User talk:Martin Hogbin/Monty Hall discussion

Moved here from my talk page.

New example

To make you understand the error in the simple solution, I give you a new example.

My son will paint the wall in the living room in his new house. He still has to decide between three colours: White(W), light green(G) or light yellow(Y). He will need 10 liters paint, costing: W: 30 euro, G: 40 euro and Y: 45 euro. I have offered to pay for the paint. How much will I have to pay? The only answer will be: if he uses W then 30, etc. What if he decides at random? Would you give a different answer? You could of course calculate the expected amount I will have to pay, but would you consider that the correct answer. But after he has painted the walls, what then? Don't you agree that the answer is still: if he uses W then 30, etc? Now what if all the colours were of the same price? Would that make any difference of 30 euro? The only difference is that I'm sure I'll have to pay 30 euro. But not because this is the expected amount!! Nijdam (talk) 01:13, 1 January 2011 (UTC)

Nijdam, I a very happy to talk to you about this subject but it would help if you did not keep refusing to answer because you claim have already told me the answer. Maybe you did not explain the right thing or maybe you did and I did not understand. Just eplain again, it is much simpler.

I am not sure which point of disagreement your analogy above refers to. Could you explain please. Martin Hogbin (talk) 11:27, 1 January 2011 (UTC)

I don't know which question I seem to refuse to answer. The example above shows that the answer is always on the condition of the chosen colour; even when all the prices are the same (symmetric case). I.e in the MHP the answer should count for the situation the player is in, even in the symmetric case, hence calculate the conditional probability. I have on several occasions shown why the simple solution (reasoning) (as per Savant, Devlin etc) is wrong. Do you in the mean time understand this? Nijdam (talk) 12:30, 1 January 2011 (UTC)

Let us go back to the MHP then. Do you agree that in the symmetric case the probability that the car remains behind door 1 is unchanged when the host opens a door? Martin Hogbin (talk) 12:34, 1 January 2011 (UTC)

We've been through this before: the probability is changed of nature, from unconditional to conditional, both with the same value for the event of door 1 hiding the car. They really are different probabilities, based on different sample spaces. This is, I think, the only point you have to understand. Nijdam (talk) 13:16, 1 January 2011 (UTC)

A HAPPY NEW YEAR to all of you! W.Nijdam: "P has changed of nature, from unconditional to conditional, both with the same value." Does that mean that the unconditional value however wasn't correct (enough)? Or that the unconditional value was just "nonproven", that means "the mathematical proof" had still been missing, and so it had to be checked and it was found to be correct? Or quite another meaning, like transubstantiation? Who can clearly expound that "difference in nature", graspable enough for the reader? Regards, Gerhardvalentin (talk) 14:33, 1 January 2011 (UTC)

OK, so the numerical value of the probability that the car is behind door 1 is unchanged by the host opening a door in the symmetrical case. Agreed?

You seem to be giving some additional properties to the term 'probability'. As I understand it, probability is a number, a scalar, it has no dimensions or components it has only a numerical value. Is this not correct?

You also seem to be saying that any value of probability has associated with it a unique sample space. I do not agree, there may be more than one sample space that could be used to determine any required probability. Martin Hogbin (talk) 14:18, 1 January 2011 (UTC)

3 apples are different from 3 pears. Two probabilities can have he same value without being the same thing. Saying that the probability the car will be behind the other door is 2/3, is different from saying that the probability the car is behind Door 2 given Door 1 was chosen and Door 3 opened is 2/3. If you are a frequentist you are referring to two thirds odd different ensembles of repetitions. If you are a subjectivist you are referring to acceptable betting odds in different situations. What Nijdam has so far failed to explain to us is why you *must* solve MHP with conditional rather than unconditional probability. I am not aware of any moral or ethical content of mathematics, so I don't understand how he thinks mathematics can dictate what we should do. However, I did offer two motivations, one for the subjectivists, one for the frequentists, as my Xmas present to you all. Richard Gill (talk) 15:37, 1 January 2011 (UTC)

We wish to calculate numerical value of the conditional probability that the car is behind door 1 given that the player chooses door 1 and the host opens door 3 to reveal a goat. We first calculate the value of the probability that the car was originally placed behind door 1. This is 1/3 we all agree.

We then show, using a symmetry argument, that the numerical value of the conditional probability that the car is behind door 1 given that the player chooses door 1 and the host opens door 3 to reveal a goat must be equal to the value of the probability that the car was originally placed behind door 1. This gives us the answer to the question that we wished to know.

So, yes, I do and have always understood that the initial probability is a different concept from the conditional probability but, as I simply wish to calculate its value, this fact does not trouble me.

We thus use the easily calculated value of one thing to calculate the value of another, harder to calculate, thing, which is what Gerhard says above. There seems to be an attempt to add a non-existent layer of complexity to this problem.

The question that we are answering is 'what is the conditional probability that the car is behind door 1 given that the player chooses door 1 and the host opens door 3 to reveal a goat'. By this is meant 'what numerical value does this probability have?', not, 'What does this concept mean?'. We are therefore free to use any method that suits us to obtain the correct answer. Martin Hogbin (talk) 16:08, 1 January 2011 (UTC)

We're going round in circles. This has all been explained to you. As you say yourself: we want to calculate the conditional probability. That's all that counts. The way this is performed is unimportant. Using the symmetry does not turn this into the simple solution. The simple solution simply does not calculate any conditional probability, and moreover uses a logically wrong argument. Nijdam (talk) 16:34, 1 January 2011 (UTC)

The simple solution does calculate the correct value for the required conditional probability. In what way is this incorrect? Martin Hogbin (talk) 16:49, 1 January 2011 (UTC)

No Martin, do not say things without verifying. I did say just 4 lines above, that the simple solution does not calculate, not intends to calculate any conditional probability. So, why then do you write this?? Nijdam (talk) 17:25, 1 January 2011 (UTC)

Important sign: symmetry. Did I get that correct:
The simple solution gives the correct value for the required conditional probability, as long as that conditional probability is correctly based on symmetry? Gerhardvalentin (talk) 17:06, 1 January 2011 (UTC)

Sorry Gerhard, you're sympathetic, but you do not understand much of probability theory, and mutatis mutandis, not much of the MHP. Nijdam (talk) 17:27, 1 January 2011 (UTC)

The simple solution tells us that (unconditional, overall) Pws is 2/3. Symmetry (in the symmetric case) tells us that the six conditional Pws's are all equal. The law of total probability tells us they must all equal the unconditional Pws, hence are all equal to 2/3. This shows us that always switching not only beats always staying (overall successrate 2/3 vs 1/3) but also that it beats any mixed strategy. If you just want to compare "always staying" with 'always switching" the simple solution is quite enough. If you want to show that "always switching" beats everything, you will need to look at the conditional probabilities. Which are not difficult to find, thanks to symmetry. On the other hand, no-one in their right mind would ever imagine that mixed strategies make any sense at all, so going the full Monty and proving optimality of "always switching" could be considered an academic luxury by non mathematicians. On the other hand, every mathematician knows that the standard and foolproof way to find the optimal solution is by conditional probability, and by symmetry, this conditional probability is child's play to find. So it is all not a big deal, as long as no-one insists either on being excruciatingly pedantic, nor on only using language which their great-grandmother who never went to school could understand. Richard Gill (talk) 19:00, 1 January 2011 (UTC)

Nijdam, here is one of the 'combining doors' simple solutions given in the article:

"Monty is saying in effect: you can keep your one door or you can have the other two doors." The player therefore has the choice of either sticking with the original choice of door, or choosing the sum of the contents of the two other doors, as the 2/3 chance of hiding the car has not been changed by the opening of one of these doors. [my bold]

No, Monty is saying: You have chosen door 1, I have shown the car is not behind door 3, would you prefer door 2?Nijdam (talk) 00:57, 9 January 2011 (UTC)

It would seem to me that the intent of this statement is to answer the question of what the probability of the car being behind the remaining door would be 'given that the host has opened either unchosen door (door 2 or door 3) to reveal a goat'. Would you not agree? Martin Hogbin (talk) 18:39, 1 January 2011 (UTC)

The combined door solution, interpreted in the right way, is equivalent to a form of the simple solution. It may be interpreted as calculating the unconditional probability and hence addressing the "unconditional" form of the MHP (if this may be called MHP anyway), i.e. the form in which the player is asked to switch, even before she has made her initial choice. I would much prefer if you react on my comment of 17:25. Nijdam (talk) 19:02, 1 January 2011 (UTC)

I do not understand what you mean by, 'The simple solution simply does not calculate any conditional probability'. Which simple solution do you mean? Martin Hogbin (talk) 21:38, 1 January 2011 (UTC)

You did read the article, did you? Nijdam (talk) 23:12, 1 January 2011 (UTC)

The article has several simple solutions. Do you mean vos Savant's solution as given in the article? Martin Hogbin (talk) 00:36, 2 January 2011 (UTC)

Martin wrote:

"I do not understand what you mean by, 'The simple solution simply does not calculate any conditional probability'. Which simple solution do you mean?"

Response from Richard

Let me try to answer.

The quizmaster will certainly open door 2 or 3 and reveal a goat. So conditioning on that is not conditioning at all. The probability the car will be behind the other door to the door opened by the quizmaster is 2/3. On the other hand, the events "quizmaster opens door 2", and "quizmaster opens door 3", are not certain. In fact they're complementary. It's true that 2/3 = P( car behind other door) = P( car behind door 3 | host opens 2 ) x P( hosts opens 2 ) + P( car behind door 2 | host opens 3 ) x P( host opens 3). That is an application of the law of total probability. We need to do some more work to find out the value of P( car behind door 3 | host opens 2 ) or of P( car behind door 2 | host opens 3 ). For instance, an appeal to symmetry and some careful thought is enough to convince us that in the symmetric case, both these conditional probabilities are 2/3. However we cannot say that "the simple solution computes them". More brain work is needed to deduce their values.

So the answer to Martin's question is "Yes" to the question of the probability given the host has opened either door (2 or 3) (conditioning on a probability 1 event), but "No" to the question of the probability given that the host has opened door 2, and also "No" to the question if the probability given the host has opened door 3.

It all depends on whether the "or" between door 2 and door 3 is part of the description of one event, or if it actually intends us to talk about two probabilities given two different events. Ordinary language can be ambiguous. That's why one has to be very careful in formulation of statements about probabilities. Unfortunately these subtleties can be completely lost on some amateurs and totally confuse others (I do not refer to present company). Writers about MHP have to know more about probability than their intended readers. Richard Gill (talk) 09:21, 2 January 2011 (UTC)

But, in practice, as demonstrated by the current Wikipedia article, the only way to communicate all this to the reader relies on changing the problem by eliminating the 50/50 host premise (!), and then arguing that hypothetically, this might in some way make a difference. What a bunch of BS. Glkanter (talk) 10:17, 2 January 2011 (UTC)

@Glkanter, we both agree that the current Wikipedia article is overweight and biased. And what I wrote above does not introduce or eliminate any controversial premises. It offers a short and I hope clear argument why both unconditional and conditional probabilities are 2/3 under the usual symmetry assumptions. Hence that conditioning on a-door being opened gives the same answer as conditioning on a specific door being opened. Do you understand? Richard Gill (talk) 08:26, 4 January 2011 (UTC)

Richard, thanks for your response. I am not sure if the point you make is the one that Nijdam is making (perhaps, Nijdam, you could say whether Richard's point is the one you are trying to make). I have suggested that we add some extra (unobtrusive) text to the simple solutions to make clear that, for the conditional probabilities, 'More brain work is needed to deduce their values'. Nijdam has not supported this proposal, so I assume the point that he is making must be a different one. Martin Hogbin (talk) 10:32, 2 January 2011 (UTC)

Actually Richard I think your point does have a weakness, which leads to the same mistake that Morgan originally made, and this is to treat the three unknown distributions (producer's initial door choice for the car, player's initial door choice, and the host's door choice) differently. By starting with, 'Let's fix the player's initial choice as door 1' you have already made the decision that the car must be randomly placed, otherwise we could not fix the player's initial choice. Having made this decision you should logically be forced to make the same decision for the other unstated distributions, specifically the host door choice. Having done this, it no longer makes much sense to refer to door numbers at all. If we can fix the player's original choice we can ignore the host's door choice. I am not saying that you are wrong but the way you have put things can easily lead you into an erroneous line of thinking. Martin Hogbin (talk) 10:44, 2 January 2011 (UTC)

I agree that I am assuming that the car's location is uniform random. This could be a frequentist's given (because we know the quiz-team hides the car at random) or a subjectivist's expression of indifference to the door numbering. I disagree whether the quiz-master's choice should now be taken as uniform random or not. This will depend on our frequentist/subjectivist stance. But in either case it doesn't matter, as Morgan et al. show and as Gerhardvalentin continually reminds us: it is never disadvantageous to switch, and often disadvantageous not to, even if there is any (unknown) host-bias (provided the location of the car is uniform random! But otherwise not necessarily).

I could have written out the argument keeping the initial choice random, and not even necessarily uniformly random; Nijdam would like that. It just costs longer formulas with six instead of two configurations to be considered. The argument goes through unchanged. Again: if there is no host-bias and the location of the car is uniform random, all six conditional probabilities of winning by switching given door chosen by player and door opened by host are equal, hence equal to their weighted average, 2/3. Note that in all six conditional probabilities we have conditioned on the player's initial door choice. So it does not make any difference how that is chosen (uniformly or not, random or not). All of the 6 conditional probabilities are determined "host-side". The player's choices do not influence them. So we only need symmetry on the host side to get equality. The 6 weights do depend on player-side and host-side probabilities together. But it doesn't matter if they're equal or not. They add up to 1 anyway.

I agree that if everything is uniform then we have full symmetry and we can argue in advance that specific door numbers are irrelevant. From that point on we can follow the simple solution and we know it is optimal since we know (and said in advance) that there is no way to improve it.

