3.

It uses nothing more than P(A) + P(not A) = 1. This is no more (and perhaps a lot less) counterintuitive than all the rest of the considerations which focus directly on choosing the switching option. I have always given perfect marks for this elegant solution and wonder why it is not mentioned in the article. Focussing on the switching strategy directly involves something like a probability tree or other construct (entailing conditional probability), fraught with potholes that can trap or fool the not-so-careful reader.

Some of the most elegant solutions to problems in mathematics arise by looking at such problems from a different perspective. This is just one such example. -- Chuck (talk) 19:29, 28 August 2008 (UTC)[reply]

Wait a minute. The expression P(A) + P(not A) = 1 refers to set complementation in a universal set - which has probability 1. And the not-switch strategy is not the complement of the set consisting of the switch strategy in the universe consisting of all possible strategies. For one thing, if there a two strategies there an infinite number of mixed strategies in that universe. While it is true that the probability of winning by switching plus the probability of winning by not switching is 1, it is something that you have to demonstrate by analyzing the problem. Its not hard but it sure doesn't follow from the tautology given. Just think, changing the name of the not-switch strategy to 'sticking strategy' sends the argument down the tubes.

In this particular game, the contestant always has the opportunity to switch so the switch strategy and the not-switch strategy always end up selecting two different doors and since the the other door is open and showing a goat, the car must be behind one of those two doors. Thus probability of winning by switching plus the probability of winning by not switching is 1. However in a variant of the game where the host did not always present an opportunity to switch then some of the time the switch strategy and the not switch strategy pick the same door, hence the sum of their probabilities of winning would be less than 1. Is that an example that shows P(A) + P(not A) = 1 is not true?Millbast5 (talk) 07:43, 18 September 2008 (UTC)[reply]

Chuck, my hat is off for you! I like your way of thinking.
"P(not switching) = 1/3 and P(switching) = 1 - P(not switching) = 2/3. It uses nothing more than P(A) + P(not A) = 1"
So simple but still so powerful! Only that well crafted sentence says more than the complete article in my opinion.
I think that reasoning in combination with the new suggested "frequency" solution would greatly improve the article--92.41.17.172 (talk) 20:53, 28 August 2008 (UTC)[reply]

I don't mean to be rude, but have you read the article? The solution section says Players who choose to switch win if the car is behind either of the two unchosen doors rather than the one that was originally picked. In two cases with 1/3 probability switching wins, so the overall probability of winning by switching is 2/3 as shown in the diagram below. The diagram then shows (visually) what happens if you switch (Chuck's point is simply the inverse of this). The killer argument against this simplistic analysis (and Chuck's) is that it completely ignores the fact that the host opens a door. Why should this same argument apply both before and after the host opens a door? The initial pick is 1/3 (surely), but then the host opens a door. Why is this any different from Howie opening a case on Deal or No Deal (or would you say never take the deal because the chances of your case being the grand prize go up every time a case is opened)? I don't teach math at a major university, but this solution simply fails to address the problem (per Morgan et al.) and doesn't seem worth more than a B- (correct answer, completely inadequate reasoning). Consider three different hosts - host 1 who opens a randomly selected "goat door" if the player initially picks the car, host 2 who always opens the rightmost "goat door" if the player initially picks the car and host 3 who opens one of the unpicked doors randomly. With host 1 the correct answer is 1/3 chance of winning if you don't switch. With host 2, assuming the player stays with the initial pick (and it's door 1), the correct answer is 0 chance of winning if the host opens door 2 but 50% chance of winning if the host opens door 3. With host 3 the correct answer is 1/2 chance of winning (assuming the host didn't reveal the car). What this means is that the argument that the initial pick results in a 1/3 chance of winning is, um, not quite correct. If you don't also consider how the host selects what door to open the answer may (coincidentally) end up with the right answer (if it's host 1) but this is fundamentally a coincidence. Given any possible host behavior the average chance of winning if you don't switch is 1/3, but if you pay attention to which door the host opens (and how the host chooses to open doors) the chances of winning by staying are anywhere from 0 to 50% depending on which door the host opens. The point is the real solution has to include a consideration of how the host chooses which door to open. If it doesn't, then the answer applies regardless of which host we're talking about (host 1, host 2, or host 3) - i.e. you arrived at the right numerical answer but your reasoning is faulty. -- Rick Block (talk) 04:45, 29 August 2008 (UTC)[reply]

Sorry, Rick, but you are being rude (which is out of place here) and it seems you have been very much caught up in the veridical nature of the posed problem. Yes, I have indeed read the article, and it beats around the bush with various possible host actions (which is another matter that I'm not addressing) ... but of course I am observing the generally agreed on rule that says the host always opens a goat door different from the door chosen by the contestant. What you don't see is that there are two and only two options which under the host assumption exhaust all possibilities: (A) be obstinate (not switching no matter what is shown) or (not A) always switch when a door (different from the one selected by the contestant) showing a goat is opened. While (not A) has all the folderal about choosing to switch), option (A) does not (if you just think about it for a moment). Moreover, (A) union (not A) is the entire event space, so P(A) + P(not A) = 1. And since P(A) is manifestly eequal to 1/3, P(not A) = 2/3.

You complain that my solution ignores the fact that the host opens a door showing a goat. Note that it makes no difference in the outcome if the host opens either of the doors not initially selected. In the obstinate case (never switch no matter what is opened) the probability of winning the car is 1/3. If the contestant has initially chosen a goat door and her strategy is always to switch, then no matter which of the other two doors is opened, she will always with the car by switching to the opened door if it reveals a car or to the other closed door if it reveals a goat.

In either of these two scenarios, the problem you and many have is in realizaing that (always switching) and (never switching) exhaust all possibilities. That is precisely where the problem seems like a (veridical) paradox ... and that is why the solution I give is so elegant and, at first glance, seems that it must be wrong - even though it is completely correct.

I am reminded how, in teaching how to find areas of regions to students, we tell them that one way is to decompose the region into simpler regions and add up the areas of the pieces. Then we watch the solutions come in for a problem such as: find the area of the shaded region shown - which is nothing more than a large rectangle with a smaller rectangle removed - and students slavishly decompose the region into four rectangles, compute the area of each, and add them up, correctly of course - but completely missing the obvious solution to subtract the area of the smaller rectangle from that of the enclosing rectangle! Too bad people can't think outside the box or (in this case) inside the box.

People's dogged insistence that the result for never switching has something to do with opening a different door from the one selected is a bit like this, the only point to the problem is to see that since the host opens a different door from that chosen by the contestant, then the obstinate and always switch options are discjoint and do indeed exhaust all possibliities.

Oh well, some people look at a sphere and see perfection - others look at one and see a snowball and duck for cover. -- Chuck (talk) 14:11, 29 August 2008 (UTC)[reply]

Sorry for the rudeness. However, see below as well. -- Rick Block (talk) 15:26, 29 August 2008 (UTC)[reply]

ha, ha. I and many with me prefer Chuck's simple solution plus 82.39.51.194 "frequency" solution anyday over that lengthy and confused outburst by Rick Block!
There are no such thing as host-1,host-2 and host-3. The only host that exist is host-1 = Random selection of "goat door". Why would I consider an example that does not apply?
With all due respect there might be a reason for why you are not teaching math at a major university. --92.41.46.220 (talk) 09:06, 29 August 2008 (UTC)[reply]

Respect acknowledged, however your reasoning is still inadequate. Yes host 1 is our host, but there's nothing in the simple solution that makes use of this fact. Assume for the moment we're talking about host 3 (makes it more like Deal or No Deal), not host 1. The same simple argument can be used, but in this case this argument results in an incorrect solution. The flaw is the implied claim that because P(not switching) = 1/3 when the player initially selects a door it remains so after the host opens a door, i.e. that how the host chooses to open a door is "extra information" that can be ignored. Not so.

You have to at least explicitly say that P(not switching) doesn't change when the host (our host, host 1) opens a door because of the constraints on the host's behavior, but then the question becomes how do you know this? Asserting it to be true with no reasoning doesn't seem sufficient.

If you have access to it, please read the Morgan et al. paper. This paper distinguishes host 1 from host 2 which is a much more subtle distinction than the difference between host 1 and host 3. With host 2, the average chance of winning by not switching (ignoring which door the host opens) is 1/3 - just like host 1. However, with host 2 there are two distinct probabilities depending on whether the host opens the rightmost or leftmost door. If the host opens the rightmost door (which he does unless the car is behind it) the chance is 1/2. If the host opens the leftmost door (which he does only when the car is behind the rightmost door) the chance is 0 (of winning by not switching). You can in some sense say the probability of winning by not switching in this case is 1/3, however the probability for any given player (with knowledge of which door the host opens) is either 0 or 1/2.

So, back to the article. I believe the solution as stated is as simple as possible without being too simple. -- Rick Block (talk) 15:26, 29 August 2008 (UTC)[reply]

I also agree with 82.39.51.194 and 92.41.46.220. I think we have consensus now to change the solution section!
--Pello-500 (talk) 13:51, 29 August 2008 (UTC)[reply]

It appears that Chuck actually doesn't agree with your change;
It's incredibly obvious that Pello-500 is just a new account created by 92.41.

If you're not interested in engaging in serious discussion here, please stop editing the article. TenOfAllTrades(talk) 14:00, 29 August 2008 (UTC)[reply]

In addition to edit warring over the article, Pello/92.41/82.39 also erased my comment calling him on it. I've restored it above. Note that he had originally (and erroneously) claimed that Chuck had agreed with his position. TenOfAllTrades(talk) 21:43, 29 August 2008 (UTC)[reply]

I also agree with 82.39.51.194 and 92.41.46.220. The frequency solution is most optimal.--84.92.246.30 (talk) 17:26, 30 August 2008 (UTC)[reply]

RfC: New or old solution section?

The new suggested "frequency" + Chuck's solution is easier to understand. Current editors refuse to accept this due to ownership. They won't let anyone edit the article !

Initial thought - if I understand correctly the question is not about what the actual solution is, that is agreed, but it is about the best way to explain it. From my experience different people have different way of looking at things. When I am trying to get to grips with a probability problem I usually find that one particular way of looking at it helps me to get it. The may not be the way that helps other people. My suggestion, therefore, is to have both solutions. Maybe the second could be added under the heading 'Another way to understand the solution'. Martin Hogbin (talk) 17:56, 29 August 2008 (UTC)[reply]

Disagree, please leave the old solution. The new proposed "frequency" solution is poorly written and referenced. Chuck's solution is just the "naive Bayes" one (uniform prior on the host behavior). As remarked by Rich Block, the "old" solution section, i.e. the one that passed the F.A. review, offers a more comprehensive treatment that is supported by the refrences and does cover more sophisticated host behaviors.The Glopk (talk) 00:29, 30 August 2008 (UTC)[reply]

My suggestion as to what should be done depends on what people believe that the purpose of this article is. For example, is the purpose purely to consider the problem academically and with some degree of rigor or is there a secondary function of explaining to as many readers as possible what the solution is and why in a way that makes sense to them. Martin Hogbin (talk) 08:53, 30 August 2008 (UTC)[reply]

I don't know if it's necessary to repeat it, but in case someone doesn't read the section above, I'll note that the proposed replacement section is formatted poorly, lacks references, and contains less explanatory text (both quality and quantity) than the proposed replacement. TenOfAllTrades(talk) 17:48, 30 August 2008 (UTC)[reply]

Can I make sure that I now understand what the current debate is about? Some editors want to keep the page as it is but recognise that there is a rather subtle problem with it. Others wish to replace it with a more rigorous approach which some feel has been badly presented and is not necessary. Martin Hogbin (talk) 09:39, 31 August 2008 (UTC) On a second look at things I can see that I got that wrong, I was confusing the current dispute with and earlier discussion that Rick pointed me to.Martin Hogbin (talk) 15:01, 31 August 2008 (UTC)[reply]

Now that I have some idea of what is going on here, I can give my views on the current dispute. I do not think that the current solution should be replaced with the proposed one. However, I do think that it is more natural to have a fixed car position and vary the pick, but I also understand the reasons that the solution is the way that it is. I do have some criticisms of the article as it is and hopefully I will have some suggestions for improvements, but scrapping what has been done and replacing it with the proposed replacements not the way to improve in my opinion. Martin Hogbin (talk) 18:08, 31 August 2008 (UTC)[reply]

I'm in favour of including both solutions. Because one method, to quote, "is formatted poorly, lacks references, and contains less explanatory text" is not grounds to remove it altogether, but to bring it up to standards. If we removed everything that was poorly written rather than improved them, Wikipedia would have about 20 articles. Besides, other topics like Zeno's paradoxes have entire articles dedicated to the various proposed solutions, so the "we already have a solution" argument doesn't seem valid here. Android 93 (talk) 06:14, 6 September 2008 (UTC)[reply]

As you will see I have made a similar suggestion here but any proposals for change get a very frosty reception. Have a look at the history of this page and you will see that an IP editor recently added an extra explanatory section only to have it immediately deleted, with no explanation and with the deletion being marked as a minor edit!

Another person tried to start a separate article only to find it set upon with a voracity I have never seen before in WP and deleted with days.

I have been trying hard to work with the existing/historical editors to find way to improve the page without compromising the integrity of the current article. So far my advice has been that I can make minor edits but if they are reverted I must immediately discuss them.

I Strongly believe that the current article fails to be convincing in respect of the basic paradox and that this must be improved. My view is supported by the fact that since the RFC there have been two people on the talk page who did not believe the article. One was eventually convinced by discussions here. As you say, there are other articles which make a better job of explaining their central paradoxes.Martin Hogbin (talk) 10:09, 6 September 2008 (UTC)[reply]

In an effort to allow this RFC to proceed unimpeded by "current editors" I have refrained from commenting, however I feel compelled to respond to this. I can see how editors new to this page might interpret responses to suggested changes as frosty, but in defense of the current editors this article is a featured article that has been through two featured article reviews [1] [2]. This certainly doesn't mean it's perfect, but it does mean there's been a significant amount of effort by multiple people spent on making the article not only mathematically correct with all major claims referenced to reliable sources but also comply with all other featured article criteria which include adherence to the WP:manual of style and a "professional standard" of writing. The recent addition by user:Technicalmayhem (a new, but not an IP, editor) was first deleted (by me) with this edit summary: revert - many WP:MOS issues with this addition. Technicalmayhem re-added it and this repeat addition was reverted by another editor also apparently watching the page. The re-addition was done with no explanation and marked as a minor edit (!), as well as the re-deletion. The separate article, "Introduction to the Monty Hall Problem", was created by user:Pello-500 (arguably the same user as 82.39.51.194 and 92.41.46.220) for the apparent purpose of including this user's preferred "solution" without gaining consensus to include it here. This is not how content disputes should be handled per Wikipedia:Dispute resolution, and since this user previously initiated this RFC he or she is clearly aware of the proper steps for handling content disputes but for whatever reason chose to ignore them. It's hard to interpret creating this article as anything other than bad faith tendentious editing.

Rick, I was not referring to your reversion but this one: 4 September 2008 Jonobennett m (48,223 bytes) (Reverted 1 edit by 68.2.55.158.)Martin Hogbin (talk) 20:24, 6 September 2008 (UTC)[reply]

From my point of view, the discussion below of Martin Hogbin's suggestions for improvement has been proceeding pretty much exactly how such discussions should go. I'm not, and I don't think anyone else is, saying no changes can be made. The caution about making a change once and if it's reverted then discussing it was not intended to inhibit changes but to encourage them (per WP:BOLD). I partially agree with the point that the article fails to be convincing (not just one, but both "doubters" have apparently been convinced) however I'm highly skeptical that ANY explanation or even combination of explanations would convince all readers. To be clear, I'm not saying it can't be made more convincing but like it says in the lead - even when given a completely unambiguous statement of the Monty Hall problem, explanations, simulations, and formal mathematical proofs, many people still meet the correct answer with disbelief. -- Rick Block (talk) 18:09, 6 September 2008 (UTC)[reply]

I agree it will not be easy but surely it is worth a go. As with similar problems, explanations from many different angles could be a good place to start. What is the best way to proceed? I do not want to start editing the current article by adding a new section, which will start off badly written, formatted, and referenced and may well get removed. Do you agree that a development version of the article might be a good place to start, with the intention of a merger when agreed, or perhaps leave it as a separate article with a link from this one? Martin Hogbin (talk) 20:24, 6 September 2008 (UTC)[reply]

Creating a new article to be merged later is definitely not the way to proceed. Proposing the text for a change, or even a new section, here (well, probably not here but in a section below) on the talk page would be perfectly reasonable. If you're talking about major structural changes, you could create a sandbox version, perhaps on a page in your user space like user:Martin Hogbin/Monty Hall problem (draft). -- Rick Block (talk) 20:54, 6 September 2008 (UTC)[reply]

I have set up a user page as described - see section below.Martin Hogbin (talk) 10:14, 7 September 2008 (UTC)[reply]

keep original. If I understand correctly the two solutions that are being offered (no one has bothered to supply diffs), then I think the current (original) version is better. having taught this problem in the past, however, I think it needs some revisions. does anyone object if I do some editing? --Ludwigs2 03:07, 7 September 2008 (UTC)[reply]

Bayes was right, but I think the logic of the existing solution is flawed

The Bayesian analysis is correct to the extent that it demonstrates that the probability of the car being behind Door 1 remains 1/3 after the decisions are made by the player and the host. That is the "decisions" by the player and the host do not affect the probability of the car being behind Door 1.

However, I believe the interpretation of the Bayesian analysis is flawed.

In particular, although the Bayesian analysis correctly accounts for the fact that the host knows the location of the car it does not reflect knowledge gained by the player when it is revealed that the car is not behind Door 3.

Bayes's Theorem can be used to take account of the fact that the player learns the car is not behind door 3 by calculating the probability of the car being behind Door 1 given that it is not behind Door 3.

Let's do it in English.

The probability that the car is not behind Door 1 given that the car is not behind Door 3

= [(The probability that the car is not behind Door 3 given it is behind Door 1) x (The probability that the car is behind Door 1)] / (The probability that the car is not behind Door 3)

= [1 x 1/3] / 2/3

= 1/2

This result is consistent with many people's intuition that if there are two doors and a car is behind one of them then there is a 50:50 chance of the car being behind either door.

That is, I think conclusions shown on the website are wrong.

Given the debate that has already taken place I guess there will be some opposition to my view. --Paul Gerrard at APT (talk) 11:51, 30 August 2008 (UTC)[reply]

Paul, I would be happy to try and persuade you that you are wrong but this is not the place. You can email me at martin003@hogbin.org. Martin Hogbin (talk) 14:02, 30 August 2008 (UTC)[reply]

The basic issue is that "the probability that the car is not behind Door 1 given that it is not behind Door 3" (which is indeed 1/2) is NOT the question posed by the Monty Hall problem. In English, what we're looking for (flipping "is not behind Door 1" to "is behind Door 1" - seems a little clearer this way) is "the probability that the car is behind Door 1 given that the host opens Door 3". With the standard rules, this probability ("host opens Door 3") is 1/2 given the car is behind Door 1 (from the problem statement: If both remaining doors have goats behind them, he chooses one randomly). This probability is also 1/2 given the car is not behind Door 1, since if the car is not behind Door 1 it's equally likely to be behind Door 2 or Door 3. Combined, what this means is the probability the host opens Door 3 is 1/2 whether or not we're also given the car is behind Door 1. This makes the Bayes expansion be (1/2) x (1/3) / (1/2), i.e. 1/3. -- Rick Block (talk) 16:28, 30 August 2008 (UTC)[reply]

Looks like it is OK to answer here. Here is a very simple way to see that the probability of getting the car if you switch is 2/3. If you switch, you always get the opposite of what you started with. You have a 2/3 probability of starting with a goat. Martin Hogbin (talk) 16:51, 30 August 2008 (UTC)[reply]

This is simple, but it relies on an assumption that when the host opens a door nothing changes (or changes the question from "what is the probability of winning after the host opens a door given the Monty Hall rules" to "what is the probability of picking a car from among 3 doors"). Under the standard rules it turns out that the host opening a door doesn't affect the player's initial chances of having selected the car (this is reflected in the Bayes expansion above with the two 1/2's that cancel each other out), but this simple explanation does not use those rules so it is fundamentally incomplete (there is much more discussion of this in the #Suggestion to replace existing solution section thread just above). Why does this explanation not apply to Deal or No Deal? -- Rick Block (talk) 17:25, 30 August 2008 (UTC)[reply]

Yes Rick, you are quite right. However if I add to my explanation the observation that, with the standard rules, no information about the location of the car is given when the host opens a door and thus the original probability of getting a goat holds after the door has been opened, then my explanation is a good one for the standard rules case. However it is then not such a simple explanation and cannot be directly applied to other cases. At least I understand what the RFC is all about now. Martin Hogbin (talk) 18:01, 30 August 2008 (UTC)[reply]

Here is rather extreme example to show what Rick means and why my original answer was incomplete. Suppose we work to the standard rules, so you would expect my explanation to apply, but, after he as opened a door, the host tells the contestant whether he has initially chosen a car or a goat (rather spoils the game) then the probability of getting a car if the contestant swaps is clearly not 2/3 but either 1 or 0 depending on what the host tells the contestant. Martin Hogbin (talk) 18:12, 30 August 2008 (UTC)[reply]

The question still remains how we know given the standard rules that "no information about the location of the car is given when the host opens a door". This is an assertion, not reasoning, and (as it turns out) showing this with reasoning is not overly simple. Again, I think the best reference for this is the Morgan et al. paper (in the references), or the current version of this article. This topic was discussed at length several months ago, shortly before the latest featured article review - starting with Talk:Monty Hall problem/Archive 6#Rigorous solution. In my opinion the ensuing discussion (that continues in the archives into Talk:Monty Hall problem/Archive 7), somewhat contentiously spearheaded by an anonymous user (using multiple IP addresses in the range of user:70.137.168.95) who originally pointed us to the Morgan et al. paper, in conjunction with scrupulous referencing improvements to satisfy concerns raised during the last FARC have led to significant improvements in this article (as well as to my personal understanding of the subtleties involved in this problem). One of the key points this user brought to the discussion (backed up by the Morgan et al. paper) was that most solutions presented to the problem are overly simplified and include but don't justify the assertion that the host opening a door doesn't change the initial 1/3 probability. Another very good reference about this is the Falk paper (also in the references). According to Falk the answer that most people present (after the host opens a door, it's a 50/50 chance) is rooted in a deeply intuitive "equal probability" assumption; however many "correct" answers commonly presented simply replace this intuition with an appeal to another one, i.e. the "belief that exposing information that is already known does not affect probabilities". Neither of these is based on particularly sound reasoning and problems can be constructed where either one leads to an incorrect answer. This problem is contentious at least in part because it's not simple. -- Rick Block (talk) 21:21, 30 August 2008 (UTC)[reply]

There certainly is more to this than meets the eye. Might a way forward be as follows? Formulate a version of the problem where a simple analysis, such as I gave, is correct. We are at this stage picking a problem to suit our desired answer. In other words construct a problem where what has been called the most attractive false solution is actually a good solution. Now clearly state our reformulated problem as an idealised version of the Monty Hall problem and give the, now correct, simple solution. My guess is that this will be what 90% of readers will want and expect. It is just a continuation of the process started by Krauss and Wang.

The discussion about the hosts behaviour and the contestants understanding of this then can become a secondary, much more complicated, and perhaps contentious, discussion for those interested in the fine details.Martin Hogbin (talk) 22:25, 30 August 2008 (UTC)[reply]

This has been suggested before - the problem is that if we do this we drift into WP:OR unless the version of the problem we present clearly meets WP:RS. I think this boils down to whether the solution currently presented is accessible to a lay person. I think the answer is yes (but I am sufficiently "into" this problem that I realize my opinion on this is basically worthless). -- Rick Block (talk) 04:14, 31 August 2008 (UTC)[reply]

Maybe we should consider the occasional exception (WP:IGNORE) here. The dividing line between OR and verifiable is rather vague anyway; not all third party references should carry equal weight and the process of assessing how authoritative any given reference is is a form of OR. With this particular problem we could probably find more references, from normally authoritative sources, that give the completely wrong answer than we could good ones. On the other hand experience with the problem has shown that our intuition is not to be trusted either. Regarding the RFC, I have continued above.Martin Hogbin (talk) 09:35, 31 August 2008 (UTC)[reply]

Hi Martin, thanks for the information.
I agree with much of what you say, but I disagree with the assertion that:

"no information about the location of the car is given when the host opens a door".

It seems to me that when the host opens a door the player is given information about the location of the car. In particular, the host's action informs the player that the car is not behind Door 3 and that the car must therefore be behind either Door 1 or Door 2.

Yes, you are right, I put it badly.Martin Hogbin (talk) 12:59, 31 August 2008 (UTC)[reply]

This information does not alter the fact that at the outset of the game there was a 1/3 chance that the car could have been behind any particular door.

That is right but also the chance a car is behind the door that I initially chose remains at 1/3 if the host must open a goat door and, if given a choice, must chose randomly, and I know all this. This is generally agreed to be true in the above circumstances but my mistake was to take this as being self evident. If the host acts differently, then the above statement is often not true. If you agree that, after a door is opened, the chances of the car being behind your initial door is still 1/3 then my argument still holds. Martin Hogbin (talk) 12:59, 31 August 2008 (UTC)[reply]

I can't see the logic of the probability remaining 1/3. If it were 1/3 and it is known that there was a goat behind Door 3 this would mean that the probability of the car being behind Door 2 had risen to 2/3. I know that is what the website says, but in my response to Glopk below I explain why I believe that probabilities can't be transferred as suggested on the website. Can you point me to any of the previous discussion about transferring probabilities? --Paul Gerrard at APT (talk) 12:23, 1 September 2008 (UTC)[reply]

Let me try a slightly different version of the problem. The contestant picks a door. We agree that he has a 2/3 chance of picking a goat. We also know that if he switches he will get the opposite of his original choice. Suppose, therefore, that this person decides, before a door is opened, that he will switch. Do you agree that he then must have a 2/3 chance of getting a car? Martin Hogbin (talk) 18:08, 1 September 2008 (UTC)[reply]

Hi Martin - I agree that at the outset there is a 2/3 chance of picking a goat, but I am not sure what you mean when you say that "if he switches he will get the opposite of his original choice". Also, if the player decides to switch why would the probability of picking a car increase to 2/3. If you extend the logic to switching a second time would you say that the probability of picking the car was 1 (i.e. a certainty). What would be the reason for suggesting that deciding to switch would increase the probability of picking the car? —Preceding unsigned comment added by Paul Gerrard at APT (talk • contribs) 00:11, 2 September 2008 (UTC)[reply]

By "if he switches he will get the opposite of his original choice" I refer to the simple fact that , if the player initially chooses a goat and the switches he must get a car(because his original choice is one of the goats and the host has removed the other one from play leaving the car as the only other door) and if the player initially chooses a car and switches he must get a goat (because there is only one car and he has it). The above always applies, without exception, regardless of the strategy of the player or the host (we are here assuming that the game rules require the host to always open a goat door and offer a swap).

Regarding a second swap, no of course, the probability does not rise to 1. The whole point of this problem is that after a door has been opened, the player has more information as to the whereabouts of the car, making the unchosen door more likely to have it. I have to say, and indeed am saying, that the current explanation is not as convincing as it should be. I am currently trying to improve it. Martin Hogbin (talk) 08:46, 2 September 2008 (UTC)[reply]

However, while the original probabilities remain unchanged the information that the car was not placed behind Door 3 effectively changes the game from a "3 door game" to a "2 door game".
The effect of knowing that the car is not behind Door 3 on the probability that the car is behind Door 1 can be demonstrated as I have done above using Bayes' Theorem.
Bayes Theorem can also be used to calculate the probability that the car is behind Door 2 given that it is not behind Door 3. (The answer is 1/2.)
Our different perspectives are as intriguing as the puzzle itself.
What am I missing? (I have tried to make sense of the references with the exception of Morgan et al which costs $14)--Paul Gerrard at APT (talk) 11:29, 31 August 2008 (UTC)[reply]

No, because the opening of a door does give you extra information about the chances of the car being behind the other door. Martin Hogbin (talk) 12:59, 31 August 2008 (UTC)[reply]

Paul, please re-read the "Bayesian Analysis" section. As others have already pointed out, the proposition of interest for the MHP is not

\qquad C_{i}|\neg C_{3}\,=\,

"The car is behind door i, given that it is not behind door 3", for

i=1,2\,

. Rather, the player is interested in the truth of

\qquad C_{i}|H_{j3},I\,=\,

"The car is behind door i, given that the host opens door 3 after the the player initially chooses door j", for

i,j=1,2\,

, and where

I\,

represents everything the player knows prior to making the first choice, including the host's behavior. The two propositions are obviously different, and therefore it is to be expected that they evaluate to different numbers. But only the latter one matters. Further, the evaluation of

\qquad P(C_{i}|H_{j3},I)\,

using the Bayes theorem shows the need to make explicit the assumptions concerning the host's behavior through the value of

\qquad P(H_{j3}|C_{k},I)\,

, the probability of the proposition "The host opens door 3 after the player chooses door j, given that the car is behind door k". All possible host's behaviors are specified by assigning values to this function of j and k. The Bayesian Analysis section in the article shows only one particular such function, the one consistent with the "standard" interpretation of the MHP, i.e. "The host shows a goat behind one of the two doors not chosen by the player. If two such doors are available, they are equally likely to be opened".The Glopk (talk) 16:46, 31 August 2008 (UTC)[reply]

Thanks The Glopk, it is good to be reminded that $\qquad P(C_{i}|H_{j3},I)\,$ is not the same as $\qquad P(C_{i}|\neg C_{3})\,$ . This a very important distinction and I agree that these expressions are very different and that there is no reason to believe they should be equal.
It is therefore important to reconsider the problem to make sure we are trying to answer the same question. My understanding of the problem is that we are trying to determine whether it is to the advantage of the player to switch after the host opens a door.
We also need to review how the expressions relate to the issue of whether it is to the advantage of the player to switch after the host opens a door.
The differences between the expressions are subtle.
My understanding is that $\qquad P(C_{1}|H_{13},I)\,$ = 1/3 shows that the rules the host follows to select a door do not alter the probabilities - they are still 1/3.
However, $\qquad P(C_{1}|H_{13},I)\,$ does not take into the fact that the host has revealed that there is a goat behind Door 3 even though the rules of the game require the host to choose a goat door.
It is really hard to find words which explain why this is a problem, but please consider $\qquad P(C_{2}|H_{13},I)\,$ . If you evaluate this expression in the same way that website evaluates $\qquad P(C_{1}|H_{13},I)\,$ you will find that:

$\qquad P(C_{2}|H_{13},I)\,$ = $\qquad P(C_{1}|H_{13},I)\,$ = 1/3

Similarly, you will find $\qquad P(C_{3}|H_{13},I)\,$ = 1/3 unless you change the probabilities to reflect the fact that it is known that a goat is behind Door 3.

However, if you change the probability of a goat being behind Door 3 to zero then you must also change the probabilities of the car being behind the other 2 doors such that their sum is 1. This raises the question of what would the new probabilities become. I see no alternative other than 50% chance for both Door 1 and Door 2.

Also, I know the existing website suggests that the host's showing of the goat behind Door 3 has the effect of "transferring" the 1/3 of probability to Door 2. The following example helps demonstrate why I believe such transfers do not occur. Imagine a similar game with 3 doors, two goats, and a car. The probabilities for each door are 1/3. Also the probability of the car being behind any pair of doors is 2/3 (e.g. the probability that the car is behind either Door 1 or Door 2 is 2/3). Now imagine that any goat door is opened - what then happens to the probability that the car is behind either of the other doors? If we assume that probabilities are transfered as per the website then for each of the remaining two doors the probability that the car is behind that door becomes 2/3 and the sum of these 2 probabilities is 4/3 whereas the sum of the probabilities must be 1. Thus, there is a problem in transferring probabilities.

Now let's consider $\qquad P(C_{1}|\neg C_{3})\,$ . It reflects everything that is known by the player. The process by which the host chose Door 3 has given no more information to the player than that a goat was behind Door 3. Thus, $\qquad P(C_{1}|\neg C_{3})\,$ = 1/2 seems to me to indicate that it is the true probability of the car being behind Door 1 and there is no advantage in switching.
Finally, lets consider what happens to $\qquad P(C_{1}|H_{13},I)\,$ if we adjust the probabilities to reflect the fact that we know the car is not behind door 3 as suggested above.
That is $\qquad P(C_{1})=P(C_{2})=1/2\,$ and $\qquad P(C_{3})=0\,$ . If you do this you will find that:

$\qquad P(C_{1}|H_{13},I)\ =P(C_{1}|\neg C_{3})\ =1/2,$

That is, the Bayesian Theorem actually gives the same result as $\qquad P(C_{1}|\neg C_{3})\,$ providing the probabilities used in the formula are adjusted to take account of the fact that a goat is behind Door 3.
Thus, it seems to me that there is no advantage in switching. --Paul Gerrard at APT (talk) 11:47, 1 September 2008 (UTC)[reply]

\qquad P(C_{i}|H_{j3},I)\,

is the probability that the car is behind door i given the host opens door 3 after the player initially selected door j (under the rules of the game). This is exactly the question we're interested in. The rules of the game ensure the host opens a goat door. You say this "does not take into the fact that the host has revealed that there is a goat behind Door 3" and admit that it "is really hard to find words which explain why this is a problem". That's good, because it is not a problem.

\qquad P(C_{2}|H_{13},I)\,

evaluates to 2/3, not 1/3, since

$\qquad P(H_{13}|C_{2},I)=1\,$ (if the car is behind Door 2 and the player picks Door 1 the host is forced to open Door 3)

Similarly,

\qquad P(C_{3}|H_{13},I)\,

is 0, not 1/3, since

$\qquad P(H_{13}|C_{3},I)=0\,$ (if the car is behind Door 3 and the player picks Door 1 the host never opens Door 3)

No "changing of probabilities" is required to get these to sum to 1. Before the host opens Door 3 the probabilities are 1/3, 1/3, 1/3. After the host opens Door 3 (under the rules of the game) the probabilities are 1/3, 2/3, 0. The effect (in this case, under these rules) is that Door 3's probability has been "transferred" to Door 2. This doesn't say anything like you can group any two arbitrary doors and probabilities transfer between them.

Our differences of opinion seem to hinge around what probabilities we feed into the Bayesian formula.

If it is correct to say that all probabilities are 1/3 then I agree with your calculations.

If it is correct to adjust probabilities to 1/2, 1/2, 0 for Doors 1, 2, 3 respectively then I believe my calculations are correct.

I believe you are right if we are talking about probabilities prior to the door being opened and I am right if we are talking about probabilies after the host opens a door. As we are trying to take account the information available to the player at the time the player must decide to switch or not not to switch, we must use the information available to the player and at this time the player knows only that there is a goat behind Door 3. The rules of the game provide no more information than this. Thus I believe the probabilities must be 1/2, 1/2, 0 for Doors 1, 2, 3 respectively rather than 1/3 for each door.

With respect to the grouping of doors I agree that one cannot group doors and probabilities transfer between them. My example was to prove that one cannot do it. However, it seems to me that your solution does exactly this. Isn't your solution efectively saying that:

The probability of the car being behind Door 1, 2, or, 3 is 1/3 for each door.

You then group the 2 doors that are not chosen by the player and say there is a 2/3 chance the care is behind one or other of those doors.

The host then selects a goat door from the two doors that were not chosen by the player.

You have then concluded that the entire 2/3 probablity that the car was behind either of these 2 doors must therefore be the probability of the car being behind the door that was not selected by either the player or the host.

I disagree with the logic for two reasons.

Firstly, in the same way as you have grouped the unselected doors you could have said that the probability of the car being behind either Door 1 or Door 3 is 2/3. Then when the host opened Door 3 I think your logic would suggest that the probability of the goat being behind Door 1 would become 2/3.

Secondly, if we go back to Bayes and consider the probability of there being a goat behind Door 2 if it is revealed that a goat is behind door 3 assuming that the probability of a goat being behind either door is 1/3. If we calculate:

\qquad P(C_{2}|\neg C_{3})=P(\neg C_{3}|C_{2}).P(C_{2})/P(\neg C_{3})\,

we get

\qquad P(C_{2}|\neg C_{3})=\,

1 x 1/3 / 2/3 = 1/2
I don't see how the rules of the game can push this to 2/3 as the constraints on the host do not reveal any information other than the car is not behind Door 3.
--Paul Gerrard at APT (talk) 02:36, 2 September 2008 (UTC)[reply]

\qquad P(C_{1}|\neg C_{3})\,

is indeed 1/2. The difference is this ignores the constraints on the host (in particular, that the host MUST open Door 3 if the player picks Door 1 and the car is behind Door 2). This would be the probability of winning if the host happens to open Door 3 choosing randomly between Door 2 and Door 3 (this is the "forgetful Monty" variant discussed in the article). -- Rick Block (talk) 16:30, 1 September 2008 (UTC)[reply]

Let's consider if the constraints of the game affect the probability of the host selecting Door 3. If we have no information about the location of the car the probability of the host opening Door 3 is 50% given that he is not going to open Door 1. This probability is true because:

There is a 1/3 chance the car is behind Door 1 and if it is there is a 50% chance the host picks Door 2 and a 50% chance the host will pick Door 3.

There is a 1/3 chance the car is behind Door 2 and if it is there is a 100% chance the host picks Door 3

There is a 1/3 chance the car is behind Door 3 and if it is there is a 100% chance the host picks Door 2

That is the probability of the host selecting Door 3 is the same as the probability of the host selecting Door 2. (This probability is 1/6 + 1/3 = 1/2)

I cannot see how the constaints on the host influence the probability that the car is behind Door 1.--Paul Gerrard at APT (talk) 03:04, 2 September 2008 (UTC)[reply]

Let's try this again. You're ignoring the effect of the game rules on the host's actions which is why saying we're given Door 3 has a goat is different from saying the host (under the rules of the game) has opened this door after the player initially picked Door 1. These are different givens, yielding (in this case) different probabilities. I'm not "adjusting" any probabilities or "grouping" any doors, only using what the probabilities are and plugging them into Bayes formula. I AGREE

\qquad P(C_{2}|\neg C_{3})=1/2\,

but disagree that this is the question we're answering.

Now it's time for you to say where you disagree with the following:

\qquad P(H_{13}|C_{2})=1\,

(probability of host opening Door 3 after the player initially picks Door 1 given the car is behind Door 2 is 1 - the game rules enforce this)

\qquad P(C_{2})=1/3\,

(probability of the car being behind Door 2 is 1/3)

\qquad P(H_{13})=1/2\,

(probability of the host opening Door 3 after the player initially picks Door 1 is 1/2 - the game rules enforce this as well and you correctly derived it, just above)

So, by Bayes

\qquad P(C_{2}|H_{13},I)=(P(H_{13}|C_{2})\times P(C_{2}))/P(H_{13})\,

\qquad P(C_{2}|H_{13},I)=(1\times 1/3)/(1/2)\,

\qquad P(C_{2}|H_{13},I)=2/3\,

Please note the big bold banner at the top of this discussion page which says the article is mathematically sound and the result has been experimentally verified numerous times. If you disagree, it's FAR more likely (I'd say certain) that you're incorrect rather than that the article is wrong. -- Rick Block (talk) 03:31, 2 September 2008 (UTC)[reply]

The player's initial choice is relevant to the extent that it restricts the choice of the host to the other remaining doors. However, is it relevant to whether there is any advantage in switching?

It is true that the player has a 2/3 chance of picking a goat and given the host must choose a goat it is very tempting to conclude that there is a 2/3 chance that the car will be behind the unselected door. That is, 2/3 of the time the rules of the game will cause the host to reveal with absolute certainty the location of the car. But, is it really true that the probability that the car is behind the unselected door is 2/3? Do the rules of the game really make this conclusion valid? It is tempting to say yes.

However, hmmmm ... BOTHER

I think you are right and I am wrong.

At the outset the player has a:

1/3 chance of selecting a door with the car behind it and if he switches he certainly loses.

2/3 chance of selecting a door with a goat behind it and if he switches he certainly wins.

That is, a switching strategy gives a 2/3 chance of winning.

This is a really good problem. Thanks for your patience in helping to me to work through this.

I WAS WRONG! THERE IS AN ADVANTAGE IN SWITCHING --Paul Gerrard at APT (talk) 13:14, 2 September 2008 (UTC)[reply]

And, to your original point, the Bayes expansion supports this as well. Please also note that you are in very good company - folks who initially got this wrong include Paul Erdős. -- Rick Block (talk) 13:54, 2 September 2008 (UTC)[reply]

Player's initial choice is irrelevant

The first choice of the player is irrelevant. The host will ALLWAYS open the door with a goat behind it leaving the player with two doors of which either one has the car behind it. The palyer then picks one of these two doors and so has a 50-50 chance of winning the car. Mathematically you should not count the hosts action because the propability that he opens the door with the car behind it is 0. So the number of actual choices are four of which two leads to the car and two to the goat. —Preceding unsigned comment added by 81.197.22.58 (talk) 21:12, 1 September 2008 (UTC)[reply]

The player's choice absolutely matters. The host can't open this door, and must open a door with goat. Yes, the host always opens a door with a goat behind it but because the host can't open the player's door and must open a door with a goat, if the player initially picks a goat (there's a 2/3 chance of this happening) the host is showing the player which of the other two doors (the unselected ones) has the car. Please read the Solution section and look at the diagram. If you're still confused we can talk about it. -- Rick Block (talk) 01:02, 2 September 2008 (UTC)[reply]

Allright, let's then ask what is the propability for the player to win the car AFTER (for the rest of the game) the host has opened the door? That must most certainly be 50%. Then we only have left the discussion about the relevance of the player's first choice. —Preceding unsigned comment added by 85.156.86.237 (talk) 09:30, 2 September 2008 (UTC)[reply]

Just because there are two doors left after the host opens one does not mean they're equally likely to have the car. Under the rules of the MH problem, they're not. Do you agree the playing card simulation is equivalent, i.e. shuffle two red twos (losers) and the ace of spades (the winner), deal one to the "player", and you (playing the part of the host) look at your two cards and turn up a red two? If so, try this, with playing cards, maybe 20 or 30 times and keep a record of how often you (the host) end up with the ace (in which case, switching would win). I suspect if you actually do this you'll discover not only that switching wins more often than not, but that you can take a shortcut which is that when you (the host) look at your hand you don't even have to turn up a two to decide if the player will win by switching. If the ace is one of the two cards (2/3 chance), switching wins. The same thing is true with the MH problem. After the player picks, if the car is behind either of the two unpicked doors then ... switching will win (and this is known before the host opens a door, based only on whether the player initially picked the car!). -- Rick Block (talk) 10:03, 2 September 2008 (UTC)[reply]

You seem to be right. Now when I get the idea, it leads me thinking that this problem would be easier to understand seen from the host's point of view. When the player doesn't switch his choice, the host can lose the car only if the player has picked the door with the car behind it in the first time. This is 1/3 chance so the host has 2/3 chance for keeping the car. By switcing his choice the player lets the host keep the car if it is behind the door he picked first (with 1/3 chance) and the player gets the 2/3 odds which the host had before switching. I think this pretty much complies with Chuck's (?) P(not switching) = 1/3 and thus we have P(switching) = 1 - P(not switching) = 2/3. I thank You very much for your time and comments. —Preceding unsigned comment added by 81.197.28.197 (talk) 22:09, 3 September 2008 (UTC)[reply]

JUST ONE MORE THING. To make it even clearer, you could present an analogyous case where there are 100 doors with 99 goats and one car behind them. First you let the player choose one door and then the host opens one door which he knows there is a goat behind it. Then the host would ask if the player wants to switch the door he picket first to all the 98 doors that are left. The most favourable choice must be obvious! The cunning thing in the original problem, is the number three, which reduces the choice to two doors in the second phase. It would though be interesting if someone could present a similar (simple and without too much mathematics) example in the three-prisoner case! —Preceding unsigned comment added by 80.186.57.76 (talk) 08:53, 4 September 2008 (UTC)[reply]

The way results are recorded in the playing card simulation makes it equivalent to a game in which the player chooses door #1, and then Monty says, “In a minute I’ll open one of the doors. Would you like to stick with #1 or switch to whichever of the two doors remains closed?”

A better simulation for the game, as stated, would be to keep track of four sets of numbers: wins by switching to #2, wins by switching to #3, wins by sticking with #1 when #2 is the remaining choice, and wins by sticking with #1 when #3 is the remaining choice. Out of all runs, the player will tend to win 1/3 of the time by switching to #2 and 1/6 of the time by sticking to #1 when #2 is the remaining choice.

Keeping track of these numbers might help the player understand not just that switching doubles ones chances, but why switching doubles one chances even when the puzzle uses a scenario that has only a 50% chance of occurring once door #1 is selected. Since the odds of winning by switching have been split into two, with one half being ignored, the odds of winning by sticking should be split as well: compare the 1/3 chance of winning by switching to #2 to the 1/6 chance of winning by sticking with #1 when #2 is the door that remains closed. Simple314 (talk) 02:10, 26 January 2009 (UTC)[reply]

Confusing text

In the Sources of Confusion section, the last paragraph is confusing.

Another source of confusion is that the usual wording of the problem statement asks about the conditional probability of winning given which door is opened by the host, as opposed to the overall or unconditional probability. These are mathematically different questions and can have different answers depending on how the host chooses which door to open if the player's initial choice is the car (Morgan et al., 1991; Gillman 1992). For example, if the host opens Door 3 whenever possible then the probability of winning by switching for players initially choosing Door 1 is 2/3 overall, but only 1/2 if the host opens Door 3.

The example of the difference between conditional and unconditional is not illuminating. Specifically, why the probability is 1/2 if the host opens Door 3. What probability are we referring to here? --dhawk (talk) 17:06, 31 August 2008 (UTC)[reply]

The probability of winning by switching is reduced in the above circumstances because the fact that the host has opened door 3 indicates a greater likelihood (1/2 rather than 1/3) that you have chosen the car. This is because, if you had chosen a goat there is the possibility that he might have had to open door 2 because the car was in 3. If, on the other hand, he opens door 2, you know it is your lucky day because the car must be in 3 as otherwise he would have had to open it. Martin Hogbin (talk) 17:55, 31 August 2008 (UTC)[reply]

Ahh. That's interesting enough that I think it should be expanded, perhaps to include the fact that you would be certain that switching would get you the car if the host opened door 2. Maybe it's just me, but that paragraph by itself doesn't make those details clear. dhawk (talk) 05:43, 1 September 2008 (UTC)[reply]

Yes, perhaps further explanation would be in order but I believe that the issue of different host strategies is a sideline that greatly and unnecessarily complicates the the main problem, which is that nearly everybody does not believe that you get a 2/3 chance of the car if you swap (conditions apply). Martin Hogbin (talk) 18:47, 1 September 2008 (UTC)[reply]

Suggested improvements

Although I agree that the solution given should not be replaced with the suggested one I do think that the proposer has a point in that the page does not do a very good job of addressing the fundamental issue that the chances of getting a car if you swap is 2/3 and not 1/2 (Morgan et al. conditions plus the constraint that the host picks randomly when he has a choice, and the contestant knows this).

I do understand that there are issues regarding verifiability in deviating from the current solution but I believe that we should make every effort to make the current solution more convincing. I am sure that there are ways in which issues of verifiability can be addressed if we try.

I think the issue of different host actions is a red herring that complicates and already exceedingly difficult problem and that this should be omitted from the main part of the solution and from 'Sources of confusion' - the last thing that section needs is another source of confusion.

Two starting suggestions are:

1 Describe the problem at the start of the solution in sufficient detail that only one answer exists. This would include describing the host's exact policy.

