User_talk:Gill110951/Archive 1

I was sorry to hear that you want to quit the mediation. Sorry, because you have made a real effort to be a conciliatory voice in the mediation. Unfortunately, while there were gains, there are also participants who are, as yet, unready to move from their own position to common interests. So I understand your frustration. You deserve a break. Meanwhile, I am not ready to give up on this group. I have one more approach I will propose. If it works, would you be willing to come back to the mediation? Sunray (talk) 00:52, 2 December 2010 (UTC)[reply]

Sure, I'd be willing. Richard Gill (talk) 20:36, 3 December 2010 (UTC)[reply]

I hope, Richard, you'll find the energy to explain to me what you (and perhaps Martin) mean by saying that the simple explanation, together with a symmetry argument, is a complete solution. If you convince me, I would accept a form of the simple solution, but till now, as I wrote before, I do not belief this is true. The simple explanation just calculates (states) the unconditional probability. What has symmetry to add there? Nijdam (talk) 22:56, 2 December 2010 (UTC)[reply]

Here's your chance to do the right thing, Richard. Is Glkanter's disposition responsible for the ultimatum Nijdam just issued to you? Will you step up and bring an end to Nijdam's endless and disruptive game of cat and mouse? Or will you further enable him in his goal to prevent any honest consensus? Glkanter (talk) 23:29, 2 December 2010 (UTC)[reply]

Let me try then. What is wrong with this? The simple solution calculates the unconditional probability P(C=1), we wish to know the conditional probability P(C=1|X=1,H=3) which, in general, will have a different value from the unconditional probability. The symmetry argument tells us that, in the symmetrical case only, P(C=1) = P(C=1|X=1,H=3). Martin Hogbin (talk) 19:33, 3 December 2010 (UTC)[reply]

I hope Martin you notice the lack of logic. It is just the other way around. You are right when you say; we wish to know the conditional probability P(C=1|X=1,H=3), and that's just the simple solution does not wish. Nijdam (talk) 22:56, 3 December 2010 (UTC)[reply]

I think there are two answers, since there are two common sets of assumptions. Is the player's first choice fixed at door number 1, or is it also uniform random and independent of the (uniform random) location of the car? For both cases, take Y to be the door to which the player may switch.

In case 1, by the simple solution and the law of total probability, 2/3 = Prob(Y=C) = Prob(Y=C|H=2)*Prob(H=2) + Prob(Y=C|H=3)*Prob(H=3 (*). By symmetry, Prob(Y=C|H=2) = Prob(Y=C|H=3). Replace Prob(Y=C|H=2) with Prob(Y=C|H=3) in (*), and use Prob(H=2)+Prob(H=3)=1, to deduce 2/3 = Prob(Y=C|H=3).

In case 2, for the same reasons, 2/3 = Prob(Y=C) = Prob(Y=C|X=1 & H=2)*Prob(X=1 & H=2) +five other similar terms. By symmetry, all six expressons Prob(Y=C|X=x & H=h) are equal while Prob(X=1 & H=2) +five other similar terms = 1. In the same way we get that all six expressions Prob(Y=C|X=x & H=h) are equal to Prob(Y=C) = 2/3.

In words, by symmetry the conditional probabilities are all equal, hence by the law of total probability they must be equal to the unconditional probability. Or in other words: by symmetry the conditional probabilities that switching gives the car are equal, thus independent of X and H. In order to decide whether or not to switch it is a waste of time to take note of the number of the door you chose and the number of the door opened by the host. Richard Gill (talk) 20:51, 3 December 2010 (UTC)[reply]

I know all this, Richard, but I wonder why you present this as a justification of the simple solution. It isn't. The simple solution does not intend to calculate any conditional probability. So, if a solution should be based on the conditional probability, it cannot be the simple one!Nijdam (talk) 22:56, 3 December 2010 (UTC)[reply]

I really cannot understand your logic here Nijdam. The simple solution calculates the unconditional probability. If you can show that this is equal to the desired probability, the conditional one, then surely you can do the simpler calculation in order to find the answer to the answer to the more difficult to answer question.

You have yet to tell us what is wrong with this argument. It is used very frequently in many branches of mathematics. The thing we want to know is difficult (or maybe impossible) to calculate but we can show that it must be equal to some other more easily calculated thing, so we calculate the easy thing to find the answer to the thing that is hard to calculate. Are all the mathematicians that use this technique wrong? Martin Hogbin (talk) 23:35, 3 December 2010 (UTC)[reply]

Really, Martin, I hope you heard the big sigh. I did explain more than once, it is not a matter of calculation. It is a matter of what is calculated. The simple solution do not allow any completion. What they say just is not wrong, but wrong as a solution to the MHP. They just state that the probability for the chosen door to hide the car is 1/3. And that's right, but not appropriate for the problem. The simple solutions do not intend to calculate any conditional probability, although that is the thing to be done. The symmetry argument you are so keen to emphasize, may be used to ease the calculation in the conditional solution, not in the simple solution, as there is nothing to ease. Please, try to understand.Nijdam (talk) 23:55, 3 December 2010 (UTC)[reply]

Nijdam can you rephrase,'The simple solutions do not intend to calculate any conditional probability' please there may be some kind of misunderstanding due to your English. Martin Hogbin (talk) 23:59, 3 December 2010 (UTC)[reply]

Richard, do you understand what Nijdam is getting at here? Martin Hogbin (talk) 23:59, 3 December 2010 (UTC)[reply]

This is a question of semantics. For Nijdam, I think, a "simple solution" is "The decision to switch or stay should depend on Prob(Y=C). We compute it as follows ... ... 2/3, so we should switch". For him, however, a "right solution" is "The decision to switch or stay should depend on Prob(Y=C|X=1,H=3). We compute it as follows ... ... 2/3, so we should switch".

It's interesting, I think, that almost no one gives an explanation for the "should". Richard Gill (talk) 07:06, 4 December 2010 (UTC)[reply]

Are there many reliable sources that advocate the 'should' and the criticisms that Nijdam and Rick so strongly champion for the MHP article? Are they as dogmatic as Nijdam and Rick? Glkanter (talk) 09:43, 4 December 2010 (UTC)[reply]

Richard, thanks for that explanation but it was not quite the point that I misunderstood. Nijdam says, 'The simple solutions do not intend...' (my italics). In normal English inanimate or abstract objects, such as solutions, do not have intentions. Although I can guess what Nijdam may mean I want to be sure exactly what he is trying to say here. Martin Hogbin (talk) 11:24, 4 December 2010 (UTC)[reply]

Nijdam, could you phrase what you are trying to say in a different way please to avoid a misunderstanding. Martin Hogbin (talk) 11:24, 4 December 2010 (UTC)[reply]

I tried to formulate it in many ways. That's why I tried the word "intend" here. But if you wish, the simple solution does not mention any conditional probability, nor leads to the calculation of the desired conditional probability. The simple solution simply states: the probability to find the car behind the chosen door No, 1 is 1/3, hence (?) the probability to find the car behind the remaining closed door No. 2 is 2/3. And that is plain wrong, as the probability to find the car behind door No.2 is 1/3, the same as for the doors 1 and 3, due to the random distribution Nijdam (talk) 19:32, 5 December 2010 (UTC)[reply]

Is that OR: car behind the remaining closed door #2 is 1/3, the same as car behind the open door #3 is 1/3? Or do reliable sources support this assertion? Gerhardvalentin (talk) 20:41, 5 December 2010 (UTC) Gerhardvalentin (talk) 09:10, 7 December 2010 (UTC)[reply]

When Nijdam talks about "the probability" he means the same probability as before. The probability of a given event never changes, in his world. When he talks about "the conditional probability" he means what other people would call "the probability after...". Conditional probabilities depend on what you are conditioning on. Typically you condition on everything you know at a certain point in time, so "the conditional probability given the past" evolves in time as "the present" moves forward. All this is just a confusion between the technical language of probabilists and the ordinary language of ordinary people. Unfortunately both kinds of people use the same words. Unfortunately the technical people are aware of subtle distinctions which they express in their sophisticated technical language in a very precise way. But ordinary people can't see what they are talking about.

To answer Glkanter, the question about the "should" on which Nijdam and Rick so emphatically insist, I know what are the good reasons behind the should, I have written them for you all several times, every professional probably knows them and has internatlized them and forgotten that there has to be a reason when you say "should". (I gave a different reasoning for people who like to use probability in different senses: subjective and objective). I have no idea if any reliable source before me has ever written these things explicitly. It's up to Rick and Nijdam to provide the sources. I am not editing the wikipedia MHP pages. I write reliable sources. Richard Gill (talk) 15:03, 20 January 2011 (UTC)[reply]

I think Nijdam is wrongly representing the simple solution here. I would write it in mathematics, as follows. Car is behind door C, player chooses door X, host opens goat-door H and offers switching to door Y. The triple X,H,Y is a permuation of 1,2,3. The prize C is equal to one of the three, but never to H, so either C=X or C=Y (but not both). If Prob(X=C)=1/3, then Prob(Y=C)=2/3. But Nijdam is right is saying that the simple solution doesn't tell us that Prob(Y=C | X=1,H=3) = 2/3. However you can derive the latter from the former by using symmetry and the law of total probability. See Gill (2011).

Well, Richard, I think you're wrongly representing the MHP. If the player chooses door 1 and sees door 3 opened, then Y=2, and Prob(C=2)=1/3. Nijdam (talk) 16:28, 7 December 2010 (UTC)[reply]

Another question is whether you *must* solve MHP by using conditional probability. From my mathematical point of view, that is the difference between showing that always-switchers win more often than always-stayers, and showing that switchers win more often than anyone else at all. It's the difference between saying that 2/3 of all the times, the switcher wins, and saying that 2/3 of all the times that the player chose door 1 and the host opened door 3, the switcher wins. The job of a mathematician is to show how you can correctly derive these two statements. The great unwashed, the consumers of wikipedia MHP page, decide which solution they like better. Mathematics professors brainwash their students that you *must* do it one way and not another. I have yet to hear any supporter of the conditional solution explaining the use of the word *must*. Is there a law about it? Is it a question of morality or religion? Is it a question of being rational or irrational? Tell me, what is irrational about liking the simple solution as an argument for switching? Richard Gill (talk) 07:43, 7 December 2010 (UTC)[reply]

Richard, you say, 'There is a continuing discussion in the literature about whether simple solutions or conditional solutions are the right way to solve MHP'. Where can I read this? Is it online? Martin Hogbin (talk) 23:44, 3 December 2010 (UTC)[reply]

I just mean the reliable sources. Jason Rosenhouse has a whole chapter. My own papers are part of this discussion. Rosenthal's paper is part of the discussion. Richard Gill (talk) 07:09, 4 December 2010 (UTC)[reply]

Is that the Jeffrey Rosenthal paper referenced in the MHP article where he gives this solution:

"Shaky Solution: When you first selected a door, you had a 1/3 chance of being correct. You knew the host was going to open some other door which did not contain the car, so that doesn't change this probability. Hence, when all is said and done, there is a 1/3 chance that your original selection was correct, and hence a 1/3 chance that you will win by sticking. The remaining probability, 2/3, is the chance you will win by switching."

Then, because he intentionally left out the 50/50 host premise, he continues:

"This solution is actually correct, but I consider it "shaky" because it fails for slight variants of the problem."

And then uses the variants in order to develop his "Proportionality Principle" solution technique? Glkanter (talk) 09:39, 4 December 2010 (UTC)[reply]

I have read Rosenthal, Rosenhouse and your papers on the subject. I think the general opinion is best summed up by Rosenthal's statement that Glkanter quotes above, "This solution is actually correct, but I consider it "shaky" because it fails for slight variants of the problem." Would you agree Richard? Martin Hogbin (talk) 11:30, 4 December 2010 (UTC)[reply]

I disagree. It's pretty clear to me that Jeff Rosnthal would agree with Nijdam that the a solution is only correct if it sets out to determine the conditional probability, and does so correctly as well. So the simple solution "coincidentally" gets the right numerical answer for the conditional probability, since the argument which is presented only delivers the unconditional probability.

""This solution is actually correct..." Just *WHAT* do you disagree with, Richard? The English language? That I correctly copied and pasted his sentences? Its the correct source. What is it, Richard? How can your personal opinion be clearer than Rosenthal's own words I posted, given in their proper context? A High Priest on His High Horse. Glkanter (talk) 14:55, 5 December 2010 (UTC)[reply]

"The conditional probability is 2/3, and you should switch", is what Rosenthal considers to be the result of the correct, conditional, solution.

"The probabability is 2/3, and you should switch", is the result of the simple solution. There is the good number there, and the advice is good too, but the reasoning has not shown that 2/3 can't be improved by possibly not switching for some configurations of door numbers chosen and opened.

Did you ever study elementary probability theory? If you haven't, you could easily misinterpret Rosenthal's words. He's writing for *teachers* of probability, not for the great unwashed. I could copy and paste sentences out of a classified NASA space shuttle engineering manual, but since I don't understand the context, I might easily end up giving bad advice to North Koreans wanting to build their own. I earlier noticed that you weren't familiar with high school algebra. Nothing wrong with that, you managed fine without it, right? You could fill a lot of encyclopaedias with all the stuff that I don't know about. And you are right to distrust the mathematicians. I still haven't heard a pro conditional solution editor explain why you "must" solve the MHP with conditional probability. It seems to be a dogma, something which people were brainwashed to believe and never thought about it since.

I will accept the implied and oft stated (by many editors) Math PhD = Godlike Knowledge, when it is explained how that statement is consistent with:

Morgan's paper having been written by 4 PhDs and peer-reviewed by countless more, yet it:

Grossly mis-quotes vos Savant relative to the language regarding doors 1 and door 3 being opened

It was 20 years before the elementary probability math error was acknowledged

It was *not* a Mth PhD who recognized and advanced the correction neeeded in Morgan's paper

I took enough elementary probability to know that Rosenthal's words:

"but I consider it "shaky" because it fails for slight variants of the problem..."

have no bearing on the validity of simple solution to solve the MHP itself, and that only an ignorant, though educated person would think differently

As English is my first language, I can recognize the lack of ambiguity in Rosenthal's statement:

"This solution is actually correct..."

Appealing to the exclusive ownership of some greater level of knowledge is a disgusting, dishonest argument. Glkanter (talk) 23:18, 5 December 2010 (UTC)[reply]

I agree that appealing to the exclusive ownership of some greater level of knowledge is a disgusting, dishonest argument. You are right to distrust the mathematicians. I still haven't heard a pro conditional solution editor explain why you "must" solve the MHP with conditional probability. On the basis of my European high school mathematics (approx the level of many US PhD's) I am 99% sure that Rosenthal believes that you *must* solve MHP by finding the conditional probability, and that that is why he finds the method of the simple solution wrong, even if the answer is right. You don't have to believe me.

I repeat: you are right to distrust the mathematicians. I still haven't heard a pro conditional solution editor explain why you "must" solve the MHP with conditional probability. Richard Gill (talk) 05:00, 6 December 2010 (UTC)[reply]

I don't distrust Math PhDs. Once again, you put words in my mouth. Amazing how you can do that, just sentences away from my actual statements. Why should I trust your ability to read Jeffrey Rosenthal's mind? You fail repeatedly to even read my typed English words properly. I have no opinion on the reliable sources Math PhDs. The Math PhDs I find objectionable are the ones who are Wikipedia editors and claim they have a superset of skills greater than my own, thus endowing them with greater cognitive skills regarding the MHP. I give these Math PhDs arguments no more value than I attribute to anyone else's regarding the MHP. That seems to put me in the minority amongst MHP editors. Figures. Further, I will trust my ability to read the English language and exercise my critical reasoning skills to understand what Jeffrey Rosenthal has actually published. I find facts more reliable than blathering conjecture when defending a position. I mention my education on my talk page in the 'Convention Wisdom' section. Glkanter (talk) 05:15, 6 December 2010 (UTC)[reply]

I have thought about it carefully, and have several good reasons, depending on which flavour of probability you like. Richard Gill (talk) 19:11, 5 December 2010 (UTC)[reply]

His "proportionality principle" is nothing new, except in name. It's just Bayes rule in the odds form and restricted to the situation that initial odds are even.

It's amusing neither Rosenthal nor Carlton thought of the symmetry argument. As educationalists they missed a splendid opportunity.

It's also strange to me that they make no effort whatsoever to explain *why* it should be thought important to compute the conditional probability given everything observed by the player. There are different reasons, tuned to your interpretation of probability. For the frequentist, whosectheory is not normative, there is a utilitarian reason: this optimizes your overall success rate. For the subjectivist it is axiomatic. The theory is normative. It says you *must* always condition on everything you know, since probability is a measure of *your* certainty. Richard Gill (talk) 20:57, 4 December 2010 (UTC)[reply]

Of course, Rosenthal didn't actually 'write' any of those words, did he, Richard? Does he write 'coincidentally'? And if he doesn't mention something at all, how can it be attributed to him in any fashion? More mind reading bs. Maybe the contestant and the host *do* exchange information via esp, after all, Richard. Glkanter (talk) 21:14, 4 December 2010 (UTC)[reply]

As Glkanter says, I only have the benefit of reading what the authors actually wrote. The concept that you must condition on everything you know is clearly an idealisation as to do so would be vastly impractical. Martin Hogbin (talk) 23:34, 4 December 2010 (UTC)[reply]

I happen to have had a similar professional education to Rosenthal and I happen to work within the same professional community. Indeed, Martin, there is an idealisation here. The idealisation is that "all we know in advance" is encapsulated in our prior uniform distributions of car-location and goat-door-opening, and "all we know during the show" is first we make our own choice (according to our lucky number, 1) and secondly Monty opens a door revealing a goat. Richard Gill (talk) 08:12, 5 December 2010 (UTC)[reply]

There could be more that is revealed during the show; the exact words spoken by the host could be significant. In the real show, Monty liked to tease his contestants a little to add some excitement, in a poker match each player listens to everything said by the others for clues as to what they may hold. As we have discussed and agreed before, people choose to 'pre-process out' events that they consider irrelevant. The MHP show that this can be misleading sometimes but I choose to pre-process out the door number opened by the host and this intuition proves correct. There is, in fact, more justification for considering the host's words to be significant than there is for considering the host's door policy to be important. The former is at least mentioned in the problem statement. Martin Hogbin (talk) 10:40, 5 December 2010 (UTC)[reply]

Sure, you can pre-process the door numbers out of the problem. And afterwards remark that by symmetry this was perfectly justifiable. Richard Gill (talk) 19:16, 5 December 2010 (UTC)[reply]

There becomes a point when the things you pre-process out of the calculation are not worth mentioning, as in the words spoken by the host. Nobody feels the need to say, 'I am taking the probability of interest to be independent of the words spoken by the host' or 'I consider the problem to be symmetrical with respect to the word "pick"'. Martin Hogbin (talk) 22:51, 5 December 2010 (UTC)[reply]

One could derive the conditional solution by tacking on the symmetry argument after the simple solution. Alternatively one could say in advance, by symmety the probability the other door hides the car is independent of the specific door numbers chosen and opened. Therefore it is sufficient to find the unconditional probability that switching will give the car.