I also have frequently pointed out that if we know nothing, and don't wish to use subjective probability, then a wise course of action is to choose our door initially uniformly at random and then switch. The simple solution tells me my (unconditional) success chance is 2/3 and the minimax theorem shows this can't be beaten. No assumptions at all are made on host behaviour nor on our knowledge or ignorance concerning host behaviour. Whatever the host does, we have our 2/3 guarantee of going home with the car. And because we chose our door at random and *not* by our lucky number, we won't feel foolish when (1/3 of the time) we switch and lose. By the way, it we are planning to switch anyway, and if we believe in our powers of extra-sensory perception, we should make our initial choice by our *unlucky* number: we should pick the door least likely to have the car behind it.

It's all written out in Gill (2011). I'm already getting fan-mail from colleagues all over the world for my fresh and refreshing take on MHP, which is of course gratifying for me! Richard Gill (talk) 09:01, 4 January 2011 (UTC)

Further response to Nijdam

Nijdam, is the point you are making that the simple solutions given in the article do not make explicit that they are intended to be solutions to the conditional problem, in particular, they do not mention door numbers. Is this a good description of your point or has Richard understood it better above? Martin Hogbin (talk) 10:45, 2 January 2011 (UTC)

Well, Richard have said almost all. Of course he did not give a complete analysis of the MHP, he just explained you what was needed for you to understand the difference between the unconditional and the conditional solution. I think he did not want to dwell by the necessity of some assumptions. Instead of trying to understand what he wants to show you, you write about the distributions, what has nothing to do with the point you have to understand. Nijdam (talk) 20:56, 2 January 2011 (UTC)

What puzzles me is that, given that the simple solutions do not compute a conditional probability without additional thought, you have not supported my suggestion to add some text to them to show how, with a little additional very intuitively obvious logic, the conditional probability can be obtained from the simple solutions. Martin Hogbin (talk) 23:41, 2 January 2011 (UTC)

Have some patience, I said already I come to this. For the moment, your deduction is logically false. The simple solution may be obtained from the conditional, but not the other way around. The conditional solution cannot be obtained from the simple solution (as such of course). The simple solution, extended with some extra argument may give the conditional solution, of course. With the right extension it may give any solution, even the proof of Pythagoras' theorem. Nijdam (talk) 10:17, 3 January 2011 (UTC)

So exactly what additional argument do you consider to be the minimum necessary to make the simple solutions solve the conditional problem? Martin Hogbin (talk) 11:31, 3 January 2011 (UTC)

Let me guess Nijdam's response: it will be that the simple solution never solves the conditional problem. The simple solution sets out to compute the unconditional probability, full stop. Nijdam is using the word "solve" in a more narrow sense than you, Martin. This is the difference between mathematician's and, say, physicist's or engineer's language. Richard Gill (talk) 09:22, 4 January 2011 (UTC)

I think you may well be right Richard, but the reason has nothing to do with mathematicians and physicists. To show that two calculations must give the same result and then use the easier method of calculation to obtain the desired result is a well known and well respected mathematical technique. In fact it is probably used by mathematicians more than anyone else. To use a symmetry to show the equivalence of two calculations is particularly common, indeed it was a mathematician, Boris, who first pointed out the symmetry. Neither solution is that rigorous, as Boris also pointed out.

To stick resolutely on this minor point is not, in my opinion, in the spirit of cooperative editing, bearing in mind that we can discuss the whole subject in detail later on in the article. There are some her who do not want any additions to the simple solution but I am trying to find an acceptable compromise. Martin Hogbin (talk) 10:34, 4 January 2011 (UTC)

I like the compromise you suggested, Martin. And anyway, wikipedia is about reporting what is "out there" in reputable published sources, not about what any particular editor believes to be the truth. But to do good editing it helps to understand the mathematical truth and to understand the different cultures of "problem solving" within professional mathematics and in ordinary recreational puzzle-solving. It would be interesting to learn what Martin Gardner thought of conditional versus unconditional solutions to MHP since he is someone who probably understands both worlds and can communicate in both worlds and between them. Richard Gill (talk) 13:04, 4 January 2011 (UTC)

I do not like the way you talk. I have repeatedly explained to you, and in fact Richard just some lines above says the same, that the simple solution SETS OUT TO CALCULATE THE UNCONDITIONAL PROBABILITY. Hence (simple logic) does not calculate the needed conditional probability. Hence is wrong. Symmetry has nothing to do with this. The way things are calculated is UNIMPORTANT. It is not about HOW things are calculated, but WHAT is calculated. Do you have difficulty in understanding this?? Cooperative editing from your side would be understanding, accepting this! And once and for all, never mention it again. Nijdam (talk) 11:56, 4 January 2011 (UTC)

But it is not clear, Nijdam, that MHP *must* be solved by conditional probability. It is not even clear what probabilistic assumptions should be made, if any. In fact, different sources make different assumptions and derive different "solutions" based on those assumptions. I am not aware of any reliable source on the mathematical statistical side, except perhaps for Gill (2011), which explicitly gives a rationale why we ought to go for a conditional solution. Different sources give different reasons for switching. As I point out in my article, the simple solution gives a good reason for switching under minimal assumptions. The conditional solutions give better reasons for switching but under stronger assumptions, hence are more restricted in scope. MHP is not the exclusive property of one particular branch of science. It is in the public domain. All interpretations and all logically correct analyses are legitimate. Richard Gill (talk) 13:04, 4 January 2011 (UTC)

Please Richard, let's not mix things up. I appreciate your comment above, and I'll try to respond to it, but has no connection to the point Martin is making over and over, although I explained him there is no point. Symmetry may be used to calculate the conditional probability in an easy way, but the intention has to be to calculate such a probability. As you said yourself does the simple solution not calculate the conditional probability. So if one wants to use the symmetry, he is not busy with the simple solution. I'm really getting tired of repeating this for the I don't what time. Nijdam (talk) 17:36, 4 January 2011 (UTC)

You are right Nijdam, we should separate the separate issues.. In my defense let me just say that Martin objected to me "fixing" the player's choice, saying that this implied I was making certain assumptions on the host-side. I needed to respond to that side-issue. Part of my response was to point out that your insistence on solving MHP by conditional probability, though certainly shared by many writers from statistics and probability, is not universally accepted either. All solutions seem to me to be relative to what is felt to be an adequate formalization of the question and what is felt to be adequate answer in different scientific cultures. Just because MHP often features in introductory statistics texts, and possibly was born there, does not mean that it is "owned" by the priests of a particular cult. And if the priests of a particular cult think they have something important to say about MHP which most people miss, they have to give good reasons why people should listen to them. (I think there are good reasons to promote the conditional solutions, but we are not doing very well in getting these across to the non-specialists. At the same time, I think the simple solutions are also of greater importance than the promoters of those solutions seem to believe, since they make weaker assumptions hence have wider applicability). Richard Gill (talk) 18:29, 4 January 2011 (UTC)

Acceptable formulation

Due to the obvious symmetry in the problem (NB we have to explain the standard assumptions before), the probability for the chosen door 1 to hide the car has the same value 1/3 before the player makes her choice and after the host has opened door 3. Hence at the moment of decision the remaining door 2 has probability 1-1/3=2/3 to hide the car. Nijdam (talk) 12:21, 4 January 2011 (UTC)

Would you object to this being put in more reader-friendly and less obtrusive language? We are talking about the 'Simple solutions' section, which is written for the general, non-expert, reader. The whole thing can be explained later in full detail. Martin Hogbin (talk) 13:18, 4 January 2011 (UTC

Is this more reader friendly?: As the car is placed randomly, the car is with probability 1/3 behind the chosen door. It may be intuitively understood, and it can be proven, that the host opening a door with a goat does not influence this value, hence after the door with a goat has been opened, the car is also with probability 1/3 behind the chosen door. As the only alternative is the remaining closed door, the car is behind this door with probability 2/3.

Martin, I do not agree with your criticism. And +1 to W. Nijdam. I fully appreciate his proposal as unequivocal and clear in its content. Reader-friendly formulation is secondary. Gerhardvalentin (talk) 13:36, 4 January 2011 (UTC)

Can we please take this to the mediation page? There's more to object to here than how reader-friendly the wording is, and I'd really rather discuss it there than here. -- Rick Block (talk) 16:32, 4 January 2011 (UTC)

Please do. And for the wording: I gladly leave this to a native speaker. Nijdam (talk) 17:25, 4 January 2011 (UTC)

Here, by courtesy of Martin, we are discussing The Truth. On the mediation page people discuss how to resolve editorial conflicts concerning the MHP page. I don't see anything wrong in side discussions which help interested editors to better understand what the sources are talking about, and the different forms which are used in different scientific cultures. Richard Gill (talk) 18:10, 4 January 2011 (UTC)

Ah, The Truth. So is the "Monty forgets" variant symmetric? My point is that symmetry by itself is not ensuring that the "obvious probability", the prior probability P(car behind player's door)=1/3, is the same as the "unconditional" posterior probability, i.e. P(car behind player's door|the host has opened a goat door) - but is rather ensuring this "unconditional" posterior probability is the same as any of the "conditional" posterior probabilities, e.g. P(car behind player's door|player picked door 1 and host opened door 3). I believe Martin's understanding of the MHP is that the crux of the issue is whether the prior probability P(car behind player's door) is the same as the posterior probability P(car behind player's door|the host has opened a goat door) but Nijdam's understanding is that the crux of the issue is whether the prior probability is the same as a specific conditional probability such as P(car behind player's door|player picked door 1 and the host opened door 3). If the reason people think the answer is 50/50 is because they know with certainty that the "probability" the car is behind door 3 is 0 after the host has opened this door, then (IMO) Nijdam's understanding is the correct one - since the posterior "unconditional" probability P(car behind player's door|the host has opened a goat door) is identically the same as the prior probability (because "the host has opened a goat door" is a null condition). With this null condition, there is no paradox - the probability of all doors starts and ends 1/3 (including doors 1 and 3 in the example case where the player picks door 1 and the host opens door 3!). Only with a condition are we faced with the prospect of looking at two closed doors and an open door showing a goat. -- Rick Block (talk) 19:46, 4 January 2011 (UTC)

Let X be the number of the door chosen by the player, let H be the number of the door opened by the host, and let Y the number of the remaining door. Let C be the number of the door hiding the car. Prob(C=Y)=2/3 because C is never H, thus the events {C=X} and {C=Y} are complementary, and therefore Prob(C=Y) = 1-Prob(C=X) = 1-1/3 = 2/3. This is a version of the simple solution, correctly giving 2/3 as the unconditional probability that switching gives the car. I did not condition on anything (in the probabilistic sense) to deduce this fact. So I don't understand what Rick means by saying "only with a condition are we faced with the prospect of looking at two closed doors and an open door showing a goat". Richard Gill (talk) 00:28, 5 January 2011 (UTC)

To make sure everything is understood in a proper way, I want to remark that this remaining door Y (a random variable) is not the same as the remaining door in the (our) MHP. In our example i.e. is door No. 2 the remaining door (a fixed number).Nijdam (talk) 13:12, 5 January 2011 (UTC)

W. Nijdam: "Y is a random variable". But you can read "Prob(C=Y) = 1-Prob(C=X) = 1-1/3 = 2/3" also as follows:
Prob(C=y) = 1-Prob(C=x) = 1-1/3 = 2/3, that means:

Prob(C=door#2) = 1-Prob(C=door#1) = 1-1/3 = 2/3 (is it that what you want to know?) – Other possible variants:
Prob(C=door#3) = 1-Prob(C=door#1) = 1-1/3 = 2/3 (if door #2 has been opened)
and so on. Gerhardvalentin (talk) 17:05, 5 January 2011 (UTC)

Sorry Gerhard, this all makes no sense. Indeed is, as Richard wrote: Prob(C=Y) = 1-Prob(C=X) = 1-1/3 = 2/3" , but what you make out of it is rubbish. On the contrary is P(C=y)=1/3 for y=1,2,3. Furthermore is P(C=2)=1-P(C=1 or C=3). So, what do you want to say? Nijdam (talk) 21:11, 5 January 2011 (UTC)

Door#3 is open now, showing a goat, and P(C=door#1) is unchanged 1/3. Gerhardvalentin (talk) 22:12, 5 January 2011 (UTC)

Well, Gerhard, believe it or not, also P(C=door#2)=P(C=door#3)=1/3. Nijdam (talk) 21:33, 7 January 2011 (UTC)

Please help me get what you are saying. Things have progressed, meanwhile the guest did choose door#1 out of three doors, and then the guest sees the host opening another door. And he sees that the host just has opened door#3 showing a goat, that's the actual situation now. Are you really saying that at this point the probability that the car is behind the open door#3=1/3? Thank you for your help, W.Nijdam. Gerhardvalentin (talk) 13:51, 8 January 2011 (UTC)

Symmetry on the host-side (location of car, door opened revealing a goat) ensures that all six conditional probabilities

Prob( Car is behind "other door" | player chose Door x, Host opened Door y ),

for x, y any pair of different door numbers are equal. Thus

2/3 = Prob(Car is behind "other door")

= sum over all x,y Prob( Car is behind "other door" | player chose Door x, Host opened Door y )

times Prob( player chose Door x, Host opened Door y )

by the law of total probability, since the six events { player chose Door x, Host opened Door y } are mutually exclusive and exhaustive. Hence, calling the common value of the six conditional probabilities q,

2/3 = q times sum over all x,y Prob( player chose Door x, Host opened Door y ) = q times 1 = q

using again the fact that the six events { player chose Door x, Host opened Door y } are mutually exclusive and exhaustive.

The point I am making is that the symmetry argument which allows us to deduce the conditional probabilities from the unconditional makes explicit use of the fact that Monty never opens a door revealing a car; note that we use the law of total probability as well as symmetry. If you want to use symmetry to solve the forgetful Monty problem, you will have to make quite a few adjustments to the computations I just gave.