2 Describe the problem in a way that eliminates the issue of host startegy, for example by including the quote from Morgan et al,'You will be offered the choice of three doors, and after you choose the host will open a different door, revealing a goat. What is the probability that you win if your strategy is to switch?'. Martin Hogbin (talk) 18:59, 1 September 2008 (UTC)[reply]

The various comments on this page since I have been here indicate to me that the solution needs to be better explained. Martin Hogbin (talk) 22:09, 1 September 2008 (UTC)[reply]

Certainly no article is ever perfect. The existing "Problem" section provides both the well-known (ambiguous) Parade version and a fully qualified version that permits only one possible (numerical) answer. Given this, I'm not sure what your first suggestion is.

Your second suggestion (to use the Morgan et al. phrasing) would be odd, since they use this as a description of what they consider to be a different problem. Their issue is not so much "host strategy", but how to interpret the question. Nearly all phrasings of the Monty Hall problem put the player at the point the host has already opened a door and ask whether switching to the other door is better. This phrasing is consistent with Morgan et al.'s interpretation of the problem as a conditional probability question, which leads to a 2/3 chance of winning by switching only if the host is constrained to randomly (with equal probability) select from among two goat doors in the case the player initially happens to select the car. The unambiguous problem statement in the Problem section includes this constraint. Without this constraint what Morgan et al. call the "unconditional" question (corresponding to the "overall" chance of winning across all hypothetical players, i.e. if 1000 players play this game and all of them choose to switch, how many would we expect to win the car) still has the same 2/3 win by switching solution - but (as they show in their paper) permits other host behaviors where knowing which specific door the host opens yields a probability of winning by switching anywhere from 1/2 to 1. This is in some sense a technicality, but it's a significant enough technicality that it needs to be mentioned here. Nearly all people have a VERY hard time understanding that these are two different questions and that they require slightly different reasoning. We had an extended discussion about this some time ago (it started with the thread I referred you to - and continued for weeks if not months) and ultimately ended up with what's in the article now. The goal is to be pedantically (mathematically) correct but to avoid introducing the concepts of conditional vs. unconditional probability until after the (usual) unconditional solution is presented.

As an encyclopedia article, I think we're certainly in the ballpark of a reasonable approach. The goal is NOT to convince everyone that the 2/3 answer is correct but to provide the "normal" answer (in a way that most people can understand) and then to expand on this to cover the more technically accurate Morgan et al. approach (hopefully without losing too many people). That it's possible to have different "conditional" and "unconditional" answers for the same problem (with the same host constraints) is (I believe) extraordinarily unusual (I've tried, but have been unable to come up with other examples). This isn't exactly the crux of the Monty Hall problem, but I think putting the player in front of two closed doors and one open door is. The "normal" solutions certainly depend on an assumption that the host does nothing that changes the probability of the player's first pick, i.e. the door the host opens is a goat so the other door has a (1 - 1/3) chance of being the car or if you pick a goat first with a 2/3 chance and then switch you now have a 2/3 chance of winning the car. These are very common, but (per Morgan et al.) mathematically insufficient ways to "solve" the problem. I don't think the slightly more rigorous approach presented in the Solution section is tremendously more difficult to comprehend.

Again, no article is ever perfect. If you have specific suggestions for wording improvements please offer them up (or boldly make them - but only once without discussing them here if they're reverted). -- Rick Block (talk) 02:32, 2 September 2008 (UTC)[reply]

I think you are missing the point of what I am trying to do. I know much thought and hard work has gone into the page and those editors responsible would not be happy to see their work replaced but there has been a criticism of this page and a suggestion that the solution is replaced with another. I do not support this, but I do think that some of the criticisms are valid and should be addressed. As I originally said in response to the RFC, my basic suggestion is to add rather than replace.

Wikipedia should reliably inform, but it cannot do this if the reader does not believe what they are being told. The current attitude seems to be, 'We have done a good and rigorous job of explaining the paradox, that is all we can do. If readers do not believe the answer that is just too bad'. In this respect some of the articles on the related problems do a better job even though the articles are of poorer quality by Wikipedia standards.

What about a separate section called, 'Still not convinced? or even a separate article. We could then try different approaches to help people understand the paradox. We could state at the top that this section ignores some of the subtleties of the problem in favour of clarity. We could also consider my suggestion 2 above. It is not uncommon in mathematics to first answer a different question from the one originally posed and then to consider how the two questions differ. This approach might lead to a better discussion of the conditional/unconditional issue.

I would like to do some serious addition to this subject in cooperation with others, but in such a way that the integrity of the current article is not compromised. Perhaps a development article called 'Explanations of the Monty Hall problem', with consideration to a possible future merger - it might fall flat on its face but that way the current article is unharmed. Martin Hogbin (talk) 09:24, 3 September 2008 (UTC)[reply]

I feel WP:NOR should be pretty much sacrosanct; what we write in Wikipedia must be verifiable. What we could do, however, is to add a page in the Talk namespace with some more explanations, or even a Wikibooks article (are there any guidelines on referring to Wikibooks?)

Also, your suggestion of "Still not convinced?" immediately made me think of Talk:0.999.../Arguments. We could follow a similar approach here for those who dispute the theoretical result. Oliphaunt (talk) 17:33, 3 September 2008 (UTC)[reply]

We may be able to achieve some of what I am hoping for without OR. For example, my suggestion to start with a different question was based on the statement of a similar question by Morgan. Regarding WP:NOR there are reasons why we should also consider WP:IGNORE and WP:COMMON SENSE. Because of the concentration on NOR, the article does not do as well as it might in terms of being convincing. Also, in this particular case there is in fact a degree of OR in the selection of references that editors have chosen to cite. A perverse editor could rewrite the article giving the wrong answer but with it still being fully verifiable, citing letters and statements from many eminent organisations.

I do not think "Still not convinced?" should be just relegated to the talk pages although I agree that a dedicated talk page is a good idea. Can we not also try to improve the article's explanatory power by adding something? Martin Hogbin (talk) 20:16, 3 September 2008 (UTC)[reply]

Article deletion discussion

A newly-created article related to this topic – Introduction to the Monty Hall Problem – has been nominated for deletion. Interested individuals may participate in the discussion at Wikipedia:Articles for deletion/Introduction to the Monty Hall Problem. TenOfAllTrades(talk) 11:59, 4 September 2008 (UTC)[reply]

It's gone now. Wow, that was fast... —Preceding unsigned comment added by 69.36.227.135 (talk) 23:45, 4 September 2008 (UTC)[reply]

User space for development of a new section

I have set up a user page user:Martin Hogbin/Monty Hall problem (draft) for the purpose of developing clearer and more convincing solutions and explanations of the problem. All editors please feel free to add your pet solutions there for discussion and futher editing.

All editors (current, historical, and prospective) please note that I hope this page will eventually be edited to a standard where all, or parts of it, can be added here under a separate section.Martin Hogbin (talk) 10:21, 7 September 2008 (UTC)[reply]

Section Aids to understanding in the article exists for this purpose, and I think it covers the best of the notable common sense explanations well. If you can find a notable explanation that is easier to grasp then more power to you, but the suggestion that we ignore rules in favor of common sense is ironical, for as a veridical paradox the very point of the problem is that common sense is wrong. The fact that some people remain unconvinced by any explanation is a testament to its excellence as a veridical paradox: common sense rebels against it. ~ Ningauble (talk) 20:23, 9 September 2008 (UTC)[reply]

I do not understand what you mean by 'notable common sense explanations'. I do not believe that many people remain unconvinced by any explanation, it may be that there is one that will do it for them. Martin Hogbin (talk) 20:54, 10 September 2008 (UTC)[reply]

By "common sense explanations" I was referring to those which sacrifice rigor in favor of clarity and ability to convince. Perhaps "informal explanations" would be a better term, but I was straining to emphasize a point about common sense being at issue. By "notable" I mean to emphasize that, since there are so many erroneous treatments in circulation, even an ordinarily WP:RS needs some vetting akin to WP:GNG. Perhaps that is an overstatement, but the situation calls for careful scrutiny.

I didn't say many people remain unconvinced, but that some do. I know several who are normal, functioning adults. Don't underestimate the cognitive dissonance that must be overcome when common sense is challenged. ~ Ningauble (talk) 20:51, 11 September 2008 (UTC)[reply]

Not ambiguous

The article says "Some of the controversy was because the Parade version of the problem is technically ambiguous since it leaves certain aspects of the host's behavior unstated, for example whether the host must open a door and must make the offer to switch." I don't think this is true. I don't think that the original version is ambiguous. It doesn't say, if host behaves every time the same, but that is not important, because the question is about a single decision. That single decision has probabilities for every choice you take. The host could behave differently at a different point in time, and the resulting probabilities might change. But that is not the question.

You can use a computer to simulate this, in which case you would have to put the car behind a different door every time, let the host open a different door all the time, and then let you decide to stick with the door you chose first or not. Then you can use the results of the simulation as an estimation of then probability of the original setup. When you simluate that with a computer, you of course have know if the host always behaves the same. But that was not the question. In the original question, the host *does* open a door, behind it is *not* the car, and he *does* let you choose again. --195.227.10.66 (talk) 17:49, 10 September 2008 (UTC)[reply]

If one assumes Monty opened a door at random then the odds are different. His intention, or the rule he follows, does matter. Vos Savant acknowledged the ambiguity by admitting in a follow up article that the constraint was implicit in her solution but had not been explicit in her statement of the problem:[3] "... remembering that the original answer defines certain conditions, the most significant of which is that the host always opens a losing door on purpose. ... Anything else is a different question." ~ Ningauble (talk) 18:23, 10 September 2008 (UTC)[reply]

Of course it is "ambiguous", or, more accurately, "incorrect". Lacking knowledge of the rules of the host's behavior introduces an independent variable into the equation. I don't see how "technically incorrect" differs from "incorrect" in mathematics. As far as I can tell, the words "ambiguous" and "technically incorrect" are media spin, trying to hide the fact that the problem and solution given by vos Santos were wrong.Apollo (talk) 16:23, 11 September 2008 (UTC)[reply]

Further ambiguity if host behaviour not clear from start: If you are in a quix programme where you suspect the producers want to avoid having to pay a car for you, you may suspect the extra choice is given because you already picked the car. Or if you suspect they WANT you to win the car - to boost the popularity of the programme, say - the extra choice may be an indication you DIDN'T pick the car to begin with. So, without info on host behaviour, this is not a pure math problem; it's one about psychology, TV company economy, viewer behaviour, and other issues way outside math. What is the information content of the fact that you are given the second choice? Only if you knew in advance that you WOULD be given this second choice is the content nil, and the math of the standard solution to the problem correct.--Noe (talk) 16:36, 11 September 2008 (UTC)[reply]

"Ambiguous" doesn't mean a little bit wrong (as in a little bit pregnant). Rather, it conveys more information about the nature of the error than "incorrect." Her answer to the problem stated was incorrect because she failed to state unambiguously what she meant the problem to be. I find it amusing how she tried to spin it as a misinterpretation by the reader rather than a misstatement by herself, for ambiguity may always be construed against the author. ...but we digress. ~ Ningauble (talk) 18:39, 11 September 2008 (UTC)[reply]

Essentially none of the objections to her answer were based on grounds that the problem statement was ambiguous (there's a quote in one of the references about this - maybe we should add this). The problem statement IS in a technical sense ambiguous, but given the context of "math puzzles" the missing assumptions are both what most people intuitively assume and highly reasonable. -- Rick Block (talk) 19:15, 11 September 2008 (UTC)[reply]

I agree. Almost any statement of any problem leaves some questions unanswered. These issues were used as a means of avoiding embarrassment by people who got the answer wrong, based on reasonable assumptions as to what the question was.Martin Hogbin (talk) 20:06, 11 September 2008 (UTC)[reply]

I agree, most of the confusion does not arise from this ambiguity. Although the objection of Morgan et al. could (arguably) be dismissed in similar manner as "a different question" than that intended, it is an analysis too important to leave out. ~ Ningauble (talk) 21:28, 11 September 2008 (UTC)[reply]

The problem as stated is not mathematically ambiguous nor incorrect - because it describes a situation which could occur and asks a question which has an answer. Yes, the hosts behavior is not specified to the degree you might wish but as a math problem, you just have to solve it without knowing more. You have to deal with the possibility that the host opened the door at random, and it happened to have a goat behind it; you have to deal with the possibility that the host knows the contestant picked the car and is merely trying to convince him or her to switch; you have to deal with every behavior that is consistent with the information given.

For all the furor, once the hosts behavior is fully specified it is a straightforward math problem but it becomes much more interesting if you assume you're supposed to solve the problem exactly as it stands. (Almost certainly, that was not the intent of the person who asked the question originally, but that is another matter.) Yes, it does have an answer and the answer is that it is better to switch - how much better doesn't have a simple numerical answer but the form it takes shows that switching is superior. And that is all thats needed to answer the question as posed.

Millbast5 (talk) 10:42, 13 September 2008 (UTC)[reply]

I agree with you that it is possible to give a solution based on reasonable assumptions. However, I think that in your statement, 'once the hosts behavior is fully specified it is a straightforward math problem', you completely miss the point. There are interesting and relevant issues concerning the host's behaviour but what makes this problem special is that even when fully specified, nearly everyone gets it wrong, including excellent mathematicians and statisticians. Some of them have tried to use the issue of the host's behaviour to cover up the fact that they have made a mistake.Martin Hogbin (talk) 12:07, 13 September 2008 (UTC)[reply]

The problem may be tricky as a word problem but as a probability problem it is one that most students in an elementary course in probability would get right. Stated as a problem in terms of the classic balls and urn it is: an urn with three balls in it, known by everyone concerned to be two white balls and one red. One person reaches into the urn and withdraws a ball at random which he keeps concealed in his hand. A second person announces that she will withdraw a white ball then looks into the urn and lifts out a white ball, as promised. What is the probability that the ball left in the urn is white? What is the probability the ball left in the urn is red?

Stated this way it is transparent is it not? The ball left in the urn is white is if and only if the first person withdrew the red ball - which happens with probability 1/3. The ball left in the urn is red with probability 1 - 1/3, or 2/3.

I can't say why the statement in terms of doors, etc., hides the fact that the initial action determines everything that follows. But here its quite clear - the guy draws a red ball, the second person withdraws one of the two white balls, white ball left - automatic. The guy draws a white ball, the second person withdraws the other white ball, red ball left - automatic.

I stand by my claim the that as a math problem it is straightforward. (Actually, a piece of cake relative to some other ball and urn problems students routinely solve.)

Now to the interesting form of the problem - where nothing about the hosts behavior is known beyond what was explicitly stated in the given problem. I did not say that the problem could be solved given reasonable assumptions - I said it could be solved as it stood. It does require methods that are accepted as good probabilistic techniques - the use of Bayes Theorem for example - but no assumptions.

How does one deal with the possibility that the host might have opened a door at random and, by chance, a goat was revealed? That part is relatively easy. Opening a door at random leaves the switch and non-switch strategies equal, whether a goat or a car is revealed, hence if the other possibilities for the hosts behavior put the switch strategy ahead it is better overall.

I should amend my statement a bit - I should not have said the problem had no numeric solution for the amount of the difference between the switch and non-switch strategies. I should have said that I can determine that the switch strategy is better but the method I used does not quantify the difference. The way I dealt with the possibility of random door opening is what stands in the way of quantification - happily the problem doesn't require it.

The rest of the solution is fairly complicated but I will glad to post the details if someone wants to see them. The overall method is to parameterize all the possibilities for the hosts behavior with two continuous random variables, p and q, where p is the probability the host will open a door he knows has a goat behind it when the contestants initial choice is the winning one and q is the probability he'll do that when the contestants initial choice is a loser. To arrive at a distribution for these random variables, Bayes Theorem is used with an equal-likelihood apriori distribution and the data that in the one known trial, a door was opened revealing a goat. Using that distribution one averages over all possibilities to arrive at a number for the probability of getting the car using the switch strategy.Millbast5 (talk) 08:01, 14 September 2008 (UTC)[reply]

You should try asking your urn version of the problem to people that you know. You will be amazed how many get it wrong. There is nothing magic about the door-and-goat statement that makes it especially difficult - the difficulty is in understanding how the probability works. Of course, once you know the answer it is easy!

It makes no difference whether the host knows where the goat is, all that matters is that he does, in fact, reveal a goat. Let me restate this in terms of your urn problem: an urn with three balls in it, known by everyone concerned to be two white balls and one red. One person reaches into the urn and withdraws a ball at random which he keeps concealed in his hand. A second person withdraws a bell at random which proves to be white. What is the probability that the ball left in the urn is white? What is the probability the ball left in the urn is red? What is your answer to the above question?Martin Hogbin (talk) 09:18, 14 September 2008 (UTC)[reply]

It is of course possible to "solve" the problem without making any assumptions about the host's behavior as User:Millbast5 suggests by assigning probabilities to all possible variants. Morgan et al. do this with the host's preference for choosing between two goat doors while keeping the other constraints intact (host must open a goat door and must make the offer to switch). This analysis says that as long as the host must open a door to show a goat and must make the offer to switch, the probability of winning by switching varies from 1/2 to 1 (depending on how the host picks between two goats), i.e. switching is always at least as good as staying but unless we further constrain the host (for example, adding that the host picks between two goats randomly with equal probability) we can't know a numerical answer.

There's a paper by Chun presenting a formal way to describe the fully generalized problem, "On the Information Economics Approach to the Generalized Game Show Problem" published in The American Statistician in 1999 Vol 53(1). Without digging out the paper I forget whether Chun uses the description to net out the bottom line for the probability of winning by switching, but in a similar fashion to Morgan et al. one could clearly assign probabilities to the variants and end up with a formula involving these probabilities. The "Other host behaviors" section hints at this, listing some of the published variants and the resultant probabilities of winning, but doesn't "sum it up". Note that given the variants listed in this section, the probability of winning by switching is between 0 and 1, and an analysis that concludes you're better off switching is going to rely on how probabilities are assigned to the variants. If there's a paper that shows an assumption-less analysis (Chun's or some other paper), we could try to present it here, but without a source to work from I think we're definitely in WP:OR territory. Another question is whether such an analysis would be appropriate to include in this article, which is often criticized for being too long already. I'd suggest that it might be appropriate to mention such a analysis (if one has been published) but not attempt to include it in the article. -- Rick Block (talk) 16:42, 14 September 2008 (UTC)[reply]

Are you saying that certain host actions (subject to host must always show a goat and offer the switch) can reduce the odds of winning if you switch to less than 2/3?Martin Hogbin (talk) 19:40, 14 September 2008 (UTC)[reply]

Depending on what you mean by "the odds", yes. This is the whole point of the Morgan et al. paper. Must show a goat and must make the offer makes the overall (unconditional) probability, across all players, 2/3. But, if the host opens the rightmost door whenever possible (or has any other identifiable preference for one door or another) there are two subsets of players. Those who switch when the host opens the rightmost (absolutely preferred) door win 1/2 the time. Those who switch when the host opens the leftmost (absolutely unpreferred) door always win. The usual presentation of the problem ("say you've picked Door 1 and the host opens Door 3") implies we're asking about the probability given the knowledge of which door the host has opened, i.e. the conditional probability as opposed to the unconditional probability. If we don't constrain the host to pick between two goat doors with equal probability, the best we can say about the chances of winning by switching if we already know which door the host has opened is that it's somewhere between 1/2 and 1, with an "average" probability (across all players) of 2/3. Morgan et al. assign the host's preference for the door he opens a probability p - switching wins with probability 1/(1+p). -- Rick Block (talk) 00:16, 15 September 2008 (UTC)[reply]

Thanks for that explanation.Martin Hogbin (talk) 08:47, 16 September 2008 (UTC)[reply]

To answer the question Martin asked: the probability that the first person draws a white ball is 2/3 and nothing can change that, though information about whether he has in fact drawn the red or a white ball in specific instances can be obtained by drawing another ball at random from the urn. When the second random ball is red, which happens with probability 1/3, the conditional probability is 1 that the first ball drawn was white; when the second random ball is white, which happens with probability 2/3, we let x be the conditional probability the first ball drawn was white. Combining the two cases we get 1/3 * 1 + 2/3 * x as the overall probability of the first ball drawn being white. Since we know the overall probability is 2/3, we can solve for x: 1/3 * 1 + 2/3 * x = 2/3 <=> x = 1/2. Happily, that agrees with our intuition.

Yes, after some thought, I agree with you. There is a difference between does draw a white ball and must draw a white ball. Martin Hogbin (talk) 08:47, 16 September 2008 (UTC)[reply]

We can get the conditional probability for the case when a white ball is intentionally extracted from the urn (my statement of the problem) as follows: when the second drawn ball is red, which happens with probability 0, the first drawn ball was white with conditional probability 1; when the second drawn ball is white, which happens with probability 1, let the conditional probability be x as before. Combining the cases: 0 * 1 + 1 * x = 2/3 <=> x = 2/3.

Did you really say "It makes no difference... he does, in fact, reveal a goat."? I keep reading it and it keeps coming up the same. You must have seen the argument above in some form and rationalized that away. How about changing the rule for the second person to be that she withdraws the red ball if its in the urn and only when there are two white balls in it does she withdraw a white. Thus whenever she withdraws a white the first person must have drawn the red ball, so the conditional probability of the first person having drawn white is 0. Hopefully this is sufficiently clear that you can see, regardless of what your intuition tells you, that the method that leads to the event of a white being drawn by the second person does make a difference in the conditional probability of the first person having drawn a white ball.

Yes you are right.Martin Hogbin (talk) 08:47, 16 September 2008 (UTC)[reply]

I think we might have stumbled onto something here explaining the population's difficulty with this problem - they somehow twist it around in their minds until it depends on grasping how two events that are exactly alike in appearance can result in different effects on a conditional probability. It seems so obvious to them that they cannot have different effects that simple, straightforward derivations of the solution are rejected, even though they can't point to a flaw.

Could be.Martin Hogbin (talk) 08:47, 16 September 2008 (UTC)[reply]

Yes, I think you are right. My mistake was to assume, without thinking, that a white ball being removed by chance has exactly the same effect as its being removed intentionally. I therefore think it is important in our explanation to cover this point in some way. In the current solution the probability of losing by switching, if the player initially chooses a car, is split into two lots of 1/6, with no explanation of why this is so and in what circumstances it would not be.Martin Hogbin (talk) 10:54, 17 September 2008 (UTC)[reply]

I don't believe we should go into this level of detail in giving a solution for the general reader - you have to be into 'philosophy' to even worry about the information aspect of various actions. I reworded my description of the game (in terms of urns and balls), into doors, goats and cars. It came out just as clear to me - clearer in fact because I avoid probability altogether. I read in one of the references that people understand frequency better than odds, so I stated it in frequency terms. Think the guy was right. I inserted it into the solutions sections above. Seven lines, no pictures, no probability and only two symbols - 'A' and 'B'. Almost anyone can read it over several times in five minute and get the whole thing in his head at one time.Millbast5 (talk) 23:49, 17 September 2008 (UTC)[reply]

I suppose it might seem a bit childish, but one can also settle this by playing the game numerous times, applying different methods for the extraction of the second white ball and keeping track of the results for the different ones.Millbast5 (talk) 01:40, 15 September 2008 (UTC)[reply]

Oh rats, I didn't answer exactly the question Martin asked - he asked about the ball in the urn and I answered as though he asked about the ball in the first persons hand. It should make no difference because I did state what I was calculating. —Preceding unsigned comment added by Millbast5 (talk • contribs) 02:05, 15 September 2008 (UTC)[reply]

I assume the statement of the problem discussed in this thread is the one from Parade, as quoted in the lead:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

As I stated earlier in this thread, it is ambiguous because we do not know whether there is any information content in the fact that the host is opening another door - did he choose to do so BECAUSE the guest either hit or didn't hit the car with his first choice? If the cost of the prize won by the guest is deducted from the host's salary, say, he may simply be trying to direct the guest away from the car, so THE FACT that he was given a second choice is an indication that he DID in fact choose the car from the start, i.e. DON'T SWITCH! Or, if the advertising income for the quiz show is boosted whenever a guest wins a car, the producers may want to direct the guest to the car, so that the fact he was given a second choice is an indication he did NOT choose the car to begin with, i.e. DO SWITCH! We are not given enough information to decide whether one of these scenarios may be taking place, so in order to attack the problem as a math problem, we need to make certain assumptions to begin with, e.g. that the host ALWAYS gives the guest a second choice after opening a door with a goat behind it. This assumption may be a natural one to make, and the fact that many people defend "wrong" solutions to the problem is NOT simply caused by this ambiguity, but still, it IS ambiguous.--Noe (talk) 07:41, 15 September 2008 (UTC)[reply]

The point user:Millbast5 is making is that the problem is not ambiguous (meaning it has two different, conflicting interpretations) but rather it simply has more uncertainties than we'd like. We can solve the problem even not knowing any of these things by assigning them each a probability and developing an equation for the overall probability of winning based on these probabilities. For example, we could start with three probabilities p1, p2, and p3 representing the probabilities that the host made the offer to switch because the player initially picked the car, because the player didn't initially pick the car, and because it's simply the rules of the game (respectively). Clearly,

p1+p2+p3 = 1

and a player who switches always loses in the p1 case, always wins in the p2 case, and may or may not win in the p3 case. So, we have (so far)

p(winning by switching) = p2 + x * p3

where x is the conditional probability of winning given that we're in the p3 case (i.e. neither the p1 or p2 cases). This sort of analysis can continue and we'll end up with the probability of winning being a function of 5 to 10 probabilities expressing the likelihood of a set of various scenarios. The range of this function will be [0,1]. If we assume each of the mutually exclusive probabilities are equally likely (for example, p1, p2, p3 above are all 1/3) we can compute a number - but this number would depend on how we break down the possibilities. Instead of the 3 possibilities for "host intent" above, we could say there are two possibilities - host "plays fair" (p3) or not (p1 or p2). Breaking it down this way and assuming equal likelihood we'll end up with p1=p2=1/4, and p3=1/2 (so our final numerical answer will be different).

The point is a conditional probability analysis is sufficient to analyze a problem with as many uncertainties as you'd like, however I'm not sure anyone other than a mathematician would think such a solution would be satisfying. -- Rick Block (talk) 14:11, 15 September 2008 (UTC)[reply]

Thanks for the assist Rick, but I meant something simpler - I meant the question of whether the switch or not-switch strategy is better does have a definite answer. (The answer is that the switch strategy is better.) Assigning probabilities in the way you're describing comes under the heading of making assumptions. Statisticians, mathematicians and others have developed methods to deal with this type of shortage of information which have been borne out as accurate and those methods are what I used - I would expect another mathematician solving this problem to do the same.

As Noe mentions one might be in a situation where the non-switch strategy is better even though the switch strategy is better over the full gamut of possibilities. (In game theory, a strategy which is better in every case than another strategy is said to dominate the other one - which isn't true here.) Of course, it takes information that is not given in this problem to know you are in a situation where you should use the non-switch strategy - but if you do have some other information, you use it. Or if you have a hunch about that evil looking host that you just can't ignore. If you got a peek behind one of the doors and saw a goat, use that too. (If you're using the switch strategy, of course your original choice will be that very door so you're certain to win if you get a chance to switch.)

Can I get this straight. There are two types of issue that can affect the conditional probability. The first is really down to the rules of the game (rather than host behaviour but there is no clear distinction) and is that the host must open a door to reveal a goat, in other words by the rules of the game the host can never open a door to reveal the car.Martin Hogbin (talk) 10:54, 17 September 2008 (UTC)[reply]

The second type of issue is the way in which the host chooses which goat to reveal when he has a choice.Martin Hogbin (talk) 10:54, 17 September 2008 (UTC)[reply]

I agree with Martin that this latter type of issue is far different than the rules of the game.

I'm of the opinion that bringing up the possibility of the host having a known preference for which door he opens to reveal a goat is deleterious to the article. Of course, its possible but nothing in the problem statement hinted at the possibility that contestants watched the show long enough to spot his preference - the original question applies as well to the very first time the game show was held as any other. It is not a question that can even come up if the doors are not identified and the doors were identified as for expository reasons - making things more concrete. The problem can be stated without such identification and statements of equivalent problems don't have anything corresponding to doors.

Taking an accidental feature of the problem statement plus adding an assumption that the process has been going on long enough for a preference to become known is basicly a different problem - at best it should be in a footnote. You could bring up the possiblity that host used his left hand to open the door more often than his right in some circumstances or blinked his eyes at a differing rates - distracting trivia is the category this sort of thing falls into.Millbast5 (talk) 06:33, 18 September 2008 (UTC)[reply]

I think I will clean up my computation of the probability of winning with the switch strategy and put it on my talk page so that anyone that wants can see exactly what I'm talking about. Everyone thats agreed with me up until now has misinterpreted what I was saying. Should be there by tomorrow night at the latest.

Maybe I've made a little progress because Noe did not mention the possibility of the host randomly opening doors as something making the problem ambiguous.

Millbast5 (talk) 01:44, 16 September 2008 (UTC)[reply]

My talk page contains the computation for the switch strategy.Millbast5 (talk) 09:17, 16 September 2008 (UTC)[reply]

@

Lead change

The two sentences added to the lead (see this diff) assume the host opening a door has no effect on the probability of initially selecting the car (which is true only because of the constraints on the host), and essentially explain the fairly trivial point that 1 - 1/3 = 2/3. How about if we state the major reason that the initially picked door remains a 1/3 chance? For example:

Because there is no way for the player to know which of the two unopened doors is the winning door, most people assume that each door has an equal probability and conclude that switching does not matter. In the usual interpretation of the problem the door the host opens is not randomly selected and opening this door does not affect the player's initial 1/3 chance of having picked the car. The car is behind one of the two unopened doors; if there is a 1/3 chance it is behind one there must a 2/3 chance it is behind the other. The player should switch—doing so doubles the probability of winning the car from 1/3 to 2/3.

The critical point is that the host opening the door does not affect the player's initial 1/3 chance of picking the car - mostly because the host does not open a random door. I think assuming this to be true, without no explanation whatsoever, does not accurately summarize the solution. -- Rick Block (talk) 02:27, 13 September 2008 (UTC)[reply]

I prefer the version previous to the diff. It gives the lead section more punch, and gives the reader pause to think. To elaborate: (1) Stating the naive perspective and the correct result, with emphasis on the magnitude of the error, is an excellent way to explain why the subject is interesting. (2) While teasers in the lead are generally discouraged, it is appropriate and arguably necessary to make an exception for articles about puzzles or paradoxes. Explanation of the solution, even in summary, ought to be deferred to the body of the article in order to accommodate readers who are inclined to try reasoning it out for themselves. ~ Ningauble (talk) 14:14, 13 September 2008 (UTC)[reply]

Another objection - moved from main page

I moved this from the main article to here. It does not really belong here but it has been suggested that we set up a space for discussion of the subject so that it is kept separate from the article itself and discussion on how to improve it. I am not sure how to do that but I think it is a good idea. Can any one help?Martin Hogbin (talk) 08:50, 13 September 2008 (UTC)[reply]

I found this answer to this question on another website and am in no way taking credit for this answer...

"The solution given is that the selection should be switched because the door you chose has a 1/3 chance of success, while the other unopened door now has a 2/3 chance of success.

This solution is surprising, counter-intuitive... and wrong.

Marilyn vos Savant was wrong. The solution given by Mark Evanier was wrong. And the solution given by these fictional math geniuses of "21" was spectacularly wrong.

What am I talking about?! How can all these real and fictional geniuses be wrong?

Switching doors can double your odds of winning, but only if Monty Hall not only knows which door has the car, but is REQUIRED under the rules of the game to open a door that (i) is not the door you chose, and (ii) is not the door that has the car behind it. This must be stated as part of the puzzle in order for the correct solution to be that there is an advantage to switching doors.

This is not merely a semantic distinction -- it is critical to the problem. If Monty Hall is operating under those rigid rules, by opening a goat door after you've chosen, he is giving you valuable information. If you guessed correctly (a 1/3 chance), he can open either of the other two doors. But if you guessed incorrectly (a 2/3 chance), he is FORCED to open up the one door that is neither your choice nor the door with the car.

Now, the puzzle, as presented by Kevin Spacey, does not state that Monty Hall is required to open up any door. He simply says that after you choose your door, Monty "decides" to open up another one that reveals a goat.

If the rules of the game are that he opens a randomly selected door after you make your choice, then the logic of switching doors break down -- there is no advantage to switching whatsoever.

If he simply opens up a door without regard to any pre-determined rules of the game, there is also no advantage to switching. You don't know why he opened a door, so you haven't gained any information.

And in fact, in the clip from the movie "21," the solution given is spectacularly wrong because Kevin Spacey's genius character explicitly posits that you don't know why Monty Hall is opening up a door -- he says that the game show host may be trying to "playing a trick on you, trying to use reverse psychology to get you to pick a goat." Despite what these fictional math geniuses say, there surely is no advantage to switching in this case, because, again, you've gained no useful information. It may be to your disadvantage to switch under these circumstances.

Let's say you play this game 100 times, and Monty reveals a goat door only when you've initially picked the car door (and does nothing when you initially pick the goat door -- just opens your door and tells you that you lost). 67 times you'll pick a goat door and automatically lose. If you adopt a switching strategy, the 33 times you pick a car door, you'll then be shown a goat door by Monty, get "tricked" into switching, and lose. The switching strategy will yield you exactly zero wins out of 100. (Refusing to switch will give you 33 wins.)

The same holds true if you play only once. You pick a door, and then Monty reveals a goat door. Unless his actions are explicitly part of the rules of the game, you have no idea why he showed you the goat door. He may have only showed it to you because your first pick was the car door, and he is in fact trying to trick you into switching. You're now in a mind game with Monty Hall that has nothing to do with the "statistics" and "variable change," that Kevin Spacey's super-student refers to.

This distinction is part of the reason the commonly-given solution is so counter-intuitive. The way it's presented, you're playing this game, and Monty Hall suddenly and without explanation reveals a goat door and gives you a chance to switch -- it doesn't seem as though it would be to your advantage to do so, and in fact it isn't. But once you say that Monty Hall is required under the rules of the game to reveal a goat door and give you a chance to switch, it makes a bit more intuitive sense that there may be an advantage in switching (and in fact there would be)."

source -->(http://gocomics.typepad.com/tomthedancingbugblog/)

The objection above results from the way that probability problems are often stated.

In the Monty Hall problem, if Monty picks a door at random, there is the possibility that he will pick the car. But, he is always seen to pick a goat (and this is said in the statement of the problem). Now it could be that Monty picked randomly but the TV company only showed the times when he picked a goat, however, it is far more likely that he knew where the goat was. In both cases the results are the same, of the screened shows the player who swapped won more often than the one who did not. The problem states that Monty reveals a goat thus we are expected to consider only the cases where this is so.Martin Hogbin (talk) 10:08, 13 September 2008 (UTC)[reply]

This objection is already addressed in the article (2nd to last paragraph in the lead, in the Problem section, in the Sources of Confusion section, and in the Other host behaviors section). The topic is already covered (with references), and this material is copyrighted (so can't be used directly, and isn't what I'd call a reliable source anyway). Is there a suggestion here for some change to the article? -- Rick Block (talk) 20:07, 13 September 2008 (UTC)[reply]

Not by me. I moved the text above from the article (where it hand been added by an IP editor) to here.Martin Hogbin (talk) 20:40, 13 September 2008 (UTC)[reply]

Language clean up

In the introdudtion, it currently states:

" Switching is only not advantageous if the player initially chooses the winning door, which happens with probability 1/3. With probability 2/3, the player initially chooses one of two losing doors; when the other losing door is revealed, switching yields the winning door with certainty."

Two things need changing in this. The first is the reduction of words. Not advantageous to disadvantageous to begin with. The second thing is that the sentence is misleading imho. The point of the Monty Hall problem is that it is a counter-intuitive probability based problem. The point is not at what point is it disadvantageous but that probability shows that in the long run it is advantageous to switch. The addition of the statement I quoted above seems to distract from the point the article is trying to make and should not be in the introduction. --Candy (talk) 18:57, 22 September 2008 (UTC)[reply]

Agreed. Rick had suggested a clearer explanation two sections previous in this discussion (Lead change) but, for reasons given there, I favor simply removing these two sentences from the lead section. Does anyone object to this proposed revert? ~ Ningauble (talk) 19:03, 24 September 2008 (UTC)[reply]

Agreed. Anything that makes things clearer.Martin Hogbin (talk) 17:49, 25 September 2008 (UTC)[reply]

I agree with the suggestion of using the word 'disadvantageous' but not with removing the two sentences - to me they are the best explanation in the whole article of why the switch strategy wins with probability 2/3 - I believe that compactness is a big plus and that pictures are a distraction. I would favor emphasizing the idea at this stage that the host is intentionally opening a door with a goat behind it. So something like 'when the host opens the other door he knows conceals a goat'. Grammatically it is incorrect as it is because it isn't the 'losing door' that is revealed but the goat behind it.Millbast5 (talk) 00:45, 27 September 2008 (UTC)[reply]

Random host variant

Regarding the variants on host behaviors, I don't find the example of:

"The host does not know what lies behind the doors, and opens one at random without revealing the car (Granberg and Brown, 1995:712)."

to be very useful. This conditional ignores the outcome that occurs 1/2 the time when the car is behind door 2 or 3 and the host must choose a random door to open - he accidentally chooses the car. Since this example of host behavior ignores these potential outcomes of the game and the idea that host chooses randomly is covered well in the previous "aids to understanding" section, I would advocate deleting this variant from the table. However if it is to be included, shouldn't the odds of getting the car by switching still be 2/3? Of course the overall odds of the game if the host does not know the location of the prize has been shown to be 1/2, with the car being prematurely revealed in 2 of the 6 overall outcomes and the win/lose status of the game to be unknown. But in this specific case, it is restricted to the condition where the host "opens one at random without revealing the car." Isn't it correct that once you are to this point in the game, the probability is exactly the same as if the host had known the location of the car and deliberately exposed a goat? This variant eliminates the option of the host exposing the car, so he must act as if he has knowledge of the car's location even though it is stated that he does not. In other words, once you have gotten to this point, you have one of three scenarios (sticking with the idea the player has chosen door number1)

car behind door1 and host has opened door 2 or 3
car behind door2 and host has opened door 3
car behind door3 and host has opened door 2

Each of these scenarios has a 1/3 chance, in 1 you loose by switching and in 2 or 3 you win by switching thus the odds are the same as with a knowledgeable host switching gives a 2/3 chance of winning. Clarence3456 (talk) 20:03, 25 September 2008 (UTC)[reply]

This is a notable variant (many references discuss the difference between this variant and the standard one), so I think it should remain in the table. The odds are 1/2 because there are 6 equally likely events:

car behind door1 and host has opened door 2
car behind door1 and host has opened door 3
car behind door2 and host has opened door 1
car behind door2 and host has opened door 3
car behind door3 and host has opened door 1
car behind door3 and host has opened door 2

but because we're given that the host happened to open a door that didn't show the car we know we must be in one of (still equally likely) cases 1,2,4, or 6. We don't know what would happen in cases 3 or 5 so we simply ignore them, but that doesn't increase the probabilities of cases 4 and 6. If we're given the host didn't show the car, we're only looking at 4/6 of the total cases and among those there's a 50/50 chance of winning by switching. -- Rick Block (talk) 13:39, 26 September 2008 (UTC)[reply]

I'm not disagreeing with your explanation Rick, but I think the best method of showing the probability is 1/2 is to compute the number from an indisputable probability - that being: the probability is 0 that the winning door was chosen if the car is revealed. Randomly opening a door entails two results - revealing a goat and revealing a car. Neither of these can possibly alter the fact that the winning door was chosen with overall probability of 1/3 but each result does alter the probability in a particular case. Since the overall probability cannot change these different results must cancel each other out. One third of the time the randomly opened door reveals a car and when that happens the probability that the chosen door conceals the car is 0. Two thirds of the time the randomly opened door reveals a goat and when that happens the probability the chosen door conceals the car is x. 1/3 * 0 + 2/3 * x = 1/3 is the equation that describes the state of affairs. Solving for x one gets 1/2.

It is generally treacherous to consider one outcome of a random event in isolation because its easy to lose sight of the importance of other outcomes. Even if you avoid the pitfalls you can often use the others to verify your computations.Millbast5 (talk) 01:41, 27 September 2008 (UTC)[reply]

You can use this same argument to show that the host intentionally opening a door he knows conceals a goat cannot affect the probability in a particular case. As before let x be that probability. The probability is 1 that he will reveal a goat so the equation is: 1 * x = 1/3, i.e. x = 1/3. Or you could say that since there is no alternate outcome to cancel whatever change you might hypothesize, the change must be zero.Millbast5 (talk) 01:58, 27 September 2008 (UTC)[reply]

Milsbast5 I see you and Rick are correct. An alternate to Milsbast's equation that helps me understand is that the total probability must of course be 1, and as you state, if the host reveals a goat, the chance that door contains the car is 0. The chance the each other two contained the car was equal at the start of the game and no actions have changed that, so call the odds that either door contains the car x. So overall: 0 + 2*x = 1. So x must be 1/2. If you guys want, you can delete this section of the discussion to keep it clean or leave for others.Clarence3456 (talk) 19:02, 27 September 2008 (UTC)[reply]

Clarence your argument is simpler and simplicity is always good. The reason for my more circuitous argument is that it ties together two things which should be tied together - the different results that come from one random event. It also explains why opening a door at random and revealing a goat by chance is not the same thing as opening a door known to conceal a goat, even though to the casual observer they are identical. —Preceding unsigned comment added by Millbast5 (talk • contribs) 01:14, 28 September 2008 (UTC)[reply]

I keep trying to think of new ways to explain this problem so that even someone who firmly believes that the two strategies are equally good can finally get it. Here is another go at it.

Imagine there are two players involved, named N and S - initially they have to guess the same door but after the host has opened a door, S has to switch to the remaining door. Also imagine, to begin with, that the host opens a door at random. That is extremely easy to analyze - each of the three participants has a 1/3 chance of having or opening the door that conceals the car. You will note that in the cases where the host opens a door concealing a goat, one of the other two wins and they win equally often, of course, since they each get the winning door 1/3 of the time.

Those who believe that it matters not a whit whether the host opens a door he knows hides a goat or he opens a door at random and reveals a goat by chance, can point out that none of the trials where the host did the latter would have come out any different if he picked the same door on purpose. Couldn't possibly get a different result in those cases - N has same door, Host opens same door so S is forced to take the same door. Case closed - it doesn't matter.

Now we consider the cases where the host opens a door at random to reveal a car, which happens 1/3 of the time. Can't forget those cases if the host is going around opening doors at random. So N has a goat, host shows car, S stuck with the other goat. Now pretend nothing changes except that the host is not allowed to show the car. He's also not allowed to take N/S's initial choice so he has to take the door S was eventually forced to take in the random play - there is no other door to consider. Well, now S is forced to take the door that hides the car. Clearly not fair - S gets every car the host showed in the random play in addition to the ones he got when the host randomly revealed a goat. N gets exactly what he got before.

So if one focuses on the cases where the host opens a door randomly and reveals a goat then the result is the same if he opened the same door because it hid a goat. On the other hand, if you consider all the cases where the host could open a door he knew concealed a goat - which is every case - then there is huge difference. S wins twice as many cars as N does.

What should you do if you don't know why the host opened the door to show a goat (as you don't know in the original problem statement), you still jump at the opportunity to switch because you never gain by not switching and you might gain big time by switching. Heck, even if you still believe that switching and not-switching always end up of the same you ought to switch because of the small chance you are wrong and the, so-called, experts are right - stranger things have happened. Of course, if you have formed a strong attachment to the door you chose at the outset or if you have a lurking suspicion the host knows you picked the right door and is trying to trick you into switching, then defy the odds and stick with it. Imagine the mental anguish if you chose the winning door originally and switched - the experts haven't factored that into their analysis!Millbast5 (talk) 00:47, 28 September 2008 (UTC)[reply]

Correction! While its true that you never gain on your 1/3 chance of winning by not switching, its possible to fall below winning 1/3 of the time by switching - whenever the host is allowed the option of not opening a door at all. He may, for example, open a door only when your first guess was correct. The general rule governing it is: if the host opens a door known to hide a goat P percent of the time when he knows you've picked the winning door and does the same less than P/2 percent of the time when he knows you've picked a losing door, then the switch strategy wins less than 1/3 of the time.Millbast5 (talk) 05:39, 28 September 2008 (UTC)[reply]

I agree that a more convincing explanation is required. Why not add your suggestion to this page user:Martin Hogbin/Monty Hall problem (draft), where it can be discussed and improved with a view to adding it here when it is ready and acceptable.Martin Hogbin (talk) 09:13, 28 September 2008 (UTC)[reply]

I see that you have.Martin Hogbin (talk) 11:05, 28 September 2008 (UTC)[reply]

Reversion of good faith edits marked as minor

I have seen at least two occasions where good faith edits to this article have been immediately reverted and the reversion marked as a minor edit. This is extremely rude and contrary to WP policy. I understand that editors wish to maintain the quality of this article but this is not the way to do it. Martin Hogbin (talk) 08:34, 3 October 2008 (UTC)[reply]

I suppose you refer to edits like this and this (perhaps also this). You're absolutely right. Except, it's maybe a little bit rude to call others extremely rude because they made a slight mistake. (And now I nearly did the same!)--Noe (talk) 11:10, 3 October 2008 (UTC)[reply]

I didn't deliberately mark my revert as minor -- looks like Twinkle did that for me, and it appears Huggle does the same. I'll change my settings. I explained my reasons for the revert in the summary line -- the edit was pure OR, something this particular article suffers from a lot. Jonobennett (talk) 12:50, 3 October 2008 (UTC)[reply]

Generally, reverting an edit occurs because of vandalism, so all software features dealing with this automatically mark it as a minor edit. This is true of the admin rollback button too. Now, whether these software features should be used to revert good faith edits is a different issue. Mindmatrix 13:38, 3 October 2008 (UTC)[reply]

OK, I accept the points made and maybe overstated my case a little, but I first came to this page in response to an RFC which claimed that the current editors were being overly protective of the page. Immediate reversion of somebody's work, with that reversion being marked as minor, is not welcoming to new potential contributors. As I have said before, I believe that the very strict enforcement of no OR might be relaxed a little to the benefit of the article. Martin Hogbin (talk) 16:45, 3 October 2008 (UTC)[reply]

Problems with the current Souces of Confusion section

The first two paragraphs give a reasonable overview of what other people have said about the sources of confusion for this and similar problems.

The first sentence in paragraph three is: "A competing deeply rooted intuition at work in the Monty Hall problem is the belief that exposing information that is already known does not affect probabilities (Falk 1992:207)." This gives the impression that exposing information that is already known does affect probabilities, while the truth is that our deeply rooted intuition, in this case, is right on target. The Falk paper gives the same (false)impression but whenever the idea was used the problem turned out to be a mistaken belief that no new information was or would be exposed. In most cases the author pointed out the new information himself but in no case did exposing only known information change a probability.

In this problem the question comes up relative to host intentionally opening a door he knows conceals a goat. The old information was that there was a door, unpicked by the contestant, which hid a goat, and the new information is the location of one of the goats - which has no effect on the probability of the contestants initial choice being right but a great effect on the probability of winning by switching. In the Falk paper referenced (2nd reference) the author does point out that in the variant games, alluded to in paragraph 3, the mistake was in failing to recognize new information was imparted, not that the deeply rooted intuition was wrong. The current section does not mention that and thus solidifies the false impression given before.

Try to imagine the computation of a probability that you would agree would be different if information you knew hadn't been given to you a second time. Yeah, it fairly boggles the mind that anyone would think such a thing is possible. (Just imagine how repeated exposures to same information might give a whole sequence of different answers - it would never be safe to compute a probability!)[This paragraph added in edit]Millbast5 (talk) 02:04, 9 October 2008 (UTC)[reply]

The fourth paragraph is all about variant games - those in which the host has a preference for which door to open when there are goats behind both doors the contestant didn't pick. In the Falk paper it is made clear that for that preference to have any effect it must be known to the puzzle solvers - which is not mentioned in the current section. Since no such preference is mentioned in the puzzle, it has no effect whatsoever - that is to say, the whole subject is irrelevant. Clearly no confusion about the puzzle this article is concerned with is going to be eliminated by the fourth paragraph but, since it fails to clearly delineate its subject matter, more confusion can be added.