The pedantic probability teacher (PPT) wants to convert the MHP into a probability exercise *before* noticing the symmetry. The intutive unwashed quick thinker (IUQT) drops the door numbers first. The PPT has got a good point, since intutive quick thinking causes most people to give 50-50 answer. His approach is guaranteed to get the right answer. The IUQT has got a good point, because his approach can also be mathematically formalized too. So why don't we all have a beer together and talk about something else? Richard Gill (talk) 05:11, 6 December 2010 (UTC)[reply]

What is meant by the simple solution? That is: what do the sources produce as a solution, that is referred to by us as the simple solution. It reads: the car is with probability 1/3 behind the chosen door, hence switching wins the car with prob. 2/3. Thar's all, no more. No mentioning of a probability after a door is chosen and one opened. I've explained this many times. Nijdam (talk) 09:46, 4 December 2010 (UTC)[reply]

No, there are many forms of simple solutions (Selvin's, vos Savant's, Combining Doors) put forth by reliable sources. These all clearly contradict your 'No mentioning...' conclusion above, rather showing indifference to which door is open. I, too, have explained this many times.

I don't suppose you have any reliable sources to support your extreme viewpoint? Glkanter (talk) 10:51, 4 December 2010 (UTC)[reply]

Simple solutions are those which focus on the unconditional probability. Conditional solutions are those which focus on the conditional probability. One of the conditional solutions is obtained by glueing together a simple solution plus a symmetry argument (and use of the law of total probability). Richard Gill (talk) 08:14, 5 December 2010 (UTC)[reply]

Right, but it remains a conditional solution, and not a simple one. Nijdam (talk) 14:09, 5 December 2010 (UTC)[reply]

Yes, that's what I said. Richard Gill (talk) 18:47, 5 December 2010 (UTC)[reply]

Is the indifference to door 2 vs door 3 conditional or unconditional? I think its a 100% condition that a goat will be revealed. Which is unlike Random Monty. But you could use the Carlton simple solution-derived decision tree to solve both. Glkanter (talk) 08:24, 5 December 2010 (UTC)[reply]

I use the word "conditional" in the sense of "conditional probability". Indifference to door 2 or 3 gets expressed within a probability framework as the symmetry of prior probability distributions. Richard Gill (talk) 09:35, 5 December 2010 (UTC)[reply]

Does the Carlton/Monty Hall simple solution-derived decision tree, with the 100% condition that a goat will be revealed, represent an unconditional solution, or a conditional solution? Glkanter (talk) 09:42, 5 December 2010 (UTC)[reply]

As per your suggestion, the door #s have been removed from the column headings. Glkanter (talk) 09:43, 5 December 2010 (UTC)[reply]

Very good! The simple solution corresponds to realising, in advance, that the door numbers are irrelevant. Richard Gill (talk) 04:53, 6 December 2010 (UTC)[reply]

The simple-solution derived decision tree produces a simple solution. There is no probabilistic conditioning on specific door numbers. Richard Gill (talk) 18:46, 5 December 2010 (UTC)[reply]

Why should we base the decision on the conditional probability given the chosen door and the opened door? The simple reason is: because this is given in the situation the player is in. Just like Martin's question about the urn with the ten balls. If we know players one and two did not draw the number 9 ball, the probability for the third player to draw number 9, is almost automatically understood as the conditional probability given the situation. The simple solution in this problem would give the answer 1/10, i.e. the average over all third players. I noticed Martin didn't have any difficulty in understanding this, but strangely enough, did not see the analogy with the MHP. And BTW, Martin, the conditional solution is not the solution that uses Bayes' formula, but just the solution that calculates, inone or another way, the conditional probability given the situation of the player. Nijdam (talk) 10:25, 4 December 2010 (UTC)[reply]

Can you provide reliable sources that support this viewpoint as being the sole 'correct' interpretation of the MHP? Or is this nothing more than unpublished OR? Glkanter (talk) 10:55, 4 December 2010 (UTC)[reply]

Nijdam, we need to start with our interpretation of (vos Savants version of) Whitaker's statement. In this we have the two phrases: 'You pick a door, say No. 1' and ' the host, ... opens another door, say No. 3'.

In normal English these phrases have a meaning somewhere in between:

1 'You pick door No. 1' and 'the host opens door No. 3'.

and

2 'You pick a door' and 'the host opens another door'.

In fact 2 is probably much closer to Whitaker's original question since vos Savant added the 'say door No 3' purely, as she later explained, to try to make clear what 'another door' meant.

Your interpretation of this question as meaning 1 above, is not the only way it can be understood. In fact it is very hard to see why Whitaker would have literally wanted to know the answer to the question only in the specific case that door 1 is originally chosen and door 3 is opened by the host. From all we know, Whitaker was most likely asking a clearly unconditional question. Thus we can say, 'the simple solutions solve the MHP' without further comment. Martin Hogbin (talk) 11:49, 4 December 2010 (UTC)[reply]

Well, all I know, is that the player is given the opportunity to switch, AFTER a door has been chosen and one been opened. It need not be doors 1 and 3, it may be doors 3 and 2, or any other combination, even in general doors x and h. But specifically this last general formulation is more difficult to picture for the average reader. So if you prefer doors 3 and 2, be my guest, but you'll see it does not really make a difference, so let's stick to the given combination 1 and 3. Nijdam (talk) 17:59, 4 December 2010 (UTC)[reply]

It is not clear that the question intends to identify the doors in any way. Would you agree that if the door opened by the host is not identified in any way the question is unconditional even if it is asked after the host has opened the door? Martin Hogbin (talk) 23:26, 4 December 2010 (UTC)[reply]

Firstly, I'm surprised you use the word "intends", as per your own analysis, a question cannot have an intention!? Then, the question does identify doors, as the player knows of course which door she has chosen and sees the door opened by the host. That is the charm of the puzzle, there you find yourself as the player in front of two closed doors, between which you have to decide. What you suggest comes down to the decision asked to an outsider, not knowing which door chosen by the player and not seeing which door opened by the host. Do you really think this is the MHP? Nijdam (talk) 21:16, 5 December 2010 (UTC)[reply]

The questioner can want to know what is best to do, for any of the six situations he might find himself in. It's a mathematical theorem that the best overall success rate is obtained by making the decision which corresponds to the conditional probability given everything known at the last moment. By symmetry, all these conditional probabilities are the same. Hence equal to the unconditional, which by the simple solution is 2/3.

In a nutshell, the simple solutions say that switching, regardless, is better, on average, than staying, regardless. The conditional solution says that switching, regardless, is better than any other conceivable strategy. The easiest conditional solution is obtained from combining the simple solution with an appeal to symmetry. Read Gill (2011).

Nijdam might consider submitting a paper himself, explaining his POV in the peer reviewed literature. Richard Gill (talk) 09:44, 5 December 2010 (UTC)[reply]

Why should I? I agree completely with you, or you agree completely with me. I never said anything different. But it seems you try to widen the meaning of simple solution. And indeed is the easiest CONDITONAL solution obtained by an appeal on symmetry. This has nothing to do with the simple solution being correct. Nijdam (talk) 14:05, 5 December 2010 (UTC)[reply]

Nijdam, if it is just the word 'conditional' that you want that I have no objection at all to using it. The argument seems to about exactly what conditions we should apply. There is always the condition that 'the host opens a door to reveal a goat'. So if all you want to do is call the simple solutions 'conditional' where the exact condition applied is not specified, I have no problem. Martin Hogbin (talk) 14:24, 5 December 2010 (UTC)[reply]

Nijdam, you wrote "Why SHOULD we base the decision on the conditional probability given the chosen door and the opened door? The simple reason is: because this is given in the situation the player is in." (my emphasis). Why do you use the would "should"? Obviously you MAY do that, but you make it an imperative. Is it a moral imperative? A legal imperative?

Clearly, it would be rational to do so, but why would it be irrational not to do so? Please give reliable sources who write in plain English for the great unwashed who have never looked in a probability textbook in their life, as well as a book on probability. I looked in Kolmogorov's treatise but did not find it. Richard Gill (talk) 04:50, 6 December 2010 (UTC)[reply]

I asked this question of a friend. Imagine you are on a game of Deal or no deal (in which there are 22 boxes only one of which contains the, randomly placed, main £250,000 prize and the contestant picks a box at random). Suppose in this (unrealistic) version of the game the host knows where the main prize is and, after you have chosen your box, opens all the other boxes but one, intentionally never opening the box with the main prize. You are then offered the option to swap your originally chosen box for the box left unopened by the host. Would you swap and why. The answer, which completely amazed me, was that my friend would stick, as it was 50/50 which box held the main prize.

This example shows just how persuasive the equal probability argument is for most people and why we must try to make absolutely clear why the answer to the MHP is 2/3 not 1/2.

Nijdam, given that the player originally chose box 15 and the host left box 6 unopened what would be your solution to this problem? In particular what sample space would you start with? Martin Hogbin (talk) 12:59, 4 December 2010 (UTC)[reply]

22 boxes? No "unacceptable, tricky game"?  Same easy problem as the MHP!  –  Serious sources deal primarily with our psychological fallacy, with our faulty "assessment skills" in reviewing the obvious "50:50(same odds)" versus "2:1(double chance)" paradox, or in your example versus "21/22 chance". In any case it's not just a simple "probability puzzle".
What do you know to be known, and what "reasonable assumptions" can be considered to be tolerated just yet as "acceptable"?  Falk (as per her "1 Million" example) would say:

Your friend may have chosen box #3, and then the host opened all boxes except box #3 and box #17, both remaining closed.
If the host should eventually be extremely biased never to open box #17 if any possible, as he didn't, then you have to admit that the chance to win by switching (Pws) is 1/2 at least (but never less!), even in that 22 boxes example. Or he might be biased to always open door #17 if ever possible, but actually he didn't.

• So Pws in any case will be within the range of at least 1/2 (but never less) to 1.

Although any "host's bias (q)" will remain unknown forever. Whether three doors, or 22 doors, or 1 million doors:

• Pws will in any case be within the range of at least 1/2 (but never less) to 1.  –  In any case, independent of the number of doors.

So tell your friend that she/he should switch anyway. And, without any "assumed host's bias (q)", Pws will on average be 21/22, in your example.

But, as said above: any "host's bias (q)" will remain unknown forever.
You can apply conditional probability theorems, that will give you exactly the same answer. Gerhardvalentin (talk) 17:29, 4 December 2010 (UTC)[reply]

I guess you also assume the contestants random draw is independent of the placement of the main price. Before I can give an answer, I need to know the policy of the host. Nijdam (talk) 18:07, 4 December 2010 (UTC)[reply]

Seriously? I suggest just to ask the host on his policy. Gerhardvalentin (talk) 19:10, 4 December 2010 (UTC)[reply]

Yes, Nijdam the two separate random events are indeed independent. The host policy is unknown. So how would you solve the problem and what sample space would you start with? Martin Hogbin (talk) 23:18, 4 December 2010 (UTC)[reply]

The host policy is unknown. Our knowledge about it says that the 22 numbers are mere labels, interchangeable. So our subjective probability that the host would open any of the doors available to him is uniform.

That would be my response, but I would like to hear Nijdam's. Martin Hogbin (talk) 10:15, 5 December 2010 (UTC)[reply]

If I do not know the policy of the host, I cannot solve the problem. If forced to, I could make an assumption about the host's strategy, and assume randomness. Then I calculate the conditional probability given the situation. Nijdam (talk) 12:36, 6 December 2010 (UTC)[reply]

To calculate the conditional probability to find the price in the box left closed by the host, I may argue it is the complementary conditional probability to find the price in the firstly chosen box, and use a symmetry argument to conclude that this conditional probability will have the same value of 1/22 as the unconditional one. Nijdam (talk) 12:43, 6 December 2010 (UTC)[reply]

Excellent, you would do just the same as I would. Or are you being facetious and you would insist on starting with a vast sample space which included a different probability for each possible box left unopened by the host (or maybe even the same probability)?

Actually I am finding it hard to see where we disagree now. You gave a simple solution which says that the probability of winning by switching is the complement of the probability of originally choosing the car which is what I said years ago. If you switch you always get the opposite of your original choice. What about the combining doors solution? Do you object to that? Martin Hogbin (talk) 21:27, 6 December 2010 (UTC)[reply]

Martin, you have to read better. Read again, and see that I wrote about the conditional probability. Nijdam (talk) 22:19, 6 December 2010 (UTC)[reply]

Please read the literature carefully, especially the recent masterpiece in MHP studies Gill (2011). Richard Gill (talk) 09:49, 5 December 2010 (UTC)[reply]

The other reason for giving this example is to show the amazing power to confuse people that MH type problems have. My example was used by vos Savant as an example of a case in which she thought it would be obvious that you should switch. I have asked other friends and none of them found it obvious that you should swap. This what the MHP is primarily about, its amazing power to confound most people and that is what we must address first and foremost in the article. I might add that I asked the question without giving box numbers (arguable unconditional) on later occasions but it did not help people to get the right answer.

By all means let us mention the conclusions to be found in that masterpiece in MHP studies, Gill (2011), along with a scholarly and balanced discussion of all the possible complications and extension but, please, after we have fully and convincingly explained the basic paradox. Martin Hogbin (talk) 10:16, 5 December 2010 (UTC)[reply]

I agree!! The simple solutions are the most convincing way to show anyone that 50/50 - so don't bother to switch, - is wrong. The conditional solutions are for connoisseurs and for students of MHP Studies. Thanks to the symmetry bridge, the conditional solution is accessible for everyone too. The MHP page must give a neutral overview of the controversy. Proponents of the conditional solution must come up with good motivation. Richard Gill (talk) 14:08, 5 December 2010 (UTC)[reply]

Let us assume the car is not hidden uniformly, but half of the times behind door 1 and the rest equally behind door 2 and door 3. What would the (a) simple solution be? Nijdam (talk) 15:59, 6 December 2010 (UTC)[reply]

No "tricky assumptions". Not the MHP. Nothing to do with the MHP-question, to "assume" that it is just nothing but a "dirty tricky sneaky question".

Always behind "door 1" or half of the times? No evidence whatsoever that exactly this very game show ever had been or will be "repeated" more than one or three times. No sneaky assumptions! Besides that, you forgot to mention that one of the three doors had already been selected by the guest, and then a non-selected door had already been opened by the host (whose policy is unknown) and who already offered to switch to the second non-selected door. You forgot to mention this very basis of any solution (whether simple, or conditional probability maths). Gerhardvalentin (talk) 18:54, 6 December 2010 (UTC)[reply]

Nijdam, we all agree that the simple solutions only apply to the completely symmetrical case. Martin Hogbin (talk) 21:17, 6 December 2010 (UTC)[reply]

No, also in the symmetrical case they do not apply. Did you read what I wrote about why the simple solution is wrong? And did you understand it? Nijdam (talk) 22:22, 6 December 2010 (UTC)[reply]

No Martin, and no Nijdam: switching gives the car with probability 2/3 if and only if your initial choice hits the car with probability 1/3. And this is true if and only if the car is hidden uniformly at random or your own choice is uniform random. The simple solution is correct with minimal assumptions. Richard Gill (talk) 03:08, 7 December 2010 (UTC)[reply]

Then please, Richard, comment on what I wrote on the "mediation" page under: Wikipedia talk:Requests for mediation/Monty Hall problem#Why the simple solution is wrong. Nijdam (talk) 16:16, 7 December 2010 (UTC)[reply]

Switching gives the car with probability 2/3 if and only if (there are plenty of other ways the probability of switching gives the car with probability 2/3) your initial choice hits the car with probability 1/3, however your initial choice hitting the car with probability 1/3 does NOT mean the probability of winning for a player who initially picks Door 1 and then sees the host open Door 3 is also 2/3. I think this apparent contradiction is the source of a lot of the disagreement about the "validity" of the simple solutions. -- Rick Block (talk) 17:21, 7 December 2010 (UTC)[reply]

Nijdam asked my comments on the following statements, which I here shorten and clarify slightly:

I take as the MHP the situation in which the player chooses Door 1, and the host has opened Door 3 with a goat, and the player, being confronted with two still closed doors, may decide which one to open. The number of the door with the car I call C, the choice of the player X=1, fixed, and the door opened by the host H. I assume C to be distributed uniformly. Hence (before the host opens a door!) P(C=1)=P(C=2)=P(C=3)=1/3. So the player has a probability 1/3 to pick the car in her first choice (before the host opens a door). If she decides to switch to door 2, the probability (before the host opens a door) to find the car there is also 1/3, because P(C=2)=1/3 and not 2/3. Most people however intuitively consider the new probability, the probability given the new situation, the conditional probability. And indeed the conditional probability, if the host chooses uniformly at random, P(C=2|H=3)=2/3, but NOT because of the argument that P(C=1)=1/3."