The symmetry argument is easier to write out when we first fix (or equivalently, condition on) the door chosen by the player. We only have to consider a partition of the sample space into two events then, instead of six. Richard Gill (talk) 00:16, 5 January 2011 (UTC)

It's not 2/3 & 1/3 if there are 4 doors, and Monty reveals 1 goat. All premises are required, equally. No more, no less. Glkanter (talk) 20:04, 4 January 2011 (UTC)

The unconditional probability that switching gives the car is 2/3 if and only if the original choice of the player is correct with probability 1/3. And for this to be true, it is necessary and sufficient that either the player's choice is uniform at random or the location of the car is uniform at random. But if you want the six conditional probabilities all to be 2/3 too, then you must assume the location of the car is uniform at random and the host-choice is uniform random (when he has a choice). So what premises you require depends on how strong a conclusion you want to get. Richard Gill (talk) 00:35, 5 January 2011 (UTC)

Speaking of 'Monty Forgets' or 'Random Monty', what kind of a game show is it when the host reveals the grand prize, ending the game prematurely? Glkanter (talk) 20:28, 4 January 2011 (UTC)

Note that Nijdam continually speaks of THE probability, at different time points. His picture is the following: random things happen in order, in time, and each time something happens THE probability distribution of the location of the car is updated (according to Bayes' rule, of course). We have: 1/3,1/3,1/3 - Player chooses Door 1 - 1/3,1/3,1/3 - Host opens Door 3 revealing goat - 1/3,2/3,0 - Player switches and Door 2 is opened revealing a ... - 1,0,0 or 0,1,0. For him we must compute THE probability distribution of the location of the car at the time the player must make his decision, and this is the conditional distribution of that location given the information which has been received so far: door 1 chosen, door 3 opened. What he didn't give us yet is a reliable source saying that this is what we MUST do and WHY we must do this. Also he didn't tell us where his assumptions of uniform distributions come from and whether they are subjectivist or frequentist. He doesn't mind how we compute the probability distribution but he insists that we do more than just conjure up some numbers from out of our hat: we have to justify those numbers. Saying that the unconditional probability that switching gives the car is 2/3, is not enough. But saying this and saying that therefore by symmetry of the host-side probability distributions and the law of total probability, the conditional probability of the same event given the specific door numbers chosen and opened is the same value 2/3, is enough.

His short argument above corresponds to splitting the "door opening" by the host into two steps: first a door is opened revealing a goat but without saying which door was opened, then the identity of the opened door is communicated. We have: initial situation - 1/3,1/3,1/3 - Player chooses Door 1 - 1/3,1/3,1/3 - Host opens a door revealing goat but keeping identity of opened door secret - 1/3,1/3,1/3 - Host divulges identity (Door Number 3) of the Door he has opened - 1/3,2/3,0 - Player switches and Door 2 is opened revealing a ... - 1,0,0 or 0,1,0.

I would rewrite Nijdam's short derivation as follows: The probability for the chosen door 1 to hide the car has the same value 1/3 before the player makes her initial choice, and after she has made her initial choice and the host has opened a door revealing a goat but without yet knowing which door has been opened: 2 or 3. By the symmetry of the host-side probability distributions, it is equally likely that door 2 or 3 is opened if the car is behind door 1, and it is equally likely that door 2 or 3 is opened if the car is not behind door 1. Hence the identity of the opened door, in our case, 3, gives no information about whether or not the car is behind Door 1 and therefore does not alter the probability the car is behind door 1: it remains 1/3. Hence at the moment of decision the remaining door 2 has probability 1-1/3=2/3 to hide the car.

At the crucial step I am here using Bayes' rule in the odds form. Just before the identity of the opened goat door is divulged, the odds on Door 1 hiding the car are 1:2. The probability Door 3 was opened given the car is behind Door 1 is 50%. The probability Door 3 was opened given the car is not behind Door 1 is 50%. The likelihood ratio is therefore 1:1. The posterior odds are therefore unchanged at 1:2. Note that the two probabilities of "50%" each come from a different component of the host-side probability distributions. In case the car is behind Door 1 we use the unbiased host premise. In case the car is not behind Door 1 we use the initial uniform distribution of location of the car, conditioned on "not Door 1". Richard Gill (talk) 08:58, 5 January 2011 (UTC)

Basics

I think, the discussion basically comes down to the question whether the simple solution, i.e. the solution without any reference to the conditional (posterior if you like ) probability, can be considered a correct solution to the MHP. Here the MHP is understood to be the version in which the player is offered to switch after the host has opened the goat door. IMO it is not; it produces the right advise: "switching is to your advantage", but based on the wrong arguments. And then: the version of the simple solution that concludes that the car must be with probability 2/3 behind the remaining door, because for the opened door the probability is 0, and the chosen door has probability 1/3 due to the random placement of the car, is completely wrong, and should be banned or at least be exposed as such. Nijdam (talk) 13:27, 5 January 2011 (UTC)

LOLZ. That's your conclusion from all that discussion in a section called "Acceptable formulation"? Glkanter (talk) 13:55, 5 January 2011 (UTC)

Questions allowed? Please who can tell me whether the following variant of the sentence above is correct or is incorrect ( I have added bold, underlined words to the aforementioned full sentence):

"the version of the simple solution that concludes that the car must be with probability 2/3 behind the remaining door, because for the opened door the probability is 0, and the originally chosen door still has probability 1/3 due to the random placement of the car and due to the unbiased opening of door #3, is completely wrong, and should be banned or at least be exposed as such."
Who can tell me the "TRUTH"? Thank you all! Gerhardvalentin (talk) 14:58, 5 January 2011 (UTC)

This sentence of yours is correct, be it that the independence or car placement and choice of door should be counted for, but it is not a variant of what is understood as the simple solution. Nijdam (talk) 21:32, 5 January 2011 (UTC)

Please, in this section, just about the basics. Nijdam (talk) 21:24, 6 January 2011 (UTC)

Topic??

@Nijdam: define C = door hiding car; X = door initially chosen by player; H = goat-door opened by host; Y = remaining door (door to which player is offered to switch). C must equal exactly one of X, H, and Y; it never equals H; therefore C must equal exactly one of X and Y. The events {X=C} and {Y=C} are therefore complementary and hence Prob(Y=C)=2/3 if and only if Prob(X=C)=1/3. What is "completely wrong" with this argument? - apart from your insistence that what we really ought to be computing is Prob(Y=C|X=1,H=3), not Prob(Y=C).

"What we ought to be computing" is a matter of opinion - editors disagree and sources disagree. Selvin himself seems to find both quantities equally interesting and equally valid as solutions. Rosenthal's argument that we ought to be computing the conditional probability is pretty feeble. Carlton gives no reason at all for his preference. Morgan et al. give no reason at all. The only source as far as I know which explicitly gives a reason behind the dogma is Gill (2011).

Are you telling me that my computation of Prob(Y=C) is incorrect? Or are you just telling us that in your opinion, MHP is solved by computing Prob(Y=C|X=1,H=3), not Prob(Y=C)? You are welcome to your opinion of course, but editors' personal opinions are not supposed to play any role in wikipedia articles. Richard Gill (talk) 22:02, 5 January 2011 (UTC)

You've shown me this several times before. What is the purpose? Note that your rv, Y, is something different than the remaining closed door in the MHP, a constant in each possible situation. Nothing wrong with your computations, I do not expect otherwise. But why do you do these calculations, that's the question. Nijdam (talk) 21:24, 6 January 2011 (UTC)

My purpose here was to give a short mathematically rigorous and at the same time intuitively transparent proof that switching gives the car with unconditional probability 2/3, if and only if staying gives the car with unconditional probability 1/3. Y is the number of the door which remains closed after the player's choice and the door opened by the host. It is not fixed in advance. It becomes fixed, once X and H are fixed. What's the problem? Richard Gill (talk) 21:32, 6 January 2011 (UTC)

May be it is the beer, but Y is a random variable and never gets fixed, Yes, as you say, when X and H are fixed, then also Y is fixed, but you may wait till the end of times, X and H are also rv's, hence never will be fixed. I know what you want to say, but you didn't. Nijdam (talk) 22:16, 6 January 2011 (UTC)

Maybe it is the beer, maybe it's the wine, but in my opinion (in my way of thinking) a random variable can become fixed in time, in the sense that as information becomes available and the probability distribution is continually updated by Bayes' rule, its probability distribution can become degenerate. This seems to me to be a common stochastic process (filtration, history) point of view. But it is just words, used to visualize the formal mathematical model. I think we agree on the mathematical facts. Richard Gill (talk) 16:12, 7 January 2011 (UTC)

Be my guest. It's a way of describing what's happening. Instead of conditional probabilities, you have to speak of updated probabilities. What's in a name. It doesn't however change a bit. The simple solution also do not calculate your updated probabilities. Nijdam (talk) 21:46, 7 January 2011 (UTC)

Let A, B and C be any three events. Note that P(A|C)=P(A&C)/P(C)=P(C|A)P(A)/P(C) and P(B|C)=P(B&C)/P(C)=P(C|B)P(B)/P(C). Divide the "A" formulas by the "B" formulas, P(A|C)/P(B|C) = { P(A)/P(B) } . { P(C|A)/P(C|B) }. Posterior odds equals prior odds times likelihood ratio, Bayes rule.

We can generalize Bayes' rule to an evolving history where events C1, C2, C3 in turn are taken account of:

P(A|C1 C2 C3)/P(B|C1 C2 C3) = { P(A)/P(B) } . { P(C1|A)/P(C1|B)} . { P(C2|A C1)/P(C2|B C2) } . { P(C3|A C1 C2)/ P(C3|B C1 C2) }.

BTW this basic idea is one of the secrets of the Kalman filter, the mathematical equations of systems theory, without which we wouldn't have landed on the moon.

Now consider the evolving history of the game: player chooses door (C1) - host opens goat-door (C2) - player learns number of door opened by host (C3). As the history unfolds the odds between any particular pair of statements must be continually updated according to Bayes' rule. Let A={car is behind Door 1}, B={car is not behind Door 1}. Initially the odds between A and B are 1:2.

The first piece of history is unveiled (the first event "C", C1): the player chooses door 1. The probabilities of A and B do not depend on C so the odds between them remain 1:2.

The next piece of history unveiled (the next event "C", C2) is that the host opens one of the other doors, as yet unspecified. He does this with probability 1 both when the car is behind door 1 and the player chose door 1, and when the car is not behind door 1 but the player did choose door 1. So the likelihood ratio for this piece of "information" is 1/1=1 and the odds remain 1:2 on the car being behind door 1.

The next piece of history unveiled (the next event "C", C3) is the identity of the opened goat-door. In scenario A (car behind door 1) we know car is behind door 1, player chose door 1, and host opened a door revealing a goat. The door he opens is Door 3 with probability 0.5 (no host-bias). In scenario B (car not behind door 1) we already know player chose door 1 and the car is not behind door 1, the host opens a goat-door. He doesn't have any choice since he has to open door 2 or 3 and the car is behind door 2 and door 3 each with probability 0.5 since it isn't behind door 1 (initial location of car is uniformly distributed). The door the host opens is therefore Door 3 with probability 0.5. The likelihood ratio is 0.5/0.5=1 and the odds remain 1:2 on the car being behind Door 1.

The difference between the simple and the conditional approaches is quite simply whether or not they explicitly take note of the last step: the revealing of the number of the door opened, after we already know the number of the door chosen by the player and that the host has opened an unspecified goat-door. The final step in the history - the final piece of information given to the player - the specific number of the door opened by the host - turns out actually not to give any information about whether or not the car is behind Door 1. (Of course it does give information about whether or not the car is behind Door 3!). The odds for Door 1 versus NOT Door 1 remain 1:2 throughout the whole history, in fact none of the steps give any information about this question, in the sense of changing the odds.

I could have gone through this story considering the odds on Car behind Door 1 : Car behind Door 2 : Car behind Door 3. They start out as 1/3:1/3:1/3. The player's choice does not alter them, and the information that the host has opened another door revealing a goat doesn't change them. However the probabilities that he opens Door 3 given the player chose Door 1 and the car is behind Door 1, 2 or 3 are 0.5, 1, 0; so the posterior odds on Car behind Door 1 : Car behind Door 2 : Car behind Door 3 are 0.5 : 1 : 0 = 1 : 2 : 0.

Considering Door 1 versus Not Door 1 is a clever simplification which brings the simple and the conditional solution very close together. But of course it gives the same final answer as considering Door1:Door2:Door3. Richard Gill (talk) 22:29, 5 January 2011 (UTC)

Richard, I agree with what you have said, particularly your statement starting, 'The difference between the simple and the conditional approaches...'. I assume that you are using 'simple' and 'conditional' a sense specific to this discussion. I reality both calculations involve conditional probability. In the first case we have the condition that the player has chosen door 1 ( he could have chosen any door) and in the second case we have an additional condition. Do you agree? Martin Hogbin (talk) 01:04, 6 January 2011 (UTC)

Simple solutions like Carlton's and Monty Hall's are conditioned on the 100% likelihood that the host reveals a goat. The decision tree that has been banned from the article shows that. They are indifferent as to which door is opened. Glkanter (talk) 02:23, 6 January 2011 (UTC)

@Martin: I was indeed using the words "simple" and "conditional" to refer to the two main classes of solutions to MHP, and moreover for me "conditional" refers to conditional probability. One can say that conditioning on a certain event is also conditioning, but I think it is better to try to restrict it to "serious" conditioning when one conditions on events which are not certain, so that at least some probabilities will change by doing so.

One can give a simple solution without conditioning on the initial door choice. Define C = door hiding car; X = door initially chosen by player; H = goat-door opened by host; Y = remaining door (door to which player is offered to switch). C must equal exactly one of X, H, and Y; it never equals H; therefore C must equal exactly one of X and Y. The events {X=C} and {Y=C} are therefore complementary and hence Prob(Y=C)=2/3 if and only if Prob(X=C)=1/3. This little solution leaves quite open whether we are thinking of X or C or both as random. If you want it to work for X fixed (and equal to 1, say), you have to assume C is random and Prob(C=1)=1/3. You can also use this argument from Monty's point of view, for him C is fixed (1, say) and the player's choice X is uniform random.

@Glkanter: indeed there is a decision tree representation for the simple solution too. And indeed the reason that the initial probability that the door chosen hides the car is not changed when the quizmaster reveals a goat (but without yet saying which door he pulled it from) is because conditioning on a certain event does not change any probabilities.