By the way, if the information were added it would not change the odds of winning by switching, it would only increase the odds in some cases and decrease them an exactly compensating amount in other cases. That is just how irrelevant the whole subject is to the current problem.68.2.55.158 (talk) 01:35, 9 October 2008 (UTC)[reply]

I'm not sure why but the preceding edit identifies me by number. In case you're wondering, my name in here is Millbast5.68.2.55.158 (talk) 01:41, 9 October 2008 (UTC) Okay, now I see... it didn't automatically log me in like it did other timesMillbast5 (talk) 01:44, 9 October 2008 (UTC)[reply]

I think the point is that if we view the action of opening a door as providing no additional information pertaining to the initial pick (i.e. we already know one of the two unpicked doors is a goat so knowing which does not change the probability of our initial choice being the car) then we're being misled, at least in some circumstances, just as much as we're misled by the "n doors means the probability must be 1/n" intuition. I forget the examples Falk uses for this, but this "no new information" intuition certainly can lead to incorrect conclusions - for example in the (Morgan et al.) case where the host has a preference for one door over another. The main point she (Falk) is making is that people have intuitions about probability, but these intuitions often interfere with the actual solutions and, conversely, solutions appealing to these intuitions are generally inadequate.

The fourth paragraph is not about variant games, but about the nature of the question. The variants are introduced only to show the difference between the two potential questions that might be what is being asked. Many solutions address the unconditional question (ignoring which specific door the host opens, as if the question is raised before the initial pick and before the host opens a door) even though the general statement of the problem clearly (to some) is a conditional question (specifically, with knowledge of which door the host has opened). This particular "confusion" is the main topic of the Morgan et al. and Gillman papers.

If you can clarify these points, please do so. -- Rick Block (talk) 02:58, 9 October 2008 (UTC)[reply]

I am more than happy to clarify matters.

Here is an example taken from Falk's article - 'second' refers to our "no-news, no-change" belief, as she calls it. It makes my point very clearly that the error is always in evaluating 'no-news' and never in the intuition/belief itself.

I agree, it is not always obvious what information is gained by a particular action and it is is easy to think that it is none when that is not, in fact, the case.Martin Hogbin (talk) 20:22, 9 October 2008 (UTC)[reply]

It is often the case, as in the problem below, that no new information is given about one thing but there is about something else. Someone focussed on the first thing might well jump to the conclusion there is no new information at all. I'm sure that happens with MH, too - host opening a door to disclose a goat reveals no new information regarding the contestants chosen door hence it contains no new information.

Now we just have to bolster your, and others', faith in your intuition that no new information invariably means no change in probabilities!Millbast5 (talk) 22:07, 9 October 2008 (UTC)[reply]

>>The second, mentioned in passing, is embedded in Dick’s attempt to induce the warden to tell him the name of one of the other prisoners who will be executed. Dick argues that he already knows that at least one of Tom or Harry is to be executed. Therefore, he maintains, he will receive no information about his own fate by getting one of these two names. If nothing new has been learned, no change in the probabilities should ensue. This is the “no-news, no-change” (or “I’ve known it all along”) belief. The uniformity belief is seldom questioned. It appears to be second nature for many naive, as well as statistically educated solvers. As for the “no-news” argument, it has convinced not only the warden, but most of the readers of the three-prisoner problem as well. It leads to a probability of pardon of l/3 for Dick (the correct answer for this particular problem)...<<

Dick tells the warden he will receive no information about his own fate and he was quite right about that, whether by accident or design. However it is quite clear that the warden did communicate some information and that information can be used to change the calculation of both of the others chances for a pardon - Harry's cut from 1/3 to 0 and Tom's raised from 1/3 to 2/3.

My point is that while we do make mistakes about whether new information has been communicated that in no way indicates that our belief that "no-news, no-change" is not precisely correct. When Falk states that the warden and problem readers were convinced by the "no-news" argument, isn't it clear that they erred in believing that no news was communicated?

When Falk says, "If nothing new has been learned, no change in the probabilities should ensue." she is entirely correct. Since a change in probabilities did ensue that meant something new had been learned and if one goes through the calculation of the new probabilities you can see exactly what that new information is - the identity of one the prisoners to be executed. (Not that it wasn't obvious beforehand.)

There are subtlties in evaluating information but our intuition that "no new information means no change in probabilities" is as solid as a rock.

While it might seem that Falk is indicting our intuition in this regard, on page 207 she correctly points out that the problem lies with the mistaken assumption that nothing new has been learned - she is discussing a different puzzle at that point but the same principle is at work. Without that I might have been concerned that she was as confused as the attributions used in the current "Sources of Confusion" make her seem.

Rick, I will promise you that if you bring forth any purported example where no new information changed probabilities, I will point out the new information. Heck, if you go through the two probability calculations yourself, its hard to believe you won't spot it right off.

Yes, we do have intuitions about probability that lead us astray but bringing up one that never does and treating it like it was a myth or an old wives tale is a terrible way to clear up confusion. Frankly, I think it would be very encouraging to the readers to point out that some of their intuitions are dead right.

I do have to agree with you that conditional and unconditional statements of the puzzle are part of the subject matter of the fourth paragraph. I should have indicted that aspect of it as well. The goal of the section is to reduce confusion about the central problem of the article and dragging in conditional and unconditional probability without a thorough discussion of what they mean is guaranteed to increase confusion - in virtually any context. Perhaps in some remote corner of the article, with warnings posted and a thorough introduction to the concepts one might profitably broach the subject - though, I doubt it.

I stand by what I said regarding the variant games - I assume their introduction was well intentioned but bringing up a variant of the game then proceeding to say and I quote, "In its usual form the problem statement does not specify this detail of the host's behavior, making the answer that switching wins the car with probability 2/3 mathematically unjustified. Many commonly presented solutions address the unconditional probability, ignoring which door the host opens; Morgan et al. call these "false solutions" (1991)." makes it sound like the solutions given in the article are erroneous.

How can that do anything but create more confusion? Solving the variant games was not hard and after doing it, I still had no idea what these statements meant. Exactly what is mathematically unjustified? "false solutions"? I suppose I could somehow dredge up Morgan et al and see what they were talking about but after doing it with Falk and finding that it did not support what "Sources of Confusion" purported, I figured that would be a similar waste of time. And on a subject that is far better left out altogether.Millbast5 (talk) 13:51, 9 October 2008 (UTC)[reply]

Of course if no new information has actually been imparted the probabilities cannot change, just as lacking other information about which of N alternatives might be the correct one makes the probability of each 1/N. These are BOTH "true" intuitions but they resonate so powerfully that they are easily misapplied.

The Morgan et al. paper is available online (for a fee) but should also be available in nearly any university library. They argue 1) the problem as generally stated should be interpreted as a conditional problem, and 2) since the problem statement (per vos Savant) doesn't specify the host opens one of two goat doors with equal probability we must treat the host's preference for one door over the other as a variable p. Given these assumptions, the general 2/3 answer is either non-responsive (answering the unconditional question, not the conditional question) or incorrectly assumes a constraint on the host that is not imposed by the problem statement. They argue the "correct" answer to the vos Savant version of the problem (granting that the host must open a door, and won't reveal the car) is that the probability of winning by switching is 1/(1+p), i.e. somewhere between 1/2 and 1 depending on the host's preference for the door that the host opens (in the case where the host has a choice). Relating this back to Falk, this is a case where the "no news" intuition (nothing the host does can affect your initial chance of winning the car, which must remain 1/3) leads to an incorrect answer.

Are there specific changes you're suggesting should be made to the article? -- Rick Block (talk) 14:11, 10 October 2008 (UTC)[reply]

Naturally I do have some suggestions for what might reduce the confusion of the readers though I do believe we can do a much better job of providing solutions and try to eliminate confusion that way. The primary source of confusion about this problem is that readers have a very strong intuition that it doesn't matter whether the host purposefully reveals a goat he knows is behind a door or he opens a door at random and by chance reveals a goat. The latter leaves the switch and non-switch strategies equally good - they can reason that out quite well - so how could the former not do the same? Anyway, I will address that source of confusion in a short while.

For the moment I will delve into the issues of paragraph 4.

Recall that the problem posed in "Ask Marilyn" did not state that the contestant picked door number 1, it said "say No. 1" which means that the particular door is not significant in the English speaking world and he's choosing a particular one just for purposes of illustration. Likewise the host opening door No. 3 is purely illustrative. That means the intent of the question is not what is the probability of the contestant winning with different strategies in the case he opened door 1 and the host opened door 3, the question is clearly what are his chances of winning the game with either of two strategies. Not only is that question answerable, it is answerable in case the host has a preference for some doors over others and the answer is the same!

Clearly Morgan et al interpreted the question to be about a particular pair of doors because to fully specify the hosts door opening preferences 3 parameters like p are needed - one for each possible pair of doors he may have to choose between, 1 & 2, 1 & 3, and 2 & 3. Why they choose that interpretation is anybody's guess, but mathematicians do things like that - take a different view and you might have another interesting problem. (I did the same thing you might recall - I chose an interpretation where the puzzle had to solved exactly as stated, without any knowledge or assumptions about the hosts behavior beyond what the puzzle statement supplies. I knew that that was not the intent of the question - for one thing its far too difficult to ask of a general audience. It was too complex to be included in this article, too - it was considered "OR", I believe.)

For illustrative purposes say the contestant picked door 1 and p is the probability the host will pick door 2 to open given the choice between doors 2 and 3. Indeed the probability of winning using the switch strategy with door 2 being opened is 1/(1+p) and the corresponding chance of winning with door 3 being opened is 1/(1+(1-p)). So what are the chances of the switch strategy winning? All we have to is multiply the two probabilities above by the probability of each of the doors being opened and add the two products together - childs play. Door 2 is opened with probability 1/3 + p/3 and door 3 is opened with probability 1/3 + (1-p)/3 thus the chances of winning via the switch strategy is: (1/3 + p/3) * 1/(1+p) + (1/3 + (1-p)/3)*(1/(1+(1-p))) - the algebra is easy and the answer is 2/3, as we knew it would have to be.

Door 1 was just illustrative and since the answer is independent of probability p, the answer will be the same for any other choice by the contestant. So the answer to the problem is that the contestants chances of winning by switching are 2/3 - versus 1/3 by not switching.

I suppose by their peculiar lights it isn't mathematically justified to say the probability of winning by switching is 2/3 when it varies according to which door is opened. Of course, if you do not know what value p has, then you only know one door may be better than the other but not which and by how much. Yet you still know the probability of winning by switching is 2/3 regardless of p - what are they going to say to that? What could they say? "Yeah, but if you don't know p you can't tell what are the chances when Door 3 is opened and thats what the puzzle asked and intended to ask." Frankly, I'd disagree with them if the puzzle had been stated without the writer making it clear the mention of doors was only for illustrative reasons - and so would almost everyone else. I suppose if one of them were contributing here we could have a philosophical-mathematical discussion about how to state puzzles so only one interpretation is possible.

Also note that the puzzle as stated here only asked if switching was better and thats true as long as p is not 0 or 1 and the appropriate door is opened with each value - door 3 when p=0 and door 2 when p=1. In those extremal case not switching is equally as good, not better. If you don't know what p is then you're clearly better off switching since you can never do worse than by not switching and you may do vastly better.

How are the readers going to be helped by going through a mess like this? None of them interpret the problem the way Morgan et al did so one would have to have a very good reason for delving into it. Yes, you could explain that if the contestant did know the value of probability p for the pair doors he did not pick, he'd have a better idea of whether he won or not depending on which door the host opened for those few seconds before the door he settled on was opened to make the probability either 1 or 0. You would also explain that there was no way for the contestant to make use of his knowledge of probability p - except if p was 1 and door 2 was opened or if it was 0 and door 3 opened - then he could not switch and win with the same probability as switching (1/2). And then you could reassure him that the contestants chances of winning by switching were still 2/3, regardless of the value of p.Millbast5 (talk) 13:23, 11 October 2008 (UTC)[reply]

Your second to last paragraph above is exactly the same as the conclusion Morgan et al. come to - i.e. if you don't know p you might as well switch since by doing so your chances of winning don't go down but may go up considerably (and, people who a priori decide to switch before seeing which door the host opens win with 2/3 probability). I don't think anyone takes "let's say" as anything other than illustrative - on the other hand, I think the point Morgan et al. make is valid. Their point is that the player is not making a decision at the beginning of the game, before picking a door and before knowing which door the host opens, but after picking the initial door (illustratively, door 1) and after the host has opened a door (illustratively, door 3). At least one of the experimental psychology papers about this problem indicates that this aspect of the problem (putting the player in front of two closed doors and one open door at the point of attempting to evaluate the probability of where the car is), is what makes it resonate so strongly with the "must be 1/N" intuition. Roughly a year ago (in the talk archives) there was an extended discussion here about this, which led to changes to the article including introducing the paragraph you're objecting to. To a large extent, the issue is not what we think but what the published sources say. We could certainly rephrase things for better clarity. Is this what you're suggesting at the very end of your post above? -- Rick Block (talk) 15:27, 11 October 2008 (UTC)[reply]

I agree with Millbast5that the issues addressed by the likes of Morgan et al are essentially a diversion from the main question. The current article has too much about 'versions' of the problem but does not provide a convincing resolution of the central paradox. Martin Hogbin (talk) 20:07, 11 October 2008 (UTC)[reply]

I'm glad to hear that Morgan et al recognized that short of knowing 'p' took the value 1 or 0, their analysis is entirely without benefit to the contestant and that in no case could he actually gain from it. Of course that means that while the contestant doesn't have to make a decision until the door is opened, he can't do any better than he could by deciding whether to switch or not before the game started. If one looks at their paper in that light perhaps there is a place for a reference to it - in case someone wondered if there wasn't some way putting off the decision might help the contestants chance of winning. However, to appear in the 'Sources of Confusion' section it needs to be enlightening to the puzzle in general, and I see the objections I made originally as still being in full force.

I just had a cool idea - a way that someone who didn't believe switching would improve his chances of winning could hedge, just in case he was wrong. Suppose he thinks door 2 is the door with the car behind it. Instead he picks door 1, intending to switch to door 2 when the host gives him the option to switch. The only way this play could go awry would be if the host opened door 2 to reveal it as a loser - and then he'd have to be glad he didn't pick it and stick with it. Of course, with door 2 out of play he has a real choice about switching. Heck, why not rank all three doors as to his subjective feel of their chances of concealing the prize and he intially picks the worst of the three so he will get to switch to one of his first two choices for sure. He could explain that rationale to his friends and family if they started to tease him about switching from the winning door - and they might even buy it. Or tell them his plan ahead of time and then he probably wouldn't panic under the gun and not switch - lets face it, switching is wussy.

I'm not sure it belongs in the Sources of Confusion section but I've been thinking that someplace in the article we should bring up the idea that people do not like to look indecisive and the switch strategy does give that appearance. Few would fail to switch on that basis if they did believe switching doubled their chances of winning but it probably is part of the reason why everyone who believes the two strategies are equally good elects not to switch.

Sorry, I'm getting off topic here.

Rick, I'm in full agreement that the contestant being faced with two closed doors stimulates the automatic response that they are equally likely to hide the prize. I can't remember if I explained that I believed one reason the urn and balls version of the puzzle is easier to understand is because that image of the contestant is not called to mind. Also it does not appear that the person who knowingly removes the white ball from the urn, after the subject blindly picks a ball, is an adversary while the host is still rather suspicious in that regard.

Published sources should attributed wherever their ideas and results are used but, outside of history and 'related topics', we should not look to refer to them simply because they exist if they do not help us achieve our goals.

I also believe that in an article that is about a popular topic we should have a goal of making the article interesting or even entertaining to read because people will not be enlightened if reading the article is so boring they stop reading. A little lattitude with OR ought to allowed, too, when the subject is just not that serious. Something publishable in a learned journal is not what I mean, of course, but 'novel and creative' enhance an article.

With popular topics, such as this, an article that one has to find the disclosing authority for and then pay thirty bucks to eyeball it, is about as close to useless as you can get - again outside of history and related topics. So if you can't quote enough of it to be helpful you should try to take a course that avoids it altogether - paraphrasing is rife with pitfalls, especially when it is done with an eye to keeping it short. Look what happened with Falk and Morgan et al in the current section - Falk didn't conclude what the section implied and the paraphase of Morgan highlighted their conclusions that made it sound like the puzzle solution was chock full of errors. As I mentioned above the important conclusion from Morgan is that knowledge of the hosts preference for opening doors doesn't affect the chances of the contestant winning, yet that part got left out.

I disagree with the tone of the article (and many of the references) in regard to people's intuition about probability - they are, in fact, generally quite good and we should try to explicitly use them in presenting solutions. Thus they will be nodding their heads and thinking "yes, I do think just that way" instead of feeling like they are being treated as dolts for thinking/believing as they do.Millbast5 (talk) 23:23, 11 October 2008 (UTC)[reply]

Isn't the first part of the solution section about as simple as you can get? Three equally likely scenarios, switching wins in two and loses in one, ergo 2/3 chance of winning by switching. The subsequent deeper analysis, using the Morgan et al. reference, addresses a mathematically significant criticism of this solution which is that the solution doesn't utilize the "equal goat door" constraint imposed on the host. Without this constraint, the simple solution may be incorrect making it at least incomplete. You might argue that no one would interpret the problem to mean that the host exhibits a preference for one door over another, but if you don't include this constraint that's exactly what a mathematician would do (find a hole in the argument). If you want the answer to be 2/3 chance of winning by switching, and you want the decision to switch to be made after the host has opened a door, you have to include the "equal goat door" constraint and the solution has to use this constraint. Anything less and you're either asking a different question (like what's the probability of winning if you decide to switch before the host opens a door knowing that the host will open a door), or 2/3 isn't always the right answer. -- Rick Block (talk) 04:30, 26 October 2008 (UTC)[reply]

Proposed substitute for the last two paragraphs of Sources of Confusion section

Just for the sake of variety, you get to be the contestant for a while.

There is no question that the most difficult thing to evaluate is the effect of the host intentionally opening a door he knows hides a goat after you've made your initial pick. Everyone can figure out the effect if the host just picks one of the two available doors at random and opens it. A third of the time the action will reveal a car and you'll know immediately that your door and the switch door each hide the car with probability 0, and two thirds of the time the action reveals a goat leaving your door and switch door each harboring the car with probability 1/2. Our intuition is very good about that. If the host jumped ahead of you and opened a door at random before you could pick the outcome is exactly the same. Our intuition is all over that one. Suppose the host didn't wait for you to pick but opened the door he knew hid a goat before you picked. That would be great, you'd win half of the time instead of a third of the time. Also the 'switch' door would hide the car half of the time. Our intuition nails that in milliseconds!

One thing that has been true in every case is that your picked door and the 'switch' door have hid the car with equal probability so it is very tempting to assume the same holds when the host opens the known goat door after you pick and your intuition sends you a strong 'buy' signal on that assumption. Still the smart thing to do is check it out before you buy.

One good way to get a handle on it is to imagine the game being played 9 times all at once. First you make your nine choices. But then you're allowed to send an observer to look behind the doors - she reports that there are nine cars altogether and 3 of them are behind doors that you picked. Now the host goes through all nine games opening a door he knows conceals a goat. The observer does her thing again and reports all the cars are just where they were before - no funny business going on. So there are 18 doors that are closed and three of the cars are behind your 9 of them - that leaves six cars behind the other nine doors. You do the math again to make sure. It doesn't help - there twice as many cars behind the 'switch' doors as behind your doors. Is that darn observer being truthful? What she says makes sense - 3 cars out nines tries is normal and you really didn't expect any more cars to move behind your doors so it sure seems like she is telling the truth. There's no way out of it - the 'switch' doors are twice as likely as yours to harbor a car.

Why does our intuition get this last case wrong? In every case the image we have in our minds is the same - we are standing before two closed doors and one open door, and leaving out the random cases where the open door reveals the car, there is a car behind one of the two closed doors and we don't know which. Generally speaking, when we don't know which door its behind its equally likely to be behind either of the closed doors. It seems only fair - heck it seems right - that it should be equally likely. Even when it is not equally likely but we don't know which door is more likely, its the same as it being equally likely as far as we're concerned. This particular case is the oddball - the doors are not equally likely and we know which one is more likely. It just isn't apparent until we exercise our imagination a bit.

We still can't really fault our intuition because it led us through the 9 games demo just fine, taking it one small step at a time.Millbast5 (talk) 04:44, 13 October 2008 (UTC)[reply]

Here is another go it it, where I focus on tracking down the source of confusion

Another possibility for the Sources of Confusion

Comment: This version directly shows "why the probability isn't 1/2" so it replaces the section by that name as well.

The source of confusion is not the puzzle itself. It arises from contemplating what would happen if the host opened one of the 'other' doors at random. The solution in that case is gotten right by virtually everyone: the contestant picks the winning door 1 time in 3, the host opens a randomly chosen door to reveal the car 1 time in 3 and the unpicked and unopened door hides the car 1 time in 3. Now comes the confusion: if you forget about the 1 time in 3 the host reveals the car then the contestant wins 1 time in 2! Of course, if you eliminate half the losing choices for whatever reason you want, the contestant would have winners half the time. Its just simple numbers. It means nothing and means no more if you phrase it this way: when the host reveals a goat by chance the contestant wins half the time. Clearly that is just another way to say you are eliminating half his losing choices. Now the source of confusion leaps into full view - if you attribute the change in frequency of winning to the revelation of a goat instead to the fact that you're disregarding half his losing choices you have the makings of a real misconception.

In the actual problem only goats are ever revealed but if one believes that the revelation of goats causes the contestants chances of winning to increase, it ought to do it even more when more of them are revealed. And so it does, according to the faulty belief - instead of the contestant winning 1 time in 2 when half his losers are removed, his original pick wins 1 time in 2 overall.

This belief gets a boost from another source - the mental image of the contestant having to pick between two closed doors, one of which hides the car. The contestant doesn't know which one, and generally this means that the car is behind either door 1 time in 2 - perfect corroboration! The belief becomes so solid that nothing can shake it. (This is not speculation as a number of the referenced sources demonstrate the absolute conviction of many believers - some of them get irate if one continues to challenge the belief.)

One way to combat the belief is to consider a number of plays of the game simultaneously so as to make the original picks more concrete. Imagine for example, 99 plays of the game in which the contestant picks 33 winners and 66 losers - which is, statistically, the most likely outcome. Now its clear that if no cars or goats are shuttled to and fro behind his chosen doors one can do anything one wants with the 198 other doors and the contestants 33 winners and 66 losers will not be altered. One can open those doors at random to ones hearts content or look behind them and expose goats till the cows come home and the contestants 33 winners and 66 losers sit there undisturbed.

Phrased in frequency terms - the contestant picks the winning door 1 time in 3 and the host purposefully revealing goats does not change that fact. It applies to the imagined contestant choosing between two closed doors, too. True, he doesn't know for certain which door hides the car but he does know his initially chosen door still hides the car 1 time in 3 so, inevitably, it is behind the other door 2 times in 3.

As is apparent at this point, the solution to the game is extremely simple if one just proceeds without doubting that the initial pick remains a winner 1 time in 3 after the host opens a door to reveal a goat. One easy logical step shows the remaining door, the 'switch to' door, hides the car 2 times in 3. This simplicity probably works against it as far as engendering belief - the problem baffles our immediate intuition so it surely it can't be as simple as subtracting 1/3 from 1!Millbast5 (talk) 11:07, 18 October 2008 (UTC)[reply]

I think you are quite right in where the confusion lies, but I am not sure that you have found a way of making that clear to the reader.

As you point out at the end, if we take what has been described here as the unconditional case, then there is a a very simple and convincing argument - you have a 1/3 chance of initially choosing the car and you always get the opposite if you swap. With the reasonable assumptions that most people make (host aways offers swap, aways opens a goat door) the same argument 'happens' to apply for the conditional case (since the host action does not change the probability that the player has chosen a car). I therefore suggest that the conditional/ unconditional issue is something of a red herring and should take a secondary place in the article. In the game, as understood by most, the conditional version of the problem has the same odds as the unconditional one. It might have been otherwise, but it was not (as most people understand the problem).Martin Hogbin (talk) 12:27, 18 October 2008 (UTC)[reply]

I inserted a new phrase into the middle of the third paragraph for increased emphasis.

I'm not too happy myself with the clarity of the what I wrote. That first paragraph is too long and I don't get to the actual statement of the source of confusion until the end of it. I've also considered starting with the other half of the false argument - the contestant faced with two doors, etc, but I believe the confusion does start with contemplating the effect of opening a door at random. I imagine it varies from person to person where it starts. In any case, if you have any thoughts about how to improve it, I'd be glad to hear them.

You know you don't have to sell me on the idea of eliminating any mention of conditional and unconditional probability from the article. Of course, I'm using the idea myself in talking about the odds under the condition that a goat was revealed but people are quite used to the idea as we phrase things in that way in our everyday speech. "If TMac stays healthy the Rockets have a good shot at the title." We understand perfectly that different conditons change the odds but developing the formalism is out of place in an article like this and using the terms 'conditional' and 'unconditional' without developement is worse. Frankly, I'm not sure at all what you (and others) mean when you use those terms in these discussions. Imagine how someone feels who is a bit shaky when the word probability is used all by itself.

Yes I agree. I was not really making my conditional/non-conditional point to you but to others who might be reading. I think this article suffers from what might be called 'wiki correctness'. In other words it keeps to all the rules but does a bad job. Martin Hogbin (talk) 23:26, 18 October 2008 (UTC)[reply]

Monty's Action Does Not Cause The Original Odds To Change.

When Monty opens a door, he doesn't tell us anything we didn't already need to know. He ALWAYS shows a goat. It makes no difference to this puzzle WHICH remaining door he shows. So it starts out as 1/3 for your door + 2/3 for the remaining doors = 100%. Then he shows a door, but we knew in advance that he was going to show a goat. The odds simply haven't changed following his action. They remain 1/3 for your door + 2/3 for the remaining doors (of which there is now just 1).

That's why on the show, Monty didn't HAVE to show a door (unlike this puzzle), and he could bribe you with cash. Otherwise, there's no show. You always switch.

I'm new to this. Is there a way to get this 'solution' onto the article page?

75.185.188.104 (talk) 17:09, 25 October 2008 (UTC)[reply]

Isn't this essentially the same as the explanation in the Combining doors section? -- Rick Block (talk) 17:48, 25 October 2008 (UTC)[reply]

It would seem that your solution is not much liked by many of the current editors of this page because it cannot be verified by reliable sources. Martin Hogbin (talk) 22:59, 25 October 2008 (UTC)[reply]

Martin - was that directed at me? If so, then I'm confused. I didn't say I didn't like it, but that it's already included in the article (with reliable sources). -- Rick Block (talk) 03:58, 26 October 2008 (UTC)[reply]

No. It was directed at all the current editors. If you remember, I first came to this article in response to an RFC which claimed that the current editors were being overly protective. Although I understand why that attitude has some justification, I think that there are possible improvements that could be made that are being resisted.

My central point is this: The Monty Hall problem is particularly notable because, even when people make the 'standard' assumptions, the great majority of them get the wrong answer and it is difficult to convince them of the right one. This article could, and in my opinion should, address that issue better.

The more complex issues of host action/ game rules etc are not part of the core problem. In particular the issue of conditional/unconditional probability only adds to the confusion for many people. It is a fact that, for the 'standard' version of the problem, there is no difference between the two, therefore I put it to you that we do not need to make a major point at the start of the article about something that could have been the case but in fact is not. I have proposed that I very simple, and hopefully convincing, explanation along the lines suggested above should be given a prime position at the start of the article, followed by a section on sources of confusion in the basic case. The more detailed analyses, as already presented, should then follow. I appreciate that this might require a more liberal interpretation of some WP policies but I suggest that it is worthwhile considering this option in order improve the article.

I have tried to be a good Wikipedian by not disrupting this featured article with experimental edits and have set up a draft page for people to attempt to improve the it; so far only one other editor has shown an interest. So, yes, my remark was aimed at getting some reaction but only so that the article might be improved in a way that is acceptable to all. Martin Hogbin (talk) 10:33, 26 October 2008 (UTC)[reply]

I agree that the presentation of this article leaves something to be desired. I would think that most users come to this page to settle an arguement, or to enlighten themselves on how they don't 'get' the solution. I know that's why I came here.

So, an elegant, simple solution would serve the readers. I'm a stats and probability geek, but I don't like having to 'prove' something by running simulations (since I don't know who created the simulation, I can't trust it). And, if there is an easier to understand explanation that doesn't require umpteen diagrams, the typical reader would prefer that, as s/he will probably try to explain what s/he read here to other people. 75.185.188.104 (talk) 12:27, 26 October 2008 (UTC)[reply]

Isn't the first bit in the Solution section about as simple and elegant as it gets, with a diagram for the more visually minded folks out there? Three clearly equally probable cases, switching wins in two and loses in one. The transitional paragraph about this solution (the one containing the reference to simulations) and the follow-on conditional analysis (with a different diagram that appeals to other sorts of folks) is there to address a significant mathematical criticism (i.e. this solution does not address the problem that's actually posed), raised first by Morgan et al. If we end the Solution section following the first diagram the solution is incomplete as it fails to address the Morgan et al. criticism. IMO, any "simple" solution will inevitably have this same issue. There's an argument that Morgan et al. are splitting hairs over trivial details that most people don't care about, but that's like ignoring negative or imaginary roots of a quadratic equation (x^2 = 4, what's x? most people would say 2, but -2 is also an answer). If the suggestion is to simplify the solution section by eliminating any mention of the conditional analysis I strongly disagree. -- Rick Block (talk) 15:03, 26 October 2008 (UTC)[reply]

You refer to "significant mathematical criticism (i.e. this solution does not address the problem that's actually posed),". If such a thing is valid, then you don't have a solution at all. The solution should be, and is, simple. 1/3 + 2/3 = 100%. And Monty's actions do not change this formula. I don't see a mathematical ambiguity.

Glkanter (talk) 16:07, 26 October 2008 (UTC)[reply]

This is the same topic discussed in more detail three sections down (about the subtly different question), and indeed Morgan et al. argue the "simple" solution is not a solution at all. -- Rick Block (talk) 16:28, 26 October 2008 (UTC)[reply]

Well, I thought we were going to discuss the solution to the Monty Hall Problem. And that the 'Solution' portion of the Wikipedia entry would provide just that. I can see now that there is some other agenda for Monty Hall Problem page. I provided a mathematical proof to a very simple problem. So far, not you, or anybody else has challanged that proof. I already pointed out an error in the VERY FIRST LINE OF THE EXISTING SOLUTION paragraph. Yet you bring up all sorts of 'subtly different questions', 'mathematical inconsisancies', and whatnot. It's your page. Do whatever you want.

Glkanter (talk) 16:42, 26 October 2008 (UTC)[reply]

We're off on the wrong foot here. It's not my page, it's openly editable by anyone. In its current form it's the result of the efforts of at least a dozen editors over the span of 7 years. I'm simply one of a number of editors who have contributed to this page for a long time (in my case, since at least June 2005). I'm perfectly happy to discuss changes, and very happy when discussions result in improvements to the article. Please delete the first sentence in the solution section (I'll wait a while and if you don't I will). Thank you for this suggestion. I think nearly anyone would agree it's an improvement (fewer words are generally better). What I mostly meant by the comment from 16:28 above is that the discussion is continuing in the other section (below), so rather than continue here we should continue there. -- Rick Block (talk) 20:48, 26 October 2008 (UTC)[reply]

Re: The 'Combining Doors' comment above. Yes, it's similar. imho, these should be a prominant part of the 'Solution', not relegated to 'Aids to Understanding'. Glkanter (talk) 15:12, 26 October 2008 (UTC)[reply]

I find the explanation in the combining doors section somewhat abstract. Here's the explanation I've been using with people who have a hard time understanding why switching is the best strategy -- I know this probably shouldn't go on the main page (I'm the only reference I know of) but I wanted to share it.

I'm going to write it in the form of a dialog because that is the shortest form of this explanation and also the form in which it has been used:

"What if I could show you a tricky way to pick two doors instead of just one. Would you agree that you have a 2/3 chance of winning if you could pick two doors?"

"Sure! If I can actually pick two doors I'll have a 2/3 chance of winning."

"OK, tell me (but don't tell Monty) which two doors you want to pick."

"OK, I want to pick doors two and three."

"Great -- now tell Monty you want to pick door number one."

"But I don't want door number one!"

"I know, that's the tricky part. By telling Monty you want door number one you'll get him to open one of the two doors you actually picked -- and the way the rules are setup he'll always pick a door that has a goat. After he opens one of the two doors you actually picked now tell him you're switching that way you'll get to pick both doors two and three."

Am I correct that this would be inappropriate for the main page? Stepheneb (talk) 17:08, 2 January 2009 (UTC)[reply]

I would say yes, you are correct that this would be inappropriate for the main page (unless you can find a reliable source for it). It is logically the same as the "Combining doors" explanation. -- Rick Block (talk) 18:53, 2 January 2009 (UTC)[reply]

A twist - same choice, different odds?

I follow the reasoning all the way, yet the following twist to the scenario puzzles me. Can someone explain what I'm missing?

Contestant chooses door 1, Monty reveals a goat behind door 3 and invites the contestant to revise his choice.
Before the contestant makes his final decision known, another contestant is whisked on stage, straight from the dressing room, with no prior knowledge of the first contestant's original choice. This second contestant is presented with the open door 3 (and its goat) and asked to choose where the car is.
The two contestants then make their choices, independently of each other. The first contestant, having read this article, switches his choice to door 2. The second contestant tosses a coin on the basis of which he also chooses door 2.

As we know from the article, the first contestant has a 2/3 probability of finding the car. But for the second contestant it's a straight choice between 2 doors, so he has a 1/2 chance of finding the car. So it appears that at the moment the two contestants make their final choices, p(finding the car | choosing door 2) has two different values simultaneously depending on who does the choosing, whereas intuition (mine, anyway) says that the probability should have a unique value, and that it should depend only on the possible permutations of the original distribution of car and goats, and the shared knowledge that door 3 revealed a goat. Help! -- Timberframe (talk) 09:25, 26 October 2008 (UTC)[reply]

The probability depends on what the second contestant knows about the situation. If they are simple told,'here are two doors, one hides a car and the other a goat', then they might as well toss a coin. If, on the other hand, they have watched the show on a monitor, or are told, 'this is the door that the contestant first chose, then Monty opened this door to reveal a goat (as he always does)', they would know that the odds are better if they choose the door that the first contestant did not. Martin Hogbin (talk) 10:42, 26 October 2008 (UTC)[reply]

Mr. Hogbin is correct. Note that the 2nd player has less likelyhood of picking correctly. 50% vs 67%. This is because he has less knowledge of how the 2 doors came to be. He can only assume randomness, but the first player knows Monty did not act randomly when he opened the door revealing a goat. 75.185.188.104 (talk) 12:17, 26 October 2008 (UTC)[reply]

Another way to look at this is to compute the second player's conditional probability of winning the car. At the time the second player is choosing, Door 1 has a 1/3 chance of being the right door while Door 2 has a 2/3 chance. Someone picking randomly between these doors has a (1/2 * 1/3) + (1/2 * 2/3) chance of winning the car. Multiply this out and you get 1/2. In words what this means is that if the second player happens to pick Door 2 he also has a 2/3 chance of winning the car (but this only happens half the time, and the other half he has only a 1/3 chance). If your question is what is the second player's chance of winning if he happens to pick Door 2, the answer is 2/3 (same as the first player). Note that this is true whether or not he watched the show on a monitor. He may well think his chances are 1/2, and they are before he flips the coin, but after he flips the coin and picks a door his chances are either 1/3 or 2/3 (just like player 1) depending on which door he picks. -- Rick Block (talk) 15:24, 26 October 2008 (UTC)[reply]

To say that the second player would have a 2/3 chance of winning the car if he picks the door 2 is not, in my opinion, a good way of putting things. It depends on who is making the call. From the second contestant's point of view the chances are 1/2 (if he knows nothing of the history), from the audience's (who have seen what has happened so far) point of view they are 2/3, and from the producer's (who knows the car is actually behind door 1) point of view it is 0. Probability is a state of knowledge.Martin Hogbin (talk) 16:16, 26 October 2008 (UTC)[reply]

On the other hand, to say the first player and second player who have picked the same door have different chances of winning seems like an extremely odd way of putting things. The probability does depend on who makes the call, but I think the same perspective should be used in both cases - that being the perspective of a member of the audience. -- Rick Block (talk) 16:35, 26 October 2008 (UTC)[reply]

I would say that it is more normal to quote probabilities from the point of view of the person who is making the choice. This does lead to different probabilities for different people doing the same thing but understanding that is basic to understanding probability. For example what are the chances of a player initially picking the car - 1/3. Suppose that we have a different player who somehow knows that car is not behind door 2 - his probability of picking the car is 1/2. Same car, same goats, same doors, but a different answer. Martin Hogbin (talk) 16:47, 26 October 2008 (UTC)[reply]

You're right on target. As more knowledge is gained, probabilty increases from 1/2 (random, or no knowlegde), to 2/3 (watched Monty open a door), to 100% (you're Monty). Hopefully, Timberframe's original concern was resolved.

Glkanter (talk) 17:02, 26 October 2008 (UTC)[reply]

I'll be more precise. The perspective is generally that of someone who knows all the givens from the problem statement. So, if we're given the unambiguous Monty Hall statement (from the article) and that player 1 had picked Door 1 and the host has opened Door 3 and then player 2 has arrived on the scene and flipped a coin and has selected Door 2, player 2 has a 2/3 chance of winning (not 1/2). This matches the chances player 1 has if switching to Door 2. The probabilities are evaluated with knowledge of all the givens. Different givens result in different probabilities, which is why we'd say Martin's player who (we're given) knows that among the three doors the car is not behind Door 2 has a 1/2 probability of picking the car initially. Same car, same goats, same doors, but different givens so a different answer. -- Rick Block (talk) 17:49, 26 October 2008 (UTC)[reply]

I guess the moral of this discussion is to make always make clear what the givens are before stating a probability, which I think none of us actually did. Martin Hogbin (talk) 20:13, 26 October 2008 (UTC)[reply]

Thanks guys, got it now. The falacy in my proposition is in the assertion that because there are two closed doors and the car is behind one of them, the second contestant has a 50:50 chance of finding it. This is only true from the second contestant's limited perspective. The conditions of the scenario is posited have in reality already shifted the odds to 2:1. Thanks again -- Timberframe (talk) 19:29, 26 October 2008 (UTC)[reply]

BTW

The following is prominant in the article:

"Solution The overall probability of winning by switching is determined by the location of the car...

This result has been verified experimentally using computer and other simulation techniques (see Simulation below)."

This first sentance doesn't mean anything, and in fact is incorrect. The probability is NOT affected by the location of the car. One instance of playing the game is affected, but not the probability.

Further, a valid proof, which I believe my entry is, does not require being "verified experimentally using computer and other simulation techniques". A valid proof stands on it's own.

And lastly, this at the top of the Discussion Page is at a minimum, off-putting:

"Please note: The conclusions of this article have been confirmed by experiment

There is no need to argue the factual accuracy of the conclusions in this article. The fact that switching improves your probability of winning is mathematically sound and has been confirmed numerous times by experiment."

As I stated above, valid proofs do not rely on 'experiments'. Yet, there are two references to 'experiments'. Me thinks the lady doth protest too much. 75.185.188.104 (talk) 13:34, 26 October 2008 (UTC)[reply]

Are you suggesting deleting the first sentence in the Solution section? If so, that'd be fine with me. The "proof" in the Solution section does not depend on being verified experimentally - that sentence is there in an attempt to further convince the many doubters about the validity of the solution. If you can think of a way to make this more clear, please do so. The banner at the top of this page is there to discourage folks from arguing that the solution is incorrect (we still get a fair number of folks claiming the 2/3 answer is wrong). -- Rick Block (talk) 15:35, 26 October 2008 (UTC)[reply]

Yes, being incorrect, the first sentance should be deleted.

Why not move the 'experiment' stuff to 'Aids to Understanding'? A Solution should stand on it's merits. Making reference, in the solution, to another technique weakens it's apparent authority.

In fact, doesn't just having the 'Aids to Understanding' section imply that the proof is lacking clarity? The salient point that I believe is missing, is that the probabilities don't change because of Monty's action. That's why it's 1/3 - 2/3 when you start, and when you finish. Which is the only point of the whole Monty Hall Problem.

Glkanter (talk) 15:57, 26 October 2008 (UTC)[reply]

Can I mention again that I have set up this page:user:Martin Hogbin/Monty Hall problem (draft) for those, like myself, who do not find the current solution clear or convincing to develop a better basic solution. The concept is to develop a very clear solution that is verifiable and which can be eventually added to this article. There is also a talk page for discussion on how we might achieve this. Martin Hogbin (talk) 16:58, 26 October 2008 (UTC)[reply]

This comes very early in the 'Solution' section:

"A subtly different question is which strategy is best for an individual player after being shown a particular open door. Answering this question requires determining the conditional probability of winning by switching, given which door the host opens. This probability may differ from the overall probability of winning depending on the exact formulation of the problem (see Sources of confusion, below)."

I have no idea what this means, and fail to see how it part of a solution to the problem. People come to this page for an explanation. The above paragraph, plus whatever follows it, are not part of the solution to the Monty Hall Problem, and, imho should not be present there. Glkanter (talk) 14:59, 26 October 2008 (UTC)[reply]

There are two different questions.

If a player decides to switch before the host has opened a door, what is this player's chance of winning? Or, what is the overall probability of winning at the start of the game?
If a player decides to switch after the host has opened a door, what is this player's chance of winning? Or, what is the conditional probability of winning given which door the host opens?

The first part of the Solution section addresses the first question. The second question is subtly different, and requires examining only those cases where the host has opened a specific door. The second question can have a different answer than the first question even given the same game setup. This is exactly the same situation as the question raised two sections above about a second player who arrives on the scene after the host has opened a door (substituting before and after this player chooses a door for before and after the host opens a door). Before flipping a coin this second player has a 1/2 chance of winning. After flipping a coin this second player has a 1/3 or 2/3 chance of winning, depending on which door he chooses. Similarly, the first player has a 2/3 chance of winning by switching before the host opens a door, but may have a different chance of winning by switching after the host opens a door depending on exactly how the problem is formulated. In particular, unless the host is constrained to pick between two goat doors with equal probability (in the case the player has initially selected the car), the chance of winning by switching after the host opens a door may be anywhere between 1/2 and 1. An analysis that ignores this constraint and concludes the chance of winning by switching is 2/3 is incorrect, since the same analysis would apply to a different version of the problem (for example, where the host is constrained to always open the lowest numbered door hiding a goat). The typical wording of the problem is arguably asking the second question, not the first. -- Rick Block (talk) 16:07, 26 October 2008 (UTC)[reply]

There is no question of the probabilities at the start of the game. It's 1/3 car vs 2/3 goat. That's actually a premise, that everything is random. The only question asked in the problem, which assumes that Monty ALWAYS shows a goat (also a premise), is what are the odds when 2 doors are left? imho, everything else is clutter. I don't know what the word 'conditional' means in the context of this puzzle.

I came to the Wikipedia page because I needed to understand why it was favorable to switch. I've since distilled the reason to a single sentance, 'Monty's Action Doesn't Change the Odds.' As a user, I can assure you that all this other stuff gets in the way of understanding.

Glkanter (talk) 16:21, 26 October 2008 (UTC)[reply]

I agree. In the 'standard' version of the problem, it actually makes no difference when the player decides to switch. What is gained by answering a more complicated question that the one most people assume? The 'basic' question is plenty hard enough without trying to make it harder. Martin Hogbin (talk) 16:31, 26 October 2008 (UTC)[reply]

The question is why don't Monty's actions change the odds? Asserting this is true because we already know one of the unchosen doors hides a goat is not mathematically sound (and it's actually reasonably difficult to prove this assertion). The standard version of the problem puts the player's decision at the point the host has already opened a door. This makes it a conditional probability question. Any answer that ignores which door the host opens is fundamentally incomplete. -- Rick Block (talk) 17:06, 26 October 2008 (UTC)[reply]

Yes, but I am advocating getting the best of both worlds. Have a simple explanation that may not be rigorously complete but that people can understand and follow it with the current explanation that is more rigorous and complete. Martin Hogbin (talk) 17:22, 26 October 2008 (UTC)[reply]

This is the exact structure of the existing Solution section. Simple solution first (3 equally likely scenarios, switching wins in 2 of them), followed by a transition to a more rigorous solution. Are we fundamentally quibbling about what the simplest solution is? The one that is presented is meant to be obviously (intuitively) correct and mathematically sound (as far as it goes), but also function as a natural lead-in to the more rigorous solution. Are you arguing that the existing "3 equally likely scenarios" solution is hard to understand? -- Rick Block (talk) 18:14, 26 October 2008 (UTC)[reply]

Yes, I suppose I am arguing that simple solution shown is not simple enough. It also has some curious statements such as,'The overall probability of winning by switching is determined by the location of the car'. From whose point of view is that statement made? Martin Hogbin (talk) 19:43, 26 October 2008 (UTC)[reply]

Mr. Block, I disagree. There is no 'mathematically sound' problem. If there is, please share it with me. And your statement relating to conditional probability doesn't seem to apply here. The contestant's choice was random, and we don't care which door Monty reveals. You may repeat endlessly your asertion that it is important, but it is not valid. Like I said earlier, you seem to be interested in something other than the clear, concise answer to the Monty Hall Problem. Until you can focus on that, and perhaps show me the flaw with my solution, I'm wasting my time here. As for Monty's actions not affecting the odds, the proof is that the odds do not change once the contestant has chosen a door. His door remains 1/3. Maybe that is a tough concept for the random reader to grasp. That's why it's a paradox! But until you focus on that, and not all this other stuff, you won't be addressing the solution to the Monty Hall Problem at all.

Mr. Hogbin, I don't see how this solution is not 'rigorously complete'. Simple, yes. Incomplete, no.

I am, for the moment, simply accepting the opinion of others regarding rigorous completeness. Martin Hogbin (talk) 19:43, 26 October 2008 (UTC)[reply]

Glkanter (talk) 18:02, 26 October 2008 (UTC)[reply]

Consider a slightly different problem which is the same in all respects except that the host always opens the lowest numbered door possible. I believe your argument still holds (player's initial pick is 1/3, other two doors are 2/3, we know one is a goat, knowing which one can't affect the initial probability, etc.), leading to the conclusion that the player's chances of winning by switching are 2/3. This is even true if the question is asked before the host has opened a door. The usual version of the problem asks the question after the host opens the door. In this variant, if the player has picked Door 1 and the host has opened Door 3 the correct answer is the chances of winning by switching are 100%. In the unambiguous problem as stated in the article, the host must pick between two goat doors with an equal probability. This constraint, and only this constraint, is the difference between the variant I've suggested and the problem as posed in the article. If the solution does not use this constraint, there's a variant for which the solution produces the wrong result (i.e. the solution is mathematically unsound). -- Rick Block (talk) 18:14, 26 October 2008 (UTC)[reply]

Mr. Block, put yourself in the reader's seat. S/he wants to understand the Monty Hall Problem. Not 'a slightly different problem which is the same in all respects except that the host always opens the lowest numbered door possible.' Just the Monty Hall Problem. Now tell me, mathematically, or with symbolic logic, where my solution TO THAT PROBLEM is inaccurate, incomplete, or un-repeatable. That's all I ask. For now.