I add to his notations X, C, H also the notation Y: the door to which the player is invited to switch. The events {X=C} and {Y=C} are complementary. This is because X, H, and Y are, by definition, all different and H is different from C, yet C has to be equal to one of the three X, H, Y.

Nijdam didn't specify a distribution for X. I took it that he means X is identically equal to 1, the player's lucky number.

No need to specify the distribution of X.Nijdam (talk) 10:13, 8 December 2010 (UTC)[reply]

So you assume X has an arbitrary distribution and is independent of C? No problem. First of all, condition on X=1. All conditions stay the same by independence, and we can run through my previous derivation again, with "Prob( ...)" thought of as "Prob( ... | X=1)". Richard Gill (talk) 11:31, 8 December 2010 (UTC)[reply]

Therefore Prob(Y=C)=2/3 if and only if Prob(X=C)=1/3. That is the required simple solution. QED.

Not a solution to what I above considered the MHP.Nijdam (talk)

I know you don't consider that the MHP, but plenty of other people do (Selvin, vos Savant, ...). Richard Gill (talk) 11:31, 8 December 2010 (UTC)[reply]

By symmetry, Prob(Y=C | H=3) = Prob(Y=C |H=2). According to the law of total probability, and because X=1 is fixed, so Prob(H=2)+Prob(H=3)=1,

Prob(Y=C) = Prob(Y=C | H=3) . Prob(H=3) + Prob(Y=C |H=2) . Prob(H=2)

Therefore Prob(Y=C) = Prob(Y=C | H=3) = Prob(Y=C |H=2) = 2/3 since according to Nijdam's assumptions, X=1 and Prob(C=1) = 1/3. That is the required conditional solution. QED

Not my assumptions. And there is no disagreement about the "conditional solution". Nijdam (talk) 10:13, 8 December 2010 (UTC)[reply]

What do you mean, "not my assumptions"? I refer to the assumptions which you wrote down. Richard Gill (talk) 11:27, 8 December 2010 (UTC)[reply]

Conclusion: there is a short mathematically rigorous formulation of the simple solution by which I mean the deriviation of the marginal probability Prob{Y=C}, and a couple lines more gives you the conditional solution by which I mean the deriviation of the conditional probability Prob{Y=C|H=3} too, under the usual conditions. Richard Gill (talk) 06:52, 8 December 2010 (UTC)[reply]

I do not understand your conclusion. Is it about the simple solution being correct, or about the derivation of the conditional probability by means of the symmetry? And what is more, you did not really comment what I wrote.Nijdam (talk) 10:13, 8 December 2010 (UTC)[reply]

I did not comment on what you wrote because there was not much to say about it except to correct some English and to make some language more precise. You wrote something irrelevant about Prob(C=2)=1/3. That is not the simple solution. The simple solution is any correct derivation of Prob(Y=C)=2/3. You think that MHP *must* be solved by adding uniformity assumptions and by computation of a conditional probability. You never said where those assumptions came from. You never gave a reason why the problem *must* be approached in the way that you choose to approach it (with conditional probability). Who said so? I didn't find it in Kolmogorov's book. I never saw it anywhere.

To be more precise: I see that Rosenthal, Carlton and Morgan and his friends also say that you *must* compute the conditional probability, but they don't tell me why. Richard Gill (talk) 11:34, 8 December 2010 (UTC)[reply]

To be sure, not all editors agree with your assessment of what Rosenthal, Carlton and Morgan and his friends all "say that you *must*" do. Glkanter (talk) 12:57, 8 December 2010 (UTC)[reply]

Indeed, that is true. Some wikipedia editors don't agree with my reading of these authors. However, it is a fact that Rosenthal, Carlton and Morgan are all mathematics professors writing about teaching mathematics. I am also a mathematics professor who reads and writes such books. I think I know things about the hidden assumptions, the standard cultural background of these people, better than people from very different backgrounds. But I am not a dictator or a high priest, everyone is welcome to their own opinion. I tell you mine, you are free to ignore it. What more can I do? What more can you do? Richard Gill (talk) 13:25, 8 December 2010 (UTC)[reply]

Well, that's really it, isn't it Richard?

"I think I know things about the hidden assumptions, the standard cultural background of these people, better than people from very different backgrounds." - Richard

That's the High Priest telling the unwashed that he knows what they can never know. I figure that Rosenthal, Carlton and Morgan have actually written down what it is they wanted me to know they were thinking at that time. Nothing more, nothing less. Just a quaint notion I have about Wikipedia's reliable sources policy. But it applies in the real world, too, I suppose. Its just that such audaciousness has never come up before in my life. Glkanter (talk) 13:50, 8 December 2010 (UTC)[reply]

I am not audacious. You don't have to accept my reading of those papers. I am neither a High Priest nor a Mikitary Dictator. I give you an objective fact (about the profession of several authors and myself) which you may or may not want to take account of. So you don't trust in my good faith or intelligence. Well, that's your problem, not mine.

By the way I am sure you are intelligent enough to pick up the language. Just read a few chapters of some introductory texts. You think that's a waste of time. I think it would be a waste of time too, since you could alternatively just trust reliable sources like myself. Richard Gill (talk) 23:17, 8 December 2010 (UTC)[reply]

I mentioned where you could find my formal probability education CV. I guess you chose not to read that. Glkanter (talk) 00:03, 9 December 2010 (UTC)[reply]

Stuff from this page, up to 1 December, is going to move soon to User_talk:Gill110951/Archive 1 . Richard Gill (talk) 10:04, 5 December 2010 (UTC)[reply]

Shall we carry on discussing the MHP here or would you prefer us to move elsewhere? Martin Hogbin (talk) 10:18, 5 December 2010 (UTC)[reply]

Here is fine! I'll keep the recent new threads here. Richard Gill (talk) 13:56, 5 December 2010 (UTC)[reply]

When I get hired as a consultant to find solutions to a business problem, I will employ probability tools. I will use them to support the suggestions I make and the course of action I recommend.

Since I need to make, rather than defer, a recommendation, I am looking for ways the tools *can* work for me. And attributing a uniform distribution to an unknown distribution is very often a proper call. Which I have done countless times, and have never been betrayed by such a decision.

If I were to try to explain to a businessperson (or engage Richard to do so) why the likelihood of the car is *not* equally 1/3 for each door, only because he used his lucky number rather than a random number generator, I (we) would be immediately and appropriately escorted from the premises.

This is why I have no use for all that nonsense. It is counter-productive to reality. Glkanter (talk) 11:53, 6 December 2010 (UTC)[reply]

It is very rational to take the likelihood 1/3 for each car and the host's choice (when he has a choice) 50-50. This is the least favourable situation you could be in.

Since all doors are equally likely you then might as well choose your own at random. And by the simple solution, switching gives you the car with probability 2/3. This is your minimax strategy. (Hiding the car uniformly at random and opening a door 50-50 when there is a choice is correspondingly the quiz-show's minimax strategy)

How can you agree, Richard? You're the primary person I disagree with! The 'a lucky number is not the same as a random selection, so I can't say its uniformly 1/3' statement is yours! You just saved several lengthy, contentious discussions on this very topic to your archive. Glkanter (talk) 05:40, 7 December 2010 (UTC)[reply]

Today I'll say yes: this morning I tend to think that for pragmatic purposes your lucky number can reasonably be thought of as a uniform random number.

But on second thoughts, does the statement "your lucky number can reasonably be thought of as a uniform random number" have any meaning at all? Is there any way to prove or disprove this? I think not. It might just as well be true as not true - I can't see the difference. Because it's empirically untestable the statement has no empirical content. It's a matter of taste, of religion, of meta-physics, of culture, whether we choose to believe it or not.

By the way, my opinion does change from time to time. I don't necessarily agree now with what I wrote weeks or months ago. I appreciate it when people stimulate changes in my thinking. Richard Gill (talk) 07:24, 7 December 2010 (UTC)[reply]

While the ability to keep an open mind is to be admired, it is problematic to those one debates:

How is one to recognize that you have, indeed, changed your opinion, and the arguing can cease?

One may wonder what is the utility of an opinion that can be swayed as the wind blows?

One can never be sure of your current POV.

Belittling other people, calling them various names and insults because they don't see your point would seem to require a retraction(s) whenever one of these changes occur.

In short, it may be of no value to argue with you, Richard, or to give your opinions much consideration, at all, in the first place. Glkanter (talk) 10:29, 7 December 2010 (UTC)[reply]

That's right. It may be no value at all. I left the mediation. You are under no obligation to give my opinions any consideration whatsoever. But if you join a conversation on my talk page then I think it is generally considered polite that you do give my opinions some consideration. But still, even this is not an obligation. Richard Gill (talk) 13:29, 8 December 2010 (UTC)[reply]

Please tell me how we could empirically test the truth of the statement "Glkanter's lucky number can reasonably be thought of as a uniform random number".

I know how to test whether a dice-and-dice-throwing-apparatus is a good way to generate uniform random door-numbers. I know who to design super-duper high quality computer programs to do the job deterministically (using notions from public key cryptography), and how to prove mathematically that they are good at their job. I know how to design quantum optics chips to do the same job quantum-randomly, and I know how to evaluate their adequacy empirically.

If you, for your own purposes, want to think of yourself as a random number generator, that's your choice. I don't object to it.

But you can't demand from me that I think of you as a good random number generator. Richard Gill (talk) 11:32, 7 December 2010 (UTC)[reply]

I don't expect you to. Just explain to the client why that single issue caused us to miss his critical deadline, before he throws us out the door.

So, did you change your position back again, because I was critical of you, Richard? Its *so* hard to tell. Glkanter (talk) 12:05, 7 December 2010 (UTC)[reply]

Read my paper. I have not changed my position on this particular point at all. But I talk about things which you refuse to know about. If you miss your deadline because you talked to me and got mixed up and angry, that's your problem, not mine. Richard Gill (talk) 06:56, 8 December 2010 (UTC)[reply]

I can't force Mr. Businessperson to read your book, though. Glkanter (talk) 07:56, 8 December 2010 (UTC)[reply]

Of course not. As a consultant it's your job to know the science and to give good advice to your business clients. Just like I do to mine. Richard Gill (talk) 08:20, 8 December 2010 (UTC)[reply]

Yep. We're talking in circles, once again. Mr. Businessperson will throw us out the minute I explain that my colleagues won't let me treat his 'lucky number' as if it was a random number. Therefore, I can't model his problem using uniformity, and suggest a solution. And the more complicate the explanation that I am told to deliver from the High Priests, the angrier he gets with me. Glkanter (talk) 08:27, 8 December 2010 (UTC)[reply]

I don't say that you can't! I say that *I* don't want to. What you do is your own business. I am not a High Priest. I am a student of science. Richard Gill (talk) 11:13, 8 December 2010 (UTC)[reply]

Glkanter, can I ask you this question about your hustler. Exactly as before, three shells, one pea, no switching, hustler places the pea before you arrive. You bet on your lucky number. What do you say your probability of winning is? What would you say the probability was for someone else who planned to bet on their lucky number. Martin Hogbin (talk) 10:08, 8 December 2010 (UTC)[reply]

'Hustler places the pea before you arrive' is a new premise, but doesn't change anything.

Glkanter's lucky number: I don't know where the pea is, I can do nothing to increase or decrease the likelihood of winning this single play, each shell is 1/3.

Someone else's lucky number: He doesn't know where the pea is, he can do nothing to increase or decrease the likelihood of winning this single play, each shell is 1/3.

This is a model of the real world, not a photorealistic snapshot of a moment in time. Even if I presume the hustler has an advantage, unless I know what that advantage is, my choices are 'fight or flight'. I can choose not to play, as the odds are indeterminable, or I can make the best approximation possible, based only on what I actually *know*. Glkanter (talk) 10:56, 8 December 2010 (UTC)[reply]

Good answer, it is a *model*. Models are simplifications. Approximations. You create something small which you can handle. The choice of model is an art. Working within the model is a matter of logic, mathematics. Everything under control. So the question about whether we should model Glkanter's lucky number in the real world, with a uniform random number within mathematics is a question of taste, experience, culture. It probably is a good approximation for a single play with a street hustler.

This is exactly what my paper is about, which Glkanter still hasn't read.

Sometimes one can test models. We might use the model of uniform random numbers from 1 to 6 for tosses of a die. We can empirically test how good this approximation is. I don't know a way of testing Glkanter's lucky number in one game with a street hustler. That's why I don't want to say anything about whether this is a good model or a bad model. I have nothing to go on! I am not a High Priest, this is just my personal opinion. Does Glkanter need some HIgh Priest to tell him what he can do and can't do? I don't think so. Richard Gill (talk) 11:21, 8 December 2010 (UTC

So I need not disappoint the client, after all? I can treat his lucky number as uniform? And no testing can, or ever will, prove me wrong? How does that differ from being reasonable, logical and valid? Which are the standards I maintain and expect from others? Glkanter (talk) 14:38, 8 December 2010 (UTC)[reply]

Glkanter, so your friend fancies a bet and asks your advice. The hustler is offering 2:1 odds and your friend asks if this is fair. I guess you would advise him that it is. He bets and you keep a note of how he did. You have a lot of friends, who visit you in turn and to whom you give the same advice. They all play and you keep records of the results. Would you expect your friends on average to win, lose, or break even? Martin Hogbin (talk) 13:20, 8 December 2010 (UTC)[reply]

While you owe me a response of some sort, Martin, I *will* answer your question.

This has nothing to do with the MHP article currently in mediation. Does everybody understand this?

Especially if the hustler places the pea before seeing the next player, if he has an edge, it must be he knows which of the 3 shells the average punter is most likely to call.

So, he really has no choice but to *always* place the pea under that shell. I only need that first play of the game to know what he will do every time.

So, I'm gonna send 100 different people to him, without him knowing they are connected to me. They will each select, and win with whichever shell hid the pea in that first game I played.

Unless that first game I played was a 'decoy' game intended to confuse people who think like me. Then I'm screwed, except when he plays other 'decoy' games.

But this is not an area I've ever given much thought to. And its certainly not relevant to the Wikipedia MHP article. Is that clear to everyone? Glkanter (talk) 14:06, 8 December 2010 (UTC)[reply]

Glkanter, if you are going to pull out of the discussion every time you think you may be forced to reconsider your opinion there is no point discussing things. You brought up the hustler scenario to prove a point. I am trying to convince you that it does not prove the point you think it does.

To clarify, I do not care when the hustler places the pea so long at is in secret. In your discussion above you are talking about game theory, how you should react to what you think is the hustler's plan. As far as I know neither of us wish to talk about that subject. The question is this: you assert that the probability that the probability the pea will be under your lucky number is 1/3. You are now sending your friends to play. You are not giving them advice on how to play, each friend will, you assume, use his own lucky number. You are just telling them whether the game is fair with 2:1 odds. What do you say? Martin Hogbin (talk) 14:40, 8 December 2010 (UTC)[reply]

I have no clue what you are talking about. You still haven't commented on my earlier response. My advise is: I guess at 2:1, you'll break even. Whether you use a lucky number or a random number makes no difference. I'm not sure why you want to get involved in some other guy's action. Glkanter (talk) 15:08, 8 December 2010 (UTC)[reply]

What response do you want me to comment on? I will be happy to do so.

So you advise your friends that 2:1 is fair and on that basis they should break even. But of course they don't, because the distribution of luck numbers between 1 and 3 is not uniform. The guy is a street hustler and he knows, as you said yourself, the some numbers are more common with others so he places the pea more often under the lest often chosen number (or maybe always under that number).

How do you reconcile these two facts? You say the probability that the pea will be under your lucky number is 1/3 but when you give your friends advice on this basis they lose their shirts. Martin Hogbin (talk) 20:19, 8 December 2010 (UTC)[reply]

Et tu, Martin? "Whether you use a lucky number or a random number makes no difference." True or false? If he insists on playing, what would *you* advise him? Glkanter (talk) 21:09, 8 December 2010 (UTC)[reply]

"Whether you use a lucky number or a random number makes no difference." false. That is what I said before.

My advice, based on my state of knowledge of street hustlers, would be not to accept odds of 2:1 but ask for better. Martin Hogbin (talk) 22:15, 8 December 2010 (UTC)[reply]

Not an option.

If better odds are not an option don't bet. The point is that you told your friends that 2:1 was fair odds and the lost their shirts. How do you explain this. Each friend bet only once yet overall they lost. Which ones had the worse odds?

My friend plays once using either his lucky number or the random number generator. What's your advice?

Using your lucky number ask for better odds or don't play, using the RNG accept 2:1 for a fair bet.

What is the distribution of the population's lucky numbers across 1 through 3? Glkanter (talk) 22:32, 8 December 2010 (UTC)[reply]

I do not know, but I believe the hustler has a better idea than I do.

Besides, Martin. My contention is in regards to: random number generator vs lucky number, without the supposition of a bias, just the lack of knowledge of, or any expectation of, a bias (including 'uniform' as a bias). Glkanter (talk) 22:54, 8 December 2010 (UTC)[reply]

It is back to the same old thing. Using the Bayesian notion of probability it is a state of knowledge.

To someone who has no idea what a street hustler is but knows only that there one pea under one of three shells with no swapping or cheating, the probability the pea is under shell 1 is 1/3 as it is for shells 2 and 3. But you and I know better, so we assess the probability of winning by playing our lucky numbers to be worse than 1 in 3.