Yes, it's this indifference (100% probability of revealing a goat) to which door is open that makes the simple solutions work. The simple decision tree demonstrates that the contestant has *not* made his choice before the goat is revealed, and that the solution is *not* an 'average', or 'overall probability' of anything, except the average probability of door 2 or door 3 being opened, which are known to be equal (each a likelihood of 2/3 of having a goat, same as the contestant's door, due to the original distribution of the car and goats). That decision tree shows that Nijdam's arguments are without merit. Brushing this off *because* of the 100% condition is unwarranted. Glkanter (talk) 10:51, 6 January 2011 (UTC)

Here I am afraid I disagree. The 2/3 of this simple solution is the average over the success chances when Door 2 is opened, and when Door 3 is opened. Those two possibilities are equally likely, but that is not the crucial point. The crucial point is that the probability Door 1 was hiding the car is not changed when we receive the information whether it was Door 2 or Door 3 which was opened. In both cases it is 2/3, and their average is 2/3. But you don't know that it is 2/3 in both cases from just studying the decision tree which you authored. You need to give an argument for this fact. One argument is symmetry, another argument comes from the final contribution to the odds ratio for Door 1 versus Not Door 1, 0.5/0.5. The 0.5 in the numerator (upstairs) is the probability Door 3 is opened given the player chose door 1 and the car is behind door 1, the 0.5 in the denominator (downstairs) is the probability the car is behind door 2, given that it is Not behind door 1. Richard Gill (talk) 20:35, 7 January 2011 (UTC)

There's nothing else I can say that I haven't already said numerous times. Except this: I disagree with your arguments and conclusions, and many reliable sources give a solution that is contradictory to the 'requirements' you seem to lay out. Glkanter (talk) 20:50, 7 January 2011 (UTC)

My point of view is that we have a number of different correct derivations of different probabilities under different conditions. It is up to the user to choose which assumptions they are prepared to make and therefore which conclusions they can buy with them. It is up to the user to decide what kind of probability notion they want to use. It is up to the seller to argue why it is worth paying the extra money for a Mercedes (conditional) when a Volkswagen (simple) seems good enough for most people.

What I like about the argument I gave with the three steps C1, C2, C3 and considering the odds on Door 1 : Not Door 1 is that you see how you the simple solution is obtained (at step 2) having used very little assumptions, while the conditional solution (at step 3) makes explicit use of both host-side uniformity assumptions. What I also like is the use of Bayes rule in the odds form. This makes conditional probability simple and intuitive! It means that one can give a full mathematically rigorous conditional solution in a couple of sentences of plain English - you don't need to work through big formula manipulations. It shows exactly what the "extra" of the conditional solution is. This is what good mathematics is about: replacing computations by ideas (Riemann, Hilbert). Richard Gill (talk) 08:45, 6 January 2011 (UTC)

@Gerhardvalentin: the C1, C2, C3 and Door 1 : Not Door 1 argument allows us to take account of host bias. We just have to change one of the two 0.5 probabilities into an arbitrary number between 0 and 1 (which of the two is changed, is important!). In the two extreme cases the final odds are 2:0 and 2:2 with the "no-host-bias" giving the usual intermediate 2:1. So switching is never wrong and therefore even if you don't know the host bias the answer is "switch". The unconditional probability you'll get the car is 2/3, the conditional probability is somewhere between 1/2 and 1. (This is the whole mathematical content of Morgan et al's work). Richard Gill (talk) 08:58, 6 January 2011 (UTC)

Different solutions

Richard, let do as you suggest above and not treat events which are certain, essentially the game rules, as conditions.

Do you agree then that, on that basis, we should start with a sample space including all the possible outcomes allowed under the game rules.

The next step should be, as you say above:

1) The player chooses door 1.

We should therefore condition our sample space on that basis first of all. Note that we cannot say that the players door choice is independent of the car placement unless we know that the car placement (or the player's choice) is random. As is most commonly done let us assume that both are uniform at random. Do you agree so far? Martin Hogbin (talk) 10:06, 6 January 2011 (UTC)

I prefer to keep all my options open. Let's agree there is a door hiding a car, a door chosen by the player, a (different) door opened by the host (revealing a goat), and a third door left over. Call them C, X, H, Y. Random variables taking values in the set of door numbers {1,2,3}. There are some structural relations between them: H is certainly different from X, and then Y takes the remaining value from {1,2,3}, different from both X and H. Also, H is certainly different from C. Everyone will agree to take C and X statistically independent of one another. That still permits either or both of them to be fixed (non-random) as well as random. The joint probability distribution of everything can be built up from the marginal distributions of C and of X (host-side and guest-side respectively), and the conditional distribution of H given C and X (host-side). Even if I originally assumed X random, I can condition (in the probabilistic sense) on X=1, say, if I want to. By independence the distribution of C does not change. I already built up my model by imagining a collection of conditional distributions over the values of h, Prob(H=h|C=c,X=x) for all c and x, so I can still put the model together with X=1 fixed.

It follows from the structural assumptions that Prob(Y=C)=1-Prob(X=C). And by independence, Prob(X=C)=Prob(X=1)Prob(C=1)+Prob(X=2)Prob(C=2)+Prob(X=3)Prob(Y=3). (Note that this relation is also true if X is fixed, e.g. Prob(X=1)=1, Prob(X=2)=0=Prob(X=3). A bit of algebra shows that if either X is uniform on {1,2,3} or if C is uniform on {1,2,3}, then Prob(X=C)=1/3. So it can be forced host-side, and it can be forced guest-side, that the initial choice is correct with probability 1/3. And when that is true, switching gives the car with probability 2/3.

This is the basis for the recommendation to the player coming from game-theory. The game-theoretic-savvy player has no interest in conditional probabilities since they involve host-side probability distributions over which he has no control. He just wants to have the best probability of coming home with a car. So he randomizes his initial choice and always switches. His success-chance is 2/3. The game-theoretic-savvy Monty, on the other hand, wants to keep as many cars as possible. So he completely randomizes the location of the car and he completely randomizes the choice of door to open when there is a choice. That way he prevents any player (e.g. a player with inside knowledge about the probabilities used in the randomizations host-side) from doing better than 2/3.

A passive player who just wants to see how the probabilities change in time as information is released will postpone making any decision as late as possible, because at least that way he is guaranteed that his decision will be optimal (optimal with respect to unconditional(!) probability of succes). This player presumably either uses subjective probability, or he is a frequentist and has been given some hard information about the procedures used host-side to hide the car and to open a door. If he knows frequentistically or knows subjectivistically that initially all doors are equally likely to hide the car he might just as well start by choosing door 1 because 1 is his lucky number. (If despite his beliefs of uniformity he is superstitious, he would be wiser to choose door number 1 because 1 is his unlucky number). Such a player just wants to see the probabilities unfold and he does this because he already "knows" that C is uniform random.

So yes, for this player it makes sense to start by conditioning on his actual door choice, X=1. By independence the distribution of C is not changed and it remains uniform. So his odds on the location of the car are still uniform, 1:1:1, after he has chosen this door.

Now the host opens a door and we hear a goat bleating, but the guest was not allowed to see which door it was yet. He just knows that it is a different door from number 1. The new information was going to happen certainly, under all three possible locations of the car, so his odds remain 1:1:1.

The host asks if he would like to switch to the "other door". If he must decide at this moment, his answer will be YES, because the odds that the car is behind "the other door" are now 2:1. As they always were. Prob(Y=C):Prob(X=C) = 1:2, both before conditioning on X=1, and after, and after also conditioning on the certain event that he would hear a goat bleating behind an as yet to him unknown door H different from X and Y.

If he doesn't have to decide now, he could say "but tell me first which door you opened". The host could then say, OK, H=3. The chance that the host would say "H=3" is different under the three hypotheses: C=1, C=2, C=3. The three chances are q, 1, 0, where q is the probability the host opens door 3 when he has a choice between 2 and 3. The posterior odds now become q:1:0 since the prior odds were uniform. Whatever q is, these odds support switching to door 2. At one extreme the car is now CERTAINLY behind Door 2. At the other extreme it is equally likely behind 1 or 2, so you may as well switch. In the case of a known unbiased host, or when your subjective assesment of the bias is completely indifferent regarding direction of bias, we get 0.5:1:0 = 1:2:0. Clearly he might just as well not have asked since he's going to switch anyway. This is Gerhardvalentines signature tune, and the content of Morgan et al's work. Richard Gill (talk) 14:28, 6 January 2011 (UTC)

My recommendation to wikipedia MHP page editors is to cut a large amount of the present maths formalism out of the page, in particular, out of the conditional solutions part of the page. Conditional probability is not difficult, and Bayes' rule in odds form is extremely intuitive and powerful. Emphasize the close connections between the conditional and unconditional solutions. There is nothing wrong with most authorities' derivations of the unconditional probability that switching gives the car. Don't criticise them, just report honestly what they deliver. Concentrate on the short sharp simple derivations which don't bring in assumptions which some people find discutable. Accept that there is a mathematical school of thought that says "you *have* to compute the conditional probability" (for instance, because only in this way is it guaranteed that your solution is optimal) but the step-by-step computation of the conditional probability shows that the last step, bringing in the information of the specific door number opened by the host, does not lead to a different decision. Just as one would expect. We know that always switching has overall success change 2/3, that always staying has overall success chance 1/3. It is pretty inconceivable that there could be anything better than 2/3. Proving it mathematically seems to me to be a bit of a luxury --- except in the mathematics class-room, where it is essential to pay some attention to that issue too. Richard Gill (talk) 14:44, 6 January 2011 (UTC)

+1. Yes, thank you. The MHP-paradox (Pws 1/2 or 2/3, odds 1:1 or 1:2) is foremost a "psychological problem" of approximation and assessment of chances and risks (YES or NO), not a flat "Mathematical probability puzzle" in the first line. Caracterized by completely contradictory positions, it is necessary to distinguish: Bare conditional probability theorems, based on differing audacious "assumptions" are used primarily for mathematical training purposes, and there they are doing quite well. Although sold under the so-called rubric "solving the MHP", it is less a "solving the paradox", but more about training opportunities in maths. And "en.WP" could finally lead the way as a pacesetter to a NPOV depiction. Gerhardvalentin (talk) 22:48, 6 January 2011 (UTC)

Based on your recommendation, Richard, do you agree with my proposal on the mediation page to eliminate paragraphs 2 & 4 from the 'Conditional solution' section? How about paragraph 1? Glkanter (talk) 16:59, 6 January 2011 (UTC)

The first paragraph is presently

The simple solutions show in various ways that a contestant who is going to switch will win the car with probability 2/3, and hence that switching is a winning strategy. Some sources, however, state that although the simple solutions give a correct numerical answer, they are incomplete or solve the wrong problem. These sources consider the question: given that the contestant has chosen Door 1 and given that the host has opened Door 3, revealing a goat, what is now the probability that the car is behind Door 2?

I would rewrite this paragraph, perhaps as the following:

The conclusion of the simple solution is that "always switching" beats "always staying": overall success-chance 2/3 versus 1/3. Some sources however are interested in something different: given that the contestant has chosen Door 1 and given that the host has opened Door 3, revealing a goat, what is now the probability that the car is behind Door 2? The reason for this more subtle and detailed inspection is that in principle it is possible that in some of the six possible situations (specific door chosen by guest, specific door opened by host) it would be better to switch and in other situations better to stay. If so, the overall succes-rate could be made better still than 2/3, the success-rate which is achieved by *always* switching.

After this I would compress all the remaining material of the section on "Conditional Solution" into one or two short paragraphs. Perhaps first of all I would report the symmetry argument, according to which, in the situation of no host-bias, all six conditional probabilities are equal to one another and hence, by the law of total probability, to the unconditional probability 2/3. After that I would perhaps do a Bayesian analysis on the lines of the discussion above: using the odds form of Bayes' rule I would show how the odds on the location of the car change as each piece of the history becomes unveiled, showing how the simple solution is obtained at the phase when we know a door is opened revealing a goat but don't yet know its identity, and how the further information of the identity of the opened door doesn't change the odds at all in the symmetric case, but anyway, cf. Morgan et al., doesn't change the decision whether it is best to switch or stay. Richard Gill (talk) 21:50, 6 January 2011 (UTC)

I would add some and write: The conclusion of the simple solution is that "always switching" beats "always staying": overall success-chance 2/3 versus 1/3, meaning that on the average 2/3 of all players will win the car. It is not clear however whether this result also holds for the player in the situation given in the problem. Some sources are therefore interested in something different: given that the contestant has chosen Door 1 and given that the host has opened Door 3, revealing a goat, what is now the probability that the car is behind Door 2? The reason for this more subtle and detailed inspection is that in principle it is possible that in some of the six possible situations (specific door chosen by guest, specific door opened by host) it would be better to switch and in other situations better to stay. If so, the overall succes-rate could be made better still than 2/3, the success-rate which is achieved by *always* switching. Nijdam (talk) 22:25, 6 January 2011 (UTC)

Yes, something like that seems to me to be the right motivation for going into the more tricky conditional probability stuff. Richard Gill (talk) 16:12, 7 January 2011 (UTC)

I don't suppose y'all would consider just removing the stuff that that doesn't apply to the 50/50 host MHP, and leaving the 3rd paragraph like it is? Glkanter (talk) 00:33, 7 January 2011 (UTC)

You're referring to the paragraph

The conditional probability of winning by switching given which door the host opens can be determined referring to the expanded figure below, or to an equivalent decision tree as shown to the right (Chun 1991; Grinstead and Snell 2006:137-138), or formally derived as in the mathematical formulation section below. For example, the player wins if the host opens Door 3 and the player switches and the car is behind Door 2, and this has probability (1/3)x1 = 1/3. The player loses if the host opens Door 3 and the player switches and the car is behind Door 1, and this has probability (1/3)x(1/2) = 1/6. These are the only possibilities given host opens Door 3 and player switches. The overall probability that the host opens Door 3 is their sum, and we convert the two probabilities just found to conditional probabilities by dividing them by their sum 1/3+1/6=1/2. Therefore, the conditional probability of winning by switching given the player picks Door 1 and the host opens Door 3 is (1/3)/(1/2), which is 2/3.