Glkanter (talk) 18:37, 26 October 2008 (UTC)[reply]

(outindent) The question is if it's advantageous to switch after we've picked a door and the host has opened one of the remaining doors. Your solution is the following:

When Monty opens a door, he doesn't tell us anything we didn't already need to know. He ALWAYS shows a goat. It makes no difference to this puzzle WHICH remaining door he shows. So it starts out as 1/3 for your door + 2/3 for the remaining doors = 100%. Then he shows a door, but we knew in advance that he was going to show a goat. The odds simply haven't changed following his action. They remain 1/3 for your door + 2/3 for the remaining doors (of which there is now just 1).

I've highlighted the problematic parts. These are assertions, not deductions based on what is given in the problem statement. They're true statements, but stating them without justification is equivalent to assuming the solution. We're not given that it makes no difference which door the host opens and we're not given that when the host opens one of the two remaining doors the initial 1/3 odds don't change - this is essentially what we're asked to show. It's like in a geometry proof using that an angle in the diagram is a 90 degree angle because it looks like one without showing why it must be one. If indeed it must be one because of what else is given it's certainly true, but using it without justification is at best incomplete. The simple solution currently in the article (3 equally likely scenarios, switching wins in 2 and loses in 1) doesn't have this problem. Do you find this solution hard to understand? -- Rick Block (talk) 19:21, 26 October 2008 (UTC)[reply]

Rick, I take issue with one of your statements: '...true statements, but stating them without justification is equivalent to assuming the solution' is too strong. Stating things without justification is assuming something, but not the solution. Even with the assumption of the statements mentioned, most people would still get the answer wrong, which is the notable feature of all this Martin Hogbin (talk) 19:56, 26 October 2008 (UTC).[reply]

Rick, also perhaps you would be kind enough to criticise the solution on my development page under Martin - Very short. Martin Hogbin (talk) 20:26, 26 October 2008 (UTC)[reply]

The very short solution (the player has a 2/3 chance of initially picking a goat, and in this case the host must reveal the other goat so this player now has a 2/3 of winning the car by switching) is essentially the "In other words" sentence just before the figure. I don't think it hurts to enumerate the cases, and the diagram definitely helps at least some people so I think this very short solution is fundamentally what's there already. The other advantage of the existing slightly longer version is that it provides a framework to discuss the more rigorous solution. Is the existing solution (up to and including the figure) any harder to understand than this very short solution? -- Rick Block (talk) 21:15, 26 October 2008 (UTC)[reply]

Thank you for your response to my question. Again, I respectfully disagree. My statement that it doesn't matter WHICH door Monty shows is true. It's not germain to the puzzle. The odds not changing? That's easy. Monty does not consider the door chosen by the contestant. Therefore, we gain no new knowledge about that selection, and it remains at 1/3. Since the probabilty must equal 100%, the remaining door now carries a 2/3 probability, (100% - 1/3). This is not really a debatable point. It is a bedrock principal of probabilty. The beauty of probability is that you only have to prove a solution one way. You can use my method, or Martin's method. They are BOTH VALID. As are some of the explanations with the diagrams, etc. And yes, the longer the explanation, the harder it is to understand, and explain to others. Why would you NOT want the most elegant solution to lead the paragraph? Any statement, diagram,etc. that is not necessary should be avoided. This is standard proof protocol. When Martin reduces it to 2 statements, that's a thing of beauty. Why interfere with it with extraneous statements?

Glkanter (talk) 21:33, 26 October 2008 (UTC)[reply]

Now, let's critique the existing Solution, much like you critiqued my proposed solution.

[Paragraph 1] The reasoning above applies to all players at the start of the game without regard to which door the host opens, specifically before the host opens a particular door and gives the player the option to switch doors (Morgan et al. 1991). This means if a large number of players randomly choose whether to stay or switch, then approximately 1/3 of those choosing to stay with the initial selection and 2/3 of those choosing to switch would win the car. This result has been verified experimentally using computer and other simulation techniques (see Simulation below).

[Paragraph 2] Tree showing the probability of every possible outcome if the player initially picks Door 1A subtly different question is which strategy is best for an individual player after being shown a particular open door. Answering this question requires determining the conditional probability of winning by switching, given which door the host opens. This probability may differ from the overall probability of winning depending on the exact formulation of the problem (see Sources of confusion, below).

[Paragraph 3] Referring to the figure above or to an equivalent decision tree as shown to the right (Chun 1991; Grinstead and Snell 2006:137-138) and considering only the cases where the host opens Door 2, switching loses in a 1/6 case where the player initially picked the car and otherwise wins in a 1/3 case. Similarly if the host opens Door 3 switching wins twice as often as staying, so the conditional probability of winning by switching given either door the host opens is 2/3 — the same as the overall probability. A formal proof of this fact using Bayes' theorem is presented below (see Bayesian analysis).

Paragraph 1 means nothing. It is gibberish.

Paragraph 2 does not respond to the Monty Hall Problem, and does not belong under the Solution heading.

Paragraph3 refers to a special case where Monty opens door 2 only. It too, is not germain to the Monty Hall Problem, which has ALREADY been proven by whatever method leads the discussion.

These three paragraphs of the existing Solution do not add anything to the reader's understanding of the solution, and should be removed. At best they are superfulous, at worst they are confusing, and maybe just wrong.

Glkanter (talk) 22:09, 26 October 2008 (UTC)[reply]

And, for the record, there are only 2 scenarios.

1 The contestant chooses a goat, which happens 2/3 of the time. In this case, as Martin so eloquently demonstrated, Monty eliminates the remaing goat leaving the car behind the closed door.

2 The contestant chose the car, in which case Monty eliminates one of the goats. This happens 1/3 of the time.

Therefore, any solution relying on diagrams, etc should have only these 2 scenarios. Anything beyond this convolutes the solution, which, of course should be avoided.

Glkanter (talk) 22:16, 26 October 2008 (UTC)[reply]

It doesn't matter which door Monty shows, but only because he's constrained to open one of the two goat doors with equal probability in the case the player initially selects the car. Again, you're asserting this without justifying why this is true. Regarding the odds not changing, are you saying Monty does not consider opening the door chosen by the contestant? This, by itself, is not sufficient to ensure that Monty's actions don't change the initial 1/3 probability. The total probability must be 100%, and before the host opens a door it's surely 1/3 player's door vs. 2/3 for the other two doors (so 2/3 of the players who decide to switch before the host opens a door will win) but the only thing that keeps it that way after the host opens a door is that pesky equal goat door constraint. The best way I know to show that it CAN change is to contrast the problem as stated with a different problem (i.e. the aforementioned "host opens lowest numbered door possible" variant).

With the equal door constraint, if the player picks Door 1 the probability the host opens Door 2 (or Door 3) is 1/2. This 1/2 is not random, but composed of a 1/3 chance that the car is behind the other door plus a (1/2 * 1/3) chance the car is behind Door 1. This fact makes the problem symmetric with regard to which door the host opens. We in effect have two identical subproblems, and it's this that makes it true that it doesn't matter which door the host opens (and prevents the host opening a door from changing the initial 1/3 chance).

In the lowest numbered door variant, the probability that the host opens Door 2 is not the same as the probability that the host opens Door 3. In this variant, the host opens Door 2 if the car is behind either Door 1 or Door 3 (i.e. 2/3 chance), and Door 3 only if the car is behind Door 2 (1/3 chance). The player has a 100% chance of winning if the host opens Door 3, but only a 50% chance of winning if the host opens Door 2. Multiplying out the composite probabilities we get (100% * 1/3) + (50% * 2/3), which totals 2/3 (meaning if you decide to switch before the host opens a door you have a 2/3 chance of winning) but in this variant the chances change to either 1/1 or 1/2 when the host opens a door (depending on which door the host opens).

The ONLY point in bringing up this variant is to demonstrate the flaw in the argument that the host's actions can't change the probability. They most certainly can. The host can't change the fact that 2/3 of players who decide to switch before a door is opened will win, but this doesn't necessarily mean that 2/3 of the players win when the host opens Door 3. This is an important aspect of the Monty Hall problem that, frankly, VERY few people understand (I didn't, until about a year ago - when an anonymous user brought it up on this talk page and provided the initial reference to the Morgan et al. paper and quite persistently argued about it). It is, in a sense, a second order issue (the first order issue being the "is it 1/2 or 2/3" confusion), but treating the problem rigorously, as a conditional probability problem, is not that difficult and leads to a much better understanding. This more rigorous solution is the topic of the paragraphs you've highlighted above. Could these be made more clear? Almost certainly. Should they be removed? I don't think so. -- Rick Block (talk) 00:24, 27 October 2008 (UTC)[reply]

Monty choosing the CONTESTANT'S door? Not in the puzzle. Therefore, the rest of your diatribe is meaningless. Equal goat door constraint? Never heard of it.

Look, the opening Wikipedia solution - once you remove the first sentance - proves that the odds are 1/3 vs 2/3. Once that mission is accomplished, the rest is either redundant, or incorrect. Is more than one proof valuable? Sure, but not the malarky that's there now.

Conditional probability? Hogwash!

I'm not familiar with a valid proof that is 'rigorously insufficient'. This isn't a beauty contest, or a political debate. It's math. It's a binary system, either it is, or it isn't a valid proof. The probabilities of those two events equals 100%. Like I said, that's a bedrock principal.

You have failed to demonstrate a weakness in the Wikipedia proof, Martin's proof or my proof. Yet you bring up all these other things. It's a simple puzzle, once you realize that Monty leaves a car on stage 2/3 of the time (thank you Martin). Again, 100% minus 2/3 leaves 1/3 for the contestant's choice.

Glkanter (talk) 01:42, 27 October 2008 (UTC)[reply]

Maybe we should take this one step at a time. Do you agree there may be a difference between the probability of winning by switching before the host opens a door and after the host opens a door? Or, at least, that these are different questions? -- Rick Block (talk) 02:09, 27 October 2008 (UTC)[reply]

I've read this entire discussion page from top to bottom. Other people have come before me, and pointed out essentially the same things I'm saying. No, thank you, I do NOT want to debate some other puzzle questions with you. It's a simple puzzle. In fact, the puzzle is almost like a Seinfeld episode. Nothing happens! The probabilities DON'T CHANGE! You seem to not understand Probability Theory. Otherwise, you would have acknowledged the validity of the proofs, and therefore the uselessness of those other issues as they relate to the Monty Hall Problem. Despite your valiant efforts at requiring 'a rigorous solution', you allowed a COMPLETELY ERRONEOUS STATEMENT to lead the solutions discussion. And have resisted multiple efforts to improve that section! That erroneous statement is at odds with everything that Probability Theory stands for. At times, you've argued with me over the premises of the Monty Hall Problem. Monty opening the CONTESTANT'S door? Give me a break.

Glkanter (talk) 02:28, 27 October 2008 (UTC)[reply]

I'm sorry you're finding this frustrating, but I assure you I understand probability theory. You perhaps missed my post in a section above agreeing that the first sentence in the solution section could (should) be deleted. The question about which door the host is opening was asking for a clarification of a comment you made (I'm well aware the host doesn't open the contestant's door). It is in fact not a simple puzzle, which is one reason why it remains so controversial. I'm very patient, and I'm willing to go through this step by step with you if you'd like. It's sounding like you might not like. That's OK, too. If you are willing, I think the first point is the one I ask above. Are there two possible questions? -- Rick Block (talk) 02:46, 27 October 2008 (UTC)[reply]

Well, we could debate your grasp of the science, but that would just get personal. I'd just like the interested reader to be able to see a couple or three simple proofs to help him/her understand the solution. That's really the only reason Wikipedia exists. There is a bunch of stuff under the Solutions heading that interferes with that. I, and people before me, have already demonstrated that. Is it possible to speak with your supervisor, or are you the final word?

Glkanter (talk) 03:13, 27 October 2008 (UTC)[reply]

Point by point form your response above...

"It doesn't matter which door Monty shows, but only because he's constrained to open one of the two goat doors with equal probability in the case the player initially selects the car. Again, you're asserting this without justifying why this is true."

False. I simply said "It doesn't matter which door Monty shows'. You added the rest. There's nothing to justify. How does Monty's or the contestant's behaviours change based on the door #s, or which goat he reveals? There's nothing in the puzzle about that.

"Regarding the odds not changing, are you saying Monty does not consider opening the door chosen by the contestant? This, by itself, is not sufficient to ensure that Monty's actions don't change the initial 1/3 probability."

Here's where you challanged the assumptions of the puzzle. Sad.

"The total probability must be 100%, and before the host opens a door it's surely 1/3 player's door vs. 2/3 for the other two doors (so 2/3 of the players who decide to switch before the host opens a door will win) but the only thing that keeps it that way after the host opens a door is that pesky equal goat door constraint."

False. The only thing 'keeping it that way' is that pesky law of probabilty that the outcomes must = 100%.

Glkanter (talk) 03:46, 27 October 2008 (UTC)[reply]

I am no more, and no less, an editor than you are - although I am an admin on the en.wikipedia site (admins are merely editors who are viewed by the community as trustworthy enough to have certain capabilities not available to most users, like deleting pages and blocking vandals - but admins distinctly do not work for the Wikipedia Foundation, so talking to my "supervisor" is a curious thought). The "final word" here is the consensus of the editors of this page. If you think there's a conflict you're welcome to follow the steps described at dispute resolution. If you'd like, you're also welcome to make whatever changes to the article you think are appropriate, although since you're apparently a new editor I'd suggest you review the 5 pillars. I'll also point out that this article is a featured article and has gone through two featured article reviews. The "stuff" you're objecting to in the Solution section was added in response to the most recent review, so it's not just me you're arguing with here but essentially the entire set of folks who have an interest in this article.

I'm quite willing to discuss this with you (if you'd like) in a less public forum such as your talk page, or mine, or email. -- Rick Block (talk) 04:30, 27 October 2008 (UTC)[reply]

Rick, thanks for your comments on my 'very short' solution. I have to say that Glkanter is not alone in being dissatisfied with the basic explanation in the current article. Myself and at least one other editor have also criticised it. It is also obvious from the reactions of editors just passing through (readers) that the article is not as convincing as it could be. Because of the FA status of this page, I am reluctant to just wade in and start editing it, which is why I set up my development page. The hope was that, on that page and it its associated talk page, we could all (old and new editors) work together to find ways to improve this article without compromising its current status. I would therefore suggest that all persons interested in the quality of this article discuss deficiencies of the current solution and proposals for new ones on that page. In particular, I will answer you questions to me above concerning what I do not like about the current article and I will look forward to your replies there.Martin Hogbin (talk) 09:24, 27 October 2008 (UTC)[reply]

My development page was set up to allow discussion rather than edit warring so I do not think that it would be right for old and existing editors to completely ignore my page but then start reverting changes to this page based on a consensus reached by those expressing their views on the development page. We need to move forwards but in the right way. Martin Hogbin (talk) 09:24, 27 October 2008 (UTC)[reply]

Proposed Changes

Several editors believe that, although the current article may meet the FA criteria, it still remains unconvincing and also concentrates far to much on variations of the basic problem that are irrelevant to the core problem.

It is therefore proposed that the current solution be replaced with a new version giving simple, intuitive, and verifiable solutions to the basic problem.

Preparation for this new section is taking place on a [[4]] . All interested editors are asked to make their comments on the suitability of this section for inclusion in the current article and to make improvements to the new section. Martin Hogbin (talk) 20:54, 6 November 2008 (UTC)[reply]

I couldn't find any discussion on the linked page, but I think the simplest explanation of the solution would be as follows (START):

Sorry, I have now corrected the link [[5]]. Martin Hogbin (talk) 09:26, 7 November 2008 (UTC)[reply]

The wording of the puzzle can be misleading because it seems to imply that after the goat is revealed, the player is now picking one of two doors. In reality, when offered a switch, the user is ALWAYS picking between their one door and BOTH of the other doors. Think about it this way. Instead of openening a door with a goat, once the player selects a door, Monty offers the player a choice, they can take whatever is behind their door, or take whatever is behind the two doors they didn't pick. Simple logic tells the player that AT LEAST ONE of those two doors MUST contain a goat, there's only one car. However, the player also knows that two doors are better than one. Since they picked one of three doors, the chance that their door contains the car is only 1/3. The chance that one of the other two doors contains the car is thus 2/3. Whether or not the user switches, we know one of the unpicked doors contains a goat. Revealing the presence of the goat before the switch offers no new information. Monty is not revealing a big secret, everyone knows one of two unpicked doors had to contain a goat. Monty just saved a little suspense and air-time by pre-revealing the goat we knew had to be behind one of the two unchosen doors. Therefore, the odds of winning the car by switching to both of the other doors (before or after the goat is revealed) are always 2/3. The key here is that the user IS switching to BOTH unchosen doors. The one we knew had a goat behind it just happens to already be open.

I think what you are suggesting is pretty much what is on the development page under 'combining doors'.Martin Hogbin (talk) 09:26, 7 November 2008 (UTC)[reply]

This solution assumes a couple things:

Monty will either randomly or always reveal a goat after the user picks a door. I.e. he knows where the car is.
Monty will either randomly or always offer the player the option to switch doors.

Monty randomly offering the trade is equal to always offering the trade, since in this particular instance, the trade was offered. The question becomes moot if Monty didn't offer a trade. If Monty has alterior motives, such as offering a trade only after you already picked the car, then the odds of you winning the car on a trade are 0, and it didn't matter which of the two doors (if either) was opened because once Monty offers the trade you know they both contain goats.

My main point is that not only are the 'assumptions' that you make above the natural ones that most people make, they are also what is clearly stated in the 'Parade' statement of the problem, which is where all the fuss started. It is my contention that the real problem is the one you state above and that this is what the article should concentrate on. Other, essentially academic, possibilities should then be discussed later in the article. Martin Hogbin (talk) 09:26, 7 November 2008 (UTC)[reply]

(END) Maybe this isn't any simpler to understand, but this is what I though of as I slowly convinced myself the answer really is 2/3. Hellpimp (talk) 06:41, 7 November 2008 (UTC)[reply]

Perhaps you might be able to improve the development article.Martin Hogbin (talk) 09:26, 7 November 2008 (UTC)[reply]

Are the doors distinguishable or not? They're numbered, so of course they are. This means there are two questions we can ask.

1) What is the overall chance of winning by switching, averaged across all players?

2) What is the specific chance of winning by switching for only those players selecting a given door (say Door 1) knowing the host has opened a specific door (say Door 3).

The general form of the problem statement asks the second question, not the first one. However, the argument above addresses the first question, not the second. In contrast, the Bayesian analysis section of the article answers the second question. I don't think this is merely an "academic" point. I know the online copy costs money, but I would strongly encourage anyone seriously interested in this problem to read the Morgan et al. paper in the references (it should be available at any university library). If we're attempting to answer the second question, Morgan et al. show the answer is 2/3 chance of winning by switching only if the host is also required to select between two goat doors (if given the chance) with equal probability. Any solution that does not use this constraint (the Bayesian analysis does), is either answering a different question (like the first question, above) or is incomplete (i.e., wrong).

The current structure of the article is to answer the first question first since its solution is easier to grasp, but then acknowledge that the second question is indeed a different question and proceed to answer it. If this structure is not clear, by all means clarify away. IMO, relegating this point to a subsequent "academic" section does our readers an extreme disservice. -- Rick Block (talk) 15:12, 7 November 2008 (UTC)[reply]

Although the doors may be distinguishable this is not relevant to the core problem where, to all intents and purposes, they are not. The distinguishably of the doors is another sideline which, in my opinion, should be relegated to an 'academic variations' section of the article.

Regarding the host behaviour on selecting between goat doors, the 'Parade' statement of the problem says,'...the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat...' it does not give us any information as to how the host might choose between two goat doors. In such a case it is not an assumption to take it that this is random but an obligation.

The problem with adding in all these complications is that it makes the article useless at one of its primary purposes which is explaining the solution to readers in a way that convinces them. Martin Hogbin (talk) 18:45, 7 November 2008 (UTC)[reply]

And the problem with avoiding these complications is that it makes the article useless at its primary purpose which is summarizing what reliable sources have to say about the Monty Hall problem. We don't avoid things just because they're hard to explain. Where the rubber meets the road here is the references. Two different references that appeared shortly after the vos Savant dust up, Morgan et al. and Gillman, bring up the difference between the two questions that may be asked. Yes, they're both academic papers, but unless there are other references that say these papers are obscuring the "core problem" this notion is simply your opinion. Treatments of the problem in non-academic sources still generally (possibly always) ignore this difference but this does not give Wikipedia an excuse to do so. As far as I know, pretty much all academic sources now include the "equal goat door constraint" in the problem statement. Many do so because it's necessary to justify the Bayesian approach which solves for the probability of winning by switching if the player picks Door 1 and the host opens Door 3. The Bayesian treatments solve for this, rather than simply the probability of winning by switching if the player picks Door 1, because it's what the problem ostensibly asks.

I'm aware that our exchange here has the form of an argument, but I actually don't think we're that far apart. As I've said repeatedly, I'm fine with presenting an "intuitive" solution answering the easier question first. And, indeed, the article currently does this (and whether the current intuitive solution is the most convincing is certainly debatable). However, I'm not fine with the Solution section including only something that one of the main academic references calls a "false solution".

I'll solicit other opinions about this at Wikipedia:WikiProject Mathematics. Most folks there are aware that this problem is a notorious quagmire, so it is somewhat difficult to get them to comment but I'll try. -- Rick Block (talk) 03:44, 8 November 2008 (UTC)[reply]

The solutions proposed properly solve the paradox using established Probability Theory. All this other stuff adds no value.

Glkanter (talk) 07:10, 8 November 2008 (UTC)[reply]

I got it! Start a new Wikipedia article called 'The Equal Goat Door Constraint Paradox'! Marilyn vos Savant can write a column about it in her general interest Sunday magazine column. People all over the world will marvel at how counter-intuitive the solution is, when all along, after a trip to the nearest university library, Morgan, et al and Bayesian statistics were there to explain it. Glkanter (talk) 16:17, 8 November 2008 (UTC)Glkanter (talk) 16:26, 8 November 2008 (UTC)[reply]

I'm responding to Rick's WT:WPM post. Any attempt to make the article clearer is a good idea, but it must be based upon the fact that Wikipedia is an encyclopedia. Its job is to document the knowledge found in reliable sources, not to synthesise explanations in the literature to provide the clearest and most convincing one. This article is not entitled "Explanation of the Monty Hall Problem" and Wikipedia generally doesn't have articles with such titles because they are not encyclopedic and typically involve taking a point of view. The job of this article is not to convince the reader, but to present them with information. That's the difference between an encyclopedia and a textbook: the latter aim to educate the reader, the former are a resource which help the reader to educate themselves.

Wikipedia does not have disclaimers, does not assert viewpoints, and does not contain original research. Geometry guy 19:48, 9 November 2008 (UTC)[reply]

The application of WP policies to this problem is full of difficulties. We do not need a cut-and-paste approach to verifiability, neither do we want opinion, but where between these two extremes to we want to be? What is a reputable source? In this subject there have been many sources that would normally be considered reputable that have got the answer wrong. Common sense is no help either. I am not saying it is easy but I believe that we can do better.

Clearly, the current article does not just, 'document the knowledge found in reliable sources' neither should it, in my opinion. What is the point of having an article the contains information if nobody believes it?

Finally, what is the justification for the current article's current concentration on academic variations of the problem. This is not what makes the problem notable. Martin Hogbin (talk) 23:24, 9 November 2008 (UTC)[reply]

Between the two extremes we want (from Wikipedia:Featured article criteria) to be factually accurate: claims are verifiable against reliable sources, accurately represent the relevant body of published knowledge, and are supported with specific evidence and external citations.

What constitutes a reputable (reliable) source is defined at Wikipedia:Reliable sources.

Per the quote from the featured article criteria above, the point is to accurately represent the relevant body of published knowledge. What is to be believed about any article is that it does this. References are included specifically so readers can verify that what is said in the article is supported by published sources.

I'm not sure what you mean by the article's current "concentration on academic variations", but (also per the FA criteria) articles should be comprehensive: it neglects no major facts or details. More on this from Wikipedia:Neutral point of view: content must be written from a neutral point of view (NPOV), representing fairly, and as far as possible without bias, all significant views that have been published by reliable sources. -- Rick Block (talk) 15:43, 10 November 2008 (UTC)[reply]

Still confused ..

The following introduction is still confused suffering from poor English, a lack of clarity and some misconceptions.

"Because there is no way for the player to know which of the two unopened doors is the winning door, most people assume that each door has an equal probability and conclude that switching does not matter. In fact, in the usual interpretation of the problem the player should switch—doing so doubles the probability of winning the car from 1/3 to 2/3. Switching is only not advantageous if the player initially chooses the winning door, which happens with probability 1/3. With probability 2/3, the player initially chooses one of two losing doors; when the other losing door is revealed, switching yields the winning door with certainty. The total probability of winning when switching is thus 2/3."

The object of the problem is not whether the person wins the automobile but how the probability shifts when a door is exposed (with a goat). If the person selects the correct door or not on the first guess is irrelevant to the problem - especially as an introduction.

A breakdown:

(1) Because there is no way for the player to know which of the two unopened doors is the winning door, most people assume that each door has an equal probability and conclude that switching does not matter.

Is fine

(2) In fact, in the usual interpretation of the problem the player should switch—doing so doubles the probability of winning the car from 1/3 to 2/3.

Does not need "in the usual interpretation of the problem" in the introdcution.

(3) Switching is only not advantageous if the player initially chooses the winning door, which happens with probability 1/3.

Should read, "If the player initially chooses the winning door then changing is clearly disadvantageous" I don't even think it is necessary as this is confusing in the introduction.

(4) With probability 2/3, the player initially chooses one of two losing doors; when the other losing door is revealed, switching yields the winning door with certainty.

Is badly constructed. It should read, "If the player has initially chosen one of the two losing doors: when one of the losing doors is revealed then switching wins the player the automobile."

However, I would argue that this isn't even needed anyway.

(5) The total probability of winning when switching is thus 2/3.

Is a non-sequitur as the paragraph is referring to two different scenarios (ref to 2 and 3). Both scenarios, incorrect first choice and correct first choice cannot lead to the same probability of winning when switching. --Candy (talk) 22:52, 24 November 2008 (UTC)[reply]

So, in summary, you're suggesting a change to perhaps something like:

Because there is no way for the player to know which of the two unopened doors is the winning door, most people assume that each door has an equal probability and conclude that switching does not matter. In fact, the player should switch—doing so doubles the probability of winning the car from 1/3 to 2/3. If the player initially chooses the winning door with probability 1/3, then changing is clearly disadvantageous. If the player has initially chosen one of the two losing doors with probability 2/3, then after the other losing door is revealed switching wins the automobile. The total probability of winning for all players who switch is thus 2/3.

or cutting the explanation after the "In fact" sentence. I'd support cutting the explanation (for reasons discussed at length in previous threads on this talk page this explanation is somewhat less than completely satisfactory). -- Rick Block (talk) 02:54, 25 November 2008 (UTC)[reply]

Good faith edits

Reversion of good faith edits should not be marked as minor. Martin Hogbin (talk) 20:27, 11 December 2008 (UTC)[reply]

Is there some specific revert you're talking about here? -- Rick Block (talk) 00:49, 12 December 2008 (UTC)[reply]

Recent reverts marked as minor:

Nov. 1 by Belinrahs - reverted humorous comment
Dec. 3 by Iridescent - reverted vandalism
Dec. 4 by Paul August - reverted change of 2/3 to 1/3 - sneaky, but possibly good faith
Dec. 11 by Rick Block - reverted additions - ~~probably good faith~~ - the edit combined an addition to an image caption (that could be good faith) with another addition (that clearly wasn't) - I think Martin missed that bit (I know I did!).--Noe (talk) 12:53, 12 December 2008 (UTC)[reply]

Sorry, I did miss that. Martin Hogbin (talk) 20:11, 12 December 2008 (UTC)[reply]

External link to ucs.edu page

I've moved the following link which was added to the article here. There's nothing on this page that can't be incorporated directly into the Wikipedia page, per WP:EL that means it is not an appropriate external link.

The Monty Hall Problem at LEEPS, a lab run by Dr Dan Friedman (a prominent economist at UC Santa Cruz).

-- Rick Block (talk) 14:39, 19 December 2008 (UTC)[reply]

--- I believe the above link should be on the main page for the Monty Hall Problem. It cites important academic journals and articles on the issue. We cannot ignore literature that discuss this problem as it was a hot topic of discussion amongst prominent economists and mathematicians. Some of the articles mentioned in the above link do form a basis to the solution that is widely accepted today. - Aadn

From WP:EL#Links normally to be avoided:

1. Any site that does not provide a unique resource beyond what the article would contain if it became a Featured article.

Also from WP:EL (Advertising and conflicts of interest):

You should avoid linking to a website that you own, maintain or represent, even if the guidelines otherwise imply that it should be linked. If the link is to a relevant and informative site that should otherwise be included, please consider mentioning it on the talk page and let neutral and independent Wikipedia editors decide whether to add it.

The first of these definitely applies here and I strongly suspect the second does as well. The LEEPS page itself does not meet Wikipedia's standard for reliable sources, so shouldn't be used as a reference either. You are perfectly welcome to add citations drawn from the LEEPS page to the history section, especially any that are considered major or important. Note that there are dozens (if not hundreds) of published papers on this problem, so clearly most of them will not be mentioned here. -- Rick Block (talk) 05:22, 21 December 2008 (UTC)[reply]

The Goat

What if I want the goat and not the car? Should this probability also be addressed, even if it's not part of the original Monty Hall problem? 76.95.124.146 (talk) 15:09, 8 January 2009 (UTC)[reply]

This question may be useful in settling the issue of how best to explain the solution. The goats are actually a distraction from the real puzzle, serving much the same purpose in this trick puzzle as "misdirection" serves a magician. If a goat is an anti-prize, worse than neutral, then it may be more difficult to grasp why the revealing of a goat behind a non-chosen door is not useful information in solving this particular puzzle, although it can be useful in solving other puzzles that resemble it. The first step to providing a convincing explanation of the solution may be to remind the solver that one is not actually required to take the goat home if the goat is chosen! (In fact, a goat is equivalent to no prize at all.) This may disappoint 76.95.124.146, but for other solvers, it may be a good first step to understanding the puzzle. Perhaps it is unfortunate if no reliable source has taken this approach.Simple314 (talk) 03:29, 11 January 2009 (UTC)[reply]

Addition of simplest explanation

I've undone this edit which adds an introductory paragraph to the Solution section with numerous WP:MOS issues discussing an alternate problem (boxes rather than doors). The addition is essentially the same as the existing "combining doors" section. The text is in the history if anyone wants to talk about it. -- Rick Block (talk) 06:01, 24 January 2009 (UTC)[reply]

Minesweeper analogy

I've deleted the addition of the following new section.

Minesweeper analogy

An analogy which might have a more intuitive solution is the computer game Minesweeper with the minor change that the game is played without numbers indicating mines. Suppose one mine (the equivalent of the car) is hidden in a 100x100 grid. The first action by the player is to randomly plant a flag on one of the squares (equivalent of picking a door at the beginning). The player clicks on a square without the mine (this is actually how many versions of minesweeper are programmed, the first click cannot be a mine) clearing the entire grid except for one square and the one flagged by the player (the equivalent of Monty opening the door revealing a goat). For anyone familiar with the game Minesweeper, the intuition here is that the mine is under the unflagged square since the player flagged the square randomly at the beginning of the game. This analogy also removes any possible motivation on Monty's part to trick the player which may be confusing in the classic example.

Although interesting, it's unreferenced and sounds like WP:OR (and requires an understanding of Minesweeper). If anyone can find a reliable source including this analogy we can talk about re-adding it. -- Rick Block (talk) 01:43, 25 January 2009 (UTC)[reply]

It's a very good analogy, though, and in fact it's a form of the "increased number of doors" explanation. But it does have a critical flaw: if the player actually manages to flag the mine, all the other squares will get cleared, which would be the equivalent of Monty revealing every single goat. So it actually simulates a very different "Monty" -- one who reveals every goat unless there's one behind your door.--Father Goose (talk) 09:41, 25 January 2009 (UTC)[reply]

The description mentions that all the squares except the initially flagged one and one more are cleared (making it not quite like minesweeper). Perhaps a truer version would be you flag one, Monty flags another, and then Monty clicks one he knows isn't the winner. This actually scales down to a 3-square version identical to the MH problem. In any event, if it's not published I don't think we should even consider using it. If it is published we could talk about it, but even then I'm not sure it really adds anything. -- Rick Block (talk) 17:20, 25 January 2009 (UTC)[reply]

Still no good: in most cases you'll be left with three unrevealed squares: your flag, the host's flag, and the actual location of the mine. Even in the three door version, it won't work: if both of you pick goats (flag an empty square), when you click on the last remaining square (the mine), the game will move the mine under either your flag or his, presumably randomly. So there's no restricted choice on his part: both his choice and yours are entirely random. It's the "host doesn't know" scenario.--Father Goose (talk) 21:21, 25 January 2009 (UTC)[reply]

Wrong explanation

The, what is often called, "simple solution", saying: the player originally has a chance 1/3 of winning, so switching increases this chance to 2/3, is theoretically wrong! This explanation is also offered directly under "Solution" , and although it offers numerically the right answer,it is not correct, as may be seen by comparing it to the statement of the problem right above it. In the problem statement the question is: Should the player switch to Door 2? And in the solution the player is allowed to switch to Door 3 as well. The point is, and I think it is mentioned somewhere in the above discussions, that the posed question can only ask for a conditional probability, given the situation, i.e. Door 1 chosen and Door 3 opened revealing a goat. This is also clear from the equivalence with the Three Prisoners problem and Bertrand's box paradox.Nijdam (talk) 11:26, 3 February 2009 (UTC)[reply]

Rather than flat out "wrong", I'd say the simple solution is a correct solution to a slightly different problem. This is the point of the three paragraphs below the large figure in the Solution section of the article. Many (possibly nearly all) people interpret the question to apply to all players unconditionally rather than only those players seeing a specific open door (and a fair number apparently cannot understand that these are actually different questions). These people would claim the wording "say you've picked Door 1 and the host opens Door 3" is saying that we are to assume all specific situations are equally likely, in which case the unconditional and conditional answers must be the same. If you can make this more clear in the article, please do so. -- Rick Block (talk) 14:49, 3 February 2009 (UTC)[reply]

This is an other way of putting it. But as the article is so precise in defining the problem, with all guarantees of no possible misinterpretation, the given "simple solution" is no solution to this precise stated problem.Nijdam (talk) 16:04, 3 February 2009 (UTC) —Preceding unsigned comment added by 82.75.67.221 (talk) [reply]

I've revised your clarification. Are you OK with the revised version? -- Rick Block (talk) 19:34, 3 February 2009 (UTC)[reply]

Any of you recent contributors may not have read this page in its entirety. It's pretty long. So, just to let you know, many, many people are very unsatisfied with the entire article. It's long winded, confusing, and outright wrong in places. But we respect the process, so here we are. Search for my signature, as one example. Glkanter (talk) 03:30, 4 February 2009 (UTC)[reply]

Let me see if I've got this right. Glkanter thinks the simple solution (the part above the large figure in the solution section) is ridiculously long winded but conveys the only essential aspect of the solution, and the part below the figure is incomprehensible and wrong and should be deleted. Nijdam thinks the part above the figure is wrong and, if not deleted, should be prefaced with something like "the following is wrong:". How about if you two talk. I'd be happy to try to moderate. -- Rick Block (talk) 05:48, 4 February 2009 (UTC)[reply]

I tried to follow the discussion above between the two of you, and I actually got confused, not about the problem or the solution, but about you. I'm not very happy with Rick's revision of my extension to the "simple solution". My extension shows in essence the flaw in this way of reasoning. I showed that the "simple solution" is not a solution to the stated problem. It's like the man, searching in the evening for his lost wallet under a streetlamp where things are well visible, allthough he knows he lost the wallet in a dark side alley. So I invite either one of you to read my extension an comment on it.Nijdam (talk) 10:49, 4 February 2009 (UTC)[reply]

I did a little more reading of the above discussion about this topic and I think I find Rick on my side, but Glkanter opposed because of the reasoning: 1/3 + 2/3 = 1, meaning: the initial probablility is 1/3 of picking the car, so whatever happens, as (!?????) this has not changed, the final probability of the alternative should be 2/3. Well as you may guess, I already indicate where it goes wrong. The "initial probability changes" under the assumed condition. Among all possible cases, the ones with the car behind Door 1 and Door 3 openend, only form 1/6th.Nijdam (talk) 11:28, 4 February 2009 (UTC)[reply]

Nijdam's version is here. My revision is the next edit in the history. I have two issues with Nijdam's version. 1) It's a little too harsh on the "simple solution". In its current form, this solution doesn't say anything that is actually incorrect it just doesn't precisely answer the question. Also, like it or not, it is what is commonly presented as the "solution". 2) The style doesn't follow Wikipedia style guidelines, in particular see Wikipedia:Manual of Style (mathematics)#Writing style in mathematics. -- Rick Block (talk) 15:02, 4 February 2009 (UTC)[reply]

Nijdam, please show me the error in the 1/3 + 2/3 = 1 (100%) solution. I'd also like to know what is unacceptable about Martin's solution, 2/3 of the time I choose a goat, and 100% of those times Monty show me the other goat? Everything else, from Morgan, et al, to Bayesian, to "We have run simulations that prove this" is wasted binary digits. Glkanter (talk) 23:12, 4 February 2009 (UTC)[reply]

I thought you were skilled enough to directly understand the problem. First: The "simple solution" is not a solution at all, i.e. not to the stated problem. As Rick puts it: "it just doesn't precisely answer the question"!! As if this is not enough reason to be harsh. 2) I'm really completely stunned about this. Let us translate the situation to an exam. Well, the professor utters: "let's not be too harsh on this "solution". Nothing is really incorrect, it just doesn't answer my question." 3) At least Rick understands it is wrong. 4) Why is it wrong? As I explained in my text, the "solution" offers the possibility that Door 2 is opened, and the statement of the problem excludes this. 5) Now about the specifc question about Martin's reasoning. It has the same flaw: it doesn't answer the posed question. Of course it is the right answer to the right alternative question, which I may present as follows: the player is blind and Monty only tells him: I opened one of the other doors and there is a goat behind it. But that is not the case in de stated problem. There the player sees that Door 3 is opened and shows a goat, and hence the conditions of the problem are accordingly formulated. 6) All this is because after some events has happened, we want to know the conditional probabilities, given these events. Compare it with throwing a fair dice and suppose it shows 6. What is the probability it will show 5? (0 or 1/6???). I hope it becomes clear to you. Sorry, forgot to log in.Nijdam (talk) 23:56, 4 February 2009 (UTC)[reply]

In a moderating kind of way, I have two suggestions here. 1) Please comment on content, not on other editors' mental facilities. WP:COOL has some good suggestions. 2) As far as possible, please support arguments with references rather than arguing what you simply know to be the Truth. There are no shortage of references about the Monty Hall problem - any significant point has almost certainly been published (at least once). -- Rick Block (talk) 02:00, 5 February 2009 (UTC) [reply]

Following my own advice - to Glkanter, the issue with both your and Martin's solutions is explained in Morgan et al. These are both variants of what Morgan et al call "false solution F1".

Solution F1. If, regardless of the host's actions, the player's strategy is to never switch, she will obviously win the car 1/3 of the time. Hence, the probability that she wins if she does switch is 2/3.

F1 is immediately appealing, and we found its advocates quite reluctant to capitulate. F1's beauty as a false solution is that it is a true statement! It just does not solve the problem at hand. F1 is a solution to the unconditional problem, which may be stated as follows: "You will be offered the choice of three doors, and after you choose the host will open a different door, revealing a goat. What is the probability that you win if your strategy is to switch?" The distinction between the conditional and unconditional siutations here seems to confound many, from whence much of the pedagogic and entertainment value is derived.

Similarly, Gillman 1992 (referenced in the article) criticizes vos Savant's solution:

Marilyn's solution goes like this. The chance is 1/3 that the car is actually at #1, and in that case you lose when you switch. The chance is 2/3 the car is either at #2 (in which case the host perforce opens #3) or at #3 (in which case he perforce opens #2) - and in these cases, the host's revelation of a goat shows you how to switch and win.

This is an elegant proof, but it does not address the problem posed, in which the host has shown you a goat at #3. Instead it is still considering the possibility that the car is at #3-whence the host cannot have already opened that door (much less to reveal a goat). In this game-you have to announce before a door has been opened [emphasis in the original] whether you plan to switch.

These are both from academic WP:reliable sources (The American Statistician and American Mathematical Monthly), and the article is (at this point) attempting to present the information contained in these sources.

Rick, can you please clarify the following paragraph? It is essential to my understanding of the criticism. Thank you.

This is an elegant proof, but it does not address the problem posed, in which the host has shown you a goat at #3. Instead it is still considering the possibility that the car is at #3-whence the host cannot have already opened that door (much less to reveal a goat). In this game-you have to announce before a door has been opened [emphasis in the original] whether you plan to switch.

Glkanter (talk) 16:45, 7 February 2009 (UTC)[reply]

Rick, I'm still waiting for your clarification of the above statement. As it is central to your claim of the non-validity of my proof. What does this mean:

'In this game-you have to announce before a door has been opened [emphasis in the original] whether you plan to switch.'?

Are they talking about the Monty Hall Problem or some varient? Or is this the 'before switching' beast you speak of? MOST IMPORTANTLY, what possible relavence can it have to my proof?

Glkanter (talk) 07:30, 8 February 2009 (UTC)[reply]

Is the view that the "simple solution" is sufficient supported by equally reliable sources? If so, which sources and what do they say? -- Rick Block (talk) 04:56, 5 February 2009 (UTC)[reply]

Well, don't be offended; take it as a complement instead, because from the discussions above I concluded both of you are well skilled at least in mathematics. Yet you keep asking for arguments, which I gave in the extension of the article and extensively above. Moreover, Rick quotes the same arguments I gave. So what more needs to be said? If the "simple solution" as a solution to the well stated problem, is supported by any source, the conclusion can only be, this source is not reliable. The only necessity is changing he article, as I suggested.82.75.67.221 (talk) 10:01, 5 February 2009 (UTC)[reply]

Just to be clear, you're arguing for reinstating this change (right?), which prefaces the simple solution with:

The following "solution" is often stated, but is not the correct solution to the problem, because in the above precise stated problem not only did the player picks Door 1, but is it also known that Door 3 is opened and shows a goat. Let us yet follow the reasoning and see what goes wrong.

and interjects a new paragraph

For the attentive reader it will be clear that the player in this solution is also confronted with the possibility to find Door 2 opened showing a goat and offered to switch to Door 3. This is not allowed in the statement of the problem. However there is a way to work around it. Looking at the pictures we see this is the case in the first and the last picture in the last row. If we restrict the analysis to the other cases (for the probabilistic skilled reader this means conditioning on these events), the (conditonal) probability of winning the car by switching is 1/3 against 1/3+1/6, which is also 2/3.

before the more technically correct solution. The current structure is to start with a commonly presented simple solution being careful to avoid saying anything that is actually incorrect (in particular, qualifying that this solution applies to all players before the host opens a door), and then transition into the more precisely correct solution. I've tried to clarify the transition, as follows:

Although the reasoning above is correct it doesn't answer the precise question posed by the problem, which is whether a player should switch after being shown a particular open door (Morgan et al. 1991). Answering this question requires determining the conditional probability of winning by switching, given which door the host opens. The difference is whether the analysis, as above, considers all possible scenarios or only the scenarios where the host opens a specific door. The conditional probability may differ from the overall probability depending on the exact formulation of the problem (see Sources of confusion, below).

I think this follows the sources fairly faithfully, leading the reader to the more technically correct solution without making a huge point out of it. I think it is true (I don't have a handy reference, but could perhaps find one) that even though the detailed problem statement is mathematically unambiguous many people misinterpret it as asking about the unconditional probability (for which the simple solution is the correct answer). As I've said, calling the simple solution wrong seems overly harsh - particularly since the unambiguous version is specifically designed to end up with the same answer (I don't have a reference for this either, but it seems fairly obviously true). -- Rick Block (talk) 15:36, 5 February 2009 (UTC)[reply]

Well, I find it unacceptable to start with a "solution" that isn't a solution. My suggestion (please improve my English, as I'm foreign):

Proposed Solution section

According to the problem statement above, a car and two goats are arranged behind three doors and then the player initially picks a door. Assuming the player's initial pick is Door 1 (the same analysis applies for any other door the player picks):

The player originally picked the door hiding the car. The game host must open one of the two remaining doors randomly.
The car is behind Door 2 and the host must open Door 3.
The car is behind Door 3 and the host must open Door 2.

This leads to the following situations:

Car hidden behind Door 1		Car hidden behind Door 2	Car hidden behind Door 3
Player initially picks Door 1

Host opens either goat door		Host must open Door 3	Host must open Door 2

This hasn't happened	This happens with "probability" 1/3	This happens with "probability" 2/3	This hasn't happened
	Switching wins in 2 out of 3 cases

Players who choose to switch win if the car is behind Door 2. This is the case in 2 out of 3 cases, so with 2/3 probability.

This means if a large number of players, after choosing Door 1 and being showed a goat behind Door 3, randomly choose whether to stay or switch, then approximately 1/3 of those choosing to stay with the initial selection and 2/3 of those choosing to switch would win the car. This result has been verified experimentally using computer and other simulation techniques (see Simulation below).

Nijdam (talk) 17:56, 5 February 2009 (UTC)[reply]

How about this? (and your English is far better than my Dutch) -- Rick Block (talk) 02:58, 6 February 2009 (UTC)[reply]

According to the problem statement above, a car and two goats are arranged behind three doors and then the player initially picks a door. Assuming the player's initial pick is Door 1 (the same analysis applies for any other door the player picks):

The player originally picked the door hiding the car. The game host must open one of the two remaining doors randomly.
The car is behind Door 2 and the host must open Door 3.
The car is behind Door 3 and the host must open Door 2.

This leads to the following situations:

Car hidden behind Door 1		Car hidden behind Door 2	Car hidden behind Door 3
Player initially picks Door 1

Host opens either goat door		Host must open Door 3	Host must open Door 2

Switching loses with probability 1/6	Switching loses with probability 1/6	Switching wins with probability 1/3	Switching wins with probability 1/3
Switching loses with probability 1/3		Switching wins with probability 2/3

Looking at the figure above, in cases where the host opens Door 3 (which is the case according to the problem statement) switching loses in a 1/6 case where the player initially picked the car and wins in a 1/3 case where the player initially picked a goat. Players who choose to switch win twice as often as players who stick with the original choice, which is a 2/3 chance of winning by switching.

This means if a large number of players randomly choose whether to stay or switch after initially choosing Door 1 and being shown a goat behind Door 3, then approximately 1/3 of those choosing to stay and 2/3 of those choosing to switch would win the car.

We'd probably need to follow this up with a discussion of the typical unconditional solution and why it is not responsive to the question that's asked. I think this structure is probably less accessible than the current structure to someone not mathematically sophisticated (indeed, as Morgan et al. comment, "the distinction between the conditional and unconditional situations here seems to confound many"). Typical Wikipedia style is to start with broadly accessible content and then go deeper (see Wikipedia:Make technical articles accessible), which is the overall intent of the current structure. As I've said the current structure is meant to present an analysis that perhaps doesn't quite address the question that's asked (but is not actually wrong) leading into an technically correct solution. I think this is an article of more interest to laypersons than mathematicians (although mathematicians frequently get this problem wrong as well), so I think anything we can do to make the solution generally accessible is worthwhile. -- Rick Block (talk) 02:58, 6 February 2009 (UTC)[reply]

Sorry, among all the almost simultaneous contributions to this page, I didn't notice this part. I'm sorry too to say I prefer my suggestion. Especially because the exact formulation of the problem, as given in the article, states Door 1 is chosen and Door 3 opened.Nijdam (talk) 12:23, 8 February 2009 (UTC)[reply]

Glkanter's objection

Note: There's a summary below, see #Request for brief clarifications. -- Rick Block (talk) 20:55, 8 February 2009 (UTC)[reply]

Monty does not pick doors randomly. He always reveals a goat. And it does not matter which goat. I'm still waiting to hear why Martin's solution is not satisfactory.