To the guy who has been watching the hustler and has noticed that he always puts it under shell 2. The probability for shell 1 is 0. As you yourself pointed out, if you played several games against the hustler, maybe in a different disguise each time, you could probably do better than 2:1. There is no 'real' probability, only a probability based on your knowledge (in the Bayesian notion).

Do you believe me yet? When you do I will explain why the MHP is still 2/3. Martin Hogbin (talk) 23:58, 8 December 2010 (UTC)[reply]

3 guys walk up to the hustler at the same time. They will pick different shells. You can pick one player after they have made their selections. 1st guy picks a shell, saying, 'that's my lucky number'. The 2nd guy picks a shell, saying 'that's my lucky number'. The 3rd guy takes the remaining shell saying, 'that's my lucky number'. Who do you pick? What odds? What if the 3rd guy said nothing? What else do you think is important to this thread? Glkanter (talk) 23:05, 8 December 2010 (UTC)[reply]

It is just a matter of what you know, or think you know. I would use my RNG then I would be sure that the hustler could not beat me. Martin Hogbin (talk) 23:58, 8 December 2010 (UTC)[reply]

Well, I'm just a model maker. Who sometimes is forced to make models with incomplete information. Nothing you guys have written has said 'assuming uniformity is wrong', which is the only point I have ever tried to make in these discussion. Are there risks? Of course. Can they be quantified in any way numerically? Probably not, else I would have that information available for my model. Why wouldn't you answer my question, and pick a shell-picker, Martin? Anyways, I'm tired of this. Anyone else? Glkanter (talk) 00:09, 9 December 2010 (UTC)[reply]

I did answer your question, 'It is just a matter of what you know, or think you know'. I can give you more detail if you like but you could work it out for yourself. For example:

If all I know is that three people pick three different shells by picking their lucky numbers then I am indifferent to the shells and the punters. I have no information that would enable me to distinguish the shells or the people.

If I also know that the pea was placed by a street hustler then the question becomes the same as before, I do not know which shell the hustler favours so I am best to use a RNG. You can ask as many questions as you like on this subject but I suspect you could work out the answers for yourself. Imagine it is your money.

The one thing the MHP teaches us is that probability is not always distributed evenly amongst the available possibilities. Remember how people argue vehemently that the answer must be 50:50 because there are only two remaining options. They forget that information has been revealed by the host opening a door to reveal a goat. This changes their state of knowledge and allows a better estimate of probabilities to be made.

It is just the same with the shells, and all probability problems. If all you know is that the pea is under one of three shells then the only options you have are to either say you cannot answer the question or apply the principle of indifference and take the probability as uniform.

If you also know that the pea was placed by a street hustler who expects punters to guess using their lucky numbers then you may me able to improve your assessment, in this case not of the probability of the pea being under any particular shell but of the likely outcome of a bet by a punter. AS we have agreed, we would expect experiment to confirm our revised estimate. Martin Hogbin (talk) 10:31, 9 December 2010 (UTC)[reply]

How many people are walking around with a random number generator?

If you don't have one available, but you, for some reason, *must* choose a punter, what do you do?

Me? I hold my nose, make a pick, know that it can never be proven wrong, and get on with life. Glkanter (talk) 13:06, 9 December 2010 (UTC)[reply]

If 'must' choose a punter and I must bet and I am offered 2:1, and I have no way of picking a random number, then I consider myself swindled, which is fair because I will join your list of fiends who will also lose on average to this hustler.

You are proved wrong, your friends have lost their shirts by following your advice.

I am really surprised that you do not see the connection with the MHP. Probability does not have to be uniformly distributed amongst the possibilities. You do accept that, don't you? Martin Hogbin (talk) 22:25, 9 December 2010 (UTC)[reply]

I'm talking about models, decision making, and the inability to quantify the magnitude of an alleged, but unproven, error. What's that word I once heard? Not 'optimizing', but 'satisficing'. It means the very best decision. That you can make. Based on what you know, or know you don't know, only. You're apparently interested in other things. No, nobody has convinced me of anything, least of all the need to expand my horizons in order to grasp these Truths you all so want to share with me. Glkanter (talk) 22:34, 9 December 2010 (UTC)[reply]

I fail to see how you cannot be convinced by this very simple argument. You think the probability the pea will be under any given shell is 1/3, you give your friends advice on this basis, they lose money. You must be wrong.

Suppose a group of your friends come to you and say, 'Hey, Glkanter, you told us that 2:1 were fair odds for the guy at the end of your street so we bet with him. When we add up all the money we have won and lost we find we have made a substantial overall loss. What gives?'. Were they just unlucky? You have lots of friends and they all say the same. Martin Hogbin (talk) 10:30, 10 December 2010 (UTC)[reply]

You're basing your whole argument on.

Which:

I don't know is true, and in 3 or 4 days, hasn't been shown to be true

If true, the extent of the skewing remains unknown

So you're busting my balls over something that may not exist, and/or may be immaterial. Which my buddies, who don't carry around RNGs, understood before they played.

[Edit]:Real life street hustlers move the shells around, than ask where the pea is. As our game is hypothetical (as far as I know), your insistence that the lucky numbers are not uniform between 1 & 3 is a contrivance. Like a host bias when choosing between 2 goats.

I asked that we leave this topic a couple of days ago. Glkanter (talk) 11:06, 10 December 2010 (UTC)[reply]

As I said before, it seems that every time you start to lose an argument you want to back out. I will leave you to it. Martin Hogbin (talk) 18:11, 10 December 2010 (UTC)[reply]

Well, if you're going to mutilate my postings, and ignore my arguments, well yes, I guess it could be argued I'm a quitter. Glkanter (talk) 18:15, 10 December 2010 (UTC)[reply]

Actually, it's a disaster there. Philosopher Karl Popper once said “If you can't say something clearly and simply, it is better to keep quiet and keep working until you can say it clearly.” Yesterday some nebulous formulation, called "Third compromise", has been added there, just atrocious pants that can never be meant for real. – Your help is urgently needed there. Please join in again. Regards, Gerhardvalentin (talk) 18:49, 8 December 2010 (UTC)[reply]

Gerhardvalentin, Richard's return would only be beneficial if the moderators were to stop or correct Nijdam's disruptive postings. Sunray does not see them as disruptive, however. AGK made comments otherwise. Glkanter (talk) 19:42, 8 December 2010 (UTC)[reply]

Gerhard, I agree it is chaos there. Why not support my zeroth compromise? Martin Hogbin (talk) 20:24, 8 December 2010 (UTC)[reply]

Time out for me. Maybe next year ;-) I like the Popper quote. Richard Gill (talk) 23:04, 8 December 2010 (UTC)[reply]

Solution

There are several approaches to solving the Monty Hall problem all giving the same result: that a player who switches has a 2/3 chance of winning the car.

Simple approach

The player initially has a 1/3 chance of picking the car. The host always opens a door revealing a goat, so if the player switches she has a 2/3 chance of winning the car. (Monty Hall, as reported by Selvin)

Conditional probability approach

Several sources (Morgan, ...) criticize the simple approach as there is no consideration of whether the probability of 2/3 to have initially picked a door hiding a goat is influenced by the specific sequence of door numbers (Door 1, Door 3) of the doors chosen by the player and opened by the host. Other sources consider this an academic diversion while many do not even see any issue. In some sense there is no issue, since by the symmetry of the problem this so-called conditional probability is also 2/3 (Morgan, Gill).

Comments, mathematical details, comparison. The criticism is that, conceivably, the probability that the other door hides the car could be different, in the situation that the player chose Door 1 and the host opened Door 3, from the situation that the player chose, say, Door 3, and the host opened, say, Door 2; and so on (four more "situations"). So, it is argued, it might even be better not to switch in some situations! To solve the problem properly, in the opinion of such sources, we have to explicitly rule out this possibility.

For the purposes of this brief discussion, we will make the following popular (but not universal) assumptions:

The car is equally likely behind any of the three doors

The player is equally likely to choose any of the three doors

If the player happens to initially choose the door hiding the car, the host is equally likely to open either of the two other doors

A moment's reflection now shows that the probability that switching gives the car is the same in all of the 6 ( = 3 initial choices x 2 then possible opened doors) possible situations or cases recognisable to the player. The reason for this is that by the mathematical symmetry of the problem as just specified, any probability or conditional probability in this problem is unchanged by an arbitrary renumbering of the doors. Since the six just mentioned cases are exhaustive and mutually exclusive, and since the conditional probabilities, given each case, are the same because of symmetry, it follows that all of these conditional probabilities must equal (by the law of total probability) the overall probability, 2/3.

In particular: given that the player chose Door 1 and the host opened Door 3, switching gives the car with probability 2/3.

To appreciate the difference between the two solutions, consider what they each say about 900 representative imaginary replications or repetitions of the game show, under the assumptions just listed. In particular, we'll imagine the player making his initial choice of door anew, completely randomly, in each imaginary repetition, and the car being hidden anew, completely randomly, in each repetition too. (The reader who doesn't like the scenario for the player's initial choice is invited to work through the arithmetic for the scenario of their choice themselves; for instance, for the scenario where the player always initially chooses Door 1 since "1" is his lucky number).

What the simple solution tells us is that in 2/3 of all the times, so 600 out of the total of 900, a player who switches (regardless of initial choice and regardless of which door is opened by the host) wins the car.

What the conditional solution tells us is bit more complicated. In the 900 imaginary replications, the player initially chose Door 1 300 times. In half of these cases, by symmetry, the host opened Door 3. Therefore in 150 of the 900 imaginary replications, the player found himself in the situation that he chose Door 1 and the host opened Door 3. Since the conditional probability of winning by switching given the Player chose Door 1 and the host opened Door 3 is 2/3, he'll win the car in 100 of this group of 150 imaginary repetitions if he always switches in this situation. The same can be said for each of the 5 other groups of 150 repetitions. Thus in each of the six cases, if the player switches, he'll win the car in 100 of the relevant group of 150 repetitions. In total, he wins in 600 out of the total of 900 repetitions, in complete accordance, as it must be, with what the simple solution already told us.

Compromise X is a moving target. I plan to modify it if the comments below bring up points which I think I can accommodate. Richard Gill (talk) 12:29, 11 December 2010 (UTC)[reply]

Martin's responses

Richard, I really do not like your 'probability in a nutshell', 'What this solution is saying...' bit. Martin Hogbin (talk) 10:33, 9 December 2010 (UTC)[reply]

Can you tell me what you don't like about those bits, specifically? You don't agree with the factual content? Or you don't like the presentation? Richard Gill (talk) 11:00, 9 December 2010 (UTC)[reply]

Richard, you say, 'What the simple solution tells us is quite simply that roughly 2/3 of all the players, i.e., 600 out of the total of 900, would win the car'.

1) I am not sure how the solution tells us the specific fact that you quote, it is clearly an interpretation.

I am sure that the solutions tell us these facts! Richard Gill (talk) 18:40, 11 December 2010 (UTC)[reply]

2) The statement asserts a frequentist interpretation of probability.

Thank you. I have rewritten this in a neutral way. Also subjectivists can think of repetitions when they interpret probability. The difference is merely whether the completely imaginary repetitions are in series or in parallel. Richard Gill (talk) 18:40, 11 December 2010 (UTC)[reply]

3) Roughly 2/3? How roughly? So is the probability only roughly 2/3?

Thank you. Instead of thinking of a random sample of size 900 repetitions or replications, I now think of a representative sample of size 900. So the true probabilities will be reflected exactly rather than only approximately in the sample. Richard Gill (talk) 18:40, 11 December 2010 (UTC)[reply]

4) It is not clear how the situation is to be repeated to get the claimed result. Do we put the car in the same place every time? Does the player always pick door 1?

I have made more clear how we should imagine the repetitions, and mentioned that readers who have other ideas can do the arithmetic themselves. Of course I am thinking of the standard assumptions, with the car being behind each door equally often, the player choosing each door equally often, the quizmaster switching equally often both ways when he has the choice. Richard Gill (talk) 18:40, 11 December 2010 (UTC)[reply]

5) The simple solutions also tell us (by using symmetry) the same as the conditional solution tells us. This is where the argument started.

Yes, add a symmetry argument to the simple solution and you get a version of the conditional argument. Alone, without further "interpretation" or "deduction", the actual explicit content in the steps of the simple solution and ending at "2/3" only tell us about a marginal probability. Richard Gill (talk) 18:40, 11 December 2010 (UTC)[reply]

Your statement is attempting the impossible, to sum up both the meaning of 'probability' and the MHP in one sentence. It hides all the issues that we have been arguing about for all these years. Martin Hogbin (talk) 22:36, 9 December 2010 (UTC)[reply]

My statements are attempting to sum up in a nutshell the difference between simple and conditional solutions, as well as the actual solutions themselves (in their most efficient form). I do not attempt to sum up the meaning of probability in one sentence, I make use of a familiar and common and uncontroversial interpretations. People who have different interpretations can rewrite the "meaning" in their own words. Moreover I am attempting to formulate a compromise. That means that everyone will have to accept a bit of give as well as take. Richard Gill (talk) 18:40, 11 December 2010 (UTC)[reply]

Rick comments on Martin's

The simple solution presented here is:

The player initially has a 1/3 chance of picking the car. The host always opens a door revealing a goat, so if the player switches she has a 2/3 chance of winning the car.

Unless the population of players (the probabilistic sample set) where the highlighted "1/3" and "2/3" come from are the same, then this solution makes no sense whatsoever, under ANY interpretation of probability (essentially the units are mismatched - as in, if I eat 1/3 of an apple that means I must have 2/3 of an orange left). The 1/3 is the player's initial chance of picking the car, i.e. the chance BEFORE the host has opened a door and the chance without even saying what door the player picked. The 2/3 therefore must be from that same sample set, i.e. not just the players who pick door 1 and see the host open door 3, but ALL PLAYERS (regardless of which door was picked or which door the host opens). Do you understand this rather basic point of probability, or not? You continue to say things that imply you don't. -- Rick Block (talk) 00:29, 10 December 2010 (UTC)[reply]

As I was the only one on ten years to notice an error in the paper by Morgan et al where they failed to make the distinction you kindly point out to me I guess I must understand it. Could there be something you do not understand perhaps?

Your comment has little to do with the subject under discussion, which is my objection to trying to sum up both the concept of probability and all the issues surrounding conditional/unconditional solutions in one sentence. It cannot be done. The sentence is not exactly wrong but it encapsulates many assumptions and views which may not be universally held. Martin Hogbin (talk) 10:40, 10 December 2010 (UTC)[reply]

I have adapted the "probability interpretation" to be, I hope, acceptable to both frequentists and subjectivists. I summarize both the simple solution and the conditional solution in a few sentences. And I summarize the controversy between simplists and conditionalists in a few sentences. I believe that my proposed text contains true facts and only true facts in a pretty neutral way. Of course, some people might not like being confronted with those facts. That's why I presented them in small print, so as not to offend or bother anyone unncessarily.

I could even publish the whole thing as a footnote to my paper in Statistica Neerlandica and it would be in a reliable source as well! Richard Gill (talk) 18:46, 11 December 2010 (UTC)[reply]

What you propose would be far more suitable for a later section in the article, where all these issues could be discussed properly and in more detail with appropriate reference to reliable sources. To bring them up I think that would be a good idea.

First and foremost, however, people want to understand why the answer is 2/3 and not 1/2 without distractions about conditional/unconditional and frequentist/subjectivist. This can all come later. My aim remains to write a good encyclopedia article. Martin Hogbin (talk) 11:20, 12 December 2010 (UTC)[reply]

Martin, I agree with you on that. Conditional/unconditional and frequentist/subjectivist are important issues for those who want to get deep into MHP. But to get started, the article has to discuss the basic paradox: why you have to switch. Why it's not 50/50.

Anyway, it is not for wikipedia editors to adjudicate on correct/incorrect, right or wrong. Their task is to report the contents of reliables sources.

The point of my text "Compromise X" is to bring home that the conditional/unconditional issue is not such a big deal at all - thanks to the symmetry argument, which brings the conditional solution into the grasp of (almost) everyone who, in the words of Glkanter, is able to get themselves dressed in the morning. Almost anyone can follow the argument, almost anyone can appreciate the difference between the results of the two kinds of solutions. It's for the readers to decide if they think the extra subtlety is worth the brain-power-expenditure.

I don't see why the conditional solution and the discussion of the difference between the solutions needs to be postponed. If it is in its own section, clearly labelled, people who don't care about this stuff can skip it.

Since many readers of wikipedia MHP are students and teachers of statistics classes, they have to be able to find "their solution" easily. Richard Gill (talk) 18:23, 12 December 2010 (UTC)[reply]

Which they will, of course, be able to do wherever it is placed, unlike the general readers, who will be put off by material they are not interested in and cannot understand.

Anyway, I have decided, like you, to give this article a break. The mediation is simply not working and the middle ground is being lost. Martin Hogbin (talk) 19:06, 12 December 2010 (UTC)[reply]

Still, aside from the quesiton where it ought to be placed (if it needs to be placed anywhere at all!) I am interested, Martin, in whether or not the text of Compromise X is now something which you could reluctantly say that you broadly agree with? I am talking about the content, not about the MHP page or the mediation! Richard Gill (talk) 19:14, 12 December 2010 (UTC)[reply]

As an outline for a section intended to explain to the interested reader what all this conditional/unconditional stuff is all about, I like your proposal better now but this is not what all the argument is about. The argument is still essentially about those who want to say, at the start, that the simple solutions are defective.