I would rewrite this as

The conditional probability of winning by switching given which door the host opens can be determined referring to the expanded figure below (is the figure needed???), or to an equivalent decision tree as shown to the right. It assumes that the player has already chosen Door 1. The switching player wins if the host opens Door 3 and the car is behind Door 2, and this has probability 1/3. The switching player loses if the host opens Door 3 and the car is behind Door 1, and this has probability 1/6. These are the only possibilities in which Door 3 is opened. The overall probability that the host opens Door 3 is their sum 1/3+1/6=1/2. Therefore, given the player initially chose Door 1, the conditional probability of winning by switching given the host opens Door 3 is (1/3)/(1/2), which is 2/3.

I would add to this paragraph a short discussion of the rationale for the conditional solution: that it explicitly makes use of all the information in the possession of the player, hence is guaranteed to generate the optimal overall success rate, which we thereby know is 2/3. This discussion could include the proof of the conditional result from the unconditional by symmetry and the law of total probability, because this elegant proof shows explicitly that the extra information (identity of door opened) does not change the chances that the car is behind Door 1. I would also have a section discussing the difference between the conditional and the unconditional solution, illustrated with the proof by step-wise updating by Bayes' rule in odds form. In that section I would show how the unconditional solution only makes use of the assumption that Door 1 hides the car with probability 1/3, and how the conditional result depends on the subtle balance of the 50/50 two host-side uniformity assumptions. At the same time I would show how even without host-unbiasedness, switching is always the right thing to do. I would add some remarks about the unconditional simple solution of Monty Hall himself. This solution follows the point of view of the host that the location of the car is fixed and his choice will be his choice, also fixed, if he has a choice. The player's initial choice is random and hits the car with probability 1/3. He wins the car by switching if and only if his initial choice misses the car (probability 2/3).

My aim throughout would be to minimize computations and algebra, but to maximize ideas, to do justice to the diversity of points of view taken in the reliable sources, and to emphasize the close relations between all the different solutions. Richard Gill (talk) 07:19, 7 January 2011 (UTC)

Really important ideas are: mathematical symmetry; the law of total probability; conditional probability; updating probabilistic information by Bayes' rule in the odds form.

I'd also like to see a small section on the relation between probability interpretations and the consequent meaning (interpretation) of the distributional assumptions and hence the meaning of the probability conclusions. Richard Gill (talk) 16:22, 7 January 2011 (UTC)

Let us not forget there is also an incorrect simple solution, or equivalently combined doors solution. I definitely want these forms of pseudo solutions to be criticized. Nijdam (talk) 17:40, 7 January 2011 (UTC)

You made that claim earlier, but offered no specific support for it, neither your own, or from a reliable source. Glkanter (talk) 18:49, 7 January 2011 (UTC)

I too would like to know which solution you refer to, @Nijdam. One in the present page? Attributed to which source? Richard Gill (talk) 20:27, 7 January 2011 (UTC)

Yes, look in the article. Take for instance Devlin's "solution", or the combined doors solution, which is also shown in a picture. Nijdam (talk) 21:54, 7 January 2011 (UTC)

And Richard, I really am fed up by having to repeat this over and over, without getting an answer. So once and for all, answer to the following, and restrain yourself to what I say. One form of the simple solution reads: Due to the random placement the car is with probability 1/3 behind the chosen door 1. As the opened door 3 shows a goat, this door has clearly probability 0 to hide the car. Hence the remaining door 2 must have probability 2/3 on the car. it is this formulation I'm fighting. Especially because it looks quite plausible and is copied by lots of school pupils and students who gets assignments about the popular MHP. Now just comment on what I wrote here. Nijdam (talk) 22:06, 7 January 2011 (UTC)

@Nijdam, you are absolutely right. "Hence" is wrong. What is said is a nonsequitur. The fact of specifically Door 3 being opened could change the likelihood that the car was originally hidden behind Door 1. So far we only used "all doors initially equally likely" and that isn't enough to get a conditional result, while this argument is about a conditional result, since it talks about how probabilities change (or don't change) on getting the "Door 3" information. There is a missing step where "no host bias" has to be explicitly used. Eg, appeal to independence: by symmetry, the number of the door opened doesn't change the probability the car is behind door 1. Or if you prefer, use Bayes' rule and show by explicit computation that the odds on Door 1 hiding the car isn't changed by the information *which* specific door is opened.

Next task, find a reliable source which explicitly makes this point, because I'm afraid that some editors could consider this observation of yours "OR". (Or write it yourself: why not submit a small note to Statistica Neerlandica?) Richard Gill (talk) 07:24, 8 January 2011 (UTC)

Morgan's misquotation

You guys are worse than Morgan, et al. They faulted the simple solutions because Whitaker/vos Savant didn't make explicit the 50/50 host premise. For a puzzle about a game show. Of course, Morgan says in their rejoinder that maybe only math professors would ever consider that it's not 50/50. Then they finally admit, 20 years later, that it's always 2/3. You guys take it a step further, and criticize the simple solutions for not explicitly repeating that premise as part of the solution. That is the very definition of pedantic. And this is of no benefit to the Wikipedia reader. Glkanter (talk) 10:07, 8 January 2011 (UTC)

Of course, Morgan had to dishonestly modify Whitaker/vos Savant's problem statement in order to make their argument, even though Morgan used quotation marks, from:

"...and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat."

to

"...and the host, who knows what's behind them, opens No. 3, which has a goat."

Maybe the exposure of this canard contributed to their revised, 'it's 2/3, period' 20 years later. You fellas have no such crutch. Glkanter (talk) 10:49, 8 January 2011 (UTC)

Here's Morgan's full misquote of Whitaker/vos Savant:

""Suppose you're on a game show and given a choice of three doors. Behind one is a car; behind the others are goats. You pick door No. 1, and the host, who knows what's behind them, opens No. 3, which has a goat. He then asks if you want to pick No. 2. Should you switch?""

That's barely coherent: "You pick door No. 1, and the host, who knows what's behind them, opens No. 3, which has a goat." What's 'them'? It's an idiotic, poorly conceived word substitution made solely to support their criticisms. Which they, 20 years later, in some continually vague fashion, backed up on. But here we are, still slogging through this nonsense. Glkanter (talk) 11:13, 8 January 2011 (UTC)

I have never noticed that before. This misquotation by Morgan et al has no justification. Martin Hogbin (talk) 11:30, 8 January 2011 (UTC)

They write bad English but it is coherent. "them" refers to the doorS. The host knows what's behind every one of the three doors. Their falsification lies in changing vos Savants parenthetical "say, Door 1", and "say, Door 3" into very definite Door 1 and Door 3. You could delete "say, door 1" and "say, door 3" from vos Savant's words and what would be left would be a completely coherent question which would be perfectly legitimately answered by a (correctly argued) unconditional (i.e, simple) probability solution.

And as vos Savant said later, that was also her semantic intention. Richard Gill (talk) 12:00, 8 January 2011 (UTC)

vos Savant's solutions by the way do not make any use of the 50/50 host choice assumption since they are only aimed at the unconditional probability, for which it this assumption is irrelevant. How many times do I have to remind you of this? Richard Gill (talk) 20:11, 9 January 2011 (UTC)

I disagree with your inference that the simple solutions are an average of all games played only. The indifference by the contestant as to which door has been opened in each single play of the game relies on the doors being equally likely to be opened. If the 50/50 host is required for that indifference, then vos Savant makes use of it. How many times do I have to remind you of this? Glkanter (talk) 20:31, 9 January 2011 (UTC)

Yes, you miss the entire point, Richard, which is that Morgan needed to misquote Whitaker/vos Savant to make their tenuous argument. That you offer your own misquote is actually quite amusing, pointless, and predictable. Glkanter (talk) 12:54, 8 January 2011 (UTC

Yes, the misquotation is important for Morgan's argument. Martin Hogbin (talk) 13:32, 8 January 2011 (UTC)

I have always recognized that Morgan et al deliberately misquote Vos Savant in order to boost their claim that her answer is wrong. Richard Gill (talk) 13:48, 8 January 2011 (UTC)

Devlin and combining doors

Nijdam, would you agree that the only things lacking in Devlin's solution are a statement that the probability of the car being behind the originally chosen door is independent of the door opened by the host and an explanation of why this is so? Martin Hogbin (talk) 11:06, 8 January 2011 (UTC)

Your way of putting this question seems to say: It is not a big mistake Devlin makes, it just needs some extra explanation. No, no,no, Devlin is flat wrong, he's making a serious logical error, he's not to be taken seriously as a source for Wikipedia. His article is just a popular story, not peer reviewed. He makes clear, he does not understand much of probability. I wrote him twice, but no response.Nijdam (talk) 00:04, 9 January 2011 (UTC)

It's hard to tell exactly where he's going with these two new sections on his talk page, but since he needs to include the statement, '...if one were to remove the 'Suppose you're on a game show' premise from the problem statement...' in his comments, he's obviously *not* discussing the MHP. Because there's no Random Monty/Forgetful Host, host bias when faced with 2 goats, or hinting to/telling the contestant where the car is, on a game show. Glkanter (talk) 11:27, 8 January 2011 (UTC)

Both of those two new sections on Nijdam's talk page seem to me to be perfectly clear and reasonable and uncontroversial, I would imagine that any intelligent and MHP-knowledgeable editor could agree with the content. Richard Gill (talk) 12:32, 8 January 2011 (UTC)

I am just trying to establish whether this is Nijdam's sticking point regarding this solution or whether there is another objection to it. I look forward to your response Nijdam. Martin Hogbin (talk) 11:33, 8 January 2011 (UTC)

@Glkanter: not all wikipedia readers use probability in the subjectivist sense. Also if you think "frequentist", and allow for host-bias, you can give a coherent solution to MHP. Your personal limitation in understanding of probability does not imply that only people like you will come to wikipedia MHP for information. And as Nijdam correctly points out, many readers of wikipedia MHP pages are students of probability and statistics and academics in those fields. The pages cannot be limited to a popular approach for the great unwashed, only. Hopefully people who are intelligent enough to edit the MHP pages will be intelligent enough to appreciate, if not to share, other points of view than their own. Fortunately we seem to have made a lot of progress in recent weeks in mutual understanding. Even if you want to focus on the simple solution for the simple people, you still must learn to understand some of the subleties which bother teachers of probability and statistics. It is not just a matter of arrogance or high-priestly-ness that just about all "professional" or specialist writers on MHP seem to prefer conditional solutions. Richard Gill (talk) 12:21, 8 January 2011 (UTC)

I don't know how to "...and allow for host-bias..." while discussing and editing the problem about a game show, where the K & W formulation states the host chooses uniformly at random when faced with 2 goats. Which is the MHP. Which is the subject of the 2/3 & 1/3 vs 50/50 paradox. Glkanter (talk) 12:54, 8 January 2011 (UTC)

And I do know how to allow for unknown host-bias! Neither K&W nor you own a monopoly on the interpretation of Vos Savant's question. Nor on the interpretation of probability. All of which is the MHP-P or meta-MHP. Which indeed includes the subject of the 50:50 versus 2/3 1/3 paradox. Have you read my paper yet? Richard Gill (talk) 13:58, 8 January 2011 (UTC)

BTW Richard, did you notice that K&W forget to mention the assumption that the placement of the car and the player's first choice have to be independent? Nijdam (talk) 00:07, 9 January 2011 (UTC)

No I didn't notice that. Of course everyone does assume that, for good reasons too. Richard Gill (talk) 08:58, 11 January 2011 (UTC)

I will have an interest in your paper only when the Wikipefia MHP article has the NPOV violating simple solution criticisms removed. Otherwise, I remain fat, dumb and happy in my knowledge that the original uniformly at random distribution of the car and goats allows for the proper application of the simple solutions conditioned on the 100% liklihood the host will reveal a goat. Glkanter (talk) 14:41, 8 January 2011 (UTC)

If you want your POV to win in the wikipedia MHP page mediation battles, you need to find reliable sources to support you. One of them is my paper! It says that the simple solution is correct, and it even gives mathematically rigorous (pedantic) support for that. A completely different point, is that the particular argument given by Keith Devlin is faulty or sloppy (or at best: incomplete). Which anybody can see who understands logic and can read English. I can't judge if you are fat and/or happy, but I do think it is dumb not to try to understand the points your fellow editors try to make. Richard Gill (talk) 14:52, 8 January 2011 (UTC)

The NPOV violations exist within the Wikipedia MHP article, not within Glkanter, or his understanding of the MHP or probability. Glkanter (talk) 15:55, 8 January 2011 (UTC)

@Glkanter: it's the job of maths professors to be pedantic. We are merely pointing out that really is a missing step in Devlin's argument. Which in the probability and statistics classroom it is absolutely vital to point out, because in other situations such a missing step could lead to a fatal error, and we want our students to learn correct methods of argumentation and computation; if they get the right answer by a wrong computation that is still wrong. Since Devlin explicitly talks about how some probabilities change when you know specifically that Door 3 has been opened, he ought to ask himself whether or not other probabilities might change too.

@Martin, I cannot speak for Nijdam of course, but in my opinion you are right: the only thing that is missing is the symmetry argument which you just alluded to yourself. The probability of the door being behind the originally chosen door is independent of the specific door opened, by symmetry, under the usual assumptions of no-host-bias or, for subjectivists, prior indifference to host-bias (ie it can equally well be any amount more than 50:50 as any amount less than 50:50. Richard Gill (talk) 11:33, 8 January 2011 (UTC)

If Nijdam would confirm that this is the only basis of his objection to Devlin's solution, we might be able to work towards finding a resolution acceptable to all. Martin Hogbin (talk)

I really do not understand why anyone would be so interested in defending a simple solution, even to the point of deducing the right conditional form from it, as if that would make the simple solution more acceptable. What is it? Does someone have explained in the course of time the MHP with the simple solution, and now in understanding its weakness, desperate tries to find a way out??? I really do not see why someone would otherwise defend the simple solution (the correct form of it of course), except the nitwits of course. Nijdam (talk) 00:17, 9 January 2011 (UTC)

You are entitled to your opinion Nijdam, but that is all it is. I recognise that there is a degree of incompleteness or lack of rigour in the simple solutions but this can, if it is considered really necessary, be corrected by additional wording. Your objection to using the simple solutions with additional clarification or explanation does not reflect any mathematical principle just your personal opinion on the subject. Martin Hogbin (talk) 10:57, 9 January 2011 (UTC)

Martin, you persist in ignoring the point that is being made here. There exist correct computations of the unconditional probability. There exist correct computations of the conditional probability. Devlin's is neither. There is a missing step (and it requires an as yet by him unstated assumption) to make. The probability Door 3 hides a car collapses to 0 when that door is opened revealing a goat. Devlin seems to think that opening Door 3 doesn't tell us anything about whether Door 1 hides a car. But in principle it could do. He fails to provide the argument (and the assumption which is needed to carry through the argument). It is sloppy, unforgiveable.