Glkanter (talk) 23:47, 5 February 2009 (UTC)[reply]

Please read the excerpts from Morgan et al. and Gillman, above. Even more directly, Morgan et al. discuss what they call false solution F5:

Solution F5: The probability that a player is shown a goat is 1. So conditioning on this even cannot change the probability of 1/3 that door 1 is a winner before a goat is shown; that is, the probability of winning by not switching is 1/3, and by switching is 2/3.

Solution F5, like F1, is a true statement that answers a different problem. F5 is incorrect because it does not use the information in the number of the door shown.

We've been through this before, but the easiest way I know of to show there are two different problems, and to figure out which one a given solution is addressing, is to change the assumptions slightly so that the answers to the two problems are different. When the answers are the same it's hard to tell which one a given solution is addressing (is it this 2/3 or that 2/3?). The slight change (also from Morgan et al) is to give the host a preference for one door over the other. To make it the most extreme, let's constrain the host to open the leftmost door if possible (still always showing a goat, but opening the leftmost door unless this would reveal the car). The question remains the same, i.e. if a player initially picks door 1 and the host opens door 3 what is the chance of winning by switching? Martin's solution still applies - the chance of initially picking a goat is 2/3 and the host most show the other goat. So, Martin's solution says the answer to this variant is 2/3 chance of winning by switching (right?). On the other hand, in this variant if the host opens door 3 (which we've said he did) we know for sure the car is behind door 2 so with the given setup (player picks door 1, host opens door 3) the chance of winning by switching is 100%. The essentially "wrongness" of Martin's solution is now making it come out with the wrong numerical answer. It is still the right answer for a different question, specifically what is the chance of winning for a player who decides to switch before seeing which door the host opens. But in both variants, this is not the question we're asking.

I think you don't like analogies but I'll try one anyway. Let's say we're asked what the square of squareroot(4) is. One solution is "squareroot(4) = 2, 2 squared is 4, the answer is 4". This is the right answer, but the solution is technically incorrect because squareroot(4) is either 2 or -2. For this particular numerical problem it doesn't matter if you ignore the negative root, but the solution is clearly wrong. We can expose the flaw by changing the question, for example "what is twice the squareroot(4)". We have a very similar issue here. No one's arguing that the answer (2/3 chance of winning by switching) is incorrect. It is correct. It's just not answering the question that's asked. -- Rick Block (talk) 02:01, 6 February 2009 (UTC)[reply]

I'll be honest. I don't understand most of what you guys are talking about. It's a simple puzzle. Do you increase your chances of winning the car by switching? It's been years since I took a Logic course, or a Probability and Statistics course, so I can't cite chapter and verse. But the goal is to present a valid proof with as little complication as possible. So you take shortcuts, which don't affect the proof. Two goats? I only care about the car. Ignore the goats in the proof. 3 doors? There'll only be one 'other one' left when I'm asked to switch. Ignore the door #s in the proof. Conditional probability? Don't know what it means. Don't see how anything is conditional here. Two closed doors when I'm asked if I want to switch, exactly one has a car. No conditions exist.

I do remember something about a requirement for proofs being valid and true. 2/3 of the time the remaining door I didn't choose has a car. Both valid and true.

That's why the real game show didn't work like this. Glkanter (talk) 03:16, 6 February 2009 (UTC)[reply]

The essential point is regarding the question that's being asked. Is it

a) "what is the chance of winning for a player who decides to switch before seeing which door the host opens"

or

b) "what is the chance of winning for a player who decides to switch after seeing which door the host opens"

I think most statements of the problem at least implicitly ask the second question (the unambiguous version in the article explicitly asks the second). Martin's solution addresses the first question. Since the numerical answers to these two questions can be different (per the variant mentioned above), a solution for one is clearly not the same as a solution for the other. Simply because a "valid and true" solution for #a ends up with the right numerical answer for #b (which it might for some versions of the problem), doesn't make it a "valid and true" solution for #b.

Putting this yet another way, if there are 3000 people who go on the show and initially pick door #1, roughly 1000 will pick the car and roughly 2000 will pick one of the goats. What Martin's solution is saying is that if all 3000 decide they're going to switch before seeing what door the host opens, roughly 2000 will win the car. However, the problem statement says we know the host opens door 3 so the divisor is not 3000. What is the actual divisor (roughly how many times will the host open door 3) and how do you know this? In how many of these cases will the player have initially picked the car? To answer what the probability is of winning the car by switching if the host opens door 3 these are the numbers you need to know. What does Martin's solution have to say about either of these? -- Rick Block (talk) 04:58, 6 February 2009 (UTC)[reply]

Okay Rick, at least I see you completely knows the ins and outs. So what to do? I cannot accept a reasoning that's logically wrong. What about the suggestion I made above, by changing the given "simple solution" in such a way that some possibilities are ruled out as not happened?Nijdam (talk) 11:54, 6 February 2009 (UTC)[reply]

There are really only two situations to consider. Not 3, not 6.

2/3 of the time I will choose a goat.

100% of those times Monty reveals the other goat, leaving the car.

=>2/3 of the time I should switch.

1/3 of the time, I will choose the car.

100% of those times Monty reveals one of the goats. It doesn't matter which.

=>1/3 of the time I should not switch.

2/3 is greater than 1/3.

Therefore, by switching, I increase my chances of getting the car when offered the switch.

I earlier said a good proof uses 'shortcuts'. A better phrasing would have been 'eliminates the unessential' from the proof. That is, a goat is a goat, and a door is a door.

Martin's proof, and the one above, address the problem as defined. To deny that is incorrect. [User:Glkanter|Glkanter]] (talk) 13:34, 6 February 2009 (UTC) 63.97.108.114 (talk) 14:19, 6 February 2009 (UTC)Glkanter (talk) 14:22, 6 February 2009 (UTC)[reply]

Please reread my responses above. Per Morgan et al., Martin's proof, and your proof above are perfectly correct if the question is "what is the chance of winning for a player who decides to switch before seeing which door the host opens". But, the actual question is "what is the chance of winning for a player who decides to switch after seeing which door the host opens". These questions are different and that they happen to have the same numerical answer in this specific case doesn't mean a solution for one is also a solution for the other. I'm sorry if you don't understand this, but I don't know how to make it more clear. Whether you understand it or not, it is what multiple Wikipedia:Reliable sources say so should be included in the article. -- Rick Block (talk) 15:29, 6 February 2009 (UTC)[reply]

Nope. I don't get it at all. My proof above very clearly is the case after showing the goat door. I think this 'before and after' dichotomy is a canard. I've said before, and you just agreed above, that Monty's action doesn't change anything. That is, he provides no new useful knowledge. The proof of this is that the probabilities haven't changed.

It takes exactly one valid proof to solve a problem. That proof has been demonstrated.

Glkanter (talk) 16:19, 6 February 2009 (UTC)[reply]

The key here is "valid", which your proof is not according to two papers written by mathematicians published in math journals. What we want in the article is what published papers have to say about this problem. I'd prefer if you understood it, but ultimately whether you do or don't doesn't actually matter. -- Rick Block (talk) 05:40, 7 February 2009 (UTC)[reply]

Thank you for the specific response to my question. I'll try to look at those papers. I suppose Morgan, et al is one of those mathemeticians, who's the other?

Do I understand your response to Nijdam correctly? Does "many published sources (possibly all popular expositions of this problem) include only an unconditional solution" mean that the solution I've been advocating is what most other references provide as the only solution?

Would you indulge me? In your own words, tell me which of the eight statements, beginning with '2/3 of the time I will choose a goat' and ending with 'Therefore, by switching, I increase my chances of getting the car when offered the switch.' is either untrue, invalid, or not germain to the question 'Now that Monty has revealed a goat, do you increase your chances of winning a car by switching'?

Thank you. Glkanter (talk) 07:06, 7 February 2009 (UTC)[reply]

The two papers I mentioned are referenced in the article, see Morgan et al. (written by John P. Morgan and others) and Gillman 1992 (written by Leonard Gillman). Both of these papers are available online for a fee and should be available at any university library. I've quoted relevant sections above.

The response to Nijdam below does indeed mean the solution you're advocating is what many popular (as opposed to academic) references provide as the solution. The two papers mentioned above were written in direct response to vos Savant's solution published in Parade Magazine. I've said all along the simple solution is correct as far as it goes, but it doesn't technically answer the question that's asked. My guess is because the numerical answer is the same, and the distinction between the two questions eludes many, even many academics are willing to let bygones be bygones. On the other hand, some academics (apparently including Nijdam) are less willing to accept the simple solution because it doesn't actually answer the question that is asked (this is like a teacher requiring you to show your work and marking your answer wrong because your work is wrong even though you got the right numeric answer).

I've tried repeatedly to point out what's wrong with your solution. I'll try once again. The problem step is

100% of those times Monty reveals the other goat, leaving the car.

This is where the solution goes astray. It is a correct statement, but "100% of those times" includes both times when the host opens door 2 and times when the host opens door 3. By including both doors the host may open, what this solution is showing is that if you decide to switch before seeing which door the host opens (knowing the host will open a door to show a goat) you'll win by switching 2/3 of the time - meaning roughly 2/3 of all players who switch will win. Absolutely correct. Check. But, the actual question is what is your probability of winning by switching after seeing which door the host opens, i.e. if you see the host open door 3, the question we're actually interested in is how many players who see the host open door 3 win by switching. I think you probably believe it can't possibly matter which door the host opens and the chances must be the same in both cases. The issue is it can matter and the chances might be different in these cases. I'm willing to bet this is where I lose you every single time.

If we're interested in the probability of winning by switching after seeing which door the host opens, there are actually two numbers we need to figure out, not just one. If we assume the player initially picks door 1, we need the probability of winning if the host opens door 2 and (separately, because it might be different) the probability of winning if the host opens door 3. These are driven by our expectations for the following:

A) The percentage of players who initially pick a goat and the host opens door 2 (i.e. 1/3)

B) The percentage of players who initially pick a goat and the host opens door 3 (i.e. 1/3)

C) The percentage of players who initially pick the car and the host opens door 2 (what is this, and why?)

D) The percentage of players who initially pick the car and the host opens door 3 (what is this, and why?)

The simple solution says nothing about these last two numbers. The probability of winning by switching for players who see the host open door 3 is B/(B+D) and, similarly, for players who see the host open door 2 is A/(A+C). If we don't know C or D, we don't know either of these probabilities. The simple solution is based on knowing only A and B, and says for all players (i.e. if we decide to switch before knowing which door the opens) we win with probability (A+B)/(A+B+C+D). We can figure this out because we know whatever C and D are, (A+B+C+D)=1.

As it turns out, the extra condition we impose on the host in the explicit problem statement (If both remaining doors have goats behind them, he chooses one randomly) allows us to determine C and D are both 1/6. This lets us conclude the probability of winning by switching if the host opens door 3 is (1/3)/(1/3 + 1/6) which works out to 2/3. Any solution that does not use this constraint is inherently incomplete.

If we vary the problem slightly by giving the host a preference for one goat door over the other, for example instead of If both remaining doors have goats behind them, he chooses one randomly we say If both remaining doors have goats behind them, he chooses the leftmost door instead of C and D both being 1/6, C=1/3 and D=0. With this change the simple solution still (correctly) says if we decide to switch before seeing which door the host opens we'll win 2/3 of the time, i.e. (A+B)/(A+B+C+D)=2/3. However, if the host opens door 3, switching wins with probability (1/3)/(1/3 + 0) which is 1/1 (not 2/3). So, although the simple solution is correctly telling us something, it's NOT telling us what the probability is of winning by switching if we know which door the host has opened. -- Rick Block (talk) 19:42, 7 February 2009 (UTC)[reply]

Your critique of the statement '100% of those times Monty reveals the other goat, leaving the car' does not show it invalid, untrue or not germain to the puzzle. You have not dis-proved my proof. I think you're hung up the doors being numbered, and that which door a goat or a car is behind makes any difference. And I know you think the goats matter, base on 'the random goat constraint'. Your arguments and frame of reference are consistant with that very first line of the solution that you agreed should be removed, "The overall probability of winning by switching is determined by the location of the car." That sentance passed two Feature Article reviews? Enough, already.

Glkanter (talk) 20:31, 7 February 2009 (UTC)[reply]

I indulged you. Can you now please indulge me? Which of the following statements do you disagree with:

1) "what is the player's chance of winning if deciding to switch before the host opens a door" is a different question than "what is the player's chance of winning if deciding to switch after the host opens a door".

2) For certain variants of this problem, the answers to these two questions can be different.

3) The question asked by the problem statement is the after question, not the before question.

4) Your solution addresses the before question, not the after question.

Thank you. -- Rick Block (talk) 22:44, 7 February 2009 (UTC)[reply]

I'll just address 3 & 4, my answer renders 1 & 2 moot.

Twice, my proof includes the words "100% of those times Monty reveals...(a) goat." I don't see how my proof can be of anything but the after-Monty-opens-a-door scenario.

Enough about before or after! A premise of the puzzle is that Monty will reveal a goat. There is no 'reveals car' scenario to consider! That would not be the Monty Hall Problem! So of my 8 line proof, 2 of those lines, for clarity purposes, repeat a premise. I'm surprised that this has come up again, we've already been around this block.

So, if that's all you got, let's declare this a valid proof and move forward. Especially eliminating that 'flawed solution' talk from the 'Solutions' section.

Glkanter (talk) 00:39, 8 February 2009 (UTC)[reply]

Here's where Rick questioned the 'Monty Reveals A Goat' premise a few months ago:

"Regarding the odds not changing, are you saying Monty does not consider opening the door chosen by the contestant? This, by itself, is not sufficient to ensure that Monty's actions don't change the initial 1/3 probability."

This goes on endlessly. People get tired of this monkey business and stop posting. But the problems haven't gone away! But you can be sure the next guy who brings this up will be in for one heck of a fight! I say, it's time for resolution escalation!

Glkanter (talk) 01:06, 8 February 2009 (UTC)[reply]

I think I've been more than patient with you. It's clear you're not thinking about (perhaps not even reading) what I'm saying, so it does seem pointless for me to continue trying to explain this to you. Is there anyone you would listen to? Martin perhaps? user:Glopk perhaps (I don't know his real name, but I think he's a math professor at a university)? user:Billjefferys (who is William H. Jefferys)? You might want to read Wikipedia:Dispute resolution. -- Rick Block (talk) 02:12, 8 February 2009 (UTC)[reply]

I'd like to weigh in here and say that I also don't understand what is different about when the player decides he or she is going to switch. In 2/3 of cases, a player who goes into the game intending to switch will win. In 2/3 of cases, a player who goes in with no plans, who decides to switch anyway, will win. Assuming the player does not communicate his internal plans to the outside world, whatever the player intends cannot possibly alter the probabilistic assessments of an external observer who doesn't know where the prize is. This seems to me like it is suggesting that the mind state of the player can somehow alter the probabilities of the outside world, which is clearly impossible. Am I misunderstanding? Maelin (Talk | Contribs) 02:28, 8 February 2009 (UTC)[reply]

Have you read the whole thread here? -- Rick Block (talk) 05:35, 8 February 2009 (UTC)[reply]

I have, but I'm still confused. Is the question "Is there a difference between between the cases in which [(the decision is made before) amd (the decision is made after)]" or is the question "Is there a difference between the probability of winning [(before) and (after)]"? In the latter case I can see there might be differences but in the former case I can't see how "at what stage the decision is made" can possibly influence the outcome. Maelin (Talk | Contribs) 10:50, 8 February 2009 (UTC)[reply]

What a cop out! Of course I read your responses. I just responded directly to points 1 through 4! So, answer the question. What is wrong with the 8 step proof? That's what I'll listen to! Come on, the charade is over. Everybody knows. It's OK to acknowledge it.

Glkanter (talk) 02:37, 8 February 2009 (UTC)[reply]

OK, I'll continue to ignore your outbursts and pretend you're responding civilly. I've told you what is wrong with your 8-step proof. It answers the wrong question. You say #1 and #2 are moot. Does this mean you agree with these statements? It sounds like you're agreeing with #3 as well. Which leaves only #4 that you apparently disagree with. So, one more indulgence.

5) Does your proof also apply to the variant I've mentioned several times where instead of If both remaining doors have goats behind them, he chooses one randomly we say If both remaining doors have goats behind them, he chooses the leftmost door?

The question remains the after question, i.e. (directly quoting from the article) Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you "Do you want to switch to Door Number 2?" Is it to your advantage to change your choice? Everything your proof relies on that's in the original problem is also in this one, so I think it does apply and I'm expecting you to say it does as well. There's only one little problem. The correct answer here is 100%. Please explain this to me. -- Rick Block (talk) 05:35, 8 February 2009 (UTC)[reply]

Against my better judgement, I will respond to this variant. To clarify, anyone who knows of this new constraint on Monty has, in essence, just been told there is a car behind door #2. Is that correct? Well, then how is that a game of chance? This differs fundamentally from the random selection of a goat, because the random method does not impart new knowledge (we know all along that Monty will reveal a goat and don't care which door). It is new knowledge that triggers a changing of the probabilities. In your variant, sometimes new knowledge is provided. That's why it goes to 100%. And all this has NOTHING to do with you dis-proving my proof.

Did you really say this?

"Everything your proof relies on that's in the original problem is also in this one, so I think it does apply and I'm expecting you to say it does as well."

You actually posit that the addition of a constraint would not change a proof? Or are you saying that by adding a new constraint, you have disproved the proof of the puzzle without the constraint? Get a clue, Jack!

Enough, already. Can we focus on the real issues? Finally? Please?

Glkanter (talk) 14:43, 8 February 2009 (UTC)[reply]

I deny the existance of different 'before' and 'after' questions. There is only one. It goes, "After Monty has revealed a goat, do you increase your chances of winning by switching?"

That is the only problem I have addressed since joining the discussion.

That is the only problem I have provided a proof for.

That is the only problem which I have asked you to find a flaw in my proofs.

And it is the only problem I will discuss on the discussion page of the Monty Hall Problem.

You're last gasp at finding a flaw was that I'm solving the 'before' puzzle. I pointed out the error in that statement, that twice the proof says "Monty reveals a goat". Which is a premise, for crying out loud!

So, whether or not there are two puzzles, I am solving the 'right' one. What's your critisism now?

Glkanter (talk) 06:21, 8 February 2009 (UTC)[reply]

Is my question above not clear? Does your solution apply to the variant, or not? -- Rick Block (talk) 06:25, 8 February 2009 (UTC)[reply]

Your question is immaterial to the Monty Hall Problem we are solving. I have no idea or interest in it's clarity. It is obsfucation. You are avoiding the Monty Hall Problem question. I have presented a proof, then sought meaningful critique. I have explained how a proof is validated. Yet you repeatedly refuse to remain engaged in a constructive discussion of my proofs.

Because it only takes one proof. And I've presented it. And it's valid. And it proves that ALL that other stuff is nonsense.

Glkanter (talk) 06:43, 8 February 2009 (UTC)[reply]

I'm trying to clarify for you what the problem is with your solution. It is directly related to the Monty Hall problem, which in fact YOU are avoiding. Let's switch to the terminology mathematicians use for the before and after questions. What I've been calling the before question is the unconditional question, while the after question is the conditional question. In this case, the unconditional question asks about the probability of winning by switching for all players without distinguishing between them in any way. The unconditional question doesn't care if the player picks door 1, or which specific door the host opens in response. The aforementioned Morgan and Gillman papers say the Monty Hall problem is NOT an unconditional problem, but a conditional problem. They each even propose wording changes that turn the problem into an unconditional problem. To save you a trip to the library (or the couple of bucks to get the online copy) I'll even quote them here for you. I'll even add some more of my own:

Morgan's version: You will be offered the choice of three doors, and after you choose the host will open a different door, revealing a goat. What is the probability that you win if your strategy is to switch?

[Rick here - note the scope of this. We're not limiting the population to those who've picked door 1 and see the host open door 3, we're asking a general question about all players.]

Gillman's version: You need to announce before a door has been opened [italics in the original!] whether you plan to switch.

Rick's version #1: Same as original, but at the very end instead of asking "Is it to your advantage to change your choice?" we ask "What is the probability across all players that switching will win the car?"

Rick's version #2: Same as original, but before the host opens a door the player is instructed to close her eyes, put her hands over her ears, and shout "SWITCH SWITCH SWITCH ..." or "STAY STAY STAY ..." and continue doing so while the host opens a door, until the host taps her on the shoulder at which point she can stop shouting, open her eyes, and the outcome of her choice will be revealed.

However you want to phrase it (unconditional vs. conditional, before or after) all of these differ from the Monty Hall problem as stated in the article, and as usually stated pretty much anywhere, by asking about all players without differentiation. The wording in the version published in Parade ("You pick a door, say No 1 ..") and the wording in Krauss and Wang version ("Imagine that you chose Door 1 ...") make it a conditional problem asking about a player who has picked a door and can see which door Monty opens in response. We don't care about all players in general. We care only about the subset of players who (for example) have picked door 1 and then see Monty open door 3. Using the typical probability notation, we're not asking about

P(win by switching)

but rather

P(win by switching | player picks door 1 and host opens door 3)

These are the two questions I've been yammering about that are different. P(win by switching) is the unconditional probability of winning by switching. It answers the before question. It tells you roughly what fraction of all players who play the game will win by switching.

P(win by switching | player picks door 1 and host opens door 3) is the conditional probability of winning by switching given that you've initially picked door 1 and the host has opened door 3. It answers the after question. It tells you roughly what fraction of all players who pick door 1 and see the host open door 3 will win by switching.

These are different questions which may or may not have the same numeric answer.

The Monty Hall problem asks the conditional question, not the unconditional one (according to two different math papers published about it).

The simple solution correctly answers the unconditional question. It says bupkus about the conditional problem. For the Monty Hall problem as stated it says the answer is 2/3. For the variant I've proposed above which you refuse to consider it also says the answer is 2/3. The unconditional probability IS 2/3 for both of these.

Although it is correctly answering the unconditional question, the simple solution is NOT addressing the conditional question. If we take its answer as the answer for the conditional question it "works" for the Monty Hall problem as stated in the article, but this is in effect a coincidence. The problem with the approach is exposed by applying the same solution to a variant (like the one I've proposed above) where the unconditional and conditional solutions are different.

This is EXACTLY the same as someone answering the question "what is 2²?" by saying "2 plus 2 is 4, the answer is 4". It's the right answer, and a true statement, but the wrong reasoning. We can expose this wrong reasoning by asking, "what is 3³?" Similarly, we can expose the wrong reasoning of the simple solution to the MH problem by asking about a slightly different variant, where the wrong reasoning produces the wrong answer.

If anyone else is reading this thread and can make this any more clear, please feel free. -- Rick Block (talk) 15:59, 8 February 2009 (UTC)[reply]

So, specifically, which one of my 8 statements invalidates the proof? Because, I'm sure you know that if each individual statement is true and valid, and the conclusion comes from the statements, then it's a valid proof.

Because it only takes one proof. And I've presented it. And it's valid. And it proves that ALL that other stuff is nonsense.

Glkanter (talk) 16:14, 8 February 2009 (UTC)[reply]

Here's where I start to wonder if you're reading what I'm saying. I tell you the math sources say there are two different questions, the unconditional question and the conditional question, and explain what these mean, and explain how they're different, and explain that the simple solution correctly addresses the unconditional question, and tell you the math sources say the problem is actually asking about the conditional question - and your reply ignores all of this and simply repeats your claim that you're right.

You said above you "deny the existance [sic] of different 'before' and 'after' questions". Do you deny the existence of conditional probability (before answering you might want to go read the article) ? -- Rick Block (talk) 17:07, 8 February 2009 (UTC)[reply]

I read every word you post. Believe me. Look, here's how it works. Someone puts forth a proof. Then everyone else looks for where there may be errors. Then when everyone is satisfied that it's correct, it is accepted. Once it's accepted, there is no saying about it: "Although the reasoning above is correct it doesn't answer the precise question posed by the problem, which is whether a player should switch after being shown a particular open door (Morgan et al. 1991)."

Step 1. Somebody puts forth a proof. I've done that.

Step 2. Everybody else looks for where there may be errors. In the proof. Nowhere else.

All you have ever done is say that I am not solving the appropriate question. To that, I again say, Hogwash! Each of those 8 lines passes. And the conclusion: 'since 2/3 is greater than 1/3, switching wins more often', answers the question being asked.

Enough, already!

Because it only takes one proof. And I've presented it. And it's valid. And it proves that ALL that other stuff is nonsense.

Why don't we admit that neither of us is going to be persuaded by the other? I've recently requested that Martin take this to the next level of Conflict Resolution. I'm tired of the circular nature of the argument, but I think I'll stick around. Glkanter (talk) 18:35, 8 February 2009 (UTC)[reply]

How it works at Wikipedia is someone finds reliable sources and uses them to develop an article about some topic. We specifically don't include original research such as putting forth original proofs, so (in articles at least) we have no obligation to examine whether proofs are correct or not. If Leonhard Euler were to come back from the grave and say this article is hogwash, here's a simple, elegant proof you should use instead - the response would be that he needs to get it published in a reliable source before it can be used here (both the proof and the claim that what's here is hogwash). That is not even close to the case here. What's happening here is you, some anonymous Wikipedia editor, are suggesting we ignore what reliable sources have to say and replace content that's based on what they say with a proof that they say does not address the question. I don't care how brilliant you think you are, but that is not what's going to happen. Trust me on this. -- Rick Block (talk) 19:12, 8 February 2009 (UTC)[reply]

And leave me out of this. Martin Hogbin (talk) 20:28, 8 February 2009 (UTC)[reply]

As usual, you're answering the wrong question. I have little interest in developing material for inclusion in the Wikipedia Monty Hall Problem Article. I'm trying to demonstrate to you and all the other editors why all that other stuff is hogwash and should be deleted! And for that, a logical proof is exactly what's in order.

Is there a Wikipedia methodolgy for determining a good source from a bad source? Maybe Morgan and Gillman are crackpots? They sure seem to have a minority opinion relative to the mainstream, based on what I've read here.

Is there a Wikipedia standard for how much prominance is given to minority opinions? If any?

It's time to escalate. Don't you want to move beyond this?

Glkanter (talk) 19:39, 8 February 2009 (UTC)[reply]

I've asked the folks at WikiProject Mathematics to comment (next step in dispute resolution is to solicit more opinions). If anyone watching this page would comment here, that might help as well. There is a methodology for determining what is a reliable source, described at the link I provided above. Academic and peer reviewed publications are usually the most reliable sources (there's a hedge about being outdated or superseded, which does not mean that popular sources published later trump academic sources published earlier). I assure you Morgan and Gillman are not crackpots and their views reflect the still current views in mainstream probability theory.

Several years ago, there was an anonymous editor (editing without a login) who initially brought up the Morgan et al. paper. I argued against him, much like you're arguing against me, but I finally "got it". I've spent a lot of time explaining this to you, not because I like to argue, but in hopes that you'll actually "get it", too. I could have justifiably said "shut up, we say what the sources say" a long time ago and dismissed you as a troll. I didn't, and still haven't, because I don't think you're a troll. -- Rick Block (talk) 20:23, 8 February 2009 (UTC)[reply]

MCDD is Morgan and three other guys. <Oh! The three other guys are the 'et al'!>Glkanter (talk) 22:00, 8 February 2009 (UTC)[reply]

"In fairness, MCDD do moderate their tone later on, writing, "None of this diminishes the fact that vos Savant has shown excellent probabilistic judgment in arriving at the answer 2/3, where, to judge from the letters in her column, even member of our own profession failed.""

http://www.math.jmu.edu/~rosenhjd/ChapOne.pdf page 48

Who is John Gault?

Glkanter (talk) 21:00, 8 February 2009 (UTC)[reply]

Who is John Gault? Perhaps John Galt's cousin? -- Rick Block (talk) 23:05, 8 February 2009 (UTC)[reply]

Right explanation

I'll do one effort to explain the problem. The situation is already conditional on the choice of Door 1 by the player. And we know Door 3 has been opened revealing a goat. Let us assume the game has been played 3000 times. What might have happened?

Player has chosen Door 1
Car hidden behind Door 1		Car hidden behind Door 2	Car hidden behind Door 3
1000 times		1000 times	1000 times
If nothing else is revealed
Player wins the car if he doesn't switch		Player wins the car if he switches
But player gets "information"
Host opens Door 2	Host opens Door 3	Host must open Door 3	Host must open Door 2
500 times	500 times	1000 times	1000 times
This case didn't happen	One of these cases happened		This case didn't happen
	Player wins the car if he doesn't switch	Player wins the car if he switches

Hence only in half the cases did the situation arise in which the player finds himself. So we see it cannot be unconditional, because only a part of all possibilities could have happened. Clear?Nijdam (talk) 18:05, 6 February 2009 (UTC)[reply]

Conditional/unconditional

Rick, I thought that it had been agreed that, for the fully defined problem (host alway offers the switch and always opens a goat door), there is no distinction between the conditional and unconditional cases. Martin Hogbin (talk) 22:56, 6 February 2009 (UTC)[reply]

In cases of mathematics the right solution is not reached by agreement. I hope I explained clearly enough why the stated problem requires a conditional solution and why this solution differs from the unconditional one (not numerically in the final answer, but that's not the issue).Nijdam (talk) 23:56, 6 February 2009 (UTC)[reply]

Martin, no, I don't think there was ever any agreement that in the fully defined problem there is no distinction between the conditional and unconditional cases. The fully defined problem ensures the numerical probabilities are the same, but these remain distinct questions requiring different solutions.

Nijdam - I have said repeatedly I'm OK with starting with an unconditional solution so long as it is clearly identified as an unconditional solution, doesn't actually say anything that's incorrect, and is immediately followed up with a transition to the more accurate conditional solution. My rationale for including an unconditional solution is because many published sources (possibly all popular expositions of this problem) include only an unconditional solution. I think not providing one here would be a significant omission. How to structure the article is an editorial decision, which is reached by consensus. We want the article to be accessible to laypersons, but also correct, which I think sometimes calls for compromises. -- Rick Block (talk) 05:03, 7 February 2009 (UTC)[reply]

Well, to be honest, I'm not. I strongly oppose starting with a reasoning that is not a solution to the stated problem. The right solution, as I suggested above, is quite accessible to the layman. If anything has to be said about the "simple solution", it should be as a solution to a (slgihtly, but essentially) different problem. It doesn't seem right to me, to have thoughts like "keep the laymen in the dark, it is no use to explain things properly to them".[Sorry, forgot to login] Nijdam (talk) 12:59, 7 February 2009 (UTC)[reply]

Nick, in what way are the conditional and unconditional cases distinct in the case of the fully defined problem? Martin Hogbin (talk) 14:44, 7 February 2009 (UTC)[reply]

Martin - see my most recent reply to Glkanter, two sections above. -- Rick Block (talk) 20:20, 7 February 2009 (UTC)[reply]

Your reply seems to me to be arguing that we can show the two cases are different by changing things so that they are different. Martin Hogbin (talk) 00:43, 8 February 2009 (UTC)[reply]

Martin - 2+2=4 and 2² = 4. Does that mean "what is 2+2" is the same question as "what is 2²"? I'm saying we have two questions about the MH problem whose answer is both 2/3. We can show they are indeed different questions by asking them about a different problem where they have different answers. -- Rick Block (talk) 05:44, 8 February 2009 (UTC)[reply]

Here's something to think about:
The presentator has a simple strategie: when possible he opens Door 3, otherwise he tries Door 2 and is that also impossible he opens Door 1. Quite easy. See what happens:

Chosen 1, opened 2: Switching wins the car

Chosen 1, opened 3: Switching wins the car with probability 1/2

Chosen 2, opened 1: Switching wins the car

Chosen 2, opened 3: Switching wins the car with probability 1/2

Chosen 3, opened 1: Switching wins the car

Chosen 3, opened 2: Switching wins the car with probability 1/2

Yet the initial prob. of winning the car is in all cases 1/3 and according to the "simple reasoning" switching would increase it to 2/3. Notice BTW that the overall prob. of winning the car when switching is (of course) 2/3. And also conditional given the chosen door. But (!) not conditional given the chosen and the opened door. Don't let it keep you out of your sleep.Nijdam (talk) 22:56, 7 February 2009 (UTC)[reply]

Firstly, I should have added one more thing to my 'fully defined problem', which is that the host chooses a goat door randomly when he has a choice.

Between the start of the show and the player choosing a door, many things happen; the host opens a door, the host talks to the player, the audience may applaud or someone in it may cough, there may be a break for advertisements. The outcomes of all these events represent conditions based upon which the players choice is made. We choose to ignore most of those conditions for one reason, we take it that they do not reveal any new information about the location of the car and thus cannot possible affect the outcome of the show. That is the criterion upon which we decide what conditions we must take into account in answering the problem.

Let me now ask this question: if the problem were stated an a form that explicitly stated that nothing occurred between the start of the show and the player making his choice that reveals any information about the position of the car, are the conditional and unconditional statements of the problem then distinct? Martin Hogbin (talk) 10:05, 8 February 2009 (UTC)[reply]

Interesting that you put is this way. Because what is the meaning of "no information is revealed"? This actually means that the possible outcomes are not restricted. And here they are: allthough the conditional probability is the same as the unconditional, the information revealed is for instance that the car is not behind Door 3. Nijdam (talk) 11:12, 8 February 2009 (UTC)[reply]

Perhaps I should have said, 'no information is revealed that would affect the probability of winning by switching'. In that case would you accept that there is no distinction between the conditional and unconditional cases. Martin Hogbin (talk) 18:42, 8 February 2009 (UTC)[reply]

Alas(?!), no. You really miss the point. The information revealed - with the proper strategie of the host - does indeed not affect the probability of winning by switching. But the meaning of such a statement is that the conditional probability given this information, is the same (numerically) as the unconditional one. And as I said before: the revealed information restricts the possible outcomes. I get the feeling that all efforts are made, against better judging maybe, to avoid rejecting the "simple solution". I sincerely hope this is not the case. Nijdam (talk) 22:58, 8 February 2009 (UTC)[reply]

You argument would apply to everything that happens between the players original choice of door and his decision as to whether to switch or not. As I said above, why do we not consider the conditional probability given that the host has spoken to the player. What I mean by this is why is the host opening a door a condition any more that the host speaking to the player? Martin Hogbin (talk) 23:30, 8 February 2009 (UTC)[reply]

Well, some things happen and do not limit the possible outcomes. Like the started rain outside or the joke the host tells in between. But, as I over and over argued, the open door does. I gave a "simulated" example further down, in reply to Gkanter. Have a look there and be convinced. I really can't think of anything more that should be said. Maybe the real issue is some discussiants are not familiar with probabilities, let alone conditional probabilities. Nijdam (talk) 00:00, 9 February 2009 (UTC)[reply]

That was exactly my point. What we include as a condition in any probability questions depends on what what believe can possibly affect the outcome. If the statement of the problem were to explicitly state that 'no information is revealed that would affect the probability of winning by switching' by a particular event then that event would not represent a condition, and the conditional and unconditional cases would be indistinguishable. 86.133.179.19 (talk) 09:28, 9 February 2009 (UTC)[reply]

I agree with Glkanter: "Everybody else looks for where there may be errors. In the proof. Nowhere else." He may not be responding to the conditional problem, as presented by Rick Block, but should he? People try to explain the conditional situation, but why do they fail to explain to him why the particular problem cannot be unconditional? Where are the errors in the simple solution, as an answer to the given question? I found some probable 'answers' to that:

- Morgan et al. : "F5 is incorrect because it does not use the information in the number of the door shown."

- Glkanter: "I think you're hung up the doors being numbered, and that which door a goat or a car is behind makes any difference."

- Nijdam: "alternative question <..> : the player is blind and Monty only tells him: I opened one of the other doors and there is a goat behind it."

- Hogbin: "Between the start of the show and the player choosing a door, many things happen <..> we take it that they do not <..> affect the outcome"

- Nijdam: "what is the meaning of "no information is revealed"? This actually means that the possible outcomes are not restricted."

The question is: should the requested chance be regarded as a conditional probability, and if so, why? Morgan et al. say it should, but their only argument or hint is that the information in the number of the door is otherwise not used. What information is in the number? Does it matter if the (blind) player has knowledge of that? What other information might be relevant? Should it be limited to events which restrict the possible outcomes? Even if it doesn't affect the outcome?

The article Conditional probability states that P(B|A) = P(B) if A and B are statistically independent. In other words: conditional and unconditional chances are the same if A does not change the probability of B. And it doesn't. Heptalogos (talk) 15:34, 10 February 2009 (UTC)[reply]

Conventional Wisdom

I've been re-reading some past postings. According to Rick, this article has been reviewed on 2 occasions as a 'Featured Article', and that much of what I find inessential actually was a (by)-product of those reviews. Rick is proud of 'shepherding' this article through at least one of those reviews.

So, in some ways, I seem to be arguing against Conventional Wisdom. But I don't feel that way. I have a few college courses on this topic, over 30 years ago, and a lifetime of being a data analyst. My viewpoint is, 'There is no possible way I am wrong about this'. To me, this whole discussion has as much to contribute as a discussion of whether the sun will rise in the east tomorrow morning.

How does a single voice effectively confront the Conventional Wisdom? This is a question not just for Wikipedia, but any societal system. In the US, a swindler set up a Ponzi scheme on Wall Street. Individual investors went to the regulatory agency numerous times, but to no avail. The guy didn't actually get caught. He turned himself in! How does a minority, but important, voice get heard?

Yes, I look at this entire article, excepting maybe 5% of it, as an elaborate hoax. I think everyone went along because they did not want to admit to limited knowledge of the subject matter. Everybody drank the kool-aid. And the emperor is wearing no clothes.

2/3 of the time I will select a goat. Therefore I should switch. Glkanter (talk) 15:32, 7 February 2009 (UTC)[reply]

You are not alone Glkanter, I agree that the page could do with improvement. My main complaint is not that the current solutions do not properly and fully address the subject; this is done well and with references to reliable sources, thus it ticks all the boxes to be a featured article. My complaint is that it lacks a section that will quickly convince the average reader of the right solution. This is unfortunately not something that appears in the FA criteria. However, the really notable point about this problem is that even when all the facts are presented in a clear and unambiguous way (and I believe if if the problem were stated in a fully unconditional form) the average punter gets it wrong. It is common in many walks of life today to find situations where you can tick all the boxes but still not do the job well. The question that should be asked of all WP articles, but generally is not asked, is, 'Does this article address the needs of the expected readership?'.

In this respect I believe that the conditional/unconditional issue is an unnecessary, and to a large degree unwarranted, complication that only serves to obfuscate the central and notable problem.

I have not pressed for the inclusion of 'my' solution because I am not totally happy with it as a convincer. I have bored many of my friends and family with the problem to try to find out what arguments are the most convincing. The solution I gave does work quite well but stumbles over the fact that the decision is not made at the start but after the host has opened a door (arghh! the very issue I claim is not important). I still stand by what I say, however, but am looking for a way to treat this issue in a different way. Martin Hogbin (talk) 11:41, 8 February 2009 (UTC)[reply]

There is a difference between a proof being valid, and being 'intuitive'. If the solution were 'intuitive', the puzzle would not be well known. I've done the same thing with friends, and generally, they're just not in the mood for this. The important point of the proof being valid, however, is that it renders all that other stuff as hogwash. Including the 2nd paragraph of the Solution section which says the just-offered solution is no solution at all! What does FA criteria say about that?

No. There's no compromise on this. It's math. Not Economics. It's time to end this horrible charade. And I am MUCH more interested in the immediate deletion of all non-value-added (wrong) (BS) stuff from the article than the elevation in prominance of any one of the elegant, valid proofs we've put forward.

Glkanter (talk) 14:07, 8 February 2009 (UTC)[reply]

And really, who wants to argue when at the top of the page it says:

Please note: The conclusions of this article have been confirmed by experiment

There is no need to argue the factual accuracy of the conclusions in this article. The fact that switching improves your probability of winning is mathematically sound and has been confirmed numerous times by experiment.

Two mentions of proof via experimentation. I'll bet not a single reference source uses 'experimentation' as their solution. Because 'experimentation' is not a statistical or logical proof. An error like this confirms my above stated concerns. I'm curious, is it common for Wikipedia discussions to have such a banner? I've never seen anything like it in any 'academic' setting before. Anywhere. Glkanter (talk) 15:54, 7 February 2009 (UTC)[reply]

And this, it's the 2nd paragraph after the 1st chart in the Solutions section. It's one of the more prominent items in the Solutions section.

"Although the reasoning above is correct it doesn't answer the precise question posed by the problem, which is whether a player should switch after being shown a particular open door (Morgan et al. 1991)."

What sort of a 'Solution' section leads off with a proof it is going to immediately (and fraudulently) discredit? And then goes on endlessly doing so? For who's benefit? The once-interested reader is by now, long gone. Glkanter (talk) 17:06, 7 February 2009 (UTC)[reply]

Here's my only edit to the actual Article. On October 23, 2008 I deleted the following erroneous line:

"The overall probability of winning by switching is determined by the location of the car."

The above statement was the very first words of the Solutions section.

Imagine. Being THAT wrong on the very first line.

Glkanter (talk) 17:57, 7 February 2009 (UTC)[reply]

So, I posted the following two days ago, at 13:34, 6 February 2009:

There are really only two situations to consider. Not 3, not 6.

2/3 of the time I will choose a goat.

100% of those times Monty reveals the other goat, leaving the car.

=>2/3 of the time I should switch.

1/3 of the time, I will choose the car.

100% of those times Monty reveals one of the goats. It doesn't matter which.

=>1/3 of the time I should not switch.

2/3 is greater than 1/3.

Therefore, by switching, I increase my chances of getting the car when offered the switch.

Eight lines. Including a premise repeated twice for clarity. Since then, there have been countless postings by Rick. Running off on tangents to and fro. But we NEVER focus on the 8 statements. It's always some BS about 'before or after', or whatever the mcguffin of the day is.

It's a valid proof, and could be shorter. Mr. Block! Tear down your wall!

Glkanter (talk) 07:46, 8 February 2009 (UTC)[reply]

This is why I say it's time to end this charade, now.

In October, 2008, Martin posted this to Rick:

"If you remember, I first came to this article in response to an RFC which claimed that the current editors were being overly protective."

So, it's not just me. It's been enough people, for a long enough time, that the offical Wikipedia RFC process was put in motion. Good. It's time for the next step in the process. BTW, who do suppose could be the 'overly protective current editors' referred to?

Glkanter (talk) 15:34, 8 February 2009 (UTC)[reply]

Do I sound frustrated? Well, I am. Here's another one from October. Sound familiar?

Maybe we should take this one step at a time. Do you agree there may be a difference between the probability of winning by switching before the host opens a door and after the host opens a door? Or, at least, that these are different questions? -- Rick Block (talk) 02:09, 27 October 2008 (UTC)

I've read this entire discussion page from top to bottom. Other people have come before me, and pointed out essentially the same things I'm saying. No, thank you, I do NOT want to debate some other puzzle questions with you. It's a simple puzzle. In fact, the puzzle is almost like a Seinfeld episode. Nothing happens! The probabilities DON'T CHANGE! You seem to not understand Probability Theory. Otherwise, you would have acknowledged the validity of the proofs, and therefore the uselessness of those other issues as they relate to the Monty Hall Problem. Despite your valiant efforts at requiring 'a rigorous solution', you allowed a COMPLETELY ERRONEOUS STATEMENT to lead the solutions discussion. And have resisted multiple efforts to improve that section! That erroneous statement is at odds with everything that Probability Theory stands for. At times, you've argued with me over the premises of the Monty Hall Problem. Monty opening the CONTESTANT'S door? Give me a break. Glkanter (talk) 02:28, 27 October 2008 (UTC)

What was that completely erroneous statement? Just this:

"The overall probability of winning by switching is determined by the location of the car."

So, yes, I'm personally invested. I'm interested. I just read that there are 'archive' pages for this discussion. So what we see here, is just the most recent discussions of this lunacy. The tip of the iceberg, so to speak.

So, enough of the charade!

Because it only takes one proof. And I've presented it. And it's valid. And it proves that ALL that other stuff is nonsense. —Preceding unsigned comment added by Glkanter (talk • contribs) 15:54, 8 February 2009 (UTC)[reply]

You know what? It's been long enough. Martin, I understand that at some time last year you were specifically requested to get involved, in order to facilitate the resolution of some conflict(s) relating to the Monty Hall Problem.

I ask Rick to invalidate my proof, he tells me I'm not addressing the Monty Hall Problem.

It is obvious these discussions have gotten nowhere, and are going nowhere. So I request that you declare this situation as 'at an impasse', and escalate this to the next level of Conflict Resolution.

Thank you.

Glkanter (talk) 17:05, 8 February 2009 (UTC)[reply]

Since October 25, 2008 I have posted on this discussion page 70 times. 66 as Glkanter, and 4 previous times as an ip address. In addition, there was a separate discussions page that was set up, and I posted there many times as well.

70+ postings over a 2 line proof! When I pick a goat, Monty leaves a car. I pick a goat 2/3 of the time.

I feel that this has gone on long enough! I honestly wonder if there isn't a bot out there generating responses from a limited selection of inaccurate/meaningless/non-relavent statements.

This has been going on with countless other editors since at least 2005. Check out the archives sometime.

To the powers-that-be at Wikipedia, I implore you, please bring this to a conclusion!

Glkanter (talk) 15:43, 10 February 2009 (UTC)[reply]

I just went to the page where Rick requested assistance. One of the people there, who described himself as 'an expert in probability' had this to say:

One probabilistic point that I observe is, (ir)relevance of the (un)conditional probability. I'd say that in this case they are equal not just by a numeric coincidence. Rather, the conditional probability (treated as another random variable) is constant (a degenerate random variable) in this case, due to an obvious symmetry. Taking into account the total probability formula we conclude that the conditional probability must be equal to the unconditional probability in this case. Thus I feel indifferent. Both are relevant in one sense or another. Do you agree? Boris Tsirelson (talk) 21:42, 10 February 2009 (UTC)[reply]

http://en.wikipedia.org/wiki/Wikipedia_talk:WikiProject_Mathematics

Does this mean what I hope it means?

Glkanter (talk) 22:37, 10 February 2009 (UTC) moved comments 76.243.187.238 (talk) 01:29, 11 February 2009 (UTC) doh! Glkanter (talk) 01:50, 11 February 2009 (UTC)[reply]

After just 4 months, I've run out of new things to say. So rather than clog up other people's exchanges, I'll just add commentary here. I don't see the point in further arguing, but I'm not giving up the quest.

I see the 'random goat constraint' has reared it's (their) ugly head again. In my proof, which I maintain is sufficient for this puzzle, the goats are immaterial. The puzzle asks about increasing the liklihood of selecting the car, not the location of a goat.

Does the probability change if instead of goats there is nothing? So would we have a 'random nothing consraint'?

My understanding of Probability Theory says if an item isn't specifically addressed, it's considered random. And anything that's random is not a constraint. And if it's not a constraint, it doesn't need to be addressed.

For those keeping score, I was indeed reported to the authorities. I, and others with my criticisms were described as a 'cadre', 'wanting to "dumb down" ' the solution by Rick.

Glkanter (talk) 14:36, 11 February 2009 (UTC)[reply]

And, the doors and their numbers are meaningless. The doors only exist to shield the contestant's view of the car and the two goats.

Let's focus on why this is such a famous puzzle. Because it's easily understood (2/3), and easily misunderstood (50/50). So, no reader really understands the solution until he has mastered the 'random goat constraint'? Pure folly.

Nobody cares about the specific case of doors 1 and 3, any more than they would care about suitcases 11 and 25 in Deal or No Deal. That's not what Probability Theory is about. Here, I disagree, until my last breath, with Rick. Probability, and this puzzle are always solving the 'over time' question. If repeated endlessly over time, on aggregate, what would the results be?