The article as it is now, has: Simple solutions, Aids to understanding (of why the probability is 2/3), Conditional solution, more stuff. This was also pretty close to your proposal, which I supported. As a starting point for 'more stuff' I think your proposal is fine. Martin Hogbin (talk) 22:28, 12 December 2010 (UTC)[reply]

Maybe one could achieve a compromise by having: Simple solutions (with no health warnings), Conditional solutions (with no criticism of the previous), Aids to understanding, Controversies (including the simple versus conditional storm in a tea-cup), More stuff. Richard Gill (talk) 13:54, 13 December 2010 (UTC)[reply]

By the way, I fail to see why anyone could not understand the intention and content of the conditional solutions. Whether people ought to be interested, and whether they would bother to study it, is another question. It seems to me that all sections on solutions, simple or conditional, need to be sanitized by presenting purely the results (assumptions, derivations, conclusions) and not the criticism of other approaches. Separate sections can *report* the various controversies, including single versus conditional, in a neutral way. Again, readers who were perfectly happy with one or other solution and don't want to do a course on Advanced Monty Hall Problem Studies will not read further anyway. Readers who are interested can read and decide for themselves.

If your notion of probability means "reasonable betting odds" then a simple solution tells you that it is reasonable to bet at odds 2:1 that the car is behind the other door, whenever the game is played. (From inspection of the specific argument, you can see what is being assumed about the initial choice of the player. Often this is taken to be fixed, or can be taken to be fixed). The conditional solution tells you that, if your initial choice was Door 1 and the door opened by the host was Door 3, it would then be reasonable to bet at odds 2:1 that the car is behind Door 2. If you think in terms of repetitions then we are talking about 2/3 of all repetitions of the game, versus 2/3 of those repetitions in which the player chose Door 1 and the host opened Door 3. The simple solution does not tell you, on its own merits, that it would be reasonable to bet at odds 2 :1 in any of the six recognisable situations which the player could find themselves in. However if you are a subjectivist then you will immediately know that by symmetry it doesn't matter, you don't need to take notice of the door numbers.

The difference between the solutions can be stated in a completely objective way which makes it clear that the conditional solution goes a bit further than the simple solution. On the other hand probably most readers are not interested in this bit of extra mileage.

The reason un-thinking teachers of probability and statistics are dogmatic about conditioning is the following: if you consider a mathematical decision problem where information comes to you in steps and at a certain point you have to make a decision, then you will always determine an optimal decision by computing the conditional probability distribution of the outcomes at the latest possible moment, i.e., in possession of the maximal amount of information. This is an easy to prove mathematical theorem which no one ever bothers to prove or even to talk about in class. So students (and these later become teachers) become merely brainwashed to believe that it *must* be done this way, they have forgotten *why it could be wise*. It is therefore not a question of "you must condition", but it is a question of "if you do condition you can be sure of getting the/an optimal result". Note, please, that I have formalized the problem as a sequence of actions: 1.) car is hidden, 2.) player chooses door, 3.) host opens door, 4.) player is given opportunity to switch. Each action is associated with "information" coming into possession of the player, namely 1) "none", 2) door number chosen by player, 3) door number of door opened by host, 3) payoff time. From the mathematical analysis we discover that the actual numbers on the doors which we obtained in steps 2) and 3) don't actually hold useful information, we need only take account of their "usage" or "role": namely the usage or role of being "chosen by the player", and the usage of being "opened by the host".

Glkanter's position and possibly yours, I think - please both excuse me if I'm wrong - , is the very practical position that you are not just interested in the final step of knowing for sure that 2/3 can't be beaten: it doesn't occur to either of you to doubt it, and you are indeed right that it can't be beaten, so you were retrospectively justified, in this case, not to have worried. Nijdam's position is that of the mathematics teacher who says that in order to solve an optimization problem it is not enough just to stumble on the optimum, you must also give a mathematical proof that it is the optimum.

The way I read it, earlier today you proposed to AGK that the criticisms of the simple solutions should be in the Conditional section:

"Part of the coverage of the conditional solutions should include a discussion of the substantive difference between the results obtained in each approach." Richard Gill (talk) 06:47, 13 December 2010 (UTC)

That would seem to contradict what you wrote above:

"Conditional solutions (with no criticism of the previous)... Richard Gill (talk) 13:54, 13 December 2010 (UTC)"[reply]

and

"It seems to me that all sections on solutions, simple or conditional, need to be sanitized by presenting purely the results (assumptions, derivations, conclusions) and not the criticism of other approaches. [User:Gill110951|Richard Gill]] (talk) 14:21, 13 December 2010 (UTC)"

And that you support nearly everything in my proposed 'Compromise 1':

"Maybe one could achieve a compromise by having: Simple solutions (with no health warnings), Conditional solutions (with no criticism of the previous), Aids to understanding, Controversies (including the simple versus conditional storm in a tea-cup), More stuff. Richard Gill (talk) 13:54, 13 December 2010 (UTC)"[reply]

Please help me understand this apparent dichotomy. And help me understand why you and I are even arguing. Thank you. Glkanter (talk) 16:07, 13 December 2010 (UTC)[reply]

Didn't I get that correctly? Who says that   to clearly show respective assumptions / derivations and showing the "difference"  should include "criticism"? No one but you, Glkanter, I think. To clearly show the difference is one thing. Can you agree that "the storm of controversies, conflict, criticism etc." is quite another agenda? Imho it is. Regards, Gerhardvalentin (talk) 17:22, 13 December 2010 (UTC)[reply]

My comment was in response to Richard's response. Glkanter (talk) 07:36, 15 December 2010 (UTC)[reply]

It seems to me that this kind of "reflection on what a solution tells you" is within the grasp of anyone who can be convinced that you ought to switch. It shouldn't get anyone upset. It is not really a big deal, except perhaps in the special context of a mathematics class-room where this kind of little detail is exactly what you have to hammer on and on about for three years, so that your students finally begin to be able to think, just a tiny bit, like a mathematician. I can't understand why people get so emotional and dogmatic about it. Richard Gill (talk) 14:21, 13 December 2010 (UTC)[reply]

Maybe because there is currently too little literature, too little "reliable source" that would have clearly presented the MHP-Problem-Problem in its "global context, with all its facets", in a clear and easily comprehensible way? I guess there is a chance that the "WP.en" could be one of the first ones, worldwide available. Gerhardvalentin (talk) 15:29, 13 December 2010 (UTC)[reply]

Maybe indeed! Thanks. Richard Gill (talk) 17:47, 13 December 2010 (UTC)[reply]

Glkanter: the apparent contradiction is because you refuse to contemplate what it means when we say a probability is 2/3. I think it uncontroversially means "2 times out of 3". I think we should ask ourselves: "3 what?" I think that 2 out of 3 apples are something different from 2 out of 3 pears. You call this OR. I call your OR labeling a formalistic excuse to avoid thinking, and I imagine that you do this because of a fear that thinking might lead you to realise that you had overlooked something in the past. We all hate to have to admit to past mistakes..

I think that the legalistic approach to resolving disputes on wikipedia by invoking OR, NPOV, etc. is disgusting and/or chiildish. It is the last resort of admins. Richard Gill (talk) 17:47, 13 December 2010 (UTC)[reply]

The contradictions I pointed out above are simply the result of you typing different things in response to the same stimulus, mere hours apart. Not due to my (lack of) understanding of probability or by following Wikipedia policies. Just inconsistencty and, perhaps, your own arrogance, as we all hate to have to admit to past mistakes. If we were perhaps writing 'The Annals Of Probability', I might agree with your tirade. However, as my first, and so far, only goal on Wikipedia has been to bring the MHP article into compliance with Wikipedia policies and make it more accessible to the reader, I think rejecting OR and NPOV violating arguments is highly appropriate. You've agreed with me that the simple criticisms receive undue weight in the current article, I believe Nijdam's proposals make the situation worse, and like your contributions, are not based on the viewpoints of reliable sources. Glkanter (talk) 07:36, 15 December 2010 (UTC)[reply]

I think the "contradictions" you pointed out stem from your reluctance to comprehend subtleties and your apparent inferiority complex with regard to maths PhDs, which is not warranted at all, since you are more than smart enough, and apparently have plenty of time to do the bit of background reading and thinking which would bring you up to maths PhD speed on this material. A bit more good faith and effort could work wonders. But of course I agree that no-OR and the insistence on NPOV are the essence of wikipedia. I also think that Nijdam's proposals make the situation worse. On the other hand, I am a reliable source and not an editor of the wikipedia MHP page. And I have consistently argued for compromise and for synthesis, and all the time done my best to make sense out of the sources, whichever side they stood on. Richard Gill (talk) 15:15, 20 January 2011 (UTC)[reply]

Gerhard's responses

Yes Richard, you brought it to the very point. Your terms     "... could be influenced by the specific choice ..."     shows what really is meant and is easy to understand. Gerhardvalentin (talk) 11:03, 9 December 2010 (UTC)[reply]

22 boxes/doors (or 3 boxes/doors) - a question:

Richard, I guess that, besides various droll variants "outside the MHP", the question answered by vos Savant isn't foremost a "probability puzzle" asking for the exact respective probability to win by switching, nor for all those imaginable variants of conceivably underlying assumptions, that might be given, or not!

Because the question is exactly whether it was "of advantage" to switch. So you just need to be aware of the situation you are facing.

Above I mentioned Falk. She correctly states concerning the MHP:

After the guest has chosen her door and the host has opened another door showing a goat, the probability to win by switching (Pws) still is 2/3 not only "because" the host always will be able to open a door showing a goat, but rather only if, in opening of his door, you haven't learned anything, have learned absolutely nothing to allow us to revise the odds of the door selected by the guest. That's the fact only if the host, if ever having got two goats, opens one of his two doors uniformly at random, not giving additional information on the actual location of the car.

Even with 1 million doors, after the host has opened all but two doors, leaving closed only the door chosen by the player and, e.g., door #777'777 (as per Falk), you  might be confronted with one of the two absolute extreme situations:

If the host never uses to open door #777'777 if ever possible, then the probability to win by switching Pws could be "only 1/2", but never less.

But if he always uses to open door #777'777 if ever possible, but actually he didn't, then Pws will be 1. And you know that on average Pws will be 999'999/1'000'000.

So that's all you need to know as a basis for your decision. You know that, and you know the overall Pws. You do not need to know more, and you cannot know more, as any of your additional "assumptions" never are to be "given". It's an illusion to take them for "given". So why bother?

In any case the range will be from (at least) 1/2 to 1!  –  So you should switch anyway.

You know that, and you know the overall Pws. You do not need to know more, and you never can know more. No need of those illusions.

But students of probability theory will be confronted with closer tasks and additional assumptions.

Is this acceptable, and do the sources say so? I know at least one reliable source does say so indeed. Your papers will notably help. Thank you.  Regards,  Gerhardvalentin (talk) 12:09, 12 December 2010 (UTC)[reply]

What you say here is entirely correct. In fact it is the main result of the Morgan et al paper. I have also reproduced it in my own paper.

Please note that how you would like to model this problem depends on your thinking about probability. If you think of probability in the personalistic way, namely as representing *your* personal uncertainty, then for you the host is equally likely to open any door when he has a choice, since for you the door numbers are arbitrary labels, completely exchangeable, so your "expectations" of what the host will do in just one game are uniform over the doors. If you think of probability as representing physical processes, e.g., in the brain of the host, then for you the probability the host would open each particular door is an unknown parameter. Fortunately, as you say, and as Morgan et al, already said, you don't need to know these probabilities in order to decide what is the best decision.

Please also note that in order to know we are making the *best* possible decision, we should work with conditional probabilities given all information at the last moment before we are forced to choose. The simple solution only shows that always switching is a *better* decision rule than always staying. Richard Gill (talk) 18:11, 12 December 2010 (UTC)[reply]

I think it's necessary to present the arguments of criticism of the "simple solution", clearly laid out in a NPOV manner,

as well as the arguments of criticism of the "conditional solution illusion" also, clearly laid out, in a NPOV manner.

Important to show what the conflict really is (was) all about.  Gerhardvalentin (talk) 13:24, 12 December 2010 (UTC)[reply]

I agree with this too. Problem is, the reliable sources do not give reasons for their criticism. I think I am the first one who took the trouble to give good reasons for the criticism. Richard Gill (talk) 18:13, 12 December 2010 (UTC)[reply]

Richard, please may I put two more questions.

1. *Before* and *After*
   Imho "simple solutions" do pay regard to the door selected by the guest, and imho it makes no difference whether she selected #1, #2 or #3. And imho simple solutions also are paying regard to the situation *After* one door has been opened by the host, and simple solutions are speaking of the remaining door offered as an alternative, too. Why do some say "they don't"?
   Imho simple solutions are indeed considering not only the door chosen by the guest (e.g. #1), but they are also considering which door has been opened by the host (e.g. #3). Some say that simple solutions "do not consider" the situation *After* the host has opened a door.
   Is it really "correct" to suspect and to say that simple solutions do not consider which one of his two doors the host has opened, just because they are not excluding any other combination besides 1+3?  Imho simple solutions do pay regard to the door opened, they just do not exclude other combinations 1+2, 2+1, 2+3, 3+1, 3+2.
   Imho it is a baseless and incorrect assertion that simple solutions do not consider which special door has been opened. Indeed they do, but for the simple solution it's just of no importance, which one.
   And, without saying, it is surely obvious and given that fears *always* could be justified. And obviously the simple solution considers even the "worst case scenario" indeed, because it is also given that, by switching, in any case you could be drawing a blank. No one says to win by switching always is guaranteed.
   But considering your state of knowledge given, and even considering your worst fears, totally independent of the original number of doors (whether 3 doors e.g., or 1 million), it is an "eternal and irrefutable iron law" that, if you have been offered to switch to the second closed door, your risk to loose can at max. be 0 to 1/2 only, *and never more*, and that the odds to win by switching will always be within the range of at least 1/2 to 1, *but never less*. You know that eternal law to be "given", (see "what if ..."), and you can pay respect to this given irrefutable fact without using conditional probability theory nor Bayes. Independent of your "approach" (with or without "conditional probability" theorems) you know already from the outset that nothing can ever be better than always to switch.
   So, who may say that simple solutions "do not consider" the situation *After*?
   
2. *Givens* vs. *speculative assumptions*
   Some assumptions of "conditional solutions":   Imho any conditional solution always gives you the correct answer, based on the underlying assumptions. A conditional solution gives you a reliable result, and yes, you should use conditional probability therefore, if you like. But one argument within the "MHP-Problem-problem" was that a conditional solution is important because the odds on the door first selected of 1/3 could eventually have been influenced by the action of the host. As said above (simple solution), if you fear it could have been influenced, you know at the same time that the odds on the door first selected anyway must have remained within the range of 0 to max. 1/2, but never more, and so the odds on the door offered to switch on must be within the range of at least 1/2 (but never less, no matter what has happened) to 1, this range being a priori "given". And, as said above (simple solution), you know already from the outset that a priori nothing can be better than always to switch. No additional "proof" is needed to find the optimal decision.
   Is it correct to say that the conditional solution is of importance because of it's correctness only, but not because of it's underlying various "assumptions, derivations and conclusions", never guaranteed to be "given", that try to help in getting "better results" for optimists and for cowards though? I think any such "deviant" conditional solution, if you think this is of importance, and if you insist on that, just is an illusion. As those various assumptions never may be "given".
   Can I say that, the *range of Pws 1/2 to 1 (but never less)* anyway a priori already known to be "given", as an eternal and irrefutable iron law, only a conditional solution with the result "2/3" can be said to be "reasonable", so any other (allegedly "better") assertion will remain a purely speculative and less helpful "illusion"? Because the eternal "fixed range of guaranteed Pws" already is known, a priori. Gerhardvalentin (talk) 13:40, 14 December 2010 (UTC)[reply]

I will try to answer these questions later - I mean in a few days, hopefully. Richard Gill (talk) 08:01, 16 December 2010 (UTC)[reply]

My point is that the "Gordian knot" has been solved easily by a stroke of the sword:  The subjectivist, seeing the host open one door showing a goat, is considering all he "knows" and says "Pws is expected to be 2/3".  The objectivist however, seeing the host open one door showing a goat  –  in considering the worst fears possible and taking into account the maximum deviation any conceivable  –  says that Pws in any case, a priori, is bound to  "always be remaining within the fixed range of at least 1/2 to 1 (but never less)", because the odds on the door first selected by the guest "could" eventually rise from 1/3 to 1/2 (but never more), and  –  on the other hand  –  could fall to 0, so a priori are bound to be remaining within the fixed range of  "0 to 1/2,  but never more". These two aspects are both eligible and justified, and are completely independent of the "simple" or the "conditional" approach. They do apply in both cases alike.  May I say so?  Regards,  Gerhardvalentin (talk) 11:53, 16 December 2010 (UTC)[reply]

I agree, Gerhard. This is one of the main messages of the paper I wrote. You say this is completely independent of the "simple" or the "conditional" approach, but I would say, this is exactly the result of the conditional approach. And it is the result of Morgan et al., too. Richard Gill (talk) 20:13, 20 December 2010 (UTC)[reply]

Thank you, Richard, for your reply, and I hope your message will help. My point: The view of this principle precedes any "approach", and any approach must be seen and must be addressed in the light of this pre-existing principle. Whether 3 doors or 1 million, the force of this "eternal and irrefutable iron principle" applies always and it precedes the decision of approach, whether simple or conditional. This principle was already in force in the creation of the world and is "ahead" of the simple and it is "ahead" of the conditional approach, no matter whether you "explicitly mention this precept", or not. It applies independently. I think this "precedence" puts the conflict into perspective and clearly helps to neutralize that conflict. What do you think of the said "precedence of that eternal precept"?     I wish you beautiful holidays, and thank you once more. Kind regards, Gerhard.  Gerhardvalentin (talk) 00:14, 21 December 2010 (UTC)[reply]

Probability theory, and conditional probability too, is just accounting, just arithmetic. Just 1+1=2. And that if you add up a table of numbers columnwise or rowwise, you'll get the same answer. What I call conditional probability is just thinking separately about the six possible configurations of door chosen, door opened. And that is what your argument does. And some arithmetic. It is not a question of dogma "you must do it this way" (or that way). It is just a question of analysis. Of inspection. Of arithmetic. So I do think that wee are talking saying the same, only you call it an eternal precept, and I call it conditional probability. Richard Gill (talk) 13:15, 22 December 2010 (UTC)[reply]

Thank you, yes. I just tried to communicate my "firm belief" that the feeble argument of illusion-flawed conditionalists:
«The odds on the door first selected "could" have been influenced by the specific choice of the host»
forever must be ineffective for your decision to accept the host's offer to actually switch to the second closed door offered, based on any simple solution. Whether one million doors or only three: Because you KNOW at the same time that Pws, independent of the host's action, in any case will be remaining within the range of 1/2 (and never less) to 1! You KNOW that in many ways, e.g. because of the most grim and sneaky assumption "what if the host never uses to open this second closed door, if any possible", that clearly tells you that Pws forever must be at least 1/2 and is to remain within the range of 1/2 to 1. You KNOW that ab initio, independent of conditional or of simple approach. Regards,  Gerhardvalentin (talk) 18:40, 22 December 2010 (UTC)[reply]

On the long run

Hello Richard, I wish you a HAPPY NEW YEAR!  And thank you for your comments.  Please one question again:  You wrote here:

distinguish the overall long run, from each of the six subsets defined by initial door choice and door opened.
Unconditional and conditional is about overall versus one of the six special cases.