But anyway, this doesn't matter at all. The wikipedia MHP page doesn't have to reproduce wrong solutions alongside good ones! Devlin's work by the way is just a little opinion-piece, it is not a peer-reviewed academic publication. If any mathematician worth their salt had reviewed it in advance of publication, they would have pointed out the gap in the argument to him. Richard Gill (talk) 20:08, 9 January 2011 (UTC)

I am not ignoring the point, neithet do I not understand it; I do not agree with it. There is no such thing as a 'correct computation', there are computations of varying degrees of rigour. Devlin's solutions give the correct answer to any reasonable interpretation of the problem, including conditional ones. It misses out a step which you may consider necessary but which others may consider obvious. It is therefore less rigorous than some other solutions may be, but the so called conditional solution given in the article is not particularly rigorous either, it does not show the other doors that the player might have chosen. So yes, maybe the conditional solution given is a little more rigorous than Devlin's solution but not much, and what it lacks in rigour it makes up for in ease of understanding and power to convince.

I've now noticed the second Devlin "paper", pointed out by Rick Block. So Devlin himself agreed that his argument was wrong! Of course it is very easy to fix his argument (by symmetry, or by explicit calculation of the Bayes factor, 0.5/0.5=1. But note: either way, it is an argument for the Conditional solution! That's the important thing to recognize! Richard Gill (talk) 08:58, 11 January 2011 (UTC)

You say, 'Devlin seems to think that opening Door 3 doesn't tell us anything about whether Door 1 hides a car. But in principle it could do'. According to what principle? Only the principle that to answer a different question we require a different method of solution. Martin Hogbin (talk) 23:56, 9 January 2011 (UTC)

No @Martin, this is according to the principle that if you condition on a non-certain event, any probability can change. And some probabilities certainly do change. As Devlin correctly deduced. The "principle" in question is the universal law P1(A|B)=P1(A&B)/P1(B), where, for us at the moment, "P1" refers to probabilities given that the player chose Door 1, A is the event "Car is behind Door 1", and B is the event "Host opened Door 3". Devlin correctly notes that with regards to the event C "Car behind Door 3", P1(C)=1/3 but P1(C|B)=0. So he sure is using conditional probability! And he sure has missed a step of the argument. That's all. Whether or not you and most wikipedia readers are satisfied with intuitively appealing but logically faulty arguments is quite irrelevant.

Just because one author screws up, doesn't mean that every author's presentation is a mess. Gill (2011) and Sevlin both present both simple and conditional solutions, both correctly. Devlin tries to do the conditional solution but does it wrong. Better not to confuse wikipedia readers by presenting faulty arguments. Of course new editors will then come along and remove or correct them. Richard Gill (talk) 09:07, 10 January 2011 (UTC)

@Glkanter: please distinguish between a) what we want to calculate, and b) whether or not our method of calculating it is correct. Quite a few lovers of simple solutions are interested in an unconditional probability and give a correct computation of an unconditional probability. Nijdam himself has the point of view that we should not be interested in the unconditional probability (and he has a good reason for this), but still, that is just an opinion. Devlin seems to have the same opinion. Devlin explicitly conditions on specifically Door 3 revealing a goat. Right now we are not discussing conditional versus unconditional, but talking about something else: the correctness of a specific source's probabilistic reasoning. Devlin's *reasoning* is just  !$#*-ing wrong, or if you would like to be more polite, incomplete. It is not a simple (unconditional) solution. It's a screwed up conditional solution. Richard Gill (talk) 14:04, 8 January 2011 (UTC)

I've just emailed to Keith Devlin about his really shocking mistake / missing proof step. He may be a mathematician but he's a bad probabilist and a dangerous science popularizer if he makes blunders like that. His response will be interesting. Richard Gill (talk) 13:52, 8 January 2011 (UTC)

You might be interested in Devlin's December 05 column, see [1]. -- Rick Block (talk) 18:03, 8 January 2011 (UTC)

This seems to be standard stuff, 2/3 if the host knows where the car is, 1/2 if he does not. Have I missed something? Martin Hogbin (talk) 18:12, 8 January 2011 (UTC)

You do miss it all. Nijdam (talk) 00:19, 9 January 2011 (UTC)

Perhaps you could be more specific. Martin Hogbin (talk) 10:57, 9 January 2011 (UTC)

Devlin is showing that the simple explanation is "correct" only under assumptions that it does not explicitly use - since the same "correct" explanation seems to apply but leads to the wrong result for what is effectively the "host forgets" variant. He then goes on to say "As sometimes arises in mathematics, when you find yourself in a confusing situation, it may be easier to find the relevant mathematical formula and simply plug in the appropriate values without worrying what it all means." - i.e. if you want to know the conditional probability then use Bayes rule to figure it out rather than some "intuitive" explanation that might or might not apply. -- Rick Block (talk) 22:26, 8 January 2011 (UTC)

Rick, so you really believe in "Host forgets" and D.? I don't, and no more in the Flat Earth. And I hope this is not my mistake. Gerhardvalentin (talk) 00:52, 9 January 2011 (UTC)

That criticism/explanation, while not original, does not seem to hold up to the standards of scientific (formal?) logic, in terms of disqualifying the solution as invalid. In other words, it's bogus. 22:36, 8 January 2011 (UTC) — Preceding unsigned comment added by Glkanter (talk • contribs) 22:37, 8 January 2011 (CEST) (UTC)

Exactly as I said, the answer is 2/3 if we adopt the universally accepted standard rule that the host knows where the car is and 1/2 for what he calls 'a slightly modified version of the Monty Hall game' in which the host does not know where the car is. We all know this. I cannot see any relevance to the current discussion. Martin Hogbin (talk) 22:33, 8 January 2011 (UTC)

No Martin. The *conditional* answer is 2/3 if we adopt ... AND if we assume no host-bias (or if we are subjectivists, indifference to the direction of host-bias). Devlin gives the answer 2/3. His derivation is clearly conditional, he talks about specifically Door 3 being opened and how the probability the car is there then drops to 0. He silently assumes the probability the car is behind Door 1 doesn't change on being given this information. He does not explain how it is that he knows that opening Door 3 doesn't change is odds on Door 1 hiding the car. So to put it charitably, he was a bit quick and skipped a step. He not only skipped a step of the argument, he also skipped mentioning the assumption which he was using to make this step. Pretty shoddy stuff for someone who calls themselves a mathematician! I have written to him about this.

The answer is always 2/3 for the standard rules provided only that we are consistent in our approach (apart from the trivial case that the problem is insoluble). See section below.

I thought that Rick was suggesting that there was something new in Devlin's Dec 05 column compared with his 2003 comments. I was just pointing out that there is not. Devlin says much the same in both articles.

Devlin does not say how we know that the probability of the car being behind door 1 does not change when the host opens door 3. This is an omission but how important it is is a matter of opinion. Some may consider the fact obvious. Martin Hogbin (talk) 19:22, 9 January 2011 (UTC)

This is not to say that every simple solution is wrong. It is legitimate, in my opinion (and in the opinion of many authorities, eg vos Savant, Selvin, ...) to be interested in the unconditional probability that switching gives the car. And there exist correct arguments that that probability is 2/3. One of the correct arguments is that the host is essentially offering you the choice between Door 1 and Door 2 + Door 3. I have written up a completely respectable formal mathematical version of that argument, many many times. Richard Gill (talk) 15:43, 9 January 2011 (UTC)

Yes, I think we all agree that the simple solutions are correct for the unconditional case, the question is, how wrong are they for the conditional case? In my opinion, not very. Martin Hogbin (talk) 19:22, 9 January 2011 (UTC)

Mathematical puzzles ought to be solved by correct mathematical arguments. There are correct derivations of the unconditional probability that the switcher wins. There are correct derivations of the conditional probability that the switcher wins, given a specific door chosen and door opened. There is no point in wasting wikipedia space with nonsense (derivations which are just plain wrong, i.e., illogical). Devlin's argument is faulty but fortunately there is absolutely no need to report it on the wikipedia MHP page. There are plenty of "correct" simple solutions out there. Monty Hall, Selvin, Gill (2011), vos Savant... Richard Gill (talk) 19:54, 9 January 2011 (UTC)

It's clearly a matter or opinion how MHP should be translated into a formal mathematical problem: what assumptions should be made, and what constitutes a solution (what is the target object to be computed). Given a particular assumption-set and particular target there exist correct solutions and incorrect solutions. Devlin's solution is incorrect. He skips an important step in his chain of reasoning. There is no way to tell if he was aware of it or not. Richard Gill (talk) 20:01, 9 January 2011 (UTC)

Lemme make sure I have this straight: Devlin's valid solution gets thrown in the scrap heap because you guys have decided his presentation has an omission, but Morgan's paper remains the 800 lb. gorilla of simple solution criticism despite countless errors, including the one Martin & Nijdam made them admit, the dishonestly changed Whitaker/vos Savant quote, and all the rest of them as Martin has cataloged? You guys crack me up. But there's nothing funny about it. Glkanter (talk) 20:09, 9 January 2011 (UTC)

Devlin's argument misses a crucial step. Can't you see that? So what is the point of refering to it on the wikipedia MHP page? Wouldn't it be a service to the reader just to concentrate on the sources who give logically sound arguments for whatever they think is the right kind of solution to the problem? Both simple and conditional?

As for Morgan et al's paper, I am not aware of anyone who wants to make it some kind of centre-piece of the page. Personally I think it is a bad paper. It falsifies history so as to boost their own interpretation of the problem; the mathematics is opaque and clumsy; and it does not explain why a conditional solution should in any way be seen as better than an unconditional solution. Fortunately there are sources which do the conditional solution (with or without host-bias) in a much more simple and transparent way, and there are sources which explain what is the "bonus" of a conditional solution. There are also sources which explain the down-side --- in order to get more detailed conclusions, you need to make more detailed assumptions, and that limits the scope of your solution. Richard Gill (talk) 20:22, 9 January 2011 (UTC)

Oh, please. "Morgan is (are?) God(s)" should be the banner for the 'simple is flawed' arguers. No other reliable source says they are 'false'. Equating Devlin's presumption of the obvious to Morgan's intentional and careless errors is just silly. And there is no 'bonus' available by doing the conditional solution to the MHP. I have shown countless methods of proving that 'switch' is always the best decision, in every single play of the game. "How many times do I have to remind you of this?" Glkanter (talk)

Richard, Devlin's argument misses a step that you consider necessary but which others may consider not to be needed. To most people it is intuitively obvious that the number of the door that the host opens cannot affect the probability that the car is behind door 1. This turns out to be correct for any reasonable and consistent set of assumptions. Martin Hogbin (talk) 20:44, 9 January 2011 (UTC)

Martin, lastly it's the old argument of extreme host's bias again, in another variant, giving additional info on the actual location of the car: Because of door#3 can be hiding a goat (to be shown) only in 2 out of 3, and in exactly those 2 out of 3 the door#1, originally selected by the guest, either (1/2) hides the car or (1/2) the other goat, so in opening of "door 3" Pws is said to be 1/2 only. Cit.: "And it even gives mathematically rigorous (pedantic) support for that." Gerhardvalentin (talk) 21:23, 9 January 2011 (UTC)

Yes, I know what it is all about, but there is no reasonable set of assumptions that gives an answer other than 2/3, even Morgan agree this now. Martin Hogbin (talk) 22:34, 9 January 2011 (UTC)

@Martin: Devlin (who claims to be a mathematician) carelessly omitted a small and simple but logically absolutely necessary step in his little chain of arguments. Whether or not you or @Glkanter or Mr and Mrs Average would lose any sleep on this, is totally beside the point. Devlin's argument, with its missing step put correctly in its correct place, is an argument for the conditional solution. That's The Truth!

What is obvious?