Glkanter (talk) 14:54, 11 February 2009 (UTC)[reply]

I was struggling for the right word earlier, but it came to me. Despite what Rick is posting elsewhere about 'physical doors' and such, all Probability can do is create a mathematical model that approximates real life. That's what a proof sets out to do. And if valid, it serves as a proxy for 'real life'. From which we can analyze and draw conclusions.

I, of course belief I have presented various proofs that meet this criteria. Rendering all that other stuff, at best, redundant and confusing, and at worst, erroneous.

Glkanter (talk) 16:19, 11 February 2009 (UTC)[reply]

Glkanter, I agree to many things you say. But you should keep one thing in mind: we need an objective way to prove ideas, even if everyone seems to agree to the logic presented by e.g. you or me. So, maybe maths fails to prove what is true in this case, but maybe (our) logic fails to understand the real causes of reality in this case. Do we have other objective systems but maths to prove our logic? Are those limited to science? I think we should each try to step outside our custom systems to see if we can meet on another level. Heptalogos (talk) 21:49, 11 February 2009 (UTC)[reply]

The Monty Hall Problem gained much (most) of its current noteriety when it was written about by Marilyn Vos Savant in Parade Magazine. I read about it there. We all know the story about how a thousand PHDs challanged her, etc., etc.

But ultimately, Vos Savant was acknowledged as having the right answer at 2/3.

Let's hear from the great lady, herself, courtesy of Wikipedia'a Marilyn Vos Savant page:

Under the most common interpretation of the problem where the host opens a losing door and offers a switch, vos Savant's answer is correct because her interpretation assumes the host will always avoid the door with the prize. However, having the host opening a door at random, or offering a switch only if the initial choice is correct, is a completely different problem, and is not the question for which she provided a solution. Marilyn addressed these issues by writing the following in Parade Magazine, "...the original answer defines certain conditions, the most significant of which is that the host always opens a losing door on purpose. Anything else is a different question." [16] In Vos Savant's second followup, she went further into an explanation of her assumptions and reasoning...

http://en.wikipedia.org/wiki/Marilyn_vos_Savant

Or this, from the Newspaper of Record:

"The problem is not well-formed," Mr. Gardner said, "unless it makes clear that the host must always open an empty door and offer the switch. Otherwise, if the host is malevolent, he may open another door only when it's to his advantage to let the player switch, and the probability of being right by switching could be as low as zero." Mr. Gardner said the ambiguity could be eliminated if the host promised ahead of time to open another door and then offer a switch.

Ms. vos Savant acknowledged that the ambiguity did exist in her original statement. She said it was a minor assumption that should have been made obvious by her subsequent analyses, and that did not excuse her professorial critics. "I wouldn't have minded if they had raised that objection," she said Friday, "because it would mean they really understood the problem. But they never got beyond their first mistaken impression. That's what dismayed me."

Still, because of the ambiguity in the wording, it is impossible to solve the problem as stated through mathematical reasoning. "The strict argument," Dr. Diaconis said, "would be that the question cannot be answered without knowing the motivation of the host."

Which means, of course, that the only person who can answer this version of the Monty Hall Problem is Monty Hall himself. Here is what should be the last word on the subject:

"If the host is required to open a door all the time and offer you a switch, then you should take the switch," he said. "But if he has the choice whether to allow a switch or not, beware. Caveat emptor. It all depends on his mood."

http://query.nytimes.com/gst/fullpage.html?res=9D0CEFDD1E3FF932A15754C0A967958260&sec=&spon=&pagewanted=3

So, even as we argue semantics on this page, it's unambiguous what problem she was solving. With all that hub bub, what proof did Marilyn and all these people use to come to the conclusion that 2/3 is right? Well, it wasn't Morgan et al, via the 'random goat constraint'.

So, if the unconditional probability proof was good enough for Marilyn, Monty, 1,000 PHDs, and 10s of millions of Parade magazine's general interest readership, why is it not good enough for the Wikipedia readers of today?

Glkanter (talk) 08:01, 12 February 2009 (UTC)[reply]

It's surely good enough for everyone with less logic. Every era has it's own conventional wisdom. At the same time science is always requesting systematical proof, which is one of the reasons that conventional wisdom improves. Of course we need both, that's what I'm saying. I also believe that scientists statistically have better wisdom than random crowds (who speak through individuals), which indeed doesn't mean that science 'truth' is always superior to any common truth. So the unconditional approach is very welcome, but it shouldn't be presented as the only correct answer. Thanks for the links, I am getting curious about the exact logic of Marilyn. Heptalogos (talk) 09:53, 12 February 2009 (UTC)[reply]

Please note, I'm not saying that something like the 'Combined Doors' solution is the only correct answer. I'm saying that two paragraphs later, the Solution section should not say that the explanation just offered is incorrect. And then spend 90% of the article expanding on that. Which is exactly how the current article reads.

I suppose there's 'Wikipedia Merit' to the Morgan arguement, after all it's been published. But to hijack the entire article from the commonly understood solution? As I've recently learned, that is not the 'Wikipedia Way'.

Glkanter (talk) 16:42, 12 February 2009 (UTC)[reply]

I've just learned that Rick's comment that being 'Right' is trumped by being 'Verifiable' for Wikipedia purposes has some merit. I further understand that being in the mainstream (my term) trumps being in the minority (my term).

We're still arguing over who's 'right'. But apparently, that's not even the salient point.

Would anyone describe the 'random goat constraint' as being in the mainstream of the published literature? How about after reading this?

MCDD is Morgan et al.

"In fairness, MCDD do moderate their tone later on, writing, "None of this diminishes the fact that vos Savant has shown excellent probabilistic judgment in arriving at the answer 2/3, where, to judge from the letters in her column, even member of our own profession failed.""

http://www.math.jmu.edu/~rosenhjd/ChapOne.pdf page 48

Glkanter (talk) 13:35, 12 February 2009 (UTC)[reply]

Oh, look! Rick agrees with me that the unconditional solution represents the mainstream.

"The "vos Savant" version is far and away the best known, but it's not only conditional (sic) but under-specified."

So, now that we agree what the mainstream is, let's agree that it is under-represented and mis-represented in the article, and that the article emphasizes the minority of the published works.

Who's going to take the first stab at proposing changes to the article? I've already proposed some ideas under the heading "Request for brief clarifications".

Glkanter (talk) 16:42, 12 February 2009 (UTC)[reply]

I've posted here nearly 80 times trying to prove my proof is valid, and that I'm 'right'. Turns out, that was never the argument I needed to win.

So why, since 2005 at least, hasn't the response been that the more meaningful (for Wikipedia purposes, anyway) arguement is, to paraphrase, 'popularity of the published material'? Not until about two days ago, anyways.

So, thanks for wasting 4 months of my time. And 4 years of various other people's time.

Now, can we start fixing the Article?

Glkanter (talk) 14:48, 12 February 2009 (UTC)[reply]

Pretty interesting exchange relating to my 'being reported' by Rick at the bottom of Rick's talk page. First, he gets spanked for crying wolf, then he writes this:

There's a current crowd (Glkanter is one of them) unhappy with the Morgan et al. approach who want it removed, commenting on the talk page to gain consensus for their desired change. I guess rather than convince them I could just let them know they will never gain consensus for this change. -- Rick Block (talk) 12:23, 11 February 2009 (UTC)

Maybe next I'll be reported for trying "to gain consensus for their desired change". Does trying to build consensus make me a bad Wikipedian?

Glkanter (talk) 17:35, 12 February 2009 (UTC)[reply]

Would you consider the following statement as an indication that an editor has claimed 'ownership' of an article?

"I guess rather than convince them I could just let them know they will never gain consensus for this change. -- Rick Block (talk) 12:23, 11 February 2009 (UTC)"

Could this be an 'Aid To Understanding' for why there have been 4 years, 7 archives, and thousands of postings calling for significant changes to the Monty Hall Problem article, all to no avail?

Let me ask the 'cadre', is there 'consensus' that this editor should be reported for violating the Wikipedia Ownership policy as it relates to the Monty Hall Problem article? If forced, I guess I could be the one to make the report.

Glkanter (talk) 18:06, 12 February 2009 (UTC)[reply]

Does anybody know what form any sanctions would take? I hope it would include being blocked from editting both the MHP Article and this talk page. Glkanter (talk) 18:45, 12 February 2009 (UTC)[reply]

Here's where Rick first asked for assistance to aid in Resolving our Conflict.

http://en.wikipedia.org/wiki/Wikipedia_talk:WikiProject_Mathematics

All these other Wikipedia Math gurus already knew about Rick's MHP article Ownership issues!

I'm a first-timer here. It's been way too long, but is has been instructive as to how horribly mishapen things get when an editor claims ownership of an article.

Glkanter (talk) 19:25, 12 February 2009 (UTC)[reply]

If you're still reading these edits, you night appreciate my very first posting. I created a new section that day, but I still don't know how to link to it. "Monty's Action Does Not Cause The Original Odds To Change."

We were all so civil then! I was all ready to make my first edits to the Article! Glkanter (talk) 19:47, 12 February 2009 (UTC)[reply]

You guys should read the Featured Article Reviews. There's a link to each of them at the top of this page. In addition to some of the inane discussions we take part in here, the FA reviewers also argue about footnotes.

And there is one recognizbale name constantly making edits. That would be Rick. More documentation as to his 'Ownership' of the article.

Now, after his encouragement to suggest edits, he says this:

"Although not a veto, for consensus purposes unless you can address reasons raised to keep the content it would not be proper to make the deletion you suggest, regardless of how many editors agree with you (please see WP:NOTDEMOCRACY). -- Rick Block (talk) 03:15, 13 February 2009 (UTC)".[reply]

Looks like an 'Ownership' issue to me.

Basically, Rick is saying that the Article is already perfect. Therefore, any changes will make it un-perfect. And he knows, because he's written most of it, nominated it for FA, 'sheperded' it through FA, and personally responded to every FA critism himself. Through his 4 years of Ownership-fueled protectionism, he also has granted the existing Article 'tenure'. I wonder if there is such a policy as WPTENURE? Glkanter (talk) 04:02, 13 February 2009 (UTC)[reply]

I wrote the above about 'tenure' because this had been posted as part of a larger paragraph:

"...Given this content and these references are fairly long standing in this article and were added in response to a featured article review it would not be proper to remove them unless there is a strong consensus here..."

Rick Block (talk) 03:15, 13 February 2009 (UTC)[reply]

Then, after my 'tenure' comment and some other comments by me, this is posted, verbatim:

"I fell for it. I let him use the old 'rope-a-dope' on me. Next thing you know, instead of argueing about my proof, we'll be argueing about 'who is more mainstream', or 'what is prominant'. We are not arguing about your proof. Your proof is WP:OR and of no relevance, no matter how correct it may be. Please realise that. Leaving out the hyperbole would be helpful. To comment on the principles: no, this is not a democracy, yes all have the same potential to contribute, no Rick doesn't have a veto, no Rick did not say that he had a veto, no the argument that the content has been in place for ages is not a compelling reason to keep it William M. Connolley (talk) 08:31, 13 February 2009 (UTC)"[reply]

William has thoughtfully taken an interest in this page following Rick's request for intervention in his dispute with me. That makes him at least the second current editor brought to this page due to an intervention request involving Rick, although I do not believe the first was at Rick's request.

Yep, William gives it to me pretty good. But don't let his last sentence slip by unseen:

"no the argument that the content has been in place for ages is not a compelling reason to keep it".

So, the next time you suggest a change, and are countered with "Given this content and these references are fairly long standing in this article and were added in response to a featured article review ..." you may properly disregard such criticism. Glkanter (talk) 12:18, 14 February 2009 (UTC)[reply]

I'll probably have to document my claim. Here's another good one from Rick:

"I generally treat the talk page of the Monty Hall problem as a sort of "virtual office hours". My intent is to respond to any and all posts in a friendly and welcoming manner, in much the same way as you presumably respond to students who drop by during your office hours."

http://en.wikipedia.org/wiki/User_talk:Tsirel

Thanks, dude. You're too generous. No, really. Glkanter (talk) 08:01, 13 February 2009 (UTC)[reply]

I'm working on a new Wikipedia policy, I call it WP:DUDE-YOU'RE-NOT-SOCRATES . Glkanter (talk) 12:21, 14 February 2009 (UTC)[reply]

"This page is 495 kilobytes long. It may be helpful to move older discussion into an archive subpage." Plus 7 archive pages dating as far back as 2005.

"That's not writing, that's typing." Truman Capote

With my edits to this talk page, I have been accused of violating WP:CIVILITY, and worse. I'm comfortable with everything I've written. As I wrote at the start of this section, it's difficult for a minority voice to be recognized as credible, when it defies the Conventional Wisdom. So, I am trying to accomplish what seems, so far, to be a difficult task. The only tool I have is my ability to post on various Wikipedia pages. Talk about a probability problem! How do I know who to turn to?

So, I post here. I can't post in that pseudo-collegial, 'oh, beg to differ, chum', style that seems so popular. And I can't out-reference or out-WP:Policy Rick. He's got like 10,000 article edits to his name. No, all a punter like me can do is write with common sense and clarity, and try to draw some attention. So I have my own 'style'. Now you know why. Glkanter (talk) 14:49, 14 February 2009 (UTC)[reply]

No, it's not 'like 10,000'. His first edit (must be all edits, not just articles), was on February 20, 2004. His 10,000th was on March 21, 2005. Let's extrapolate: 10,000/ 13 months * 60 months = 46,153.8. I'm at about 100+. Little-known fact: it only SEEMS like all 46,153.8 are on this talk page. Glkanter (talk) 15:18, 14 February 2009 (UTC)[reply]

Let me ask it this way, in the 1/3 of the time when I select the car, how is Monty's motivation or actual actions going to change the fact that the car was my choice? And that by switching, I would give up the car? How is the 'equal goat door constraint' the solution to that question? It isn't. For probability proof purposes, like the most simple 2 or 3 line proof, we never even need to go there. I reject anew Morgan et al. Glkanter (talk) 16:13, 14 February 2009 (UTC)[reply]

Here's my new proof.

1/3 of the time I will pick a car.

Since 1 - 1/3 = 2/3, 2/3 of the time I will pick a goat.

I increase my likelihood of winning by switching.

Look at that. No Monty. No Monty's behaviour. Thank you, thank you very much.

Glkanter (talk) 16:21, 14 February 2009 (UTC) Glkanter (talk) 16:37, 14 February 2009 (UTC)[reply]

I think this is even better.

1/3 of the time I will pick a car.

Since 1 - 1/3 = 2/3, 2/3 of the time I will not pick a car.

I increase my likelihood of winning by switching.

No Monty, and now no goats. Glkanter (talk) 16:53, 14 February 2009 (UTC)[reply]

This is the simple solution in it's simplest form. Nothing new. Vos Savant did the same. Some people raised objections because they assumed that the problem is a conditional one. They forgot to explain why. They only said something about a numbered door being opened, maybe assuming that this number was already there before it was opened. This would reveal new information. But even then it would surely not affect the 1/3 chance of the picked door, so you are still right. But some people here insist that if part of the problem is conditional, the whole problem is. I am still waiting for the mathematical rule by which a problem is defined as conditional. I think I found it at Conditional probability where is stated that there should be statistical dependency, which means that the asked probability is affected, which is surely not the case here. But I don't get any answers to that. They just stopped explaning there. Heptalogos (talk) 14:00, 15 February 2009 (UTC)[reply]

I don't know...without Monty or the goats even mentioned, I don't see where they come into play. What did you think of Boris' comments?

Here's a different approach: My new solution is valid because the entire discussion about Monty and the goats is a carnival distraction. And doesn't affect the probabilities. It always end up a contestant, two doors, one car, and the cars haven't been moved since the original selection. It's still a 1/3 likelihood that you picked the car, so... Glkanter (talk) 14:21, 15 February 2009 (UTC)[reply]

Sorry, I just corrected 'unconditional' in my text above into 'conditional'. About your solution: it's not new and I think there's no use in repeating it all the time. What we need is solid proof/explanation that this problem is exclusively conditional. Without it, I think the article has to be changed. Let's give them some more time. You can call me optimistic, but I think something's moving. ;: Heptalogos (talk) 16:12, 15 February 2009 (UTC)[reply]

Those of you who reject the existance of Original Research are excused, and free to resume cataloging references to Bigfoot, Unicorns, etc. Glkanter (talk) 17:17, 14 February 2009 (UTC)[reply]

I just realized that any answer other than 'never switch' inproves your liklihood of winning. The worst you can do is not switch, and go 1/3. Do anything else, I'm just thinking 'go random', and yours odds increase, in this case to 50%. Glkanter (talk) 23:13, 14 February 2009 (UTC)[reply]

Request for brief clarifications

I came here from a request on WT:WPM. But the conversation is too long to read. Could each person post a 1-paragraph summary of what she or he thinks are the main issues under discussion? thanks, — Carl (CBM · talk) 20:30, 8 February 2009 (UTC)[reply]

In the thread above (#Glkanter's objection), user:Glkanter insists an unconditional solution is sufficient for the Monty Hall problem. I claim the specific wording used both here and in nearly all sources makes it technically a conditional problem asking about

P(win by switching | player picks door 1 and host opens door 3)

rather than

P(win by switching)

My stance is that the specific wording used in the article here ensures the numeric value is the same (2/3), but presenting an unconditional solution to a conditional problem (even one that happens to produce the correct numeric answer) is not sufficient. More specifically, the last two paragraphs at Monty Hall problem#Solution are mathematically accurate and important to include in the Solution section. -- Rick Block (talk) 20:42, 8 February 2009 (UTC)[reply]

In September or October I used the Monty Hall Problem Article to further my understanding of the puzzle. I found about 1000% more editorial matter than I would have expected. It's roughly 1% useful, 99% BS. Since then, I've been trying to establish why I believe it's 99% BS in order that this be deleted. I have chosen to use Probability Theory as the (only) appropriate method to convice the other editors. I have proferred what I am confident is a valid proof, and have requested that other editors find any errors. This has gone on for 4 or 5 months now. I am not unique in this. There is about 4 years of history attempting to do what I have set out to do. Rick says I'm not addressing the right problem. I insist that I am. Here's the proof. It's longer than it needs to be for clarity:

There are really only two situations to consider. Not 3, not 6.

2/3 of the time I will choose a goat.

100% of those times Monty reveals the other goat, leaving the car.

=>2/3 of the time I should switch.

1/3 of the time, I will choose the car.

100% of those times Monty reveals one of the goats. It doesn't matter which.

=>1/3 of the time I should not switch.

2/3 is greater than 1/3.

Therefore, by switching, I increase my chances of getting the car when offered the switch.

Because it only takes one proof. And I've presented it. And it's valid. And it proves that ALL that other stuff is nonsense.

There's another dozen points of disagreement. There's no point getting to those until we resolve the question of whether or not the 1st Solution (or something like it) is correct, and that all that other stuff is BS.

But, we're basically at a standstill. Neither Rick or I is going to pursuade the other. I want to see the page improved, finally, and requested that our Conflict get some Escalated Resolution help. Glkanter (talk) 21:06, 8 February 2009 (UTC)[reply]

Glkanter I totalt agree with you the current article is Bull crap! 99% useless and 1% useful. It does not do a good job communicating a simple and straightforward unconditional solution! I totaly agree with you and I want you to change it!! I have however started to lose interest because RickBlock is married to the current article and he interpret any attempt of changes as an attack on him. I just wonder why is his name RickBlock by the way? I thought the whole point was to collaboration?! I am actually a bit insulted by his choice of name!!--92.41.214.24 (talk) 22:14, 8 February 2009 (UTC)[reply]

Just FYI - it's my real name. -- Rick Block (talk) 22:50, 8 February 2009 (UTC)[reply]

I'm still not up to speed on this, but I have a few general comments.

92.41.214.24: the content of the "solution" section of this revision is far too fragmentary to be the actual content of a WP article. It would need to be turned into actual proseto be readable by an untrained reader.

ha, ha what do you mean by an "untrained reader"...My solution is the most simple one! (No BS, No waffle, like a punch in the stomach. Even though I have a PhD I dont expect anyone ells to have the same traning or understanding..:-) --92.41.112.145 (talk) 00:01, 9 February 2009 (UTC)[reply]

Glkanter: if your proof is correct when written out, this does not mean that "And it proves that ALL that other stuff is nonsense." There are always multiple proofs of any result, and different people will likely find one or another more pleasing.
Rick Block: the difficult with the Monty Hall problem is always the specific wording of the problem (hence the issues with Marilyn vos Savant). In the end there will never be a perfect wording. I think it will be possible to model the natural-language problem successfully either as a conditional or unconditional probability. It may be that the article could explain this in a footnote.

I also have a few questions if none of you objects:

Glkanter: when you talk about there being 99% extraneous stuff, do you mean just the "solution" section, or the entire article?

Glkanter has a point and that is that we should focus on the critical stuff. The more waffle you introduce the harder is it to understand. The solution should be max 5-6 sentences, all other stuff is just NOISE....bla...bla...blabbla....blabblallaaa...The current article is located at the pinnacle of confussion, waffle and BS. --92.41.41.161 (talk) 23:35, 8 February 2009 (UTC)[reply]

Rick Block: would there be a clear way that the article could explain the conditional/unconditional thing, so the article can handle it thoroughly at some point and then move on from there?

— Carl (CBM · talk) 23:03, 8 February 2009 (UTC)[reply]

Carl - having lived with this for several years, the least contentious approach has seemed to be to use quoted versions of the problem description (and a fully explicit version constraining the host to pick randomly between two goat doors is necessary to justify the Bayes treatment). The "vos Savant" version is far and away the best known, but it's not only conditional but under-specified. The conditional/unconditional thing is mentioned in the Solution section, and again under Sources of confusion. If we're to expand it (which I suspect Glkanter and others will argue against), I think the natural spot would be the "Sources of confusion" section. -- Rick Block (talk) 23:31, 8 February 2009 (UTC)[reply]

Of course, various styles of proofs would be included. You're two questions seem like one...What would be gone specifically from the Solution section is pretty much everything. I think the existing solution is overly complex. I'd obviously like to see one of the proofs offered here in the last few months be considered. Then a few other proofs. All the other current stuff in the Solution section should be deleted.

I'd like to see the 'combined doors' method elevated to the Solution section from the Aids to Understanding.

The rest of the stuff? I don't care. Just as long as each and every reference to the Solution being 'incomplete' or 'lacking rigour' is deleted. Variants? Sources of Confusion? Morgan et al? Bayesian? That stuff's above my paygrade.

Oh, yeah. And let's get rid of the banner at the top of this page. I've never seen anything like it. The two reference to 'confirmed by experiment' are awful.

Glkanter (talk) 23:29, 8 February 2009 (UTC)[reply]

Upon further review, I would eliminate the Sources of Confusion section. Anything in there that can't be moved to Aids for Understanding is itself a source of confusion and should be eliminated.

Glkanter (talk) 03:03, 9 February 2009 (UTC)[reply]

@Glkanter: look at what happens in 18 realisations of the game;
I sorted the outcomes for a better overview:

Choice  1 1 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 3 3
Car     1 1 1 1 2 2 2 2 3 3 3 3 1 1 1 1 2 2 2 2 3 3 3 3 1 1 1 1 2 2 2 2 3 3 3 3  
Open    2 2 3 3 3 3 3 3 2 2 2 2 3 3 3 3 1 1 3 1 1 1 1 1 2 2 2 2 1 1 1 1 1 1 2 2

6 times there was the same situation as in the player's case; here they are:

Choice  1 1 1 1 1 1 
Car     1 1 2 2 2 2   
Open    3 3 3 3 3 3

The conditional (!) probability to win the car by switching is (surprised?): 4/6 (BTW: the unconditional probability is 24/36, different, but numerically the same!) After all that is allready said, what needs to be said more? If you're not convinced, I guess you don't want to be convinced, or worse.Nijdam (talk) 23:13, 8 February 2009 (UTC)[reply]

Nijdam...what are you doing?? This is not quantum physics!! There is no need for simulation. It is a straight forward argument that does not need simulation. The more complex you make it the higher I will scream ! —Preceding unsigned comment added by 92.41.251.196 (talk) 23:45, 8 February 2009 (UTC)[reply]

It is not the general relativity theory either, yet the above figures explain it all. Please scream, if it make you feel better, I'll go to sleep.Nijdam (talk) 00:03, 9 February 2009 (UTC)[reply]

ha, ha you are funny ! yes, I think I will go to sleep also...I have had enough of this —Preceding unsigned comment added by 92.41.112.145 (talk) 00:06, 9 February 2009 (UTC)[reply]

One thing that would make this article better

What would make this article better would be a rearrangement of content that puts like things together, progresses from simple to complex, and uses headings that clearly define the issues. Here is a suggestion:

1 History of the problem (This would not be about the Three Prisoners problem or other “pre-history” problems, but about what happened when the problem appeared in Parade and in follow-up columns. This satisfies the needs of the casual reader.)

2 Explanations clearly consistent with the original solution (This would include 2.1 Name for the first explanation, 2.2 Name for the second explanation, etc.)

3 The common objection (This would be about the difference between asking the more general question whether one should switch, versus the more specific question whether one should switch when, say, door #3 is opened.)

4 Explanations that meet the common objection (This would include 4.1 Name for the first explanation, 4.2 Name for the second explanation, etc. Bayes would be here.)

5 Another objection (This would be about the objection that one does not know if the host is as likely to open one wrong door as another when the player chooses the right door, and the Morgan, et al. solution.)

6 Similar problems (The Three Prisoners problem would be moved into this section.)

7 Theories about the problem

Omit objections that clearly contradict the problem as stated, or put them into a final section 8: Other objections and why they are irrelevant

This proposal, or something like it, meets the needs of both the casual reader who has heard of the problem and wants to know what the deal was, and the reader who is interested in a greater understanding of the issues.Simple314 (talk) 01:10, 9 February 2009 (UTC)[reply]

Can anyone answer these questions:

1. Why is the Parade quote included in both the unnamed introduction and the first named section, “Problem”?

2. What is the significant difference between the introduction and the “Problem” section?

3. What is the purpose served by including both as separate sections?

Seriously, the article looks like a bunch of things added onto each other, one after another. The design begs for pointless arguments. Simple314 (talk) 02:05, 12 February 2009 (UTC)[reply]

1) The unnamed introduction is the "lead", which according to Wikipedia:Manual of Style must exist and (according to Wikipedia:Lead section) "should be able to stand alone as a concise overview of the article".

2) The intro contains an overview of the entire article, the "Problem" section describes the Problem statement in detail.

3) Whatever the first section is after the lead must (because of local style guidelines) repeat content from the lead. -- Rick Block (talk) 02:27, 13 February 2009 (UTC)[reply]

Thanks for your response. However, it is not completely consistent with the way things actually happen, and deciding what counts as the "Problem" section is happening far too early. Simple314 (talk) 04:06, 13 February 2009 (UTC)[reply]

I don't understand what you mean. Can you please explain? -- Rick Block (talk) 06:12, 13 February 2009 (UTC)[reply]

The reason many people are confused is that they wonder what happened to the odds that the car was behind Door #3, once Door #3 was opened. Did the odds disappear? Did they go somewhere else in the problem? According to the usual answer, they joined the odds that the car is behind Door #2. But why? Why did they go there, instead of going with the odds that the car is behind Door #1? Or splitting themselves in half, one half going with Door #1 and one half with Door #2?

Did they scurry over to hide behind Door #2 because Door #2 was closer? What if the player had picked Door #2 to begin with?

Instead, the answer tends to be, they joined the odds behind Door #2 because of some inexplicable, impossible to follow reason that avoids the question, where did the odds go?

They stay within the parenthesis of 1/3 + (1/3 + 1/3) as per the 'Combined Doors' solution. More correctly, the 1/3 is distributed randomly (evenly) across the remaining relavent doors, of which there is exactly 1. So it now reads 1/3 +(2/3 + 0). Glkanter (talk) 14:21, 15 February 2009 (UTC) Glkanter (talk) 15:29, 15 February 2009 (UTC)[reply]

A better understanding is based on this story, a metaphor, and what is talk of probability but metaphor? When the host opened Door #3, the odds that the prize was behind Door #3 were immediately banished to the Land of Things that Did Not Happen, where they live in a village called “the Host Opened Door #2.”

What is not so easy to realize is that some of the other odds on stage were also banished to that land and that village when Door #3 was opened: namely, “the odds that the prize was behind Door #1 WHEN the Host Opened Door #2.” We take it on faith that those odds are exactly ½ the original odds that the prize was behind Door #1.

That leaves on stage ½ of the original 1/3 odds that the prize was behind Door #1, and the original 1/3 odds that the prize was behind Door #2: 1/6 v. 1/3. The good wizard Ratio of Fractions magically changes those odds into 1/3 v. 2/3, and the moral is, “Switch, because it doubles your odds”

There is another version of this tale, in which a scheming bad wizard casts a spell that makes it impossible to tell what portion of the original odds the prize was behind Door #1 were odds the prize was behind Door #1 WHEN the host opened Door #2, but the people who promote that version are members of a tiny secret order, a portion of those who believe the Monty Hall Problem is fundamentally a probability problem. Their motto is “If it isn’t explicitly stated, don’t assume it is true!” The moral of their morganian story is, “Switch, because it won’t decrease your odds, but it might increase them!”

The rest of us know that the Monty Hall Problem is fundamentally a cute trick designed to fool us into thinking that the odds are even when they definitely aren’t. Our motto is “If you must assume something to make the ruse work, then the assumption must be true! But we may forget to say that.”

I dare anyone to demonstrate this is more pointless than the other points made here. Rearrangement of the article into History, Objection 1, Objection 2, etc., would help avoid silliness and endless arguments. Simple314 (talk) 02:51, 12 February 2009 (UTC)[reply]

A brilliant metaphor, is the story of the missing odds published anywhere? -- Rick Block (talk) 02:27, 13 February 2009 (UTC)[reply]

A clever example of sarcasm, is there any indication above that the author of the story is suggesting that it should be included in the article? Simple314 (talk) 04:06, 13 February 2009 (UTC)[reply]

In all sincerity, I do like it. I realize you were not intending the phrasing above to be added, and a published version would almost certainly not be quite so whimsical, but in all sincerity if there is a published analysis of this sort I think we should consider adding it. I think it is at least as convincing as what is currently in the "Why the probability is not 1/2" section. -- Rick Block (talk) 06:12, 13 February 2009 (UTC)[reply]

Sorry for the misunderstanding.Simple314 (talk) 21:22, 13 February 2009 (UTC)[reply]

Apples and oranges

The Monty Hall Paradox is only a paradox for people who like to compare apples to oranges. If a player were to pick one door out of three, two out of three times that player would be wrong. So a player should always switch if they get a second chance. With respect to the POSSIBLE OUTCOMES of this game, there are more scenarios of a player who did switch won a car. But that is not to say that the act of 'switching' improves probability. The probability of picking one door out of two is ALWAYS 1 to 2. This is true by definition. Trying to say othrwise is simply meaningless, and you can not prove it with a computer simulation. Just as you can not prove that the probability of flipping heads of a fair coin is 1 to 2. Try it, it is only true by definition. —Preceding unsigned comment added by 68.198.159.238 (talk • contribs)

If I take 3 cards, say two red twos and the ace of spades, shuffle, deal 1 to you, keep two, look at my two cards and throw out a red two (that I show you), is this equivalent to the Monty Hall problem? Are you saying you now have a 50-50 chance of having the ace? How about if I take an entire 52-card deck of cards, deal one to you, keep the rest, look at them and throw out 50 with none of them being the ace of spades (keeping only 1 card)? Still 50-50? Please try this a few times and let me know how it goes. -- Rick Block (talk) 05:24, 9 February 2009 (UTC)[reply]

Rick, is that experiment you describe above the equivalence of the conditional or the unconditional problem? Does it make a difference?

Glkanter (talk) 08:46, 9 February 2009 (UTC)[reply]

Glkanter is referring to this discussion, above. In a technical mathematical sense, since the cards are individually distinguishable and since the question of the probability of having the ace is asked at the point the discards are visible, this is a conditional problem. If you want a deeper understanding of this problem, when playing the role of the dealer as your one remaining card always keep the ace if you have it and if not then keep the two of diamonds. Keep track of how many times you try altogether and then (in the three card version) separate tallies for how many times you have the ace when the dealer shows you the two of diamonds and how many times you have the ace when the dealer shows you the two of hearts. I.e. fill out the following table:

Total tries	Total times you had the ace

Number of times dealer showed the heart two	and you had the ace

Number of times dealer showed the diamond two	and you had the ace

If you do this enough (maybe 60 times), you should see your table start to converge (see Law of large numbers) to these numbers

Total tries	Total times you had the ace
60	20 (1/3 of total tries)
Number of times dealer showed the heart two	and you had the ace
40 (2/3 of total tries)	20 (half of these tries)
Number of times dealer showed the diamond two	and you had the ace
20 (1/3 of total tries)	0 (all of these)

What this is showing you is that you (the player) have an unconditional (meaning ignoring which card the dealer discards) 1/3 chance of having the ace even after the dealer only has one card left. However, if the dealer doesn't discard randomly (when he has the ace) your conditional chances of having the ace might be different depending on which card the dealer discards. -- Rick Block (talk) 15:51, 9 February 2009 (UTC)[reply]

This is a good improvement on the card simulation as usually described. A version of the card simulation that even more closely approximates the problem as presented involves discarding a card without showing it, and works like this:

Show the ace of spades, the 2 of hearts, and the 2 of diamonds; shuffle them; then deal them in a row, into positions first, second, and third. The card in first position is the card chosen by the player. The dealer (host) looks at the other two cards, and discards the 2 when he sees both the ace and a 2, and discards the 2 of diamonds when both cards he sees are 2s. (Choosing the 2 of diamonds in such cases does indeed make the choice between discarding the second card and the third card equally likely, when both are 2s.)

Keep track of how many times the ace is in the second position, how many times the ace is in the third position, how many times the ace is in the first position and the second card is discarded, and how many times the ace is in the first position when the third card is discarded. After many repeated experiments, it will be clear that of all possible cases, the player wins by keeping the first card when the third card was discarded 1/6 of the time, and wins by switching from the first card to the second card 1/3 of the time—thus doubling the chances of winning by switching from the first card to the second card.

Why is it better that the card is discarded without being shown? The player doesn't know the difference between the two goats, but does know the difference between hearts and diamonds.Simple314 (talk) 02:52, 10 February 2009 (UTC)[reply]

Your last paragraph confuses me. Isn't it a premise of the puzzle that the dealer discards a two (reveals a goat) from among his two cards? There's never anything random about this action. It may be random how he chooses which of the two two's, but the fact that he's going to discard a two is a premise. Let's switch back to Monty Hall. Unless the contestant knows something about how Monty chooses 'which goat', he gains no new knowledge when Monty reveals the goat. If he knew something about how Monty chose goats, then that's a new premise or constraint. Which makes it a completely different puzzle.

We have a saying in my business. "If it looks like a duck, quacks like a duck, and smells like a duck, it's a duck."

At the risk of showing my ignorance, I don't think there is a difference between the conditional and unconditional problems. It's already agreed that they have the same statistical probabilities. Now, it looks to me like you're using the exact same simulation that I would use to 'aid in the understanding' of my proposed proof. Since the probabilities are the same, I would suggest that there is no difference in the total knowledge (useful information) on the contestant's part. (That is, the premises and constraints are identical, unlike the 'Monty always chooses the left-most goat-door' variant). Lastly, In order for the two puzzles to be different, yet have the same probabilities, it seems that those probabilities would diverge at some time, yet co-incidentally return to the same values. I don't see that happening. They just both start out at 2/3 and never go anywhere.

Glkanter (talk) 17:04, 9 February 2009 (UTC)[reply]

To summarize: in response to my question, you found it necessary to add a constraint, in order to make it a conditional probability problem. That makes it a different puzzle than the Monty Hall Problem. Further, one would not say that the probabilty for this new puzzle is 2/3. Going back to the Monty Hall Puzzle, since this new constaint (left-most goat-door) is known by the contestant (or else, what's the point?), there are scenarios when he knows exactly where the car is. So the answer would be stated "100% when Monty reveals door #3, 100% when I've chosen door #3 and Monty reveals door #2, etc." So this contestant makes more informed choices than the contestant in the real puzzle. So the contestant will win MORE than 2/3 of the time.

Glkanter (talk) 17:50, 9 February 2009 (UTC)[reply]

You're constantly missing the point. The player who is offered the option to change, is in the situation of having chosen a door and seeing a door opened with a goat. For different players these doors may differ from the ones for "our" player. But "our" player finds himself in a specific situation, and in this situation he is asked whether he wants to change. In the normal case the answers are the same for all the possible combinations, but all the answers involve conditional probabilities. I suggest you study the explanation I gave above, with all the possibilities summed up. In fact it is a representation of the sample space of the problem. Study it and ask me if you do not understand anything. Nijdam (talk) 18:12, 9 February 2009 (UTC)[reply]

No Nijdam I think you are missing the point.If you want an ice-cream you say "I want an ice-cream" you dont say I want a barn, I want animals, I want a cow, I want ice, I want an machine..... The closes way between two point is a straight line and your argument is not located on that line. You argument is just confusing and your intention are at best selfish! Secondly why is "Rick Block" discussing card games?! It does not have any thing to do with the Monty hall problem !! Some of you are all so deluded and confussed that I am just amazed how you manage everyday life. Further, "Rick Block" are you the Evangelist Preach "Rick Block" because if you are everything falls into place. Your stubborness and unwillingness to accept the truth must be a high in demand trade in religious circles since it can not be easy when you have invested your whole life in god to realize that GOD DOES NOT EXIST.--92.41.3.118 (talk) 18:55, 9 February 2009 (UTC)[reply]

Easy things first - I don't see the relevance, but I'm not the evangelist who comes up if you google my name. I'm bringing up card games because they are a useful simulation of the Monty Hall problem for those of us who don't have television studios with doors big enough to hide cars and enough money to hire Monty Hall. And please go read WP:NPA.

Glkanter - yes, it's a premise the host discards a red two and the randomness is how he chooses between two of them if he has both. Switching back to the Monty Hall problem, we assume the contestant knows the rules of the game we're telling her. When the host opens a door, she sees the specific door the host opens. What Nijdam is saying is that it's always a conditional problem (at least in anything like the normal presentation of the problem). If the problem statement doesn't say how Monty chooses between two goats, it's still a conditional problem and the mathematically correct answer is we don't exactly know what the probability is of winning by switching but it is something between 1/2 and 1. This is the entire point of the Morgan et al. paper.

Many people don't care about being this mathematically correct (although mathematicians certainly do!) and rather than answer the actual question mentally shift to a different question - specifically, across all players who switch, how many will win by switching (the "unconditional" question, i.e. the all 60 trials in the table above). We can get an answer to this question even if we don't know how the host chooses between two goats, but this isn't the question that is actually asked (go read it - does it say anything about "all players" anywhere?).

In the fully explicit version of the problem we avoid the unsettling "we don't know, something between 1/2 and 1" answer by saying If both remaining doors have goats behind them, he chooses one randomly. This constraint allows us to figure out a single number (2/3), but we're actually saying "given any door the host opens, the player's chance of winning by switching is 2/3". If the chance is 2/3 given any door the host opens, then of course the answer to the "unconditional" question (the probability of winning for all players who switch) is 2/3 - but not the other way around (A -> B does not mean B -> A).

In the left-most goat-door variant of the fully explicit version we can also figure out an exact, single, numeric answer, but there are two answers depending on whether the host opens the left-most or right-most door. In the problem as usually stated the host is opening a right-most door, so (for this variant) we'd say the player has a 100% chance of winning. The more general answer (for this variant) is "the chance of winning by switching is 50% if the host opens the left-most door, and 100% if the host opens the right-most door". For this variant, the answer to the "unconditional" question is still 2/3.

Please pardon my intrusion here. In the next-to-last sentence above, you state "the chance of winning by switching is 50% if the host opens the left-most door". No, it's 2/3. That's because the contestant's original selection has a 2/3 chance of being a goat. Therefore, the remaining door has a 2/3 chance of being the car. Assuming everything else you wrote is correct, your conclusion is therefore wrong. The probability would be somewhere between 2/3 and 100%. Which by definition is not "still 2/3." So it's not equivalent mathematically to the actual puzzle. That's because Monty's added constraint sometimes gives us additional information. Like when it's 100% right to switch. We would expect our chances of success to increase above the previous level of 2/3. This would appear to seriously disrupt your following statements about B=2/3. So, what's the point of the left-most goat variant, anyway? It somehow disproves my proofs in what way?

Glkanter (talk) 07:01, 10 February 2009 (UTC)[reply]

50% is correct. The cases are as follows: player selects a goat and the car is behind door 2 (1/3), player selects a goat and the car is behind door 3 (1/3), player selects the car (1/3). In the left-most variant, if the host has opened the left-most door (door 2), we can only be in the first or third of these cases. Players in these cases have selected a goat or the car with equal probability. For the players who have seen the host open the left-most door, switching wins with 50% probability. -- Rick Block (talk) 17:09, 10 February 2009 (UTC)[reply]

With the left-most door constraint added, Monty either tells us exactly where the car is, or he doesn't. So, compared to the actual puzzle, he either adds to our knowledge, or does not. But he CAN NOT, under any circumstance, subtract from our original knowledge that 2/3 of the time the contestant chose a goat. Therefore, the contestant always has at least a 2/3 chance by switching, and sometimes 100%. I'd be interested to see the proof that this varient results in a 2/3 overall probability by switching. I don't believe it. And, BTW, this variant (more correctly, 'different puzzle') doesn't matter one bit. It doesn't dis-prove my proof.

Glkanter (talk) 17:29, 10 February 2009 (UTC)[reply]

If we call the answer to the conditional question A, and the answer to the unconditional question B, what your "proof" is saying is that A = B. This is true for the "random goat" variant (fully explicit version in the article) variant. However, it's not true if we say nothing about how the host selects between two goats (in which case .5 <= A <= 1, and B=2/3) or if we say the host picks the left-most goat (in which case, A=.5 or A=1, and B=2/3).

I'll freely admit this is VERY confusing. But, it is mathematically correct. Anything else is simply a fuzzy approximation. -- Rick Block (talk) 20:51, 9 February 2009 (UTC)[reply]

Rick, would you accept that, if the statement of the problem were to explicitly state that 'no information is revealed that would affect the probability of winning by switching by anything that happens between the players original choice of door and his decision to switch or not' then the conditional and unconditional cases would be indistinguishable. Martin Hogbin (talk) 21:10, 9 February 2009 (UTC)[reply]

You asked me somwhere above about the same question. And I replied it is not a matter of not changing probabilities (meaning unconditional and conditional being numerically the same?), but of not limited outcomes. If the information does not limit the possible outcomes, yes, then there is no actual condition. But in the problem, the information limits the outcomes considerably. I.e. only one of the doors may be chosen, what allready excludes 2/3 of the outcomes. Look at my "number example" above. Nijdam (talk) 00:57, 10 February 2009 (UTC)[reply]

But the problem is symmetrical with respect to doors.{inserted by Martin Hogbin}

You might think this is kind of a cop out, but it might help some folks understand how things are supposed to work here. Would this change make the conditional and unconditional cases indistinguishable? I really don't care. I would accept it, but only if you (or someone) can provide a significant number of reliable sources, in particular academic ones, that include some condition like this in their statement of the Monty Hall problem. I say "significant number" and "academic" because I've personally read all the references in the current article, many of them academic, and several dozen more and I don't recall anyone ever stating the problem this way. On the other hand, lots of references fully qualify the host's behavior (including constraining the host to choose between two goats randomly). The article needs to be about the same problem the sources are about. Inventing our own version would be original research. -- Rick Block (talk) 03:06, 10 February 2009 (UTC)[reply]

I know how things are supposed to work and this is the discussion page, where we are allowed to think for ourselves. All I am trying to do here is establish the fact that what is and what is not a condition is a matter of choice. A condition is anything which is considered might affect the probability. The point I am making is that on this discussion page, you and others use the issue of conditional probability as a indisputable fact whereas it is in fact a decision made by certain authors of academic papers. I agree that they are generally to be considered reliable sources but that does not prevent us questioning their decisions in some matters. Martin Hogbin (talk) 09:53, 10 February 2009 (UTC)[reply]

A condition is a restriction imposed on the possible outcomes. The question of affecting "the" probability is a popular way of speaking about the unconditional and the conditional probability. A specific probability is never affected. So if someone speakes about "affecting the probability", he inherently speaks about conditioning. In the problem there is a heavy conditioning: firstly restriction to the chosen door and further restriction to the opened door. Please consider the example given above (with all the numbers) and be aware of the unconditional situation and the restricted conditional one. Nijdam (talk) 11:21, 10 February 2009 (UTC)[reply]

The following thought crossed my mind in trying to understand the lines along which some discussiants think. If someone says: no information is revealed, he should complete his statement with "in respect to some event", because in the problem information is revealed about the car being behind Door 2. It is hence tricky to decide that no information is revealed in general just by the wording of the problem. And if one wants to know that no info is revealed about some event: well, he has to compare the unconditional and the conditional prob. So let us consider Door 2. Show me the "unconditional reasoning" for it. Nijdam (talk) 12:08, 10 February 2009 (UTC)[reply]

For ease of editing I have answered in a new section below. Martin Hogbin (talk) 21:29, 10 February 2009 (UTC)[reply]

Rick, here's one of the areas where you and I fundamentally disagree.

You just wrote, "We can get an answer to this question even if we don't know how the host chooses between two goats, but this isn't the question that is actually asked (go read it - does it say anything about "all players" anywhere?)."

My belief system in Probability Theory says that 'all players' is ALWAYS the question being asked. Isn't that the question being asked when someone asks 'what's the probability of a coin coming up heads?' Why would this be different?

Glkanter (talk) 21:34, 9 February 2009 (UTC)[reply]

Different questions are different. We have to pay attention to what they say. The Monty Hall problem says "Imagine [or say] you pick door 1 and the host picks door 3. Is it to your advantage to switch?" If you want to generalize this to a bunch of players it would be "for any player who picks door 1 and the host then opens door 3, what is the chance of winning by switching?". All these players would find themselves in the same situation as our given player. Generalizing this to "for any player who plays the game, what is the chance of winning by switching" ignores aspects of the problem that may, or may not, be significant. The stated conditions are "player picked door 1 and host opens door 3" (and host always opens a door, and host always shows a goat, and host picks which goat to show randomly if he has a choice). Ignoring ANY of these conditions may make the solution mathematically unsound. I keep bringing up the example of how the simple solution (which ignores several of these conditions) results in the wrong numeric answer for the question actually asked by the left-most door variant. It produces the correct numeric answer for the fully qualified version stated in the article, but right answer sometimes and wrong answer other times is pretty much the definition of mathematically unsound. -- Rick Block (talk) 03:06, 10 February 2009 (UTC)[reply]

Two things. In the specific door 1, door 3 question, why isn't it 2/3? Doesn't the contestant have a 2/3 chance of picking a goat? Then Monty shows the other goat, leaving a car. That's all we know at this point, right? That fits in perfectly with my new and improved, customized Super Proof! (Tip of the hat to a certain MH, who I shall keep out of this.)

Every time this contestant picks a goat, Monty will reveal the other goat, leaving the car. (As you just wrote above, it is a premise that Monty will always randomly reveal a goat behind one of the remaining doors).

[When selecting door 1], 2/3 of the time this contestant will pick a goat.

Therefore, when it is offered, this contestant should switch door 1 for door 2 to improve his/her chances of selecting the car.

The second thing is, I reject the door 1, door 3 literal interpretation. (Not that it makes any difference, as I just proved.) That's not how Probability Theory works. And that's not what makes this an internationally known paradox. Heck, the real Monty didn't even use doors. He used curtains.