Please help me to distinguish the "overall probability" from "on the long run". I can distinguish (as a frequentist eg) "all possible situations on the long run" (unbiased host or host biased to any extent) - as said, all possible outcomes - and, on the other side, "one of the six special cases", where the car can be behind anyone of two nonchosen doors, "on the long run" (again unbiased host or host biased to any extent). In my 100 million example, variant #2 with extreme host's bias to open the door with the largest possible number, the result is:

Probabilities shown on complete positive distribution of any possible combination / outcome   Variant #2: Host is assumed to have been biased towards always having opened his door with the largest possible number
Total number of games	Guest selects door	Host opens door	car behind door	# switching wins		# staying wins
10 933 265	1	avoided 2	3	10 933 265	always		never
22 400 240	1	preferred 3	1+2	11 200 220	1/2	11 200 020	1/2
11 199 709	2	avoided 1	3	11 199 709	always		never
22 133 059	2	preferred 3	1+2	11 199 762	1/2	10 933 297	1/2
11 200 172	3	avoided 1	2	11 200 172	always		never
22 133 555	3	preferred 2	1+3	10 933 187	1/2	11 200 368	1/2
100 000 000				66 666 315	2/3	33 333 685	1/3

(see also there).

Can you please give me support to exactly distinguish, to understand the difference you addressed? Maybe on those examples? Would be fine. Thank you, and kind regards. Gerhardvalentin (talk) 16:51, 1 January 2011 (UTC)[reply]

You have done it yourself, in the table above. Overall there were 1 US billion repetitions. In 2/3 of those repetitions, the player won by switching. If we look at the six subsets defined by "number of door chosen by player, number of door opened by host", we see different probabilities of winning by switching. All of them at least 1/2, and of course, with a weighted average of 2/3 (weight according to relative frequency of each of the six cases). That identity of the overall succesrate with the weighted average of the success rate.s over the six cases distinguishable to the player is what is called the law of total probability. The conditional argument tells us what is a best strategy for each oft be six cases (switch), and hence what is the optimal strategy overall (always switch). By consideration of the six cases separately we see that no mixed strategy (sometimes switching, sometimes not, depending on the information available to us -door number chosen by player, number of door opened by host) can achieve a better overall succesrate than 2/3. Richard Gill (talk) 18:08, 1 January 2011 (UTC)[reply]

I think terms like 'in the long run' and 'overall probability' are not particularly helpful in discussing the MHP and have caused confusion to some editors. What happens 'in the long run' is one of the (frequentist) definitions of probability rather than a special case that might be contrasted with an 'individual probability'.

Far better, in by opinion, to refer to the probability (using any accepted definition) given, or not given, certain conditions. Although I understand that, strictly speaking, all Bayesian probabilities are conditional it is an accepted convention (certainly here) not to consider anything that happens with certainty to be a condition. Thus the game rules are not generally considered conditions. On the other hand, we might choose to apply conditions such as 'the player initially chooses door 1' or 'the host opens door 3', or 'the host says the word "pick" ' to the problem depending on what we consider the question to be asking.

We are free to apply whatever conditions we choose based on the question that we are trying to answer. Martin Hogbin (talk) 18:09, 8 January 2011 (UTC)[reply]

Well, I think it is useful to use hypothetical repetitions to understand probability statements. Whether these hypothetical repetitions are imagined to occur in time or space or in parallel worlds is not important. The important thing is what is supposed to be fixed, what is allowed to vary. Also Bayesian probabilities can be visualised in terms of repetitions. In the MHP problem it is useful to distinguish overall probabilities (relative to repetitions when we "start all over again" e.g. at the hiding of the car and/or choosing of the door) and to sub-case probabilities (repetitions in which we take half of the history already to have unrolled in a particular way - door chosen, door opened). Some people like to talk about "the long run". Just because some people have a subjective view of probability doesn't mean that wikipedia editors are obliged to restrict their imagery in getting across the conceptual difference between an unconditional and a conditional probability. The same difference applies whether we are subjectivist or frequentist. Actually, the frequentist's repetitions (in the long run) are equally imaginary as the subjectivist's repetitions (in parallel worlds). Richard Gill (talk) 15:19, 9 January 2011 (UTC)[reply]

procedural method of the host in opening of his door

Richard, the table above is based on a 100 million run) with an unbiased (50/50) host, and the entire protocol-list, game 1 to game 100 000 000, exists. Checking this protocol you will repeatedly find long sections that apparently pretend evidence of some more or less strong host's bias, although there actually is *none*. So such a bias can never be "discovered" (detected) for sure in real life, it will forever be *unknown*. If s.o. debates such a bias, it always will be a matter of "pure assumption", and that confirms the result of the simple approach "for the real life". So, in my opinion any argument for the "conditional" approach (as "door#2 or door#3" don't matter) should explicitly be based on the assumption that the host could be "assumed to be biased". But, if you "assume" him to be biased, you could "solve" that facet as Ruma Falk did: "Suppose he always uses to leave that door closed, as he did now" (Pws 1/2, but never less), or "suppose he never uses to leave that door closed, but now he did" (Pws 1/1) (to judge on this aspect you don't even need conditional maths). In clearly presenting that aspect in the MHP could help to understand why some like to present conditional probabilities for the "simple" MHP. And it should clearly be shown that the "assumption" of a (never be known) "host's bias" is a complete side-show, never addressing the "real-life-MHP", but are of a quite other nature: Just to show mathematical capability to address and to handle even this aspect. My position: For the reader, to understand what this all really is about, those "offside MHP-assumptions" should be made explicit in the MHP article.   Gerhardvalentin (talk) 14:52, 8 January 2011 (UTC)[reply]

Whether or not a bias could be detected is a different issue from whether or not it could exist. And yet another completely different issue is whether we think of probability in a subjectivist or an objectivist sense (is it out there in the real world, or is it in our minds, in our knowledge?). But, really I have said everything I can think of saying on how these issues impinge on MHP in my articles! As to the wikipedia article: I think it should be written in a layered way so as to be accessible to readers of all different kinds. It should start simple and move slowly to more complex. Given our current understanding and current tools ("symmetry", Bayes' rule in odds form) it does not have to get very technical or abstract or formal at all. And every intelligent person would be wise to understand some more about probability and conditional probability - it is always understood wrong in the newspapers, it leads to disasters in law courts, it leads to disasters in medicine and medical research. Since I have now recently published on MHP I cannot be an unbiased editor of the wikipedia MHP page. I love to participate in discussions about content and about presentation, but I am only an interested observer now. And the research continues. Next step for me is to continue discussions with some of the major authors (Carlton, Rosenhouse, Rosenthal, Devlin, Morgan) to find out how their current understanding stands, and whether they agree to some of my readings and criticisms of their work.

What really annoys me is arguing with people who can't distinguish between the logical correctness of some arguments to get to some result, and the correctness of the result itself as answer to the question as interpreted by the writer of the argument. It is a total waste of time. How can you collaboratively edit wikipedia pages about the correct solution of famous brainteasers, if your collaborators can't distinguish logical sense from logical nonsense? That's another reason why I don't want to have anything more to do with the mediation. I am not interested in arguments based on dumb waving of rule books. MHP deserves to be treated intelligently by people who have the intelligence to appreciate subtleties and appreciate differing points of view. Richard Gill (talk) 15:31, 9 January 2011 (UTC)[reply]

I'm with you, Richard. Now, all we have to decide is do we go with your preferred version of anarchy, or Nijdam's? Please advise promptly. Glkanter (talk) 17:05, 9 January 2011 (UTC)[reply]

I just wrote "I think [the wikipedia article] should be written in a layered way so as to be accessible to readers of all different kinds. It should start simple and move slowly to more complex. Given our current understanding and current tools ("symmetry", Bayes' rule in odds form) it does not have to get very technical or abstract or formal at all". You call that anarchy?

But I no longer aspire to be an editor of the wikipedia MHP page --- I've recently published two peer-reviewed scientific papers on the topics so this would be improper. Those papers do contain the absolutely latest word in comprehensive and unprejudiced mathematical analysis of MHP, offering a sythesis which has never been seen before, though all the maths is absolutely elementary, and all the components in the papers already existed in the literature. Many were learnt from my co-editors on wikipedia (who are collectively thanked in the acknowledgements). If my ongoing correspondence with Devlin, Rosenthal or anyone else leads to anything of interest I'll let you all know. The recent discussions on Martin's talk page brought up some interesting new insights, I thought. As far as I can see, just about everyone is progressing in mutual understanding and moving towards concensus; @Glkanter seems to be the only one who is totally impervious to new insights (largely because he cannot distinghuish between the logical correctness of an argument, and the logical correctness of its conclusion).

I take it you still haven't read my own work on MHP. There really is no point in talking to one another as long as that state of affairs persists. Richard Gill (talk) 19:46, 9 January 2011 (UTC)[reply]

Nijdam's responses

1) In response to Richard's

A moment's reflection shows that the probability that switching gives the car is the same in all of the 6 ( = 3 initial choices x 2 then possible opened doors) possible situations or cases. The reason for this is that by the mathematical symmetry of the problem, any probability or conditional probability in this problem is unchanged by an arbitrary renumbering of the doors. Since the six just mentioned cases are exhaustive and mututally exclusive, and since the conditional probabilities, given each case, are the same because of symmetry, all these conditional probabilities must equal (by the law of total probability) the overall probability, 2/3.

This is all very nice, but the simple solution, does not mention any of this at all.Nijdam (talk) 16:10, 9 December 2010 (UTC)[reply]

I did not say that. So: so what?

No, you didn't, but you give the strong suggestion, the simple solution is this way justified. But that's not the case. Nijdam (talk) 20:46, 11 December 2010 (UTC)[reply]

I reported a range of opinions which can be found both in the sources and among the editors, including you. What more can one do? This is wikipedia, not the Annals of Probability. Write an article in the professional literature, if you want to change the balance of opinions on wikipedia, in some distant future. Richard Gill (talk) 05:07, 12 December 2010 (UTC)[reply]

2) In response to Richard's

What the conditional solution tells us is bit more complicated. Of the 900 players in total, about 300 initially chose Door 1. In roughly half of these cases (by symmetry) the host opens Door 3. Therefore 150 contestants find themselves in the situation that they chose Door 1 and the Host opened Door 3. Since they all switch and the conditional probability of winning by switching is 2/3, about 100 of this group of 150 win the car. The same can be said for each of the 5 other groups of 150 players: in each case, if they all switch, about 100 of each 150 will win the car. In total, 600 out of the total of 900 will win.

Not clear however is, how many of these 900 players initially have chosen Door 1 and see Door 3 opened by the host, and moreover how many of these players will win the car if they switch. Nijdam (talk) 16:06, 9 December 2010 (UTC)[reply]

I thought it was pretty clear. I have tried to make it clearer still. What is wrong? I am working with the full, standard assumptions. Richard Gill (talk) 18:47, 11 December 2010 (UTC)[reply]

Well, the standard assumptions do not contain the (complete) random choice of the player. No need to. Nijdam (talk) 20:46, 11 December 2010 (UTC)[reply]

No need, but if you want to give a concrete numerical illustration it"s convenient to take so. My revised text suggests the reader do the arithmetic with another distribution, if he don't like this one. Finally, many reliable sources *do* assume a random choice. It's not original. Richard Gill (talk) 05:07, 12 December 2010 (UTC)[reply]

You know, Richard, but maybe it didn't got through to you, what I'm actually fighting is the following reasoning, that I've seen, read, many times. Given the situation of Door 1 chosen and Door 3 opened. "We know the probability for Door 1 to hide the car is 1/3. As Door 3 clearly does not show a car, the probability for Door 2 to hide the car must be 1 - 1/3 = 2/3." This reasoning is what I call - because it is found this way in the sources - the simple solution. I sincerely hope you also find this unacceptable. Nijdam (talk) 13:57, 14 December 2010 (UTC)[reply]

Why, Richard, do you not comment on this? You need not be a mathematician to recognize the error.Nijdam (talk)

No comment, because I agree that the reasoning you talk about is incorrect. But that is not the reasoning that I call "the simple solution". By "the simple solution" I mean something like the following. Let C, X, H, Y be the door hiding the car, the door chosen by the player, the door opened by the host, and the other door left closed. I hope you agree with me, Nijdam, that Prob(X=C) = 1/3 if and only if Prob(Y=C) = 2/3. In words, if the unconditional probability that your initial choice hits the car is 1/3, then (and only then) is it also true that the unconditional probability that switching gives the car is 2/3. The simple solution corresponds to ignoring the actual door numbers but only working with the "function" or "role" of each door. And I hope you also agree that by symmetry, it is true that Prob(X=C)=1/3, hence it is true that Prob(Y=C)=2/3, and from this by symmetry (irrelevance of the actual door labels) Prob(Y=C|X=1,H=3)=2/3 too. Richard Gill (talk) 11:42, 16 December 2010 (UTC)[reply]

Glad to notice we have no mathematical differences. What's left, is to decide whether the (simple) solution some sources present, is your interpretation or mine. At least it may be clear, that your interpretation is not appropriate to solve the MHP where the player is asked to review her decision after the host has opened the goat door.Nijdam (talk) 14:14, 16 December 2010 (UTC)[reply]

I disagree with your claim "not appropriate". Please explain to me, why you think you *must* look at the conditional probability. Mathematics is not a religion. It doesn't tell you what you *must* do. Richard Gill (talk) 07:33, 17 December 2010 (UTC)[reply]

In general you're not sure whether you take the right decision, if you do not base it on the latest info you have. Although in this case it is different, but the simple solution does not give arguments for it. A simple example is Martin's urn with 10 balls numbered 0 - 9. The third draw without replacement has a 1/10 chance to be number 9. But if you know the outcome of the first two draws it is different. Most ordinary people will give the answer 1/8 (the conditional probability) if the first two draws did not result in ball 9.

How do you define "the right decision"? Richard Gill (talk) 15:48, 17 December 2010 (UTC)[reply]

Perhaps you could explain exactly what is wrong with this reasoning, in the case that the problem is completely symmetrical with respect to door number. In other words, whether the host opens door 2 or the host opens door 3, makes no difference to the probability of finding the car behind door 1 or the remaining door. Martin Hogbin (talk) 22:59, 14 December 2010 (UTC)[reply]

Martin, I have explained this, and it should not need much explanation. It would be better if you just would understand this. Nijdam (talk) 23:22, 14 December 2010 (UTC)[reply]

It would seem from some of Richard's remarks that I am not the only one who would like a detailed explanation. Just thumping the table and saying it is true because I say so is no good enough.

I think you underestimate the intuitive power of symmetry. You may remember that, some time before our letter, I asserted that Morgan's ln(2) result was wrong. How did I know this without doing any calculations? By symmetry. A simple symmetry argument shows that this answer must be wrong. Martin Hogbin (talk) 20:24, 15 December 2010 (UTC)[reply]

Symmetry, or intuition, has nothing to do with the error in the simple solution.Nijdam (talk) 20:48, 15 December 2010 (UTC)[reply]

Why the great mystery? Why not explain exactly what you think is wrong with the argument above? Martin Hogbin (talk) 09:03, 16 December 2010 (UTC)[reply]

I don't think there is any mystery. Nijdam thinks that the correct solution of MHP involves a) recognising that you want to compute the conditional probability, b) computing it. He is not alone in this, that is also the opinion of Rosenthal, Carlton, Morgan et al. What interests me is that none of these authorities give any reason for this, it seems to be an article of faith.

It seems to me that if there is a reason, it will depend on your notion of probability.

I will give two reasons, one for each of the main streams of thought.

Subjectivist: for the subjectivist, information comes in and beliefs about the state of the world are continually updated. If we have modelled the MHP as a four stage process going on in time (car hidden, door chosen, door opened, reconsideration of choice) and if you are a rational subjectivist then your beliefs are updated in each phase: car hidden -> 1/3,1/3,1/3 -> door 1 chosen -> 1/3,1/3,1/3 -> door 3 opened -> 2/3,1/3,0 The theory is normative. Your rational degree of belief at any stage is *the* conditional probability distrbution given *all* that you know at each stage.