Which steps in any given calculation are absolutely necessary and which are not is a matter of opinion. I believe that the step that Devlin omitted is obvious and therefore not absolutely necessary to include. Martin Hogbin (talk) 15:45, 10 January 2011 (UTC)

The missing step is not obvious, and it requires an appreciation of the difference between a conditional and an unconditional probability. Devlin admits himself that it's important (thanks @Rick Block). Devlin is doing the conditional solution and he does it wrong, full stop. Richard Gill (talk) 08:58, 11 January 2011 (UTC)

Maybe you do not find it obvious but I do. That may surprise you and you may consider that I have no right to find it obvious but it is obvious to me and in this particular case my intuition proves to be correct. There really is no formal or even universally accepted definition of obvious, you do know the story don't you? Martin Hogbin (talk) 22:49, 14 January 2011 (UTC)

On a more conciliatory note, I have nothing against changing the combining doors solution so that it is seen to attempt the unconditional problem. No one will notice if we do it well. Martin Hogbin (talk) 22:57, 14 January 2011 (UTC)

I don't see that at all. There is only one MHP, and it has the contestant choosing door 1 (Box B in Selvin's paper) and the host opening door 3 (box A in Selvin's paper). There are a variety of solutions that are given various names. Some of the solutions are simpler than others, but given the problem statements from Selvin and K & W, they are valid and proper shortcuts for solving the door 1 and door 3 problem. Every reliable source, except Morgan's intentional misquote of Whitaker/vos Savant, uses the same problem statement. Categorizing these solutions as solving different problems is artificial, improper, and not derived from the sources themselves. Glkanter (talk) 23:51, 14 January 2011 (UTC)

I think the wikipedia MHP editors would be wise to take account of this by some minor editing. That way they will end up presenting the full conditional solution in a completely transparent and intuitively delightful way, thereby totally taking the wind out of the sails of all those annoying people who want to go on a crusade about the enormous difference between the two kinds of solutions. Richard Gill (talk) 08:51, 10 January 2011 (UTC)

Here I agree with you. I have proposed that we do some minor editing and was trying to negotiate something with Nijdam. However, as you said, most people will not be interested in this rather pedantic point so we should be careful not to compromise clarity for the majority in the cause of pedantry for the few. I stress that this is only for the 'Simple solutions' section. After that we can go into the subject in as much detail as necessary to give a full and accurate understanding. Martin Hogbin (talk) 15:45, 10 January 2011 (UTC)

Martin, it would help us all if you would only open your eyes! Devlin has introduced a 3rd party to the MHP, in the form of a 2nd contestant. It is only in this way that the Random/Forgetful host of a game show denying the contestant the opportunity to win the car makes sense. And as another editor pointed out today, Selvin does not 'own' the MHP. So, this new 3 player, reduced sample space problem is *also* the MHP! Selvin naming the problem the MHP, stating the host will always reveal a goat, has no bias between goats, giving a simple solution only in his first letter, and profusely praising Monty Hall's simple solution are obviously besides the point. Wake up and smell the coffee, Martin! Glkanter (talk) 22:51, 9 January 2011 (UTC)

Selvin profusely praised Monty Hall's simple solution in his second letter, after doing the full conditional solution, in contrast to his first letter. Monty Hall's simple solution does not make any host-side assumptions at all, but takes the host-side point of view that it is the player's initial choice which is completely random. Selvin's two letters contain both simple and conditional arguments, with just about all possible combinations of answers. In fact his paper is similar to Gill (2011) in that he offers the reader a spectrum of decent mathematical formulations of the problem and decent solutions thereof, each with their own strengths and weaknesses. Richard Gill (talk) 08:51, 10 January 2011 (UTC)

Devlin and not combining doors

@Rick Block is of course completely right: in Devlin's second internet posting (neither of these items can be called "papers") he has noticed that his original argument was faulty. Moreover he explicitly admits that he was after the conditional solution, too. There is however an alternative to "going back to first principles and doing it the hard way with Bayes' formula". We here know two "quick fixes" to Devlin's short argument.

Fix 1: symmetry. The conditional probability that the car is behind Door 1 given the player chose 1 and the host opened Door 2 is equal to the conditional probability that the car is behind Door 1 given the player chose 1 and the host opened Door 3, by symmetry (since we take the full K&W conditions on board). Therefore, given the player chose Door 1, whether or not the car is behind Door 1 is independent of whether the host opened Door 2 or Door 3. And that means that the probability the car is behind Door 1 given the player chose 1 and the host opened 3 equals the probability the car is behind Door 1 given only that the player chose 1, and we already know that that equals 2/3.

Fix 2: Bayes rule, odds form. The odds on the car being behind Door 1 given the player chose Door 1 are 1:2 against. If the car is behind Door 1 (the door chosen by the player) the probability the host opens Door 3 is 0.5 (no host-bias). If on the other hand the car is not behind Door 1 (the door chosen by the player) the probability the host opens Door 3 is 0.50, because the car is now equally likely behind Doors 2 and 3, and he is forced to choose the other one. The Bayes' factor or likelihood ratio is therefore 0.5/0.5 = 1, and the posterior odds remain at 1:2 against.

Both fixes to Devlin's argument are easy, but both are subtle, neither is completely obvious. They are especially not obvious to people who are not familiar and comfortable with the difference between conditional and unconditional probabilities. Precisely such persons will on the other hand not even notice the missing step, let alone come up on their own with a rigorous argument which fills in the gap.

We should put both corrected versions of Devlin's argument into the conditional solutions section, where they belong.

On the other hand, Devlin's remark about the host offering you the choice between Door 1 and Doors 2 + 3 is spot on: it's a perfect short simple solution to the problem of determining the unconditional probability. Richard Gill (talk) 09:28, 11 January 2011 (UTC)

So, contrary to Nijdam's recent proposition that the Combined Doors simple solution is flat out wrong, you've determined that:

It is correct

It is conditional

Devlin's logic also supports a perfect simple solution

I'm pretty much OK with the above. I've argued with Rick in the past that the Combined doors solution is clearly after door 3 has been revealed. I think this was when he first trotted out his Solution section re-write that omitted it, about July-ish. Glkanter (talk) 11:10, 11 January 2011 (UTC)

No @Glkanter, my opinion is that a combined doors simple solution is correct and unconditional ("the host is offering you the choice between Door 1, and Door 2+3: therefore the unconditional probability that switching gives the car is 2/3"). Devlin's solution is a botched conditional solution, as he realised himself in his second writing. Richard Gill (talk) 17:14, 11 January 2011 (UTC)

That's interesting, Richard, given that you wrote this above:

"We should put both corrected versions of Devlin's argument into the conditional solutions section, where they belong."

and you 'correct' me by writing:

"No @Glkanter, my opinion is that a combined doors simple solution is correct and unconditional..."

Maybe you could explain this contradiction? Of course, the main point is that once again, Nijdam's cries of 'this simple solution is wrong' are unsupported. Glkanter (talk) 17:28, 11 January 2011 (UTC)

Maybe, @Glkanter, you could read my paper and work a bit to try to pick up some subtleties? There is no contradiction. There are reliable sources which treat MHP as an unconditional probability problem. There are reliable sources which treat MHP as a conditional probability problem. Each of the two problems can be solved correctly, and each can be solved incorrectly. If you cannot distinguish between the two problems, and if you cannot distinguish between a correct argument and an incorrect argument, then of course everything I write will appear to you to be full of self-contradictions. That's your problem, not mine.

I agree with you and with Martin and I think Gerhard, that the wikipedia MHP page should start with good (correct, appealing, intuitive) simple solutions, i.e., solutions to the unconditional problem. It would be wonderful if in later sections someone like you could come to learn the difference between the unconditional and conditional problems and also come to appreciate good (correct, appealing, intuitive) conditional solutions. It's like wine. It takes a bit of practice and sophistication to come to appreciate really good wines. Is it necessary? No, it's a luxury, decent inexpensive wine is fine for most purposes! Does it enrich your life? Yes it does. Richard Gill (talk) 07:58, 13 January 2011 (UTC)

Richard, you say 'They are especially not obvious to people who are not familiar and comfortable with the difference between conditional and unconditional probabilities. Precisely such persons will on the other hand not even notice the missing step, let alone come up on their own with a rigorous argument which fills in the gap'. This is pretty much what I have been saying all along. After the vos Savant's solution it was around a decade before anybody, including statisticians and mathematicians noticed the potential conditional aspect of the MHP (although Selvin had addressed this aspect originally).

The really amazing thing about the MHP is that people resolutely believe the answer is 1/2. There is no evidence to suggestion that this as has anything whatever to do with whether the problem is conditional or not. As you quite rightly say, nearly everyone does not even notice the difference. Also, I believe that the symmetry with respect to door number would be obvious to most people if they were asked, for example, on being asked 'Would it have made any difference to your answer if the host had revealed a goat behind door 2 rather than door 3?', I am sure most people would reply, 'Of course not!', but I guess this is OR. Obviousness is an interesting topic in mathematics, I am sure you know the well known story on the subject.

So, Richard, why not give the simple solutions first, without complication, because there are many justifications for doing this, and then, for the small pedantic minority, discuss the more complex issues in proper detail? Martin Hogbin (talk) 12:16, 11 January 2011 (UTC)

@Martin: yes, I do recommend you MHP page editors give the simple solutions first without complication! I am saying that Devlin's pseudo-mathematical solution is a botched conditional solution, so it would be wise to leave it out entirely. Do include the "Door 1 versus Doors 2+3" simple solution. Why can't we *all* have our cakes and *all* eat them too? Present first sound simple arguments for the unconditional solution, and after that build on them to present sound and hardly more complicated arguments for the conditional solution, in particular, using both of the fixes which I wrote out to Devlin's argument. That way we are both respecting The Truth and the needs of two kinds of readers. Richard Gill (talk) 17:14, 11 January 2011 (UTC)

We currently give a combining doors solution in the 'Simple solutions' section and I believe that we should continue to do so, as it is on of the most convincing ways of showing that the answer is 2/3. I have suggested several time before that with some subtle changes to wording we can make the simple solutions technically correct. The lead does a good job of this. I think it may be best at the start to leave it open as to whether the conditional or unconditional formulation is being addressed. Missing out a very convincing solution because of a technical difficulty which is insignificant for most people is a bad idea. So is loading it with off-putting warnings. A little subtlety, on the other hand, might do the trick. What do you think? Martin Hogbin (talk) 13:51, 12 January 2011 (UTC)

Yes, @Martin! That's exactly what I think. It just requires collaborative editing in a constructive spirit of good faith and civility; mutual respect. We must embrace our differences. Richard Gill (talk) 08:08, 13 January 2011 (UTC)

Martin - your claim that "There is no evidence to suggest that this as has anything whatever to do with whether the problem is conditional or not." is simply wrong. According to Krauss and Wang, fully 97% of their test subjects drew a picture where door 3 is open showing a goat. These subjects are clearly trying to solve the conditional problem where it is given the player has picked door 1 and the host has opened door 3 (not the player has picked door 1 and the host has opened "a door" showing a goat). It is this problem, where the "probability" (the conditional probability) the car is behind door 3 is obviously 0 and the car must be behind one of door 1 or door 2, that nearly all people try to solve. This problem leads to the erroneous 1/2 answer, but 1/2 door 1 and 1/2 door 2 which are conditional probabilities (not 1/2 door 1 and 1/2 "door 2 or door 3, whichever one the host didn't open"). -- Rick Block (talk) 14:24, 11 January 2011 (UTC)

Is that after hearing a problem where 'another door, say #3' was opened to reveal a goat? So what? 97% of the people tested have good hearing. How about vos Savant saying in over 10,000 letters, it never came up? Glkanter (talk) 15:41, 11 January 2011 (UTC)

I, and many others, always figured it was 'the car is either behind the door I chose, or the other one remaining'. Same condition, with indifference. Glkanter (talk) 15:45, 11 January 2011 (UTC)

No, @Glkanter. Suppose you've chosen Door 1. Then there is actually a subtle difference between knowing that another door is opened revealing a goat and you have the option of switching to the third, and knowing that Door 3 is opened revealing a goat and you have the option of switching to Door 2. Vos Savant intended (she has said so, and this also corresponds to her solutions) that you could delete the words "say, Door 3" - this is a parenthetical remark to make things clear for you: the door opened would have to be Door 2 or Door 3 if your own choice was Door 1. You were also intended also to delete the words "say, Door 1".

vos Savant's intention was *not* that you should think about the situation *after* Door 1 is chosen and Door 3 is opened. However as Kraus and Wang point out, most people do create this picture in their minds, and indeed, that is one of the roots of the error which most people make. What you see in this picture is namely only a part of the information you learnt. You also got information in the way those doors got to be opened or closed.

If you are a player with a problem with short term memory and you choose your door initially at random, then step aside, then Monty opens a/the (different) goat door, and you forget which door you had first chosen, then all you see are two closed doors and one open. The chance truly is 50/50 given only this information, which door hides the car! The "wrong" intuitive solution is actually completely correct for a super-forgetful player who only sees what is in front of him, and who must decide at a moment when he has moved away from his initially chosen door. Richard Gill (talk) 17:14, 11 January 2011 (UTC)

It's too bad you cannot perceive your own writings the way they come off to others. You reached a new level of absurdity with the above paragraph. Kudos, Richard. I didn't think it possible. Glkanter (talk) 17:21, 11 January 2011 (UTC)

It's too bad you never actually carefully read what other people write, @Glkanter. You just jump to some conclusion or other and start ranting and raving about other people's stupidity.

Of course this attempted put down is absurd, unsupported, and wrong. I read everything you post on Wikipedia regarding the MHP. As I demonstrated above by showing your contradiction between 'conditional' and 'unconditional'. Which only came about because I summarized what I thought I had just read, and gave you a chance to confirm that. That's a pretty good reading and comprehension technique, wouldn't you say? So, it's a very weak personal attack. Ill-advised, too. There's no way you can comment intelligently on what I do or do not read. Your self described powers of esp, as usual, have failed you horribly. The rest of your insult is gibberish, as well. I *am* beginning to wonder why I read and respond to your postings, though. With all your absurdity, flip-flopping viewpoints, gibberish, and your various statements that you have no further interest in the mediation or editing the MHP article, I really see no further point to it. Glkanter (talk) 17:59, 11 January 2011 (UTC)

Did you read my paper yet? A lot of people told me they liked it very much. Richard Gill (talk) 17:32, 11 January 2011 (UTC)

Rick, I said '...people resolutely believe the answer is 1/2. There is no evidence to suggestion that this as has anything whatever to do with whether the problem is conditional or not'. In other words there is no evidence to suggest any connection between the fact people get the wrong answer and whether they regard the problem as conditional or not. As has already been pointed out, people show door 3 opened by the host because that is what they have heard. The question is whether they would have come up with a different answer in they had been told that door 2 had been opened, or that one of the other doors had been opened. I very much doubt that. Martin Hogbin (talk) 18:13, 12 January 2011 (UTC)

The simple problem is too easy

The whole point of the "simple" solution is that it recasts the problem from the highly counterintuitive one where you know for sure you're looking at two closed doors and an open door, to something else. I think phrasing the problem in a way that unambiguously addresses the "unconditional problem" is difficult, however regardless of how it is phrased it is mathematically equivalent to "a car is hidden behind 3 doors, you've picked one, do you want to keep your pick or do you want to trade it for the other two doors" - since no conditioning happens, all original probabilities remain exactly the same (the probability of each of the doors starts 1/3 and stays 1/3 - we don't know specifically which door was opened so we have no basis to update the probability of ANY of the doors). This problem has an intuitively obvious solution (pick two doors, duh), which is (mathematically) what the simple solution says is the solution to the MHP.