Glkanter (talk) 05:16, 10 February 2009 (UTC) Glkanter (talk) 06:14, 10 February 2009 (UTC)[reply]

The third point is, the 'left-most door variant' is a different puzzle. It adds a new constraint on Monty. Which, in some cases, causes Monty to reveal the exact location of the car. My proof was never intended to solve that problem. I wish you would stop using this as your 'proof' that my proofs are flawed. I think it's this action on your part that I react the most strenuously against.

Glkanter (talk) 05:24, 10 February 2009 (UTC) Glkanter (talk) 06:14, 10 February 2009 (UTC)[reply]

The left-most door variant is not my example, it's what is used in the academic literature to show the exact difference we've been talking about all along. The point is in this variant the player still has a 2/3 chance of initially selecting a goat, and Monty must still open a door revealing the other goat. Which means your "proof" should still apply, since these are the "givens" that your proof uses. The way math works, a proof has a "givens" section and applies to any instance of a problem satisfying those givens. Your proof does not distinguish this variant from the fully specified problem given in the article because it does not use the "given" in the fully specified version that says the host picks randomly between two goat doors. If it's a valid proof, it must be valid for any instance of a problem meeting its "givens". In effect, your proof changes the fully specified problem statement, replacing If both remaining doors have goats behind them, he chooses one randomly with If both remaining doors have goats behind them, he may choose one however he'd like - randomly or not.

The problem statement doesn't have to be taken exactly literally (meaning player picks door 1 and host opens door 3), but it clearly means we're talking about a case where a player has picked some door and the host has opened some other door in response. We're talking about 6 possible cases: player=1/host=2, player=1/host=3, player=2/host=1, player=2/host=3, player=3/host=1, player=3/host=2. The problem as worded in the article (as worded nearly anywhere) is asking about the probability of winning by switching in only one of these cases, not the probability of winning by switching in the aggregate. We could list the six cases and say, "for a player in any one of these six cases, what is the probability of winning by switching". Your solution does not apply to any of these cases individually but treats them all as one undifferentiated clump. The probability in each of these cases (the conditional probability) doesn't have to match the probability in the aggregate (the unconditional probability). -- Rick Block (talk) 16:40, 10 February 2009 (UTC)[reply]

I stated once before that I was unsure of your grasp of the subject material. The above posting increases my concern. It's unpleasant to write that. It's important that I write it. Because it's my reason for not addressing your posting point by point. I will never do that again.

Monty's 'motivation' is immaterial, unless it is known to the contestant. At which point that becomes a new premise/constraint of a new puzzle. Following your theory, we should also concern ourselves with the 'motivation' of the stage hand who originally arranged the 3 items behind the doors. For probability purposes, we call those actions 'Random'. And they deserve no further scrutiny.

Glkanter (talk) 17:02, 10 February 2009 (UTC)[reply]

And I stated before I didn't think you were a troll. You have now convinced me otherwise. I'll enlist uninvolved admins to enforce this if necessary, but in my view your posts have now reached the point of disruptive editing. If you continue in this manner I will report you to WP:ANI and suggest you be blocked from further editing. -- Rick Block (talk) 17:54, 10 February 2009 (UTC)[reply]

I've wondered how you would respond when the inevitable happened, and your arguments were proven to be meritless. Now we know. You attack the messenger, and start posting that for Wikipedia purposes, the Truth (being right) is less important than having been published.

I've asked for Wikipedia intervention for days now. I welcome any actions you take to bring closure to this festering sore.

Glkanter (talk) 18:35, 10 February 2009 (UTC)[reply]

Work around

There may be a kind of solution that comes close to the "simple intuitive solution", but is really a solution to the problem. Suppose the player is in the situation of having chosen door 1 and being opened door 3. We indicate this situation as [13]. Of course other players may find themselves in [12], [21], [23], [31] or [32]. Together all these 6 situation represent the unconditional one. With the "normal" strategie of the host, there is no fundamental difference between these 6 situation, hence the (conditional) probability of winning the car by switching is the same for each. But the overall probability, the unconditional, is 2/3, as we know from the intuitive (simple) solution. But then all the conditional probabilities must aldo be 2/3. Will this be satisfactory? Nijdam (talk) 15:38, 10 February 2009 (UTC)[reply]

Given no original research and verifiability are fundamental Wikipedia policies, how would you reference this? The bottom line is we care less about the Truth™ than accurately representing what reliable sources say. From an academic viewpoint, what we do is summarize published literature - and that's all. -- Rick Block (talk) 17:28, 10 February 2009 (UTC)[reply]

Somehow I unintentionally deleted this:

Does anyone else find the above paragraph troubling? I certainly do. And after 4 years, and thousands of postings on the Monty Hall Problem, I find it incredibly ironic.

Glkanter (talk) 18:22, 10 February 2009 (UTC)[reply]

Glkanter (talk) 18:24, 10 February 2009 (UTC)[reply]

Glkanter (talk) 20:36, 10 February 2009 (UTC)[reply]

Yes, I find it troubling also! I just hear bla bla bla bla bla.. there are so many word but no insight and understanding at all. WAFFLE WAFFLE WAFFLE --92.41.187.141 (talk) 21:02, 10 February 2009 (UTC)[reply]

What's ironic is that you think it's ironic. I've consistently been pointing you to sources that you have consistently ignored, while you have offered up not a single source to support any of your views. What I've been trying to do on this talk page is help you understand what the sources say, in an effort to get you to understand that the changes you'd like to make to the article would turn it into something other than an accurate summary of the sources (which is what it is supposed to be). Most folks tend to believe what peer reviewed academic sources have to say is correct, and if faced with a conflict between what such sources say and their own internal beliefs are at least willing to consider the possibility that their internal beliefs may be incorrect. -- Rick Block (talk) 19:13, 10 February 2009 (UTC)[reply]

It has been a fact for a long time that "Rick Block" is a burden for other wikipedia users when it comes to producing a high quality Monty hall article. That "Rick Block" has started to threaten other wikipedia editors such as Glkanter is even a bigger concern! You think you own wikipedia and no one has the right to question your arguments. I suggest you tone down your involvement because there are a lot of people that is starting to get very tired of your lenghty, emotional, unintelligent, stubborn and confussing posts. When you have realize that the best answer is a simple one then you are welcomed back.......--92.41.187.141 (talk) 20:56, 10 February 2009 (UTC)[reply]

Personal threats and accusations are undesirable from all contributors to Wikipedia. Martin Hogbin (talk) 22:12, 10 February 2009 (UTC)[reply]

Hey Rick, which "peer reviewed academic sources" did you get this gem from?

"The overall probability of winning by switching is determined by the location of the car."

That was sentence #1 of the Solutions section until I deleted it last year. It is 180 degrees counter to every bedrock principle upon which Probability Theory rests.

Glkanter (talk) 21:46, 10 February 2009 (UTC)[reply]

ha, ha that is a good one! That is just bad!! I am a bit scared actually! The person that wrote that has no clue...--92.41.187.141 (talk) 22:46, 10 February 2009 (UTC)[reply]

This afternoon I posted a reply above the big black table which I falsely assumed to be the end of the page. Now I also read the rest of it and I will post some of it again, with added content: my opinion is that this whole issue is about one simple question: should the requested chance be regarded as a conditional probability, and if so, why? I agree with Martin Hogbin that the issue of conditional probability is regarded "as an indisputable fact". Morgan et al. only give us a hint: "F5 is incorrect because it does not use the information in the number of the door shown." What information is in the number? Nijdam writes: "what is the meaning of "no information is revealed"? This actually means that the possible outcomes are not restricted." and: "The question of affecting "the" probability is a popular way of speaking about the unconditional and the conditional probability." However, the article Conditional probability states that P(B|A) = P(B) if A and B are statistically independent. In other words: conditional and unconditional chances are the same if A does not change the probability of B. And it doesn't. Heptalogos (talk) 21:51, 10 February 2009 (UTC)[reply]

yes that is correct when A and B are independent then P(B|A) = P(B). Good observation! :-) --92.41.187.141 (talk) 22:42, 10 February 2009 (UTC)[reply]

When is conditional not?

I am trying to get agreement on a hypothetical question. Suppose the question posed was this:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door and the host opens another door which has a goat and he does this in a way the does not affect the probability that your original choice was a car. He then says to you, "Do you want to pick the other unopened door?". What is your advice?

I know that this is not the question actually posed but in the above case do you accept that the conditional case is identical to the unconditional one? Martin Hogbin (talk) 22:09, 10 February 2009 (UTC)[reply]

I think that the statement of Boris Tsirelson confirms the point that I am making.

This makes the result of the conditional case the same as the result of the unconditional case, but does not make them "the same". The question is still, formally, a conditional question. -- Rick Block (talk) 05:23, 11 February 2009 (UTC)[reply]

In that case, why do we not consider all the other actions that occur between the original choice and the decision to swap as conditions? 09:39, 11 February 2009 (UTC)

What I am trying to achieve

What I would like to have is an addition to this article of a very simple and convincing explanation of the essential Monty Hall problem.

I believe that, for this purpose, it is acceptable to consider the unconditional problem for the following reasons:

1 The inability of most people to get the right answer, however the problem is formulated, is what makes it notable.

2 The academics who have studied the problem have to some degree (understandably) obfuscated the central problem with unwarranted and unjustified complications, including making it necessarily conditional.

3 Even if the conditional version is considered to be the best one, it is not uncommon in mathematical expositions to start by answering a related, and often simpler, problem and then to explain how this relates to the full blow problem.

The more academic discussion can remain as it is for those who are interested. Martin Hogbin (talk) 23:09, 10 February 2009 (UTC)[reply]

Martin - we've discussed this before and I think we're on basically the same page. The problem is notable because most people get the answer wrong, and people who assert the wrong answer are typically certain that they are correct (which makes for a large number of popular references, and a fair number of academic references including psychology papers exploring why the problem is so difficult). I'd rephrase point 2 somewhat - the academics who have studied the problem have shown even people who think they understand it and get the "correct" answer are often using mathematically unsound approaches. I view this as layers of understanding. We have people who will believe to their dying day it must be a 50-50 choice because (as any idiot can see) there are only two doors! Any of the simple solutions (treating the problem unconditionally) are more sophisticated answers. However, if we look closely at the problem, we see it is a conditional problem, not an unconditional one, so even the people who get that it's not a 50-50 choice aren't exactly correct either. Some of these people will apparently believe there is no difference between unconditional and conditional problems until their dying day because (as any idiot can see) 2/3 of the players will initially pick a goat and will win if they switch. Explaining that it is actually a conditional problem is not "obfuscating the central problem with unwarranted and unjustified complications". It's actually solving the problem as its given. -- Rick Block (talk) 05:53, 11 February 2009 (UTC)[reply]

I take it that we more or less agree about the first point.

I think you are wrong about people who get the right answer using unsound approaches. As I hope to show below Morgan et al do make the solution unnecessarily complicated by failing to take account of the natural symmetry of the problem and inventing doubts in the formulation of the problem when in fact there are none.

Even if one accepts that the conditional answer is the right one, what about my point 3. This approach is not at all uncommon in published papers. Martin Hogbin (talk) 11:12, 11 February 2009 (UTC)[reply]

Well, isn't this the exact structure of the current solution section? It correctly solves the unconditional problem, and then goes on to show a solution for the conditional. -- Rick Block (talk) 12:34, 11 February 2009 (UTC)[reply]

Proposed simple solution

We can start by observing that, if the player switches, he always gets the opposite of his original choice. The are no conditions involved, no probabilities, no dependence on host action - this is a certainty. Obviously of he does not switch he keeps his original choice.

All we have to show now is that the action of the host does not affect the probability the he had originally chosen a car, or to put it another way, that no information is revealed that might indicate the players original choice was.

Let me ask this question, looking at the Parade statement, what is the information that is revealed? Martin Hogbin (talk) 23:23, 10 February 2009 (UTC)[reply]

The information that is revealed is the specific door the host opens. It might not be a random choice if the player's initial pick is the car, in which case the probability of winning by switching might be anywhere from .5 to 1 (per the Morgan et al. and Gillman papers). Regardless, isn't the text currently in the solution section above the figure almost exactly this "simple" solution? -- Rick Block (talk) 03:51, 12 February 2009 (UTC)[reply]

Yes, I forgot, there is an explanation along these lines buried in the article. Perhaps my only complaint is that it is not clear or prominent enough. That section seems to try to answer several questions at once.

However, you are not answering the question, you are merely speculating on what the question might have been. Do you see my point? We both agree that you must answer a probability question on the information given in the question. Looking at the Parade statement, what is the information that is revealed? To put it another way, does the question tell is that it is more likely or less likely that the player originally picked a car after the host has opened a door. It clearly does neither, that is all the information that we have.

You are free to speculate on what might be the case, but that is not answering the question, it is changing it. Martin Hogbin (talk) 09:22, 12 February 2009 (UTC)[reply]

The question of wheter any information is revealed or not, is precisely where it is all about in this discussion. But here Martin merely states that no information is reavealed and hence the conditional prob. is the same as the unconditional, and also he has to admit that it is the conditional prob. which is of interest. The only thing is, he says I know on forehand what it's value will be. But the arguments why no information is revealed are missing. And the arguments may come by computing the conditional prob. or by considering the symmetry in the problem. But it is always the conditional prob. that has to be considered, and always some argumentation is needed. Nijdam (talk) 18:40, 14 February 2009 (UTC)[reply]

Experts

For clarity in the discussion: I will remove any comment of someone else besides RickBlock and Martin Hogbin given between this header and the corresponding trailer and move it above this header. Also I put this section allways at the end of this page.

@RickBlock. It's sad and frustrating all these people, hardly knowing anything about probability, but yet thinking they just know all the ins and outs of this problem. From now on, I'll only discuss matters with you (as one expert), if necessary, and with MartinHogbin, who seems to me at least responding in an adequate manner. Others may profit by reading the comments, I will no longer respond to them. Also I think it is about time this discussion ends. My suggestion above under "work around" (maybe not so well chosen title) is a sound reasoning, in the line of the "simple solution". It lookes if you crititizised it. It is (seemes?) easy to proof and may be helpfull in understanding the conditional nature as well. Nijdam (talk) 22:55, 10 February 2009 (UTC)[reply]

You say above With the "normal" strategie of the host, there is no fundamental difference between these 6 situation, hence the (conditional) probability of winning the car by switching is the same for each. I think this is the same endpoint Martin is trying to get to. I'm not sure if you're asking what I think about the math of this, or if you're suggesting including something like this wording someplace in the article. I'll answer both questions.

As math, if by "normal strategy of the host" you mean the behavior as prescribed in the fully explicit version including the "random goat" constraint I think "no fundamental difference between the 6 cases" needs to be a conclusion rather than an assumption. The "random goat" constraint is what forces the 6 situations to be symmetric - it is (at a fundamental level) why this constraint must be included implicitly or explicitly in a Bayesian analysis.

As for including this in the article, I think we might add something like it as an observation following the conditional solution, or in the transition between the unconditional and conditional solution. I know you object to the unconditional solution, however there are certainly plenty of reliable sources that present unconditional solutions. Would you be better with "so and so's solution, which says this, shows the unconditional probability is this", followed with an explanation of conditional vs. unconditional, and then a conditional solution? This is meant to be the current structure. I said this elsewhere (not sure if you noticed) but if you are an academic I think the right way to think about the article is that it is supposed to be a detailed summary of the popular and academic literature. How would you summarize the popular literature that treats the problem in an unconditional manner? -- Rick Block (talk) 13:07, 11 February 2009 (UTC)[reply]

Like you. I also strongly object the 'unconditional' non-solution, that's the whole point in this discussion. Maybe I didn't express myself clear enough. The "work around" idea is to incorporate the basic idea of this unconditional reasoning as to explain the right conditional explanation. Of course in the 'random goat' strategy, only then there symmetry in the 6 basic cases. And this symmetrie is not a conclusion of the calculation of the condtional probs. The suymmetry leads to equalness of the conditional probs. Hence it can be used ([123] means chosen 1, opend 2 car 3) as follows: for all a,b,c all different:

{\tfrac {1}{3}}=P({\text{winning by switching on forhand}})=

P([123]\lor [132]\lor [213]\lor [231]\lor [312]\lor [321])=

\!\,\sum _{[ij*]}P(...|[ij*])P([ij*]=P([abc]|[ab*])

Hence all conditional probs equal 1/3. But it remain (are) cond. probs.

The 'random goat' strategy is part of the premisses and I think the 'unconditional' non-solution doesn't find support in other cases. or is this the point you are making, that people reason in the same way even when the host has another strategy? Nijdam (talk) 12:26, 12 February 2009 (UTC)[reply]

Yes, the point is various host strategies make the problem symmetric or not and the unconditional non-solution (as you put it) ignores the host's strategy. If you assume the problem is symmetric, the conditional probabilities are all the same as the unconditional probability but how do you justify this assumption? The "random goat" constraint is the only thing that makes the problem symmetric, but it does so by making the conditional probabilities all identical which makes the problem symmetric. It's not the symmetry that makes the conditional probabilities the same, it's the other way around. Try to solve the problem with a Bayesian analysis without using the "random goat" constraint. I think you'll find you have to introduce the "random goat" constraint to get the 2/3 win by switching answer. This is the exact same issue.

BTW - please don't move the section (it's fine to add to a section that is not at the bottom of the page). -- Rick Block (talk) 03:26, 12 February 2009 (UTC)[reply]

Okay! Indeed is the problem symmetric only with the 'random goat' strategy. But this strategy is part of the formulation of the problem. The symmetry doesn't stem from the equality of the probs, but from the fact that the problem is then invariant under permutations of the doors. And hence all the conditonal probs must be the same. And this implies they are equal to the unconditional 1/3. But they are of course conditional. So I don't say the 'simple reasoning' is correct, but a kind of 'simple reasoning' may be used. And yes, I know, without the 'random goat' there is no symmetry. And also then it are the conditional probs that had to be considered. Nijdam (talk) 12:26, 12 February 2009 (UTC)[reply]

How does the random goat strategy make the problem invariant under permutations of doors? Note that the host is in some cases constrained by the players first pick (when the player initially picks a goat) but in other cases not (when the player initially picks the car)? I can see an argument that the problem must be symmetric if the host always randomly picks (that would be the "forgetful host" variant), but that's not the case here. If you can provide a reference for this, or some sound reasoning based on probability theory, that would be great. I.e. treat it like a theorem:

Theorem: In the Monty Hall problem, the random goat constraint makes the problem invariant under permutations of doors.

Proof: ??

From the perspective of including this line of reasoning in the article, if the proof is not simply "see <so and so>, published <here>" you're doing what Wikipedia calls original research (which is disallowed - making Wikipedia in a sense the inverse of academia where original research is one of the primary goals). -- Rick Block (talk) 14:54, 12 February 2009 (UTC)[reply]

What is there to be proven? In the 'random goat' strategy there is no reference to any specific door, hence no door plays a different role than another one. Hence all the situations on stage chosen a and opened b are equivalent. Is this original research? I think, this is what the advocates of the 'non-solution' really have in mind wthout being able to formulate it. Nijdam (talk) 17:07, 12 February 2009 (UTC)[reply]

@Martin Hogbin. I repeat the following points:

The precise stated problem can only be solved by determining the conditional probability given the situation the player is in, i.e. given the number of the chosen door and of the opened door.
The simple solution is therefore not a solution to the stated problem.
The simple solution lookes very attractive and its line of reasoning may be used as follows.
The conditioning will be on one of the 6 possible situations: [12], [13], [21], [23], [31] and [32], (I repeat here what I stated above), where for instance [21] means door 2 is chosen and door 1 opened. They form a partition of the complete sample space. But from the symmetry of the problem it is also clear that conditioning of either of them leads to the same probability of finding the car behind the chosen door. Hence these conditional probabilities are (numerically) equal to the unconditional one of 1/3. In the article this should be phrased in a proper way as to be understandable to the reader.
You keep asking about conditioning, so I give you (again) the sample space and derivations.

chosen  1 1 1 1 2 2 2 2 3 3 3 3

car     1 1 2 3 1 2 2 3 1 2 3 3

opened  2 3 3 2 3 1 3 1 2 1 1 2

prob    1 1 2 2 1 2 2 1 2 2 1 1 /18

The columns form the outcomes with their probability (/18 as indicated)
In which situation is the player? That is which event has occurred? Well in the stated problem [13], which is:

chosen  1 1

car     1 2

opened  3 3

prob    1 2 /18

consisting of 2 outcomes, and really a proper part of the sample space. In the second column the car is behind the not chosen door, and the well known conditional (!) probability of winning the car by switching is: (2/18)/(3/18) = 2/3.
Nijdam (talk) 23:41, 10 February 2009 (UTC)[reply]

End of experts

Morgan's formulation of the problem

A am suggesting that the formulation of the problem by Morgan et al introduced unnecessary uncertainties. Here is the Parade statement for reference:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

What does it say about the host offering the choice, it says: 'He then says to you, "Do you want to pick door No. 2?" '. Note that it does not say 'sometimes says to you' or 'may say to you' but 'says to you'. There is no mention in the statement of the problem that the host may not offer you the choice. That possibility is an invention of Morgan.

It says the host, 'opens another door'. It does not give the way that the host decides which door to open. We must therefore take this as random. This is perfectly normal in probability problems, in cases where the method of selection is not given we take it as random, this is not an assumption it is an obligation.

Does the host always reveal a goat. The statement says, 'which is a goat'. It is therefore pure invention to assume that it might not be.

There are an infinite number of perverse actions that could occur in the problem - what if sometimes there are three goats? We do not address these possibilities because they are not part of the stated problem.

Martin Hogbin (talk) 11:34, 11 February 2009 (UTC)[reply]

I think Morgan accounts that one cannot define chances in a single situation; one should always have in mind a statistically justified experimental situation. Hence, a statement could be made about the consequent behaviour of the host, making it possible to repeat the show. But I agree with you, that if you simply repeat (theoretically) the exact behaviour of the host, you shouldn't need to fantasize about behaviour rules, let alone other options. I guess Morgan is trying to make it better understandable, and actually Morgan's article seems to be written for an average public, which may make it less reliable as an academic source. Heptalogos (talk) 12:08, 11 February 2009 (UTC)[reply]

Martin - I think you may be missing the point Morgan et al. are making. I believe their primary point is that the problem is about the 6 cases of player pick and host door Nijdam has enumerated above, not the aggregate. They then solve the problem as literally stated in Parade (granting the assumption that the host always opens a goat door and always makes the offer to switch - which plenty of others, not them, criticized vos Savant for). The problem as stated in Parade doesn't say how the host decides to pick between two goat doors, so in their solution they assume nothing about this. The "natural" assumption might be the host picks randomly, but there's no mathematical justification for this. Using Nijdam's notation (i.e. [12] means player initially picked door 1 and host opened door 2), their solution could be reformulated as a theorem of the form:

There is every justification for taking it that the the host's choice of door as random and that is that we are given no information about how the host decides. We therefore musttake this as random, this is not an assumption or a supposition it is the only thing that you can do when you have no information, it is the standard thing to do in all probability questions. To invent a number of perverse scenarios and then calculate the probabilities on that basis is not maths, it is speculation. I have added a section below to demonstrate this. Martin Hogbin (talk) 21:43, 11 February 2009 (UTC)[reply]

Given <Parade description plus host must open a goat door and must offer the chance to switch>, the probability p of winning by switching for [PH] is .5 <= p <= 1 for [PH] in ([12], [13], [21], [23], [31], [32]).

In the "given" of this theorem we've said nothing about how the host decides to pick between two goat doors, so we can't assume anything about this. I think they'd be perfectly happy to solve the same problem again with a different "given", but their approach would be the same - i.e. what the question is asking is the probabilities of the individual cases, not their aggregate. I don't really know, but I think they might have been surprised by the answer they got. It's surprising enough that Glkanter apparently flat out refuses to believe it. -- Rick Block (talk) 13:46, 11 February 2009 (UTC)[reply]

Rick, Martin writes that "the host may not offer you the choice. That possibility is an invention of Morgan." I will quote Morgan: "Before proceeding, another point not addressed by the question needs to be considered. If the player's original choice does not contain the auto, does the host has the option of revealing the door that does? That is, in the playing of the game, does the host always give the player another chance, or does he sometimes end the game immediately by opening the door with the car?" Heptalogos (talk) 14:12, 11 February 2009 (UTC)[reply]

Yes, Morgan et al. are acknowledging the controversy about this. My point is this controversy preceded their paper. To be thorough, they show the effect of assuming this or not. -- Rick Block (talk) 15:09, 11 February 2009 (UTC)[reply]

OK, apologies to Morgan, I blame whoever first proposed the idea. Martin Hogbin (talk) 21:43, 11 February 2009 (UTC)[reply]

The Parade statement: "You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3" What does this mean? It seems to me that after you pick a door, it is labeled No. 1. After the host opens another door, it is labeled No. 3. Nothing is said about every door having it's own identity on beforehand. If that is true, I do not agree with these possible outcomes: ([12], [13], [21], [23], [31], [32]). So is there really a conditional restriction, or do we initially present more outcomes than possible anyway? This may seem strange, but I think it's the exact problem: if we cannot distinguish doors, we should see them as one possible outcome all together.Heptalogos (talk) 14:21, 11 February 2009 (UTC)[reply]

I don't think it matters when we label them. We're talking about three physical doors that have separate (constant) physical identities. If we want to prevent any possibility of the host's choice affecting anything, I suggested a version some time ago where after the player picks but before the host opens one of the other doors a curtain is dropped in front of the two unchosen doors. The host then, behind the curtain, opens one of them and leads a goat on to the stage from behind the curtain, and makes the offer to switch to the remaining closed (but hidden) door. There are any number of ways to turn it into an unconditional problem, but I think the essential "oh my god, how can it not be 50-50" aspect of it requires putting the player squarely in front of two doors, the one she initially picked and one more and then asking the "want to switch" question. I think this makes it inherently conditional. -- Rick Block (talk) 15:09, 11 February 2009 (UTC)[reply]

There are indeed three physical doors, but the problem has a natural symmetry. The doors can be described as the initial choice, the one the host opens, and the unopened one, there is no need to number them, in fact to do so hides a natural symmetry of the problem. Martin Hogbin (talk) 21:43, 11 February 2009 (UTC)[reply]

The curtain idea is nice, but not necessary, because there is actually only one property by which the player can distinguish door 2 from door 3, which is the fact that one is open (No. 3) and the other is closed. There is only an open door and a closed door, and the open door is always No. 3. Why do you assume that all three doors have separate physical identities? Or why do you even assume that the player notices the differences (which differences)? Heptalogos (talk) 15:23, 11 February 2009 (UTC)[reply]

I understand your point. On the other hand, I think it's fairly clear we're modeling this on physical reality - in which case the doors aren't simply abstract shields behind which we've hidden two abstract goats and an abstract car but real, physical doors which makes them distinguishable whether we label them or not (for example, if they're physical they have a physical location by which they can be distinguished even if they're not labeled). I'm not sure I've ever seen a published version of the problem that treats it as abstractly as you're suggesting. Have you seen one? -- Rick Block (talk) 16:02, 11 February 2009 (UTC)[reply]

There are indeed three physical doors, but the problem has a natural symmetry. The doors can be described as the initial choice, the one the host opens, and the unopened one, there is no need to number them, in fact to do so hides a natural symmetry of the problem. Martin Hogbin (talk) 21:43, 11 February 2009 (UTC)[reply]

The doors may be on moving platforms, thus having no location. We are inventing reality, even Morgan et al. There may indeed be a curtain in front of the doors, the player may be blind. When we use playing cards we even skip the doors, because we may understand that 'doors' are indeed abstract tools to limit our knowledge. We hide cars and goats by turning them around. The three doors may be lighting up one by one very quickly, which is stopped by the player hitting a button. He may be throwing balls at doors behind him. We only know that he picks a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3. So the first door is surely identified after picking, and the second door is surely identified after opening. Morgan makes the mistake to ask for the host's strategy, creating realities around the facts. This is a mistake, because by experiment in every situation the host will open No. 3 revealing a goat, just that. Heptalogos (talk) 20:03, 11 February 2009 (UTC)[reply]

Very metaphysical. Again, are you aware of references that treat this problem in this manner? -- Rick Block (talk) 20:14, 11 February 2009 (UTC)[reply]

Please quote any metaphysical issue mentioned by me, in your interpretation. My approach may seem unusual to you, and others, but you'd rather react to the content instead of the shape. What is surely not physical is maths. As far as I know, the only references to this are mathematical references. Two things are not (exclusively) supervised by that discipline:

1. The translation of (supposedly) real situations into mathematical problems.

2. Logic itself.

Above I argumented that this problem really has two ways of approaching: logical and mathematical. It might even solve most of the disagreement here. Heptalogos (talk) 20:43, 11 February 2009 (UTC)[reply]

Btw, Devlin, a mathematician, uses the logical approach instead of the mathematical. This is the way I would like the main explanation in the article to be, understandable for everyone. Nijdam already wrote him to say he's wrong. :) We might introduce him here. Heptalogos (talk) 21:28, 11 February 2009 (UTC)[reply]

Devlin is already referenced in the article (in the "Combining Doors" section). The combining doors solution is another unconditional variant. Are you saying the first few sentences in the existing solution section are difficult to understand? Or are you only concerned about the part that starts "Although the reasoning above is correct ..."? -- Rick Block (talk) 03:36, 12 February 2009 (UTC)[reply]

The solution part now is a mixture of easy to understand logic and mathematical objections to that. I would suggest to make two solution sections. The simple one should be just logical. I like the graphics, but I do miss an explanation like "There are two doors you did not choose, and the probability that the prize is behind one of them is 2/3. I'll help you by using my knowledge of where the prize is to open one of those two doors to show you that it does not hide the prize. You can now take advantage of this additional information. Your choice of door A has a chance of 1 in 3 of being the winner. I have not changed that. But by eliminating door C, I have shown you that the probability that door B hides the prize is 2 in 3." Heptalogos (talk) 10:22, 12 February 2009 (UTC)[reply]

A simple problem?

To support a point that I make above I ask people to answer this question:

A ball is chosen from a bag containing only red and white balls. What is the probability that it will be red?

Please answer the question as posed. Martin Hogbin (talk) 22:11, 11 February 2009 (UTC)[reply]

Unknown to me. Heptalogos (talk) 21:58, 11 February 2009 (UTC)[reply]

That is not an answer. Martin Hogbin (talk) 22:11, 11 February 2009 (UTC)[reply]

I suspect you are looking for an answer like: if R are red and W are white, R /(R+W) (assuming they're otherwise identical, and the bag is opaque, yada, yada, yada). -- Rick Block (talk) 01:16, 12 February 2009 (UTC)[reply]

That is unknown, because 'chosen' may mean that a red ball is always the target. Even if you replace 'chosen' by 'randomly chosen', what is a bag? Are all balls equal, apart from the color? One cannot strictly define chances for two things:

1. Single specific events.

2. Physical events.

If the source problem is physical, it should be translated to a similar mathematical problem. From then on, it's math discipline. Heptalogos (talk) 10:29, 12 February 2009 (UTC)[reply]

No not at all. As I understand it, according the usual rules of statistics, there is a numerical answer to the above question. I guess that it is now up to me to demonstrate that. Martin Hogbin (talk) 08:55, 12 February 2009 (UTC)[reply]

You're aiming at the answer 1/2. Probability is here to be understood as relative frequncy in all possible realisations of the experiment. The symmetry in the colours brings the answer. Nijdam (talk) 12:38, 12 February 2009 (UTC)[reply]

and, 1/2 in this case has the same mathematical meaning as Heptalogos's "unknown". -- Rick Block (talk) 14:21, 12 February 2009 (UTC)[reply]

Not really. If there were red, green, and blue balls the probabilty of choosing a red would be 1/3. Martin Hogbin (talk) 20:57, 12 February 2009 (UTC)[reply]

To the question, as posed, I believe that 0.5 is the correct answer. To say that I am aiming at that answer suggests that ther may be more than one answer. I think there is only one correct answer to my question and that it is 0.5. Are we all agreed on that or not? Martin Hogbin (talk) 20:26, 12 February 2009 (UTC)[reply]

No, I said deliberately 'aiming', because it's (again?) a matter of wording. It starts with what is meant by probability. It is good practice to interpret this as the relative frequency. Then the next question is, which experiment is to be repeated. Here it is also good practice to repeat the action of choosing and keeping the rest unchanged. So I could have assumed that the same bag i.e. with the same content, was used over and over after replacing the chosen ball. The next question is what is meant by 'a ball is chosen'. Nothing is said here about whether it would be a random choice. If the strategy would be to pick a red ball, and in the wording of the problem is stated there is at least one, the answer would be 1. In the 'random' case there is no numerical answer, only the formula Rick gave. Agree on this? Nijdam (talk) 21:36, 12 February 2009 (UTC)[reply]

'What is meant by probability'? Surely we are not meant to be discussing that here.

In addition, this problem obviously has the form of an urn problem, which is such a classical form that there is even an article here about it. I assume where you're going with this is that lacking other information, "random" is the best assumption and you'll proceed to argue that this means that in the Parade wording of the Monty Hall problem we should assume that the "random goat" constraint from the explicit wording is in effect. -- Rick Block (talk) 01:35, 13 February 2009 (UTC)[reply]

Yes exactly that. That is precisely what you should do. It is absolutely normal practice to take what we have no information about to be random, it is not something that I have just made up. Martin Hogbin (talk) 09:16, 13 February 2009 (UTC)[reply]

One will make necessary assumptions, and even best assumptions, if there is such a thing. The assumptions should create the missing sample space, but many times create a missing sample space. For every experiment we create like that, we can define the chances mathematically. Judging which physical problem fits into which experiment is not maths discipline. It's a statistician's good practice because he has to be pragmatic in some way. Heptalogos (talk) 14:17, 13 February 2009 (UTC)[reply]

What this talk page is for

This talk page is actually for talking about changes to the article, not for talking about the ins and outs of the Monty Hall problem or probability theory or the meaning of what a "bag" is. There are really only three legitimate topics for discussion here.

1) Things not in the article that should be added.

Suggest away. Please provide references in accordance with reliable sources and no original research. This article is already a Featured article and has had a fairly recent featured article review (May 2008) so the standard for its content is quite high (in particular, any significant point needs to have a high quality reference).

2) Things in the article that should be deleted.

Also, suggest away. The content is referenced, in detail, so if you're suggesting something be deleted please provide a reason.

3) Things in the article that should be changed.

Yet again, suggest away. This could range from minor wording changes (which I'd strongly suggest you just make, see WP:BOLD) to structural changes such as suggested above.

The 0.999... article has a subpage of its talk page called "Arguments", see Talk:0.999.../Arguments, and a FAQ. I've set up such a structure here, in the hopes that it might be helpful. If anyone wants to move a section they created here to that page, please go right ahead. I'd appreciate everyone's help in moving sections from here to there when it seems appropriate. Thanks. -- Rick Block (talk) 01:35, 13 February 2009 (UTC)[reply]

See Talk:Monty Hall problem/Arguments and Talk:Monty Hall problem/FAQ, both linked from the header at the top of this page. -- Rick Block (talk) 02:07, 13 February 2009 (UTC)[reply]

Suggestion to delete portions of the existing Solutions section

Yes, I would like to see the paragraphs in the Solutions section that says the first solution offered does not solve the Monty Hall Problem be removed. I believe is it not consistant with what the majority of the published references says.

I would also like to see the references to the minority point of view as espoused by Morgan et al moved out of the Solutions section.

I believe both of the changes are consistant with Wikipedia policy regarding regarding authoritative sources.

Rick, how will we know when it's proper to edit the Article? Would that be when a consensus is reached? Would any editors reserve a veto power?

Glkanter (talk) 02:19, 13 February 2009 (UTC)[reply]

Just to be clear, you're suggesting this specific content and the references (both Morgan and Gillman) be completely removed (correct?). It would be proper to remove these given a consensus on this page, where what is meant by consensus is described at WP:consensus. In the consensus process, all editors are equal to all other editors. Note it is not a vote, but a discussion where points raised for and against are considered. Given this content and these references are fairly long standing in this article and were added in response to a featured article review it would not be proper to remove them unless there is a strong consensus here.

Google Scholar (perhaps not the most authoritative citation index in the world, but it is at least a free one) says the Morgan paper is cited by 49 other works (see [6]), and the Gillman paper is cited by 30 other works, which says to me these are not simply crackpots. Deleting this content and these references, would leave the Solution section completely unreferenced as well. In my opinion, the article and this section in particular complies with Wikipedia:Make technical articles accessible (which, as it is included in the WP:MOS, is one of the criteria for being a Featured article) by presenting an easily understood solution followed by a more technically correct solution as described in the Morgan and Gillman references. For these reasons, I think this content should not be deleted.

Although not a veto, for consensus purposes unless you can address reasons raised to keep the content it would not be proper to make the deletion you suggest, regardless of how many editors agree with you (please see WP:NOTDEMOCRACY). -- Rick Block (talk) 03:15, 13 February 2009 (UTC)[reply]

I fell for it. I let him use the old 'rope-a-dope' on me. Next thing you know, instead of argueing about my proof, we'll be argueing about 'who is more mainstream', or 'what is prominant'.

Ricks comment about his veto power is not supported by the NOTDEMOCRACY paragraph. Basically, it's all give and take. And nobody 'Owns' the article.

Rick, are all editors equal, but some are more equal than others?

Glkanter (talk) 03:45, 13 February 2009 (UTC)[reply]

I fell for it. I let him use the old 'rope-a-dope' on me. Next thing you know, instead of argueing about my proof, we'll be argueing about 'who is more mainstream', or 'what is prominant'. We are not arguing about your proof. Your proof is WP:OR and of no relevance, no matter how correct it may be. Please realise that. Leaving out the hyperbole would be helpful. To comment on the principles: no, this is not a democracy, yes all have the same potential to contribute, no Rick doesn't have a veto, no Rick did not say that he had a veto, no the argument that the content has been in place for ages is not a compelling reason to keep it William M. Connolley (talk) 08:31, 13 February 2009 (UTC)[reply]

First Step Towards Resolving Alleged Ownership Issue

I guess I'll have to understand and follow Wikipedia protocol if I'm going to be successful in ending Rick's Ownership of this Article. This is from Wikipedia:Ownership of articles:

If the ownership behavior persists after a discussion, dispute resolution may be necessary, but at least one will be on record as having attempted to solve the problem directly with the primary editor.

http://en.wikipedia.org/wiki/Wikipedia:Ownership_of_articles

Rick, as you are aware, I believe you have improperly taken ownership of the Monty Hall Problem article to the detriment of the article. I estimate this has been the case for nearly 4 years. I would like to find a solution to this, in order that I, and others could edit the article, without fear of either your veto, or entering into a reversion war.

I lay out my reasoning beginning with my post of 17:35, 12 February 2009 (UTC), which comes in the section title Conventional Wisdom.

I look forward to your response to this serious issue.

Thank you Glkanter (talk) 07:36, 13 February 2009 (UTC)[reply]

I emphatically deny that there is an ownership issue. I have never claimed I own this article, have never edit warred over content, and, if you check the history you'll see numerous editors have contributed content. I have done my level best over many days now to help you understand why the change you seek won't happen. Your response has been a continuing stream of personal attacks and incivilities, including your own personal rant section above. There is definitely a problem here, but it is not ownership. -- Rick Block (talk) 08:50, 13 February 2009 (UTC)[reply]

"I have done my level best over many days now to help you understand why the change you seek won't happen" (The Dictator aka Rick Block, 2009) '

This to me implies that "Rick block" actually belives that he can decide by himself what changes should be made to the Monty hall article !!!! Rick Block should decide everything and if someone does anything that "Rick Block" does not like then "Rick Block" is going to mobilize enough support to force such a person out of the game by telling people to revert any changes to the article. Abuse and Bullying on a higher level !! Definitely ownership with capitalized 0 --92.41.74.138 (talk) 21:36, 13 February 2009 (UTC)[reply]

William M. Connolley if you remove my comment another time (just cause you cant stand the truth) I will remove ALL your comments :-) --92.41.74.138 (talk) 21:36, 13 February 2009 (UTC)[reply]

It is worse than I first thought.

There is another source of unnecessary complication introduced by the academic papers on this subject and it is this:

The Parade statement is clearly written from the point of view of the player, 'Suppose you're on a game show...What is your advice?'. The questioner is obviously asking for advice as to what they should do if they were a contestant on the show.

Some academics, however, have reformulated the question from the perspective of some imaginary third party. Someone who may know, for example, that the host has a preference for a particular door. This is something that the player does not know.

As a result of this reformulation, there have been countless arguments on this page over whether it matters if the player knows about the host's preferences. This reformulation is a major obstacle to understanding the solution. Martin Hogbin (talk) 09:30, 13 February 2009 (UTC)[reply]

Generally speaking you're right. There are many variants, involving those where the player doesn't know the rules etc. But let us stick to the problem stated in the article. There it says: these are the rules, and I suppose it means the player knows them. Even this quite restricted version brings so much confusion. Actually also Marilyn, yes she, gave the "simple explanation", and being attacked for it, and of course realizing she was wrong, tried in all possible ways to interpret the problem in such a way that she wouldn't have to admit it. Nijdam (talk) 09:46, 13 February 2009 (UTC)[reply]

I mentioned this several times. If 'we' define the chances, no player and no host have to know anything; it's all in their behaviour which is simply described. The same mistake is made describing a host who knows where the car is. It doesn't matter what he knows, as long as he does this or that. We are all inventing and assuming realities, but it all comes down to the same issue again: the specific physical problem should firstly be described as a statistical experiment or reality, which can then be translated to maths. Heptalogos (talk) 13:35, 13 February 2009 (UTC)[reply]

Martin - is this thread about the article, or about underlying mathematical issues relating to the problem? If the latter, please relocate this thread to the /Arguments subpage. Thank you. -- Rick Block (talk) 15:01, 13 February 2009 (UTC)[reply]

It is about how to improve the article, which is the purpose of this page. You know my position, which is that the concentration on certain academic papers serves only to introduce unnecessary complications, obfuscate the central and notable problem, and confuse the reader. One way in which those papers confuse the issue is by changing the viewpoint from that of the contestant (as it is in the well known Parade statement of the problem) to that of a third party observer. Martin Hogbin (talk) 20:14, 13 February 2009 (UTC)[reply]

Conditional or unconditional, once again

I have an alternate suggestion regarding the conditional vs. unconditional question that keeps coming up. How about if we add a more comprehensive discussion about this, perhaps like the following (a slightly edited version of an explanation I originally posted above that might have been buried before many people noticed it). If we don't add it to the article, I suggest that it be included in the FAQ linked at the top of this page. The wording may not be quite Featured Article quality, but I think it is close. -- Rick Block (talk) 05:28, 14 February 2009 (UTC)[reply]

Please comment in the appropriate subsection below, #Why we should or should not add this content, #Does the suggested wording fairly summarize what these articles say, or #I don't think they're correct. -- Rick Block (talk) 19:24, 14 February 2009 (UTC)[reply]

Feel free to try to improve the wording as well. -- Rick Block (talk) 19:32, 14 February 2009 (UTC)[reply]

Suggested wording

Morgan et al. and Gillman claim the Monty Hall problem must be solved as a conditional probability problem. As usually stated, the problem asks about a player who has picked a door and can see which of the remaining two doors the host opens in response, and is then given the opportunity to switch doors. The question pertains not to all players in general, but only to the subset of players who have picked a door (for example door 1) and then have seen the host open a different door (for example door 3). The point concerns the difference between

P({\text{win by switching}})\,

and

P({\text{win by switching}}|{\text{player picks door 1 and host opens door 3}})\,

$P({\text{win by switching}})\,$ is the unconditional probability of winning by switching. It is roughly what fraction of all players who play the game will win by switching. Gillman suggests this would be the probability of interest in a revised version of the problem where "you need to announce before a door has been opened whether you plan to switch." [italics in the original!]

$P({\text{win by switching}}|{\text{player picks door 1 and host opens door 3}})\,$ is the conditional probability of winning by switching given that the player initially picked door 1 and the host has opened door 3. It is roughly what fraction of all players who pick door 1 and see the host open door 3 will win by switching.

These are different questions which may or may not have the same numeric answer. As usually stated, Morgan et al. and Gillman both say the Monty Hall problem asks the conditional question, not the unconditional one.

An unconditional solution such as "a player has a 2/3 chance of initially picking a goat, all of these players who switch win the car, so 2/3 of all players who switch will win the car" correctly answers the unconditional question, but doesn't necessarily say anything about the conditional problem. For example, this solution says the answer for both the Parade version and the Krauss and Wang version of the Monty Hall problem is 2/3. The Morgan and Gillman papers both introduce a different variant, identical to the Krauss and Wang version except instead of

If both remaining doors have goats behind them, he chooses one randomly

they specify

If both remaining doors have goats behind them, he chooses the leftmost door

In this variant, an unconditional solution also correctly answers the unconditional question. The player still has a 2/3 chance of initially picking a goat, and in this case the host must reveal the other goat, so the solution above still applies and 2/3 of all players who switch will win the car. However, in this variant, if the player initially picks door 1 and the host opens door 3 the player knows the car is behind door 2 and has a 100% probability of winning by switching. If the player initially picks door 1 and the host opens door 2, either the player's initial pick was the car (a 1/3 chance) or the car is behind door 3 (also a 1/3 chance). This means, given that the host has opened door 2, the player's chance of winning by switching is 1/2. The point of introducing this variant is to show the difference between the unconditional and conditional questions. In this variant, these questions have different answers exposing the difference between unconditional and conditional solutions.

Using an unconditional solution produces the correct answer for the more exact Monty Hall problem as stated by Krauss and Wang, but this is in effect a coincidence. The problem with the approach is hidden because in this version the conditional and unconditional answers are the same. Applying the same solution to a variant where the unconditional and conditional solutions are different, like the "leftmost door variant", reveals the problem.

This is exactly the same as responding to the question "what is 2²?" by saying "2 plus 2 is 4, the answer is 4". This is the right answer, and a true statement, and related to the right reasoning, but not actually right. This wrong reasoning can be exposed by asking, "what is 3³?" Similarly, the problem with using an unconditional solution to answer the Monty Hall problem can be exposed by asking about a slightly different variant, where the wrong reasoning produces the wrong answer.

Initial reactions from Glkanter, Martin Hogbin, Heptalogos, and 92.41.236.242

I think it's perfect. Long overdue. I've always said, "What the Monty Hall Problem article needs is another 5,042 bytes dedicated to Morgan, including the final three paragraphs which refer to no references or footnotes." What's the word for that? Come on, someone help me. Oh, right! It's called Original Research! Us Wikipedians are Big Fans of OR.

Oh, wait. No mention of the 'random goat door constraint'. The Rosetta Stone of the MHP. Sorry, back to the drawingboard.

Glkanter (talk)

Glkanter (talk) 06:13, 14 February 2009 (UTC)[reply]

I do not accept that the problem is necessarily conditional, and I have given reasons why. Boris Tsirelson put it like this, ' who said I'd say that in this case they are equal not just by a numeric coincidence. Rather, the conditional probability (treated as another random variable) is constant (a degenerate random variable) in this case, due to an obvious symmetry. Taking into account the total probability formula we conclude that the conditional probability must be equal to the unconditional probability in this case'. Martin Hogbin (talk) 09:23, 14 February 2009 (UTC)[reply]

Exactly! Thank you! There is no reason why the Monty Hall problem should be conditional! A simple unconditional solution is the best solution! This will also limit "Rick Block" abilty to introduce NOISE inorder for him to increases his leverage to make sure that no one understand anything (included himself) by OVER COMPLICATING THINGS)! --92.41.236.242 (talk) 12:37, 14 February 2009 (UTC)[reply]

Would it be fair to infer that the the next sentence from Boris Tsirelson would be 'In this case then, one may, by proving the (easier) unconditional probability, also prove the conditional probability?' Glkanter (talk) 11:27, 14 February 2009 (UTC)[reply]

What is the mathematical rule for a problem 'being' conditional? Heptalogos (talk) 12:41, 14 February 2009 (UTC)[reply]

Interesting position, Martin. How do you reconcile it with WP:BECAUSE-MORGAN-SAID-SO?