Objectivist: for the objectivist, there are not normative rules coming from probability theory. But in a decision problem like MHP it is agreed that the problem is solved by finding the *optimal* strategy, and by *proving* that it is optimal. The overall probability of success is the target quantity which we want to optimize. The simple arguments of the simple solutions tell you that always switching has overall (unconditional) success probability 2/3. We have not finished "solving" MHP if we haven't mathematically proven that 2/3 can't be beaten. The only way it could be beaten, would be if there was some situation recognisable to the player (door chosen, door opened) in which the conditional probability that switching gave the car was strictly less than 1/2. This follows easily by consideration of the law of total probability: for any strategy whatsoever, the overall success chance equals the sum over the six possible situations of the probability of each situation multiplied by the conditional probability of success with that strategy, given you are in that situation.

What is so difficult to understand about this? For mathematicians, the notion of "solving a problem" is different than for lay-persons. Nijdam is right and Martin is right. Some sources are mathematicians, some sources are lay persons. Richard Gill (talk) 09:36, 16 December 2010 (UTC)[reply]

It has nothing to do with mathematicians and lay persons. I have no problem in recognising that one possible interpretation of Whitaker's question involves calculating a conditional probability (although others seem not to recognise that what Whitaker may well have wanted to know was an unconditional probability). But even given that it is a conditional probability that is required there is no need to use any specific technique to calculate it. It is a well known and perfectly respectable technique in mathematics to show that the value you wish to calculate must be equal to something else and then calculate the other think. Using a symmetry argument to do this is also perfectly correct mathematically. Do you not agree?

I agree entirely, Martin! But there is an important difference in the culture of mathematical problem-solving among mathematiclans and among lay persons. The difference lies in the customary standard of precision in logical reasoning. This includes precision in problem mathematization. If you think the question should be solved by finding a conditional probability you say so and do so, e.g. by glueing a symmetry argument before or after the simple solution. Richard Gill (talk) 03:40, 18 December 2010 (UTC)[reply]

There is no universal standard of rigour for mathematicians. As I think Boris said, the most rigorous techniques in mathematics are impenetrable to all but specialist mathematicians.

I have suggested that a symmetry argument could be added to the simple solution, so long as it is done in a simple, non-off-putting way but no one seems interested. I have also proposed that the question and simple solution are carefully worded so that they could be taken to be referring to the unconditional problem. Again, I would accept this provided that it is done 'quietly'. For example if we say things like 'players who switch...' we are talking in a way that might be referring to the unconditional case. For the average reader it is not a problem since they are unlikely to distinguish between 'all players who switch' and 'players who switch given that door 1 has been chosen and door 3 opened'. What I object to are crude and unnecessary statements along the lines of 'the solution only applies to ...' or this solution is 'incomplete' or 'does not answer the question'. These statements are more wrong than the simple solution. Martin Hogbin (talk) 16:55, 18 December 2010 (UTC)[reply]

I agree, Martin. The history of mathematics is a history of great discoveries, but also of great disasters ... where people discovered that intuition had led to terrible self-contradiction. So new rigour was sought, the paradoxes were resolved, people built up a new intuition on firmer foundations, and mathematics prospered again ... till some time later new contradictions were discovered. No reason to suppose that this process will ever stop. And I surely do hope it will never stop. Every editor who thinks there is one and only one correct answer to MHP should be banned from editing any wikipedia pages on the subject. This is exactly what my recent publications try to make clear. Richard Gill (talk) 20:19, 20 December 2010 (UTC)[reply]

Perhaps Nijdam would now give his answer. Martin Hogbin (talk) 23:40, 17 December 2010 (UTC)[reply]

[12/9/2010 Edit: For 4 months I was told the simple solutions were wrong, and that I didn't understand probability, or else I would know why it 'must' be solved conditionally. Which is exactly what the MHP article reflected at that time. This argument, of course, ignored all the reliable sources that give simple solutions. No matter, I was the dumbass. Check out my talk page for more. Glkanter (talk) 02:29, 10 December 2010 (UTC)][reply]

No, nothing's changed. All the facts are the same. No one's opinions has changed. The reliable sources have grown slightly. Still no conditionalist has taken the trouble to explain why you *ought* to condition. Only I did, in my recent papers. That is to say: I presented the argument which the conditionalists should have given straight away. By computing the conditional probability given all your information at the last moment before you must make your decision, you are guaranteed to make the optimal decision. That does not mean that there are not other ways to find the optimal decision. And some people might not even care about optimality. They just want something which is obviously damn good (2/3). It is a mathematical luxury to add to the simple solution the argument why 2/3 is not only good but optimal. Richard Gill (talk) 18:55, 11 December 2010 (UTC)[reply]

I wrote this a long, long, long time ago...

I wrote this after 4 tedious months of discussing the MHP on Wikipedia. None of the people I was arguing with had mentioned to me that the only thing that mattered was the reliably published sources. Armed with this knowledge, I have revised my estimate. Maybe 10% of the current argument article fits within Wikipedia policy, not the 5% I originally estimated. Not a single other word requires change.

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)

original diff

posted on my talk page Glkanter (talk) 20:12, 10 August 2010 (UTC)[reply]

Wikipedia is not the place for showing you are "right". If conventional wisdom is stupid and no "reliable source" says this loud and clear, wikipedia will unfortunately go with conventional wisdom. Richard Gill (talk) 18:55, 11 December 2010 (UTC)[reply]

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

[1/29/2010 Edit This is my first posting to the article's talk page on 10/25/2008...]

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)

The odds haven't changed, indeed. How do you like my Compromise X? The simple solution says that the odds don't change. The conditional solution says that the odds don't change either. The conditional solution does say a bit more than the simple solution, but possibly many people are not interested. But some reliable sources do care and are interested, so I'm afraid wikipedia will have to pay attention to them. Richard Gill (talk) 18:55, 11 December 2010 (UTC)[reply]

I am continually amazed by the lack of correlation between your responses and the words I write, Richard. I have made no claim at being 'right about the math' as to editing the article since I *finally* grasped that its all about the reliable sources. Despite that the tenor and tone of the article and the talk pages was *all* about the simple solutions being 'wrong' when I joined the fray. I am doggedly trying to eliminate the 'simple is wrong' from being the article's POV. That you bust my chops, erroneously, yet pander to Nijdam, who makes that very math claim, is very frustrating. I'm voluntarily playing Wikipedia's game, which is reliable sources, and at a minimum, a significant minority view. If you, and Nijdam were playing by the same rules, rather than each promoting your own POVs (via Gerhardvalentin) on the mediation page, things would have a chance at success. Glkanter (talk) 19:11, 11 December 2010 (UTC)[reply]

Glkanter, please stop to complain that I have asked Richard for help. Anyone who tries to read the sources but seems to not consider their particular specific angle of perspective, to consider what they are really trying to say, is in danger of misinterpreting the sources, and that exactly characterizes the existing conflict. A little more care by all editors would be appropriate and advised. Gerhardvalentin (talk) 21:38, 11 December 2010 (UTC)[reply]

I gave up promoting my own point of view, which is neither "simplist" nor "conditionalist", long ago. More recently I have given up taking part in any way in the mediation and editing of the MHP page. Why you guys spend time hanging around on my talk page is a mystery. Gerhard Valentin is his own man, if he finds some of what I wrote here attractive then he's free to use it elsewhere. And you no doubt will attack it, there. Richard Gill (talk) 19:26, 11 December 2010 (UTC)[reply]

Richard, you wrote, once again:

"The conditional solution does say a bit more than the simple solution..."

For the K & W (symmetrical) formulation, what, exactly, is this 'a bit more'? Glkanter (talk) 19:18, 11 December 2010 (UTC)[reply]

Read what I wrote in "Compromise X" and while you are at it, read my recent papers. I don't say that you have to be interested or impressed in the "extra". Richard Gill (talk) 19:22, 11 December 2010 (UTC)[reply]

We've restarted the mediation on the Monty Hall problem, with me leading the mediation alongside Sunray, and we're now making some progress. But we're being held back, partially by my own relative unfamiliarity with the dispute (although I've mostly caught up), but mostly by the inactivity of most of the parties to the dispute. Please take this as a request for you to resume participation in the mediation, which you vacated a few weeks ago. It is in all of our interests to get this dispute resolved, to the satisfaction of as many of the participants as is possible. Regards, AGK 12:30, 10 December 2010 (UTC)[reply]

I'm taking some time out. The mediation is dominated by two conflicting POV's, and especially by the extemists at each end of the spectrum, who are so certain they are right, that they do not try to understand the other POV, let alone admit to any compromise.

My own POV is another. I've said what I had to say on the mediation pages, and in two recent papers, which now belong to the "reliable sources". . Maybe I'll come back next year ;-). Richard Gill (talk) 12:21, 11 December 2010 (UTC)[reply]

As one of the extremists you repeatedly refer to to, I'd like to summarize clearly my extreme views based on the reliable sources:

(1) The criticisms of the simple solutions are given UNDUE WEIGHT in the current article, causing blatant NPOV violations

(2) The criticisms of the simple solutions do not belong in or adjacent to the Solution sections

You knowingly perpetuate this False Equivalency to Nijdam's contributions, Richard, which is intellectually dishonest, and should have ended long ago. Glkanter (talk) 12:40, 11 December 2010 (UTC)[reply]

What false equivalency are you talking about? There is a true similarity that you and Nijdam have the most extreme views among all the editors in the mediation, and in that neither of you seemed to me in the past interested in any kind of compromise. You are also similar in being intelligent educated human beings. You are similar in having absolutely no patience for the opposing point of view. You Glkanter even said that you refused to study the literature references which other editors think are relevant, because you are so certain of your own correctness.

No doubt you and Nijdam are similar in many other ways, as well as being different in many other ways too.

Since I am not in the mediation any more I don't have an opinion about anyone's currect stance on finding a compromise.

Regarding the two views which you wrote above:

(1) I agree entirely with you that the criticisms of the simple solutions are given UNDUE WEIGHT in the current article.

(2) I disagree with you that the criticisms of the simple solutions do not belong in or adjacent to the Solution sections. The "criticism" is simple and easy to understand and reliably sourced. Don't see it as criticism, see it as explanation of why some sources offer more complicated solutions. The reader doesn't have to put any weight on this, if he or she is not impressed. When motivating and explaining a conditional solution, for instance simple+symmetry, it is a service to the reader to explain the difference in what it delivers with what a simple solution delivers. The reader should be the arbiter, for him or herself, of the question whether the *difference* should be construed as *criticism*. Richard Gill (talk) 19:06, 11 December 2010 (UTC)[reply]

(1) Wonderful!

(2) This is interpretation of Wikipedia policy. Nothing more. Why have a 3rd party interject into some other reliable sources' writings? That strikes me as promoting a POV.

(3) As long as Gerhardvalentin keeps copying Compomise X into the mediation, you're involved.

(4) As long as logic and facts support my conslusions, I am under no requirements to seek out other methods.

(5) I refer copiously to reliable sources. Nijdam does no such thing. Which is the basis of Wikipedia. And the basis of my 'False Equivalency' claim. As you can't understand that, I now, once again, realize I am wasting my time with you. Beyond that, the rest of your insult/explanation is of no utility whatsoever, and is not appreciated at all. Glkanter (talk) 19:27, 11 December 2010 (UTC)[reply]

(1) Good!

(2) Say that on the mediation pages! I am not following the mediation any more.

(3) I am spending my time writing reliable sources, both on MHP and on other topics. Compromise X was an experiment, stimulated by Gerhard's remark about taking time out, rethinking, and then coming back and trying to say things in as few words as possible. There will be no Compromises from me Y, Z, ... X', X

(4) Suppose logic and facts support your conclusions, and suppose logic and facts support other people's different conclusions. Then it could be wise, since you are not the sole editor on wikipedia with interest in MHP, to try to understand them.

(5) I didn't say you are similar to any other editor in all possible respects. Yes, you do refer copiously to some reliable sources. As you know, it's my opinion that you misunderstand some of them, too. But that's just my personal opinion based on 40 years experience in learning, teaching and doing applied and theoretical statistics. Richard Gill (talk) 19:49, 11 December 2010 (UTC)[reply]

All Wikipedia editors are created equal. You ignore the False Equivalency, still. Nijdam makes no effort whatsoever to follow the reliable sources requirement, rather, he, much like yourself writes endlessly about OR. Which may explain your myopia on the issue of reliable sources. Glkanter (talk) 20:25, 11 December 2010 (UTC)[reply]

Indeed, all wikipedia editors are created equal. I have quit the mediation and the editing of the MHP page. However, my day-job is to do Own Research and to write reliable sources, in particular, on MHP. I'm concentrating on my day job at the moment.

I think Nijdam should do the same - he keeps repeating that the simple solution is wrong and the conditional solution is the only right solution, but he gives no substantive reason for this, he just reports the opinions (fatwas, papal decrees) of writers in statistics education journals, who don't seem to know the reason themselves, but were just brainwashed by their own teachers to "know" this for a fact.

I do know the reason, and I have written it up in my paper, the one which you, Glkanter, do not want to read since you don't want to know anything new. Richard Gill (talk) 18:46, 12 December 2010 (UTC)[reply]

Richard, I'm trying to find out on which points editors do agree. Please see Wikipedia talk:Requests for mediation/Monty Hall problem/Starting points. Nijdam (talk) 14:05, 13 December 2010 (UTC)[reply]

I've posted a few very brief sections on Martin's talk page that I think can actually bring about Boris' call for peaceful co-existence. They start here.

After you've looked them over, I'd appreciate it if you would, please comment in a separate section, and, once again, explain to me why I must understand the difference between frequentist and subjectivist probability in order to edit the Wikipedia article skillfully. Thank you. Glkanter (talk) 18:18, 15 January 2011 (UTC)[reply]

There's nothing I could say there which I didn't say before. I also have always called for peaceful coexistence of the two main classes of approaches to MHP, and explained many times exactly how symmetry together with the law of total probability does enable one to deduce that the conditional probability of winning by switching is 2/3, from the fact that the unconditional probability is 2/3. See for instance my "two ways to fix Devlin's solution". I have also repeatedly tried to explain exactly what the bonus is of finding the conditional probabilities, and have given several sound motivations for doing that too. At the same time I have repeatedly pointed out that most of the simple solutions given in words by ordinary writers correspond to perfectly rigorous mathematical derivation of the unconditional probability that switching wins the car. I have also repeatedly said that I felt it was a matter of taste whether one wants a conditional probability or an unconditional probability solution. However if editors aren't ready to envisage the difference between a conditional and an unconditional probability then it is pointless to talk about it.

Some reliable sources use probability in a frequentist sense. That's why editors would be well-advised to have some feeling for the difference between these two ways of thinking. It's a discussion which has been going on as long as civilization exists.

I have withdrawn from the editing of the MHP page: (1) for the obvious reason that it is inappropriate given that I just published Own Research on the topic, and (2) for the reason that I haven't any more to say to present company. I think it'll be more useful if I try to convert my insights on MHP into a popular encyclopaedia over on Citizendium, where Boris is one of the most active maths editors. When that's done, maybe the wikipedia MHP editors might find it useful. Richard Gill (talk) 18:45, 15 January 2011 (UTC). User talk: Richard D. Gill at Citizendium.[reply]

This is the last time I will bother you about the MHP. If a problem said nothing more than there are 3 doors hiding 1 car and 2 goats, from my State of Knowledge, will each door be hiding the car with a likelihood of 1/3? Glkanter (talk) 20:33, 15 January 2011 (UTC)[reply]

I remember now. As long as I use a random selection method, the door I select will have a 1/3 chance of hiding the car. So, how does that model? Are the 3 door's values undetermined until I've made my selection? After I've chosen, are the 2 other door's values still undetermined, or are they also 1/3?

What if I'm charged with making a decision, rather than calculating a probability? Can I then use 1/3 for all three doors regardless of my selection method? Will using a random method allow me to assign 1/3 to each door before I choose? How about after? Thanks. Glkanter (talk) 23:19, 15 January 2011 (UTC)[reply]

My thoughts on MHP are written up in Gill (2011), html version at my homepage.

In particular, different ideas of what probability is could make different model assumptions more or less reasonable. Moreover, the idea of probability which goes in, defines the meaning of the probability statement which comes out. Subjective in, subjective out. Frequentist in, frequentist out. But I've said all this before. I will pay attention to this kind of issue in my citizendium project. Richard Gill (talk) 12:14, 16 January 2011 (UTC)[reply]

So glad to hear that.  But in your paper (though 9/8 and 10/24) it still reads:
(1 Introduction): "His second paper Sevin (1975b) gave ..."
(2.1 What probability theory tells us, Proposition 3.): "more satisying route"
(2.2 What game theory tells us): "minimax stategy"
(Proposition 4.): Is "succesful indeed" correct?
At the end of (3 Which assumptions?): "the player has two opportunites to act"

Kind regards, Gerhardvalentin (talk) 04:32, 16 January 2011 (UTC)[reply]

Thanks Gerhard! These spelling mistakes got corrected at the page-proofs stage with Statistica Neerlandica. Richard Gill (talk) 12:11, 16 January 2011 (UTC)[reply]

Richard, I really wonder why you so firmly defend this solution. The one that reads: the player who always switches wins the car with unconditional probability 2/3. In all similar situation one has to base one's decision on the conditional probability given all known information. The unconditional solution may only be defended because the numerical answer is the same. But is it correct? The player in the situation chosen door 1 and opened door 3, can only switch to door 2, with unconditional probability 1/3 on the car. The point is that the event "switching gives the car" has different meaning for the general player (={C<>X}) and the actual player (in our case ={C=2}). It's the old song: has the player to decide before anything happens or after the host opened a door. As Morgan cs put it: it solves a different problem.Nijdam (talk) 00:38, 21 January 2011 (UTC)[reply]

Yes, it solves a different problem. But who is to say which problem is the Monty Hall Problem? The sources don't agree how vos Savant's words should be converted into a formal mathematical problem. Personally, I like the simple solution without assumptions about the host-behaviour. But don't preach to me. Explain to the simplists why you *must* use a conditional probability. More importantly, find reliable sources from probability theory which do this. Richard Gill (talk) 17:11, 21 January 2011 (UTC)[reply]

Sorry, but preaching a little seemed necessary to me, because once you wrote that the simple solution not only solved the "unconditional Problem", but also the "conditional one". Now this point is cleared, what the mathematics concerns you're in line with me, Rick, etc. The difference lies in what you like to consider the (main?) MHP. If I'm right, please make this clear to Martin, because just like you, as I think, I'm getting fed up with the ongoing nonsense I have to read. Nijdam (talk) 23:34, 21 January 2011 (UTC)[reply]

You're right, please tell Martin from me. I don't believe I ever can have said the simple solution solved the conditional problem. Simple solution plus law of total probability and symmetry does. Simple solution shows that always switching beats always staying in terms of overall (unconditional) success rate. Conditional solution shows that always switching beats every conceivable (mixed, randomized....) strategy in terms of overall (unconditional) success rate. If only we could find an alternative simple argument why you can't do better than 2/3 ... then we wouldn't need the conditional solution at all. Richard Gill (talk) 03:36, 23 January 2011 (UTC)[reply]

So, according to you, Richard:

The conditional MHP where door 3 has been opened, is solvable with: a simple solution, plus the law of total probability plus symmetry.