However, the "conditional" MHP is NOT the same as this trivial problem. It starts with this unconditional problem (there are 3 doors, you pick one leaving two) but now the host opens one or the other of the unpicked doors and you can see which one (meaning any idiot can tell its probability which started as 1/3 is now 0). It's perhaps abstractly possible to keep this as an unconditional problem (the host now opens one of the unpicked doors - keep in mind that we don't know which door, it might be one or it might be the other, you simply can't tell which door the host opens - got it? - it's like a Heisenberg uncertainty thing). As usually phrased, and I'd bet as interpreted by nearly anyone, the whole question is what happens to the probabilities of the two unopened doors in any specific case where we know for sure the probability of one of the doors is 0 - which is a conditional probability question. Asking about the door 1 and door 3 case makes this a very concrete problem. We could as well ask about any of the other cases (door 1 and door 2, etc.). The point is not that this is the only case we care about, but that we care about a specific case - not the "average" of all cases. Any specific case will do, but making it about a specific case is what (mathematically) changes the problem from "pick a door, do you want to switch it for the other two" to "pick a door, see an OPEN door, do you want to switch your door for the ONE other door". In a specific case you're not trading your one door for two doors, you're trading your one door for one other door. The only way you get to trade your one door for two doors is if you decide to switch BEFORE seeing which door the host opens. AFTER the host has opened a door there are only two doors left in play (your's and one more). At this point, the simple solution simply no longer applies because it always talks about all three doors. -- Rick Block (talk) 23:21, 12 January 2011 (UTC)

The title of this sections says what I understand you to be saying. You describe the unconditional problem as trivial, suggesting that most people would get it correct. Unfortunately there is no literature that I am aware of that tackles the question of how hard the unconditional version of the problem is, K&W come closest, so this discussion is really about our personal opinion and OR.

It is quite easy to phrase the problem unconditionally. Treat it as a request for advice for prospective players on the show, after all this is probably what Whitaker actually wanted to know. Morgan give a unconditional statement along these lines. A person is going go be a contestant on a game show with the standard MHP rules. When offered the swap what, should they do, and how much is it to their advantage to do so? If you were to ask people that question what percentage do you think would give the following answers? My answers are shown.

1) Always take the swap, you will have twice the chance of winning. MAH 20%

2) Wait and see which door the host opens then make your decision based on that information. MAH <1%

3) Do not swap, it is not to your advantage. MAH 80% Martin Hogbin (talk)

You didn't capture the unconditional question and you're ignoring the point I'm making - which is that the unconditional question is merely an obfuscated way of asking whether you want to choose one door or two. The reason the simple solutions are simple is because they completely ignore the conditional case. Look at vos Savant's solution for example. The car is equally probable to be behind door 1, door 2, or door 3. If you pick door 1 and switch, you win if the car is behind door 2 or door 3 (two cases) and lose if the car is behind door 1 (one case) - and vice versa if you stay. This solution says nothing whatsoever about the example given in the problem statement where the host has opened door 3 and we know with certainty that the car is not behind this door. In this case, the car is NOT equally probable to be behind door 1, door 2, and door 3 - so this solution does not apply. And, yes, it's personal opinion (or OR if you prefer) but I strongly believe the fundamental reason most people argue that the "simple" solutions are wrong is because they don't address the example case from the problem description. This disconnect occurs precisely because the the simple solutions address the "unconditional" problem.

As to whether the "unconditional problem" is easier or not, I think that's basically a moot point. As understood by nearly anyone reading it (K&W's 97%), the MHP is NOT the "unconditional problem". -- Rick Block (talk) 17:15, 13 January 2011 (UTC)

The sources that give simple solutions present the exact same problem description as those that offer conditional solutions. (Well, not the same one as Morgan, et al. They bastardized it to support their argument, misquoting vos Savant's column in a peer-reviewed professional journal. Aren't there any sanctions or punishments for pulling crap like that? I see in England the phony autism report guy is in hot water.) Many times they begin by saying, 'In essence, Monty is offering...', or, 'This is effectively the same as...' Your argument is of no more value than you guys parsing the difference between a 'solution' and an 'explanation'. Every one of the previous arguments has been refuted, so you're down to this non-existent trifle. Glkanter (talk) 19:45, 13 January 2011 (UTC)

Rick, what do you mean by, 'You didn't capture the unconditional question'? Are you saying that to ask for advice as to what to do should you go on the show is not an unconditional way of formulating the question?

You also say, 'unconditional question is merely an obfuscated way of asking whether you want to choose one door or two'. Yes, you are quite right, but people still get the answer wrong. That is what is so surprising about the problem. It is no great surprise that most people cannot answer a problem in conditional probability.

Finally you say, 'most people argue that the "simple" solutions are wrong'. They don't. A small minority of the more pedantic statisticians do? Martin Hogbin (talk) 20:30, 13 January 2011 (UTC)

What I mean by you didn't capture the unconditional problem is your description makes it clear we're talking about a game show (where the doors are known and visible to all concerned) and the point of decision is AFTER the host has opened a door - in context any reasonable person would be assuming that you know both which door you originally picked and which door the host opened, so there is a revised probability of the door the host has opened (whatever door it is) and it is 0 and the question pertains to the revised (i.e. conditional) probabilities of the other two doors. Nearly everyone (probably the same K&W 97%) will see this as a conditional problem. They will think through a specific example - you might as well identify the case for them and say the player picks door 1 and the host opens door 3 (we could renumber the doors to make this the case if necessary anyway). The unconditional problem changes the point of decision to BEFORE the host has opened a door, or (more artificially) requires the contestant (not just the puzzle solver) to not know which door the host has opened. BEFORE the host opens a door, all doors still have a 1/3 chance of hiding the car and you may end up switching to either of the doors you didn't choose. Yes this is probably still confusing, but it's confusing in an entirely different way (what's confusing is what the question is asking). AFTER the host opens a door, the opened door has probability 0 and your options are only the door you originally picked or the other door. The odds are still 1/3:2/3 but it's between two fixed choices (not three choices). My claim (my opinion) is most people are intending to answer the conditional problem, not the unconditional, and that this is the answer they get wrong. Given a completely unambiguous unconditional version they might still get the answer wrong, but we don't know that (and your assertion that people are getting the unconditional problem wrong is as unfounded, actually considerably less founded given K&W's 97% number, than my claim they're getting the conditional problem wrong).

The bit about people thinking the simple solution is wrong is not talking about the pedantic few, but the many, many, many people who find the simple solutions unconvincing and argue with the simple solutions saying the answer is 1/2. As far as I know, nobody puts it this way but IMO the reason is the simple solutions don't address the problem they're thinking about. -- Rick Block (talk) 02:20, 14 January 2011 (UTC)

Isn't there some sort of 'reasonableness test' that says if an explanation requires 2,591 characters, and it's really just an opinion anyways, the disputed point it supports may not be necessary or helpful in the Solution section for readers who simply want to know why it's not 50/50? Glkanter (talk) 06:32, 14 January 2011 (UTC)

You point about before and after is fine (although I believe not strictly correct). That is why I suggested phrasing the question as advice to a person who is about to go on the show. I gave three options essentially, always switch, never switch, decide later on the basis of which door is opened.

The first two answers accept the question as unconditional, hence the 'always'. The third option, argues that the problem must be treated conditionally because the door number opened by the host is important. Very few people would in my opinion choose this last option. Martin Hogbin (talk) 10:48, 14 January 2011 (UTC)

What doors are we talking about

Three doors. Important to see and to name which one we are talking about, esp.:

• selected door#1
• goat shown behind door#2
• offered alternative: door#3

The MHP-question is talking about doors:

• door selected (name it "say 1", if you like, or just "door selected", or even "door X", if you like)
• door opened, showing a goat (name it "say 3", if you like, or just "door opened", or even "door H", if you like)
• door offered as an alternative (name it "say 2", if you like, or just "the other door", or "door offered", or even "door Y", if you like)

The situation the contestant is in:
He sees two closed doors, and he knows which one is "his chosen door", and he knows which one is "the host's other door", offered as an alternative.
As said, you can name the doors if you like. And the simple solution does name them, eg "door selected" of "contestant's door" or "my door", and the other one "host's closed door" or "door offered by the host as an alternative".

So the "simple" as the "conditional" are talking about the same specific doors, the same specific information on the doors, and both are talking about the same time in the specific course of action.

After the host has opened one door of his two doors and has offered his second door as an alternative, the characteristik of the "simple solution" is that it does not matter indeed which one of his two specific doors the host just has opened showing a goat, no matter whether door "say B" or door "say C", if you like, because the situation will be exactly the same, as a given basis for the decision required, in the specific case.

Now everyone can see (eyes open!) that, in the given situation and under the given conditions, Pws never can be below 1/2.
Anyone can see that, whether "mathematician" or "non mathematician".

The simple solution concludes that Pws "actually" is 2/3, and at the same time the simple solution concludes that Pws "actually" would be 2/3 also, in case the host should actually have opened even the other door. The simple and the conditional actually give the same result: It makes no difference which of his two doors the host just has opened, Pws will be 2/3, in the given situation and under the given conditions.

• And the simple solution says "2/3 in the actual situation", same value as the overall Pws, as long as no additional info will be revealed.

• And the conditional solution says "2/3 in the actual situation", same value as the overall Pws, as long as no additional info has already been revealed ("or assumed").

So please no inappropriate confusion of terms. Gerhardvalentin (talk) 23:56, 13 January 2011 (UTC)

2/3 is the answer - What is the question?

It astonishes me that you write "the answer is 2/3". The MHP does not ask for something this could be an answer to. So what does 1/3 answer? During more than 2 years we are discussing, and one point of discussion is exactly what the number 2/3 means, and yet you come up with such an sentence. Nijdam (talk) 10:37, 9 January 2011 (UTC)

Of course on any other forum I would consider it necessary to explain what I mean but here the question of what 2/3 is the answer to is well known. To be more specific, 2/3 is the probability (based on any generally accepted meaning of the term) that the player of the MHP game, with the standard rules, will win by switching in the following cases:

1. The unconditional formulation, that is to say the player picks a door and the host opens another to reveal a goat.
2. The conditional formulation, given that the player picks door 1 and the host opens door 3 to reveal a goat.
3. As 2 with any specified legal door numbers.
4. The host is known to have a door preference but the player has initially picked a door uniformly at random.

Do you dispute any of these? Martin Hogbin (talk) 11:45, 9 January 2011 (UTC)

And you are surprised I ask what 2/3 is an answer to?? Nijdam (talk) 22:34, 14 January 2011 (UTC)

Yes, because the answer of 2/3 is correct for any reasonable, self-consistent, and soluble interpretation of the standard-rules MHP. Martin Hogbin (talk) 12:02, 15 January 2011 (UTC)

Birds

Hi Martin. Please note that it is a Bird Project convention that all bird species are fully capped, thanks Jimfbleak - talk to me? 07:06, 1 January 2011 (UTC)

Who am I to argue, but this a new one to me. How widespread (outside WP) is this convention? Martin Hogbin (talk) 11:30, 1 January 2011 (UTC)

Starting points

I urgently ask the participants to respond to Wikipedia talk:Requests for mediation/Monty Hall problem/Starting points. This may show us where we still differ in opinion. Nijdam (talk) 11:03, 9 January 2011 (UTC)

Who can give an example of when "The contestant should not switch in this particular play of the MHP"?

Anyone? Because the misstatement of the simple solutions as (1)'the average of the outcomes of always switching' is still comprised of (2)the aggregate of infinite 'in each single, unique play of the game, switch'. #2 would have to be false for a better strategy to exist. Glkanter (talk) 23:02, 9 January 2011 (UTC)

" 'the average of the outcomes of always switching' is still comprised of the aggregate of infinite 'in each single, unique play of the game, switch' " is gibberish to me. What are you talking about, @Glkanter? Richard Gill (talk) 13:48, 10 January 2011 (UTC)

In fact, who can even give a reasonable example in which the probability of winning by switching is not 2/3 (with the standard rules, as always). Martin Hogbin (talk) 23:58, 9 January 2011 (UTC)

Of course, the whole thing remains pointless. Sadly, as I wrote nearly 2 years ago, it's just a bunch of gibberish. A game show problem about 3 equally likely doors. Why is the solution so complicated, requiring 'new inquiries' into the MHP? Whatever happened to '1/3 <> 1/3', anyways? Glkanter (talk) 04:17, 10 January 2011 (UTC)

Discussion is indeed pointless as long as participants don't recognise the difference between a correct argument and a correct answer. Nobody is disputing 2/3, nobody is disputing that the contestant should switch. Everyone is happy to accept the standard and full collection of familiar premises.

If one imagines many repetitions of the game in which the true location of the car varies completely at random and the initial door chosen by the player varies in any way you like (or not at all), then (A) always switching will give the player the car in 2/3 of the times. Those many repetitions can be separated into up to six subgroups of repetitions according to the initial choice of the player, and the door opened by the player. (I say "up to 6" because if for instance the player always chooses door 1 there will only be 2 subgroups). If the host's choice of door to open is also completely random, when he has one, then also (B) in each of those up to six subgroups of repetitions, a switching player gets the car 2/3 of the times.

Surely both @Martin and @Glkanter can agree that it must cost a little more thought and/or logic to convince a nit-picking sceptic of the truth of the more refined conclusion (B) than of the rather basic conclusion (A).

Right now, no one here is fighting about whether one should solve MHP in the form of a statement like (A) or like (B). Right now, no-one here is disputing the host-side assumptions of completely random car-hiding and door-opening. Right now, all we were doing was discussing a little error in the presentation of one of the sources (Devlin): a source which pretends to give the conditional solution, not a simple solution at all! It is an internet page, an opinion piece by a Big Man. A couple of sound-bytes for his fans. It's not an academic, peer-reviewed publication.

Nijdam has correctly pointed out that the specific argument given by Devlin misses a key step. Devlin silently uses the fact, but does not explain why it is true, that the probablity Door 1 hides the car doesn't change on revealing a goat specifically behind Door 3. That's a real blunder when made by someone who claims to be a mathematician