Glkanter (talk) 10:10, 14 February 2009 (UTC)[reply]

So, how does this work in practice? There's no vote taking, and consensus doesn't necessarily rule the day. But Rick can still change his article, right? Glkanter (talk) 10:14, 14 February 2009 (UTC)[reply]

Yes he can! and from "Rick Blocks" perspective as long as he agree with the change then consensus has been reached, ha ha !! It is like the fucking wild west !! The thing is even though we would agree that an unconditional solution (consensus, as of now 4 against 1 in favor for an unconditional solution) is the best way forward then I promise you that anal "Rick Block" is going to enter into an edit war by mobilizing his "I want to become an administrator" grupies.........--92.41.236.242 (talk) 12:52, 14 February 2009 (UTC)[reply]

Actually I think we now should change the conditional solution to an unconditional one. Concensus has been reached (4 against 1)

Glkanter, Heptalogos, Martin Hogbin and 92.41.236.242 against "Rick Block" which means that the unconditional solution is the winner

What do you want to include in the unconditional solution guys? --92.41.236.242 (talk) 12:52, 14 February 2009 (UTC)[reply]

To help structure this discussion, I'm adding subsections. Please stay on topic within the subsections. -- Rick Block (talk) 18:18, 14 February 2009 (UTC)[reply]

Why we should or should not add this content

The reason I am suggesting adding this content is because the basic point made in these two papers is clearly not widely understood. If the article includes a comprehensive discussion of what they say, the wording can be improved by anyone who may be able to express the point better. We should not add the content if the point is insignificant, or mathematically discredited. It is my belief that these two papers, nearly simultaneously published (Morgan et al. was first by 2 months), are the first mathematically rigorous treatments of the Monty Hall problem which makes them crucial to include in this article. I also believe these papers were published specifically in response to the nationwide furor over this problem that followed the Parade articles published in 1990-1991. The Morgan et al. paper is cited 49 times according to Google scholar. The Gillman paper is cited 30 times. As far as I know, the central point they make is not reflected in non-academic literature, but this doesn't mean we shouldn't include what they say. -- Rick Block (talk) 18:18, 14 February 2009 (UTC)[reply]

I am not thoroughly against the conditional solution. I think that in fact there are two problems:

We have the simple, fully defined, problem that is a direct equivalent to the three prisoners problem in which the host acts simply as the agent of chance. For the reasons given by myself and Boris Tsirelson this is not a conditional problem. From what I can see from the free bit, even Morgan et al do not claim that it is. This problem, in my opinion is the most interesting and notable one in that it is a simple problem that most people get wrong. This problem is obviously not very well covered in peer reviewed journals because it is a simple probability problem whose solution has been well understood for years.

The second problem is the real world one concerning a contestant on a TV game show. In this the problem is very poorly defined. What does the contestant want? What outcome do the TV company want? How will the host act? What things could possible happen? and many more. This is the problem that has been addressed by many published papers. It is a conditional problem and it is very complicated but in my opinion it is not very interesting or notable. It is a complicated probability and psychology problem that has no clear answer and which experts have argued about for years.

So what I would like is to have more on the 'basic', paradoxical problem. In particular a solution that will convince readers of its correctness. I have no desire to get rid of the, academic conditional problem, leave it there for those who are interested but I am convinced that this article fails its readers in that it does not give a clear enough answer to the 'basic' problem. It is not that the published papers are wrong, just that they, obviously, do not cover the basic , interesting and notable unconditional problem

I have made some changes, see above, and let me know what you think. I appreciate that this article is a FA and accept the we should not go bashing it about. Let us all try to work together to improve the article. Martin Hogbin (talk) 18:34, 14 February 2009 (UTC)[reply]

You started by saying the fully defined problem is directly equivalent to the 3 prisoners problem. This is somehow contradictory to your opposition to accept that it is about conditional probability. The 3 prisoners problem is! Nijdam (talk) 21:10, 14 February 2009 (UTC)[reply]

Does the suggested wording fairly summarize what these articles say

If you think the suggested wording doesn't fairly summarize what the papers are saying, please say how or where (the analogy at the very end is original, but everything else is directly from the papers). The papers are short, but I could add references to specific paragraphs if that would help. A point they both make, not in the proposed wording above, is that viewing the problem as a conditional probability problem means the Parade version of the problem lacks a condition necessary to justify the 2/3 answer. Specifically, they BOTH mention this version does not say the host picks between two goat doors randomly. BOTH of these papers make this point, and provide a solution where the host's preference for one door or another is left as a variable q. They BOTH show the probability of winning by switching is then in the range of .5 to 1, exactly equal to 1/(1+q). I think this additional point should likely be added as well, but first things first. -- Rick Block (talk) 18:18, 14 February 2009 (UTC)[reply]

I don't think they're correct

Arguing that what these papers say is wrong in a mathematical sense has no bearing on whether the content should be included or not. If you would like to discuss the points they make, please post at /Arguments or Wikipedia:Reference desk/Mathematics. -- Rick Block (talk) 18:18, 14 February 2009 (UTC)[reply]

I, for one, am not arguing this. I am now becoming more and more convinced that there are two Monty Hall problems. The simple one, and the complicated one (which could be treated as several). My point would now be that the published statistical papers are not about the simple problem, although they do mention it in passing, but they cannot comment about it in detail because there is nothing new to say. There clearly is a simple version of the problem, as insisted on by vos Savant. So the statistical papers on the problem are not wrong but irrelevant to the simple problem. There is one other issue that academics do cover, which is why people get the simple problem wrong. Perhaps more on that would be interesting.

I think making the two versions of the problem clear in the article might prevent a lot of argument. Martin Hogbin (talk) 20:34, 14 February 2009 (UTC)[reply]

Recent changes to lead.

Rick thanks for making the players plural in my recent addition. I agree that is is a better mathematical representation and it is more PC.

I have restored the emphasis on 'always' to indicate that there is no doubt, uncertainty, or probability other than 1 for a swap to change a car to a goat and a goat to a car. I think this needs to be spelled out to the reader so that there are no lingering doubts. I was going to add either a footnote or a picture to make this absolutely clear. What do you think.

I have also put back the probabilities of swappers and non-swappers getting a goat. You may consider this obvious and superfluous but to a newcomer to the problem this may not be so. Martin Hogbin (talk) 09:05, 14 February 2009 (UTC)[reply]

Thinking about this some more, I am not sure that this solution should even be in the lead section, which is meant to be a summary of the article as a whole. Would it be better to make a very brief reference in the lead and move this explanation into the start of the 'solution' section, maybe with pictures? What we have now (and had before) is one simple solution in the lead and a different one in the body. Martin Hogbin (talk) 17:14, 14 February 2009 (UTC)[reply]

I agree with your reasoning here, although I (still) don't understand how what you're suggesting adding to the start of the Solution section is different from what is already there. -- Rick Block (talk) 18:47, 14 February 2009 (UTC)[reply]

I guess that it is just the way it is written and presented. I understand that it is done the way that it is to lead in to to a discussion of the 'academic' problem but I think this just spoils it simplicity of the solution. Can we agree that there are two problems - see above. Martin Hogbin (talk) 19:09, 14 February 2009 (UTC)[reply]

I agree there are two problems, however the conditional/unconditional distinction pertains to both. Lots of people get the "simple" (not the real world) version of the problem wrong (and say switching does not matter). They're told, "Well, you see, it's like this - you have a 2/3 chance of initially picking a goat, if you switch you get the other thing, so if you switch you get the car. Understand?" OK. Got it. Then vos Savant publishes the "host forgets" variant. Now what? You still have a 2/3 initial chance of picking a goat. Switching still inverts your initial choice. But now you have a 1/2 chance of winning by switching? WTF? The issue is these are BOTH conditional probability problems, and if you approach them both this way you get the right answer both times. IMO (and, to be clear, this is only my opinion) encouraging an unconditional approach for the "standard" variant is doing the public a disservice. -- Rick Block (talk) 20:37, 14 February 2009 (UTC)[reply]

I think you are wrong. There are many variants which must be conditional but the simple problem need not be. Any problem can be made superficially conditional by picking an event that occurred during it and making it a condition. But if that event cannot in any way possibly affect the outcome then it is a null condition, a irrelevant condition, basically not a condition at all. The fact that the conditional and unconditional solutions to the MH problem have the same value is not a coincidence 'the conditional probability (treated as another random variable) is constant (a degenerate random variable) in this case, due to an obvious symmetry', as Boris, who seems to know about these things, puts it. I am not sure that even Morgan et al claim that the basic problem is necessarily conditional. I do agree, however, that all sorts of, apparently minor, changes to the problem can make it necessarily conditional but the 'basic' problem does not need to be treated that way any more than the 'three prisoners problem' does. Martin Hogbin (talk) 21:09, 14 February 2009 (UTC)[reply]

Correction - I have just checked Morgan and they do claim the basic problem should be treated as conditional. Martin Hogbin (talk) 23:32, 14 February 2009 (UTC)[reply]

Yes, but they seem to take it for granted. There's no explanation. It's one of the disadvantages discussing within a small group of people who agree in the first place; there is no stimulus to think in any other way. And let us not forget that they spent something like "a whole afternoon" at this problem. We know how much that is. Heptalogos (talk) 10:13, 15 February 2009 (UTC)[reply]

I was not supporting Morgan in any way, just admitting a factual error in what I had said above. I agree with you, the Morgan paper is just an exercise in arrogance. Martin Hogbin (talk) 23:47, 15 February 2009 (UTC)[reply]

Quote: "But if that event cannot in any way possibly affect the outcome then it is a null condition". Martin, which rule says so? Heptalogos (talk) 22:02, 14 February 2009 (UTC)[reply]

Sorry Heptalogos, you are right. I meant, but did not say, outcome of the probability calculation, not physical outcome of the show.

If not, the why should not everything that happened between the players initial choice and his decision whether to swap or be treated as a condition. What is it that makes the opening of a door to reveal a goat a condition? Martin Hogbin (talk) 23:32, 14 February 2009 (UTC)[reply]

To explain further. Why do we ask, 'What is the probability of getting a car, given that the host has opened a door to reveal a goat?', but we do no ask, 'What is the probability of getting a car, given that the host has said the word "pick" to the player?'? In formulating any problem we have to decide what the significant conditions are, in other words define our sample space. This is not automatically done for us, it is our choice. Martin Hogbin (talk) 10:02, 15 February 2009 (UTC)[reply]

Not everything that happens reduces the sample space, like the word "pick". Opening a door with a goat does. So this could be a significant condition? Heptalogos (talk) 10:05, 15 February 2009 (UTC)[reply]

Why? How does one decide what to take into account? Martin Hogbin (talk) 10:36, 15 February 2009 (UTC)[reply]

What to take in account: every given event that reduces the sample space? Why: because it's a rule of conditional probability cq statistical dependency? Heptalogos (talk) 11:01, 15 February 2009 (UTC)[reply]

But, to be perverse, use of the word 'pick' might do just that. The host might, for example use the word 'pick' when he knows that you have chosen a car and the word 'choose' when he knows that you have chosen a goat. My point, which I think that you agree with, is that what might be a condition is more a matter for imagination than mathematics. (Also see my added comment above) Martin Hogbin (talk) 23:36, 15 February 2009 (UTC)[reply]

I do agree with you, and actually I was hoping for you to say that 'reducing sample space' is not a rule for conditional probability, but 'affecting probabilities' is. Somehow nobody reacts to that statement. Can you please, on the Arguments page? Heptalogos (talk) 10:50, 16 February 2009 (UTC)[reply]

Have you guys had a chance to see my newest solution (it's not a 'proof', after all)? If I'm right, it shows the unconditional solution solves the conditional problem. And, I won't quote anybody here, but if you follow my 'contribs', you will find an editor-of-apparent-authority saying that the solution is a valid unconditional proof of the MHP. All that would remain is to find a published source, which, since it's right, must be out there. But, see for yourself. Make your own judgements... Glkanter (talk) 12:37, 15 February 2009 (UTC)[reply]

Where is your newest solution? You'd better ask your question there. Heptalogos (talk) 13:11, 15 February 2009 (UTC)[reply]

I've copied it from near the end of the Conventional Wisdom section.

1/3 of the time I will pick a car.

Since 1 - 1/3 = 2/3, 2/3 of the time I will not pick a car.

I increase my likelihood of winning by switching.

Note that it has no Monty, and no goats. Un-conditional! And please, since I'm currently being accused of mis-representing other editors' statements, please follow my 'contribs' from roughly 16:53, 14 February 2009 for what looks to me like unqualified (that is, without reservation), corroboration. Glkanter (talk) 13:26, 15 February 2009 (UTC)[reply]

WP:Ownership Allegation Update

Rick went to http://en.wikipedia.org/wiki/Wikipedia_talk:WikiProject_Mathematics today to drum up support for his proposed changes.

I followed him by informing them of my allegations of Ownership by Rick. I suggested that today's behaviour alone shows this problem. Glkanter (talk) 22:21, 14 February 2009 (UTC)[reply]

Above you claimed that the project members have verified Rick's supposed ownership issue ("All these other Wikipedia Math gurus already knew about Rick's MHP article Ownership issues!"). That is false. We are all familiar with Rick and this article. What we are familiar with is that occasionally people unable to understand simple probability come to this article, argue for ages on the talk page, and accuse Rick of a bunch of stuff. Then he argues and discusses in good faith with the person, very patiently and correctly, while enduring a verbal assault. --C S (talk) 03:23, 15 February 2009 (UTC)[reply]

I just posted this exact post at the Math page:

This is the original post from the Monty Hall Problem talk page, verbatim:

Here's where Rick first asked for assistance to aid in Resolving our Conflict.

http://en.wikipedia.org/wiki/Wikipedia_talk:WikiProject_Mathematics

All these other Wikipedia Math gurus already knew about Rick's MHP article Ownership issues!

I'm a first-timer here. It's been way too long, but is has been instructive as to how horribly mishapen things get when an editor claims ownership of an article.

Glkanter (talk) 19:25, 12 February 2009 (UTC)

You are 100% correct. Two days ago, before I ever brought the topic of Ownership to your attention, I was guilty of pre-judging you all. I apologize for that. When I took a leap from your being aware of Rick's fondness for the Article, to the conclusion that you would therefore have already identified a WP:Ownership situation, that was wrong on my part. I'll post my apology on the MHP talk page immediately. Glkanter (talk) 03:41, 15 February 2009 (UTC)

-end of exact post

Does your past experience excuse you from researching this claim, but still commenting on it? Have you asked among the other respected editors who regularly post here for their opinions? Or done your own readings recently? Glkanter (talk) 03:53, 15 February 2009 (UTC)[reply]

Have you stopped beating your wife? Your presumptions are arrogant. I am familiar with all past discussion on this page and related discussions on the WikiProject Mathematics talk page. I saw nothing new in what is going on in this current discussion, and yes, I did read through this whole page before making my comment. It's the same thing again. I have noticed no disagreement of my assessment versus others in the WikiProject. --C S (talk) 04:34, 15 February 2009 (UTC)[reply]

Rick proposed a major addition to the article today. Just hours after about 4 of us were discussing the merits of a 'formal' proposal for deleting stuff. Then he goes to your page to drum up support. What's up with that? Who's stirring the pot? Who's not showing good sense? I know of one highly respected math editor who today described 'the discussion is so turbulent'. And Rick proposes a major addition that he flat-out knows his primary critic is dead-set against? No. That's not normal.

Here's a probability: if you don't look into it, you will not find what I'm describing. However, If you look, you might find I'm right. Glkanter (talk) 04:54, 15 February 2009 (UTC)[reply]

Since you like to misrepresent Boris' quotes, perhaps we could press him on whether the turbulence is caused by your refusal/inability to understand basic probability. --C S (talk) 05:12, 15 February 2009 (UTC)[reply]

Nice. Blame the innocent editor who's trying to play by the rules. And such a professorial tone. Rick complained about my civility. You are way beyond the customs of the past 4 months on this talk page. Don't the rules apply to you? Glkanter (talk) 13:07, 15 February 2009 (UTC)[reply]

And you knew that your smear was false. Because I didn't say it. Why not include a link? So you either intentionally smeared my good name, or you were careless with your research and accusation. Look, if you're not going to be part of the solution, you are part of the problem. Glkanter (talk) 19:11, 15 February 2009 (UTC)[reply]

You might even ask me if I have any documentation to support my claim. Glkanter (talk) 05:04, 15 February 2009 (UTC)[reply]

What documentation is relevant to your lack of mathematical understanding? --C S (talk) 05:12, 15 February 2009 (UTC)[reply]

With your question, are you arguing that being 'Right' is germain to the discussion of whether to edit the article? Have you seen my newest proof? Seen the corroborating unambiguous opinion? Nice civility, again. Glkanter (talk) 13:07, 15 February 2009 (UTC)[reply]

Here's a pretty interesting exchange relating to my 'being reported' by Rick at the bottom of Rick's talk page. He writes this:

There's a current crowd (Glkanter is one of them) unhappy with the Morgan et al. approach who want it removed, commenting on the talk page to gain consensus for their desired change. I guess rather than convince them I could just let them know they will never gain consensus for this change. -- Rick Block (talk) 12:23, 11 February 2009 (UTC)

The only 'evidence' of my WP:Ownership claim I will find will be Rick's own words and actions. Re-read the above paragraph. Is there a problem with concensus building among editors? He says, "I could just let them know they will never gain consensus for this change." I'm telling you, this isn't right. Glkanter (talk) 05:24, 15 February 2009 (UTC)[reply]

I don't think your understanding of the statement you quote assumes good faith. By its plain language, the statement you object to is a prediction that consensus for your change is not going to arise. Right or wrong as it may be, Rick is offering this prediction as a reasoned guess based on his experience with the editors of this article, and which kinds of edit ordinarily seem to achieve consensus. Do you have any particular reason to believe that this prediction actually masks an intent to forbid such a consensus from forming (assuming, arguendo, that this was within Rick's powers)? –Henning Makholm 05:41, 15 February 2009 (UTC)[reply]

Without question, I am reading into it that Rick is stating that he will not allow it. I refer to it as his 'veto' power. And, it could be interpreted that there exists a consensus for some changes contrary to Rick's wishes. But I'll let the other editor's recent and future edits speak for themselves. Glkanter (talk) 05:49, 15 February 2009 (UTC)[reply]

On which evidence do you base this assumption of bad faith, particularly given that a good-faith interpretation of the statement is readily available (that being the straightforward English meaning of what he said)? –Henning Makholm 05:53, 15 February 2009 (UTC)[reply]

Get your facts straight. I was not responsible for the previously discussed quote attributed to Boris.

And now you've accused me of being a wife-beater and mis-representing Boris' quotes. Both lies. Forget it. I do not need help like yours. Glkanter (talk) 05:30, 15 February 2009 (UTC)[reply]

LOL. Wife-beater comment was in response to your own wife-beater style question. I'm not here to help you push your ignorant points or false accusations either. Misrepresentation of quotes? Maybe I see it that way because I'm more familiar with his style of comment, whereas you want to read any nuanced, non-committal response as being in your favor. Like I said, if you believe your repeated quoting of him is justified, we can always ask him to clarify.

As for Rick's supposedly damning words, he's merely stating the obvious. Ignorant people come by Wikipedia articles all the time, trying to change article to crap. Usually they don't try that on a Featured Article. In any event, doing so is just going to fail. Rick is correct. You are so wrong you don't even know why you aren't going to garner any consensus. A less patient editor than Rick wouldn't have even bothered wasting all this time on you. --C S (talk) 05:47, 15 February 2009 (UTC)[reply]

This is 2009. In the United States, anyway, wife-beating is not generally associated with 'LOL'. I would refer you, at last, to WP:Incivility. Glkanter (talk) 13:07, 15 February 2009 (UTC)[reply]

If you're going to be dense about my reference to a standard loaded question, there's nothing I can do to help you. Learn what a loaded question is and not to use them, like you did before. --C S (talk) 15:58, 15 February 2009 (UTC)[reply]

As the former leader of the free world once said, 'Fool me once, <pause>, uh, shame on you. Fool me again...won't get fooled again!'

I fell for the 'rope-a-dope' a second time! Turns out in C S's case, my apology was unnecessary. He, at least, had already made up his mind. Hostile mother.

And don't be coy. Tell me why I am 'so wrong you don't even know why you aren't going to garner any consensus.' I'll change my posting techniques if it's futile. Just tell me the rules. Glkanter (talk) 13:07, 15 February 2009 (UTC)[reply]

As Will Connelly pointed out to Rick Block, further discussion is pointless. You have no idea what you're arguing about, and you can't even understand expert comments you solicit, instead you misinterpret them as justifying your ignorant comments. Like it or not, I am a 3rd party subject matter expert. And I clearly see you have no idea what you're going on about, nor have you even bothered to understand any of Rick Block's points. He has given more than enough explanation. There are no "rules" here that will make you right. --C S (talk) 15:58, 15 February 2009 (UTC)[reply]

Apparently it's valid for you to interpret other editors' postings to fit your POV, but not appropriate for me to do so. Which rule book do you have that I don't have. Because William Connelly said nothing of the sort. How about a link? Where does William Connelly's 'opinion' end in that paragraph directly above, and where does your's begin? Glkanter (talk) 19:11, 15 February 2009 (UTC)[reply]

I am referring to [7] where Connelly does in fact tell Block there is no reason to continue discussion. So how am I interpreting other editors' postings to fit my POV? You've been doing that, so you seem to feel it's a good tactic to accuse me of kind, but I'd like to see a single example. By the way, you like to accuse people of breaking WP:CIV (apparently partly based on your inability to understand English idiom), but you know, stopping a comment short like "Hostile mother" doesn't fool anyone. Consider this an NPA warning and tone down your language in the future.

Host-prefers-known-door behavior

The article says:

The host opens a known door with probability p, unless the car is behind it (Morgan et al. 1991).	If the host opens his "usual" door, switching wins with probability 1/(1+p). If the host opens the other remaining door, switching wins with probability p/(1+p).

The latter formula cannot be correct. In the extreme case, set p=1 and assume that the host does not open his usual door. This can only have been because the car is behind the usual door, so switching in this case wins with certainty. However p/(1+p) is only 50%.

According to my own analysis, the 1/(1+p) is correct, but the other formula should be 1/(2–p). (It is okay for the two probabilities not to add up to 1; conditional probabilities under mutually exclusive conditions have no necessary relation).

Is this an error in the source, in the reporting of it here, or in my understanding of what the behavior is supposed to be? –Henning Makholm –Henning Makholm 04:46, 15 February 2009 (UTC)[reply]

I think this situation needs to be hammered down more precisely. In the Morgan et al article, the situation considered is when door 1 is picked; host will always open a door, and always one without a car behind it; avoiding contradicting previous assumption, he opens door 2 with probability p and door 3 with probability 1-p. Then they arrive at your formulae.--C S (talk) 05:36, 15 February 2009 (UTC)[reply]

OK. I have tries to edit the description accordingly. It is not particularly elegant, though -- feel free to simplify the language if you can. An interesting observation here is that (a) it is never a disadvantage to switch no matter which door and what p; (b) the "always switch" still wins with probability 2/3, independently of p; (c) therefore the assumption that Monty flips a fair coin if he has a choice is not necessary for the standard answer to be correct (though it does simplify the analysis). –Henning Makholm 06:13, 15 February 2009 (UTC)[reply]

The error is not in the source - they say nothing about the probability of winning if the host opens his "non-preferred" door. They do say, as C S mentions, if the host opens one of the doors with probability p (and must open a door and must not reveal the car) he must open the other one with probability 1-p. I haven't chased down where the error came from, but it may be my mistake. -- Rick Block (talk) 02:29, 16 February 2009 (UTC)[reply]

What Do I Need To Demonstrate?

Is there a procedure for investigating WP:Ownership allegations? I'll be happy to share everything I can.

Otherwise, read the Conventional Wisdom section. That's where I lay everything out. —Preceding unsigned comment added by Glkanter (talk • contribs) 06:03, 15 February 2009 (UTC)[reply]

There is no such procedure, because one is not necessary. There is no formal sanction for pretending or believing that one owns an article. The principal sanction is that other editors will consider the would-be owner childish and annoying, and will be less likely to assume that his opinions about the article have a rational basis. We don't need a formal finding of ownership to start doing that; our individual reasoned opinions will suffice.

What is "punishable" is actual disruptive editing of the article, e.g. if the would-be owner attempts to enforce his veto by repeatedly reverting against a consensus of other editors. In that case it is not important whether the disruptive editor's motive for getting into a revert war is a misplaced sense of ownership or something else. –Henning Makholm 06:24, 15 February 2009 (UTC)[reply]

I hope your right. But that's not what was just implied by a recent poster. I got the impression from him that the standards are much higher than that, especially for a Featured Article. Just going by what I just read. Glkanter (talk) 06:34, 15 February 2009 (UTC)[reply]

I'm curious, who usually wins the revert wars? Glkanter (talk) 06:50, 15 February 2009 (UTC)[reply]

They are a strange game. The only way to win is not to play.

Who lose revert wars? The one who tries to force through his minority opinion through against an actual consensus of several other active editors. The dissenter's reverts will usually be re-reverted quickly. In the short run it will be a matter of who hits the three-revert limit first. Here numbers matter. In the longer run the loner tends to tire of the game before all the defenders of the consensus do -- and if he fails to tire quickly enough, eventually someone will summon an admin to apply her discretion (three reverts a day is a limit, not a right) to shut down the troublemaker. –Henning Makholm 07:16, 15 February 2009 (UTC)[reply]

That's very informative. Thank you. I wonder what other rules actually have sanctions attached. I also wonder if there is a 'super set' of editors. You know, the ones who really make the decicisions when the children are done banging their drums and making noise. Glkanter (talk) 07:43, 15 February 2009 (UTC)[reply]

Answered on User talk:Glkanter (welcome template, links to relevant guidelines). –Henning Makholm 08:58, 15 February 2009 (UTC)[reply]

Proposed truce

Rather than argue about unconditional/conditional issues we could agree, for the moment at least, to continue what is already done in the article. This is what I have done in my changes to the lead.

I think it is agreed by all that if we consider the overall probability of getting a car of players who swap compared with players who stick there is no conditional issue. All that is needed is to talk in terms of players who swap and players who stick (in other words we consider players to be either stickers or swappers). I think it is generally accepted that, for any reasonable interpretation of the problem, stickers have a 1/3 chance of getting a car and swappers a 2/3 chance and that no conditional probability is required to show this.

The two problems with this approach that I see are firstly that I do not believe that it is actually necessary, and secondly that it detracts from the true nature of the problem as a game show in which the player is presented with a choice after a particular action of the host. This makes an overall probability solution justifiably less convincing to some, nevertheless it may suit others.

However, by being careful with our wording it might be possible to present a very simple and clear solution that is acceptable to all. As far as I am concerned the overall solution already given is fine in principle, it just needs improving in terms of words and pictures. Martin Hogbin (talk) 10:21, 15 February 2009 (UTC)[reply]

It seems attractive to speak in terms of stickers and swappers. But thus it apply to the situation? Considers for a moment the situation that the host opened both other doors. Would you still speak of stickers and swappers having probability 1/3 resp/ 2/3 of winning the car? Nijdam (talk) 13:35, 15 February 2009 (UTC)[reply]

No, that is why I said, 'for any reasonable interpretation of the problem'. Note also that this approach is already used in the article in the lead section. Martin Hogbin (talk) 14:58, 15 February 2009 (UTC)[reply]

As far as I can tell, where we may be headed here (rare in math articles) is to treat this as a case where different sources (not different editors!) have different opinions and scrupulously follow WP:NPOV. It is not within Wikipedia's purview to take sides. This would suggest a longer solution section, structured like:

Various published sources treat the Monty Hall problem in different ways and present different solutions.

An easily understood simple solution (need a specific high quality reference) is ...

vos Savant's solution (reference Parade) is ...

A rigorous mathematical solution (reference both Morgan et al. and Gillman) is ...

A solution based on Bayes theorem is ... (and we'd move the Bayes solution back to the expanded Solution section).

These should probably flow from easiest to understand to harder to understand, and say what the sources say (without adding our own editorial spin). The question then becomes how many solutions are worth including and what sources would we use. -- Rick Block (talk) 17:26, 15 February 2009 (UTC)[reply]

I think I agree with this general strategy, but we should be careful not to imply a disagreement that isn't there. In particular we should scrupulously stress that these approaches are all valid and all lead to the same result. Otherwise a less mathematically sophisticated reader can easily get the impression that we're reporting on an ongoing academic dispute about who is right and who is wrong. (That is what we would do with similar language in the humanities, for example). I suggest language like

The answer, under the stated assumptions, is that the player doubles his chance of winning the car if he switches doors.

There are various ways to see this. Though they are all mathematically valid (and lead to the same end result), different people may find different explanations more compelling.

vos Savant's solution (reference Parade) is ...

A rigorous mathematical solution (reference both Morgan et al. and Gillman) is ...

A solution based on Bayes theorem is ... (and we'd move the Bayes solution back to the expanded Solution section).

I don't think it is possible to order the solutions by ease of understanding consistently -- every solution that is worth presenting is so because there is somebody for whom this solution is easiest. A better metric might be to start with the solution that we think most readers will find easiest. –Henning Makholm 17:53, 15 February 2009 (UTC)[reply]

(edit conflict) I'm late to this party, and the preceding discussion is very long and sprawling. Let me just dump my opinion, more or less from first principles, on what seems to be the matter:

The classical problem, as stated, is one where the plain and orthodox way to turn it into a mathematical question is to ask about a conditional probability. Calculating this conditional probability in the most by-the-book, pedestrian, not-trying-to-be-smart way will yield the correct answer. On the other hand, a mathematically mature reader will be able to see that because of symmetry, the same correct answer can be arrived at with considerably less calculation by imagining that the player makes his choice to swap or not swap in advance; then all one has to do is calculate an old-fashioned non-conditional probability. All of the foregoing is fact, and hopefully not in dispute; otherwise we're worse off than it seems.

Now, different readers will have different trouble with those facts, and we must try to satisfy all of them, even if that means that some readers need to read explanations that are not directly relevant to the understanding which that particular reader finds easiest. The choice between the two approaches is a trade-off between knowing more theory (conditional probabilities), or using less theory but applying more problem-specific cleverness (invoking symmetry). What one prefers is subjective, and not something that is objectively inherent in the problem.

The sticker-swapper argument is both popular and extremely attractive to readers who are capable of seeing that it is a valid approach, so it would be a bad encyclopedia that did not present it at least somewhere. On the other hand, seeing that it is valid is not entirely trivial, and there will be other readers who are more easily convinced by a straightforward conditional probability. So this must also be presented.

However, finding the right answer is not the end-all. The more curious reader will also want to investigate not only what is right, but also how he managed to be wrong when he initially thought it would be 50/50. And in order to discuss that -- an interesting and now famous perceptional problem that this article must address -- I cannot see how that can be discussed without the machinery of conditional probabilities.

What we then have to decide, from an editorial viewpoint, it not whether to present one or the other raw solution to the original problem, but merely which one to present first. Personally I would gravitate towards starting with the conditional probabilities, the one that requires less ad-hoc cleverness, but reasonable arguments for the opposite order can also be made. In the end I think it is more important the both explanations are good than which order they appear in.

Now have I misunderstood everything? –Henning Makholm 17:29, 15 February 2009 (UTC)[reply]

I couldn't have summarized things better. Nijdam (talk) 17:37, 15 February 2009 (UTC)[reply]

I do not agree with everything said especially that about conditional/unconditional and Morgan, but as I suggested, let us have a truce on that. I do not think the changes required need be that dramatic. The overall probability solution was already there in the lead, I just expanded it a bit. I would like to expand it some more and add pictures, but that would mean moving it to the solutions section. There are no references at present but I am not sure that there is a rule which says that if you add pictures you must add more references. On the other hand, I am sure that we can find some, even Morgan agree about the overall probability.

I agree with the concept of different sources having different opinions, particularly as I do not like the Morgan paper. I think we should not say that it is a rigorous solution or that it is a better solution than others unless that view can be sourced. If not, we should say something strictly factual like 'a solution published in a peer reviewed journal'. I agree with Henning's point that we should not suggest that there is any doubt about the answer, the doubt is really about the question. In a way that was my original point. Martin Hogbin (talk) 18:53, 15 February 2009 (UTC)[reply]

3rd Party Input

The policies say to resolve a dispute, 3rd party input from subject matter experts may help.

So, Rick and I have solicited this input.

Now, I've been criticized for mis-representing that feedback.

So, absent an edit to this talk page from the subject matter expert, how does one incorporate their contributions into the larger discussion?

I think one expert has unambiguously defended my proof. I think another has said he doesn't know what Morgan's problem with the unconditional solution is, so he'll research it further.

But I don't want to put words in people's mouths. I don't need to. It's out there in the edits.

As far as I can tell, the only tool available to me is to cut and paste quotes and links into new edits.

How do I incorporate those views expressed into this discussion, so as to build a consensus? Glkanter (talk) 15:08, 15 February 2009 (UTC)[reply]

None of these people are saying what you think they're saying. And yes, I read their comments. They're certainly not backing your proposed revisions to the article either. If you truly believe they are supporting your changes, ask them to come here and support your proposed removal of the "wrong" explanations. --C S (talk) 16:05, 15 February 2009 (UTC)[reply]

The Monty Hall Talk Page

The feeling I got from discussing this page with subject matter experts is that they see it as a black hole, which they wish to fervently avoid. Please go to this page to determine for yourself the validity of this statement. http://en.wikipedia.org/wiki/Wikipedia_talk:WikiProject_Mathematics

I believe, then, that any efforts to encourage them to post on this page would be counter-productive.

Nobody has answered my question. 3rd party subject matter experts HAVE made comments, suggestions and opinions. How do we incorporate those into this discussion?

We've referred to Boris. I think he has a doctorate in Mathematics. Here's his page. Read the last section. I think it's reasonable to conclude that he's agreeing with me. But you decide! http://en.wikipedia.org/wiki/User_talk:Tsirel Glkanter (talk) 19:24, 15 February 2009 (UTC)[reply]

As I said, I already read that. And nobody disagrees with what you and Boris wrote in that section anyway. Everyone here (including Rick) agrees that is the solution to the "unconditional" MHP. So you see here how you don't even understand what is going on here. --C S (talk) 19:48, 15 February 2009 (UTC)[reply]

If Boris agrees with me, and my point is that the unconditional solution satisfies the Monty Hall Problem, then how is the 'equal goat door constraint' and Morgan still necessary for proper understanding of the problem? You know, like in the Solution section where it says the unconditional solution just provided does not address the actual problem? And then the rest of the article goes off on the the conditional solution? So does any 'two answer article' avoid saying the unconditional solution is inadequate?

You don't know me. This is the only point I have been discussing since October. Glkanter (talk) 20:05, 15 February 2009 (UTC)[reply]

Ok, I didn't want to get into this, but since you keep thinking Boris is somehow in perfect agreement with what you're saying, let me make a comment here. First, you'll note that Boris didn't regard your "proof" as a proof, only as a solution, and indeed he comments it is even "too concise". This is not a pedantic distinction because it actually is relevant to your queries. A solution is just an answer. If you say the probability is such and such, and it's correct, then you gave the correct solution. Your reasoning might be bogus, or only partly bogus, or somewhat confused, or even correct. Correct reasoning, and enough spelled out to be an explanation to someone else, that is what makes a proof.

If you truly understood the problem well, you would have no problem understanding Rick Block's comments above. The crux of the matter is that there are different ways to interpret a problem like Monty Hall; one distinction is what's been called conditional vs unconditional in the above discussions. If you interpret the problem in a conditional manner, i.e. assume particular choices of doors, the host's behavior on how he chooses between two goat doors becomes important (assuming he always opens a door and always shows a goat). This choice is not actually important when you consider what you've been calling the "unconditional" problem. If you don't understand this, then I don't think you can say you really understand the problem.

The point about the unconditional solution not addressing the problem: it's a common mistake to read the problem as conditional, but then give the unconditional solution. That's just wrong. It happens that numerically the answers are the same if you assume the host chooses between two goat doors with equal probability, but that's hardly correct mathematical reasoning. Now, as alluded to before on Boris' talk page, a mathematician worth his salt would be able to use the symmetry of the host's choosing to realize that the conditional probability should work out to be the same as the unconditional probability. But there is something going on here, some reasoning based on the symmetry of the situation. --C S (talk) 20:43, 15 February 2009 (UTC)[reply]

Thank you, Dr., for the thoughtful response. I appreciate your efforts to help me better understand the inadequacies of the unconditional solution. And your right. It's over my head. I do not see where the Monty Hall Problem as told via Marilyn vos Savant has to be a conditional problem. Glkanter (talk) 20:58, 15 February 2009 (UTC)[reply]

CS, thank you for helping us out here. You say, as many others, that the problem should be read as conditional. Can you please simply explain why? Preferably at the arguments page, on which I ask this question. Thanks in advance. Heptalogos (talk) 21:30, 15 February 2009 (UTC)[reply]

Ok, I gotta get going soon (but if you respond, I'll try to respond later too). Looking over some of my responses above, I see I've been short-tempered, for which I apologize. But it was partly due to having read the whole page (I know you keep thinking I haven't read the above, but I did), and being peeved at, I think, rather unfair treatment of Rick Block.

Actually, I don't believe this material is over any reasonably intelligent adult's head. So let's give it a go. Let's address why people might think Marilyn's statement is conditional. By the way, I hope I haven't given the impression that I think the problem must be understood as conditional. As for Marilyn's exact statement, I think it's ambiguous enough on several fronts, and there's no reason to state with absolute confidence it should be conditional.

Marilyn's wording:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Note there is a lot of stuff being left out here. Let's just be naive here, taking nothing for granted. First, we have no idea what the host is up to. Why did he even offer a switch? Is it because you picked the car? Or does he always offer a switch? Now, since we can't expect to really ever understand the mind of Monty, most people assume something like, Monty is fair and always offers the switch. And he always shows a goat, not a car, as that would make the game silly. These might be considered pretty reasonable assumptions, none of which were explained by Marilyn. But how does he decide which door to open if he can choose between two goat doors?

At this point, you might decide, why should it matter how he chooses? And here's where the conditionality assumption starts coming in. The question, as stated, seems to be asking what you would do in this situation that you picked door 1, Monty showed door 3 had a goat, and you are offered to switch to door 2. As stated above, you can assume he always shows a goat and offers the switch, but that in this instance he opened that particular door instead of door 2. So if Marilyn was asking what would you do in this situation, you really need to know how he is choosing doors. Let's say that he always likes to open door 2 when there isn't a car behind it. You would be sure to win if you switch: probability of getting a car is 1.

Now the way Marilyn stated the problem, she said, "...say No. 1..." So a reasonable interpretation would be that these door numbers don't matter, since people often say "...say" to mean "for instance" and not to specify an exact instance. On the other hand, she didn't have to mention the door numbers, so a straightforward reading would be to assume that any information given in the problem is actually necessary. The funny thing is that in most types of problems it wouldn't be necessary to make this distinction. But in probability problems, it's really important. Did Marilyn intend these door numbers to be useful, or were they just extraneous bits she didn't think to edit out? Of course, to further confuse the situation, remember Marilyn never specified the host's choosing between two goat doors. If you assume the unconditional problem, this information is not necessary, since as you pointed out, you only had a 1/3 chance to begin with of getting the car, so if you can switch, you should. On the other hand, a reasonable person might assume that Monty picks either door with equal probability. And with that, you can indeed solve the conditional problem. Amusingly, you end up with the same answer either way. But that only further confuses things for people. --C S (talk) 21:52, 15 February 2009 (UTC)[reply]

Conditional probabilities are for morons! We know from optimal control that the closes way between two points is a straight line. The unconditional solution is located on that line. The conditional solution is not on that line. Occam's razor forever!! !--92.41.77.75 (talk) 22:09, 15 February 2009 (UTC)[reply]

CS, thanks for the explanation. It's all very obvious what you say, and actually that's not what I was looking for. That's why I referred to the Arguments page; here's where I ask the common rule for a problem to be conditional. Most of us are discussing the Krauss and Wang (2003:10) statement of the problem anyway, so there need not to be much discussion about what was meant by Marilyn. Even if so, it's another discussion outside maths. My question is what exactly makes the Krauss and Wang statement a conditional problem, following which rule. Heptalogos (talk) 22:25, 15 February 2009 (UTC)[reply]

I'm trying to eat my humble pie quietly, but I just gotta ask, what do you mean when you say: "By the way, I hope I haven't given the impression that I think the problem must be understood as conditional."? I'm afraid that is what I think you're saying. I've admitted defeat here precisely because I thought you said that it must be. And I'm not arrogant enough to argue math with a math PHD. Glkanter (talk) 00:16, 16 February 2009 (UTC)[reply]

This may help me understand. You wrote: "If you assume the unconditional problem, this information is not necessary, since as you pointed out, you only had a 1/3 chance to begin with of getting the car, so if you can switch, you should." What is it about the conditional problem that makes that statement not applicable to it's solution as well?. Monty's motivation, behaviour and actions don't affect that 1/3 chance, do they? —Preceding unsigned comment added by Glkanter (talk • contribs) 01:26, 16 February 2009 (UTC)[reply]

Glkanter, please try to understand what is meant. CS is not sure about Marilyn's problem to be conditional or not, because information is missing. But he thinks the Krauss and Wang 2003:10 statement (or similar) of the problem is conditional. Heptalogos (talk) 11:03, 16 February 2009 (UTC)[reply]

Thanks for your thoughts. As you can see by my question, I'm trying to demonstrate that, imho, the unconditional solution also solves any conditional statement of the Monty Hall Problem. Glkanter (talk) 14:09, 16 February 2009 (UTC)[reply]

Suggestion For New Emphaisis of the Article

I think an emphais on the unconditional solution for the Monty Hall Problem artcile is appropriate.

This would mean de-emphasising dramatically much of the remainder of the existing article. Glkanter (talk) 17:08, 16 February 2009 (UTC)[reply]

Steady on chaps

I know that I favour a simple solution but we do need to recognize that this is a featured article and we should tread carefully. The last few changes make the page much worse in my opinion and should be reverted. Let is try to reach a consensus on how to proceed. before we make such dramatic changes. Martin Hogbin (talk) 23:13, 16 February 2009 (UTC)[reply]

Door numbers

Leaving out the door numbers in the wording of the problem, essentially changes the problem and takes away not only its charm, but also changes it to a form it never was intended to. I.e. it no longer would be equivalent to the three prisoners problem, from which it seemes to be derived, but especially it no longer reflects what happens in the Monty Hall show. Nijdam (talk) 00:31, 17 February 2009 (UTC)[reply]

In Conclusion

It's been 24 hours since I posted this:

This may help me understand. You wrote: "If you assume the unconditional problem, this information is not necessary, since as you pointed out, you only had a 1/3 chance to begin with of getting the car, so if you can switch, you should." What is it about the conditional problem that makes that statement not applicable to it's solution as well?. Monty's motivation, behaviour and actions don't affect that 1/3 chance, do they?

Haven't heard a peep back yet.

So, it's over. You know it, and I know it. And I was right. From my very first post.

Rick Block, I commend you, sir. You were led, badly, by someone who you trusted, with good reasons. You were always, always considerate and professional. I hope someday you and I are on the same team.

C S, you are garbage.

I posted some fun stuff on my talk page. Check it out! http://en.wikipedia.org/wiki/User_talk:Glkanter

So, now that it has been proven that the alleged conditional statements of the MHP can be solved using the unconditional solution, there really is no 'conditional problem'. Just so much noise. Because Monty can't change the past, so it's still a 1/3 chance I picked the car.

Let's focus on improving the Wikipedia reader's experience with this article.

pick (No. 1)	show (No. 3)	outcome
car (1/3)	goat (1)	lose (1/3)
goat (2/3)	goat (1)	win (2/3)

You choose a door
You choose a door with a goat		You choose a door with a goat		You choose a door with a car
You Stick	You Change	You Stick	You Change	You Stick	You change
You get a goat	You get a car	You get a goat	You get car	You get a car	You get a goat

@@ Line 3,813: / Line 3,813: @@
 ==Door numbers==
 Leaving out the door numbers in the wording of the problem, essentially changes the problem and takes away not only its charm, but also changes it to a form it never was intended to. I.e. it no longer would be equivalent to the three prisoners problem, from which it seemes to be derived, but especially it no longer reflects what happens in the Monty Hall show. [[User:Nijdam|Nijdam]] ([[User talk:Nijdam|talk]]) 00:31, 17 February 2009 (UTC)
+== In Conclusion ==
+It's been 24 hours since I posted this:
+:This may help me understand. You wrote: "If you assume the unconditional problem, this information is not necessary, since as you pointed out, you only had a 1/3 chance to begin with of getting the car, so if you can switch, you should." What is it about the conditional problem that makes that statement not applicable to it's solution as well?. Monty's motivation, behaviour and actions don't affect that 1/3 chance, do they?
+Haven't heard a peep back yet.
+So, it's over. You know it, and I know it. And I was right. From my very first post.
+Rick Block, I commend you, sir. You were led, badly, by someone who you trusted, with good reasons. You were always, always considerate and professional. I hope someday you and I are on the same team.
+C S, you are garbage.
+I posted some fun stuff on my talk page. Check it out! http://en.wikipedia.org/wiki/User_talk:Glkanter
+So, now that it has been proven that the alleged conditional statements of the MHP can be solved using the unconditional solution, there really is no 'conditional problem'. Just so much noise. Because Monty can't change the past, so it's still a 1/3 chance I picked the car.
+Let's focus on improving the Wikipedia reader's experience with this article.

Not Switch			Switch

Pick	Show	Outcome	Pick	Show	Outcome
1	2	lose	1	2	win
2	1	lose	2	1	win
3	2 or 1	win	3	2 or 1	lose

Revision as of 01:39, 17 February 2009

Very important topic

Summary and Solution

Repetitive new solution section

Discussion

Suggestion to replace existing solution section

RfC: New or old solution section?

Bayes was right, but I think the logic of the existing solution is flawed

Player's initial choice is irrelevant

Confusing text

Suggested improvements

Article deletion discussion

User space for development of a new section

Not ambiguous

Lead change

Another objection - moved from main page

Language clean up

Random host variant

Reversion of good faith edits marked as minor

Problems with the current Souces of Confusion section

Proposed substitute for the last two paragraphs of Sources of Confusion section

Another possibility for the Sources of Confusion

Monty's Action Does Not Cause The Original Odds To Change.

A twist - same choice, different odds?

BTW

This comes very early in the 'Solution' section:

Proposed Changes

Still confused ..

Good faith edits

External link to ucs.edu page

The Goat

Addition of simplest explanation

Minesweeper analogy

Wrong explanation

Proposed Solution section

Glkanter's objection

Right explanation

Conditional/unconditional

Conventional Wisdom

Request for brief clarifications

One thing that would make this article better

Apples and oranges

Work around

When is conditional not?

What I am trying to achieve

Proposed simple solution

Experts

End of experts

Morgan's formulation of the problem

A simple problem?

What this talk page is for

Suggestion to delete portions of the existing Solutions section

First Step Towards Resolving Alleged Ownership Issue

It is worse than I first thought.

Conditional or unconditional, once again

Suggested wording

Initial reactions from Glkanter, Martin Hogbin, Heptalogos, and 92.41.236.242

Why we should or should not add this content

Does the suggested wording fairly summarize what these articles say

I don't think they're correct

Recent changes to lead.

Suggested Unconditional Solution

WP:Ownership Allegation Update

Host-prefers-known-door behavior

What Do I Need To Demonstrate?

Proposed truce

3rd Party Input

The Monty Hall Talk Page

Suggestion For New Emphaisis of the Article

Suggested Unconditional Solution Again

Steady on chaps

Door numbers

In Conclusion

Leave a Reply