Is this similar to the 'tool kit' answer I offered on Martin's talk page yesterday or the day before? Glkanter (talk) 04:38, 23 January 2011 (UTC)[reply]

Richard - The mediation page is very much the wrong page for this contribution. The mediation has been bogged down for months, and (IMO) the conflict has been nearly completely rooted in individual editors insistence that the they know The Truth better than existing published sources. Perhaps that might be true in your case (I know you are an expert and publish in this field), but by fundamental Wikipedia policy articles are to be written according to verifiable published reliable sources. Can you please limit your comments (within the mediation) to sources that have already been published? Imagine prefacing anything you say with "<Somebody> in his paper published <here> says ...." This is the right mindset to have when editing Wikipedia. Wikipedia articles are akin to the literature overview section (chapter 1) of a thesis, not the original contribution part that follows. -- Rick Block (talk) 15:58, 23 January 2011 (UTC)[reply]

Honestly, I realize this is an intrusion, and I'm trying hard to be guided by the "better angels of our nature". But to call out Richard for being disruptive to the mediation, and regular discussions because of OR, while saying absolutely nothing to Nijdam regarding his lack of ever referring to a reliable source a single time (other than to denigrate them) for over 2 years is, at best comical. Shocking is what I'm really thinking, and a lot of things you'd just call personal attacks, so I won't say them. But believe me, Rick, I'm thinking them. Glkanter (talk) 16:17, 23 January 2011 (UTC)[reply]

Thanks Rick ... well when discussions take place "off" the mediation page, people complain too... Now Glkanter brings up a good point... when is someone who favours a conditional solution going to come up with good reasons *why* one "must" solve problems like this by computing a conditional probability? I am not aware of mathematics ever being a source of moral imperatives. Of course mathematics often provides reasons why it is rational to do things one way or another. It seems to me that none of the sources who (in my reading) think that the conditional probability approach is the only proper way to solve MHP give good (and reliably sourced!) reasons for this.

Glkanter's only point at this time is that the criticism of the simple solutions does not comprise even a significant minority viewpoint. And should not dominate the article as it currently does. And that arguing to the contrary for over 2 years is not based on reality. Glkanter (talk) 23:23, 24 January 2011 (UTC)[reply]

Richard, I continue to be confused how my reading of your conclusion [paraphrasing: conditional solutions are *not* required] to your 2010 paper on this topic is incorrect. Glkanter (talk) 23:44, 24 January 2011 (UTC)[reply]

That paper and its sequel argue both argue that there is not a single true correct mathematization of vos Savant's words. It is an opinion, a point of view, whether vos Savant is asking for a conditional probability or not. Moreover the "extra" of the conditional solution is merely an answer to the question whether some kind of mixed strategy (switching or not, depending on specific door numbers) could be better still, in terms of overall success rate, than the strategy "switch, regardless" with its overall success rate of 2/3. An idea which is so far-fetched that non-mathematicians may be pardoned from not paying it the slightest attention. Anyway, there are different ways to argue that such strategies cannot exist. The symmetry of the problem can be used "in advance" to say that by forgetting specific door numbers we do not lose anything.

Both papers, like Selvin's, offer a spectrum of different solutions based on different assumptions. In contrast to Selvin though, I'm careful to make a mininimum of assumptions in order to derive each solution. This shows that the different solutions are definitely incomparable, none can be said to be better or more complete than another. The simple solution requires far less assumptions so has far wider applicability than the full conditional solution. Moreover the meaning of the assumptions and hence also of the conclusions depends on how you interpret probability. People who like to think of probability in different terms will find different solutions appealing. Personally I like the approach of the economists. Richard Gill (talk) 07:05, 25 January 2011 (UTC)[reply]

As Glkanter is a repeatedly demonstrated dumbass, all that complicated stuff in your paper is over his head. But, still he's trying to demonstrate that there are very few actual critics of the simple solutions. Somehow, both you and Rick agree that statements like Rosenthal's "...is actually correct, but fails for a different problem...", supports Nijdam's POV about the 'wrongness' of simple solutions. And despite Morgan's comment about symmetry in their rejoinder (and the things about 2/3 they say in their response to Martin & Nijdam), Rick insists Morgan finds fault with simple solutions, but *not* the conditional solutions, when *given* the 50/50 host bias. There's similar confusing logic in Rick's interpretations of the only other 3 so-called 'critical' sources. So I find it (unfortunately) necessary to 'cling' to the most unambiguous statements I can find in each source. Your conclusion is one such statement, and *still* Rick, and maybe you, Richard (I've never been clear about it, even today), says your paper supports Nijdam's POV. Hence my comment. Until you say directly, 'Nijdam is wrong, and he's unsupported in the literature, or by Boris or by myself', Rick will continue with his insistence that Glkanter, being the dumbass, doesn't understand what you all are saying.

I don't support any one point of view. I believe one should take a higher view and see how the different points of view interrelate. I don't think there is anything wrong with a correctly argued simple solution. There is also nothing wrong with a correctly argued conditional solution.

I think I understand Rosenthal's and Carlton's and Morgan and buddies's and Nijdam's point of view and I believe that they all think that simple solutions are wrong because, in the opinion of those sources and that editor, they don't address the right question. I believe that Boris too thinks that MHP has to be solved by finding the conditional probability that switching will give the car. I agree with Boris and other editors that the conditional solution can be derived from the simple solution by making use of symmetry and the law of total probability. I do that myself in my two papers.

I think that Nijdam and co. ought to give references to reliable sources which give understandable motivations for their insistence (that we must compute a conditional probability). Personally I know a couple of good reasons but I have yet to see a reliable source which spells them out.

The article currently casts doubts on the simple solutions, Richard. In many ways, many times. I'm trying to focus on ending this POV from its inappropriate undue weight and NPOV violations. Your response(s) is so vague & off topic, it adds no value to that effort. Glkanter (talk) 13:41, 25 January 2011 (UTC)[reply]

I have always supported simple solutions. I also do that in my papers (now peer reviewed and published in reputable places). I'm sorry, they were written for maths PhDs, not for the public at large. My next project is to do something in the latter direction. That's a very hard job, and because I have the ambition to do it, I stay around here, talking with you folk, hoping to learn. Richard Gill (talk) 14:44, 25 January 2011 (UTC)[reply]

I know you don't care about the conditional solutions since you "know" that the simple solutions are correct and are all anyone needs to know about MHP, but still it would help the level of the discussion if you would realise that there is a difference in showing that the switcher wins two thirds of all the times, and that the switcher wins two thirds of the times that he chose door 1 and the host opened door 3. Do you agree that both statements are true? Do you agree that the second statement costs more work to verify than the first?

HOLD IT RIGHT THERE!!!!!!!! That is a false and scurrilous statement. I understand that the conditional solution is valid, applicable to more situations, and is represented fully by the reliable sources. I am discussing the criticisms of the simple solutions, I have *not* commented on the conditional solutions, other than Morgan calling them 'false' in over a year. Knock it off, Richard. Now I will read the rest of your posting. Glkanter (talk) 13:29, 25 January 2011 (UTC)[reply]

Sorry if I misunderstood you! I stand corrected. I play with open cards, I tell people how I am reading them, so that they have an opportunity to correct me. Now please answer the questions: Do you realise that there is a difference in showing that the switcher wins two thirds of all the times, and that the switcher wins two thirds of the times that he chose door 1 and the host opened door 3. Do you agree that both statements are true? Do you agree that the second statement costs more work to verify than the first?

I have corrected you countless times on this specific item. I do not understand why you keep repeating it.

I cannot answer your question as I believe the simple solutions are always given in response to the symmetrical conditional problem statement where a door has been opened to reveal a goat. I have no interest, and have never had any interest in the first problem you describe, as it is not the MHP per any source. And as I, being a dumbass, can't even use the word 'probability' correctly, it's pointless for me to even answer your question.

ps Thanks for nothing for answering my "tool kit" answer in your usual vague, tangential, non-committal fashion on a talk page far away from the actual question. By answering it this way, you have assured that your response is of no value to my efforts, and that any reference I make to it will be rebutted by Rick Block, with "Glkanter does not understand Richard's response". I should have expected this. And nothing less. Glkanter (talk) 15:12, 25 January 2011 (UTC)[reply]

Right now I am thinking about another way to bridge the gap between simple solutions and conditional solutions. Once anyone of average intelligence has caught on to the simple solution, they feel absolutely no need to go into mathematical contortions in order to do something more. One has a strong instinct that this should be the end of the story. I share this instinct. However as a mathematician it's my job always to question instinctive solutions. The instinct is that you cannot do better than an overall success chance of 2/3; any mixed strategy of sometimes switching, sometimes not, depending on the door number chosen and the door number opened cannot be as good. The instinct is that the specific door numbers are irrelevant, all that counts in making a decision is the role or function of the three doors (chosen by player, opened by host, left un-opened by host).

Well, as a mathemtician I can say that this instinct is completely correct. The relation between the door hiding the car and the three doors identified by their role (chosen, opened, not opened) is independent of how the latter three are labelled. This is intuitively obvious and mathematically true. Therefore an optimal solution can be found by considering only the probabilistic relation between the doors identified by role (hiding car, chosen, opened, not-opened). And that is what the simple solution does.

In short, instead of giving the simple solution and then using symmetry to derive from it the conditional solution, one can go the other way round: use symmetry to argue that specific door numbers are irrelevant, and then give the simple solution, which (as Boris so carefully argued) is indeed optimal among all strategies which ignore door numbers.

Yes, I wrote about this approach last week, referring to it as 'reducing the problem'. I thought that maybe that was the 'one and only acceptable appropriate sequencing of words' that the simple solutions were missing. Apparently not. Glkanter (talk) 13:41, 25 January 2011 (UTC)[reply]

Yes, you can call this "reducing the problem". I agree that that it is a smart way to "upgrade" a simple solution to as full a solution as anyone could wish for. I wish I had worked it out before, written it up, and included it in my papers. I always knew it could be done, but never sat down to figure out the details. To a good mathematician it is obvious that one should be able to tackle MHP this way. Pity no one ever did it in a reliable source before. Richard Gill (talk) 14:37, 25 January 2011 (UTC)[reply]

However, this is OR, hence useless for wikipedia. But I will take care to include this alternative justification of the simple solution in my citizendium article. On citizendium, articles are written by acknowledged experts and are subject to expert peer review. So once that article has gone through the review process it will be a reliable source. I am looking forward to discussing this alternative approach with Boris. Richard Gill (talk) 13:22, 25 January 2011 (UTC)[reply]

Would you *please* comment on my "tool kit" description on Martin's talk page? Thank you. Glkanter (talk) 08:56, 25 January 2011 (UTC)[reply]

I don't know what you are referring to. Could you give me a link to the page and topic? Richard Gill (talk) 13:22, 25 January 2011 (UTC)[reply]

My bad. It's on the mediation page. Glkanter (talk) 13:41, 25 January 2011 (UTC)[reply]

It's in this section.

Ah, you mean sand-box, not tool-kit! Right. I have answered all questions now. Richard Gill (talk) 14:37, 25 January 2011 (UTC)[reply]

In United States English, a 'sand box' is a non-production (private) environment where one tries out new ideas, or practices existing skills. A 'tool kit' is what a professional takes on the job to solve a problem. So, no, I do *not* mean sandbox. Not in any way, shape, or form. Would it be too much to ask you to discuss the merits of my comments that include the term 'tool kit' immediately below those comments, as is customary? Glkanter (talk) 15:22, 25 January 2011 (UTC)[reply]

Anyway, right now I am searching my mind for ways to use symmetry in MHP which avoid the bother of introducing conditional probability. I'd like there to be a simple solution to MHP which is intuitively appealing and mathematically rigorous and does not involve technical notions which most ordinary people have no feeling for, and which "reconciles" the simple and the conditional solutions in the situation where we assume no host bias.

I think I'm getting very close. Consider the following.

There are two ways to refer to the three doors: by their names written in advance on them, "1", "2", "3"; and by their function or role: "door chosen by player" X, "door opened by host" H, "door left closed by host" Y. One of which is the "door hiding the car" C. By symmetry, the names written in advance on the doors are irrelevant, that is to say, an optimal strategy depends only on the role of the doors, not on their names. We know that when we list the doors by role, the probability each hides the car is 1/3, 0, 2/3 respectively. Hence the optimal strategy is to switch. Full stop (as far as the layman is concerned, I should think).

But the mathematical purist might like to see some further elaboration of this way of thinking. That's what my remark over on the mediation page does. Taking the player's choice as uniform random, by symmetry (X,H,Y) is a uniform random permutation of (1,2,3) while (C,H,G) equals (X,H,Y) or (Y,H,X) with probabilities 1/3 and 2/3 independently of (X,H,Y). Here, G, the "goat-door not opened by the host", is the third door alongside of C and H.

This says exactly (that is to say, in precise and formal mathematical terms) what everyone feels intuitively - you cannot do anything with the specific door numbers. The relationship between the (unknown to the player) door hiding the car C and the three doors identified through their known roles X,H,Y (I mean, known to the player) does not depend on what numbers, names 1,2,3, they happen to have on any particular occasion.

I can't be an editor of the wikipedia MHP page any more, but I remain a seeker after the truth. "Truth will out" - eventually the ring of truth will prevail. Richard Gill (talk) 23:12, 24 January 2011 (UTC)[reply]

1. Nijdam is *correct* that 'all simple solutions fail to solve the symmetrical conditional MHP'.

Correct. The simple solutions do not solve the symmetrical conditional MHP, by which he means the problem to find the so-called conditional probability, not the unconditional probability, that switching will give the car. Richard Gill (talk) 15:04, 25 January 2011 (UTC)[reply]

Therefore all reliable sources that use a simple solution are *wrong*.

That's what Nijdam thinks, because he thinks that you *have* to solve MHP by finding the conditional probability. I disagree. Richard Gill (talk) 15:04, 25 January 2011 (UTC)[reply]

The 5 critics of the simple solutions are *correct*.

Their criticism is poor. They don't give much reasoning, if any, why the problem should be solved by computation of a conditional probability. They tend to prefer complicated ways to doing the job to elegant ways. Fortunately we now know a multitude of elegant ways to getting the conditional solution, all of them exhibiting valuable mathematical or probabilistic ideas (symmetry, odds form of Bayes' rule, and so on). Richard Gill (talk) 15:04, 25 January 2011 (UTC)[reply]

2. Nijdam is *wrong* that 'all simple solutions fail to solve the symmetrical conditional MHP'.

Therefore all reliable sources that use a simple solution are *correct.

Not all published solutions, whether aimed at unconditional or conditional probability, are logically sound. For instance, Devlin seems to be trying to find the conditional probability but leaves out a tricky step in his proof. Later he seems to have caught on. Oh well, he works in artificial intelligence and linguistics, not in statistics and probability. Richard Gill (talk) 15:04, 25 January 2011 (UTC)[reply]

The 5 critics of the simple solutions are *wrong*.

They present their opinions. They don't present arguments for the moral or rational imperativeness of their opinions. Richard Gill (talk) 15:04, 25 January 2011 (UTC)[reply]

3. Nijdam is *wrong* that 'all simple solutions fail to solve the symmetrical conditional MHP'.

Therefore all reliable sources that use a simple solution are *correct*.

Not all published solutions, whether aimed at unconditional or conditional probability, are logically sound. For instance, Devlin motivates his computations by an essentially sound simple solution ("the host is offering you the choice between one door and two doors together") but then goes on a different tack by apparently trying to deduce the conditional probability. Richard Gill (talk) 15:04, 25 January 2011 (UTC)[reply]

The 5 critics of the simple solutions are *not really critics at all*.

They all have the POV that MHP is not solved by a simple solution. That's their POV and they are all maths PhD's or even professors too... not that that means much, in this day and age. Richard Gill (talk) 15:04, 25 January 2011 (UTC)[reply]

Your response is full of shit, Richard. Nijdam's POV is unambiguous and rigid. For the expressed purposes of discussing and editing the MHP, my 3 options are wholly appropriate in their binary form and lack of nuance.

Thanks, once more, for wasting my time for well over a year. I now know *not one single thing* more about how best to edit the MHP from reading and discussing this with you at length. Of course, that's my fault, for being a dumbass. I am amazed and now enlightened. Glkanter (talk) 15:34, 25 January 2011 (UTC)[reply]