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]

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 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).

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 completely neutral way. I could publish as a footnote to my paper in Statistica Neerlandica and it would be a reliable source, as well. Richard Gill (talk) 18:46, 11 December 2010 (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]

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?

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]

[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?

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]

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

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 o 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:

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

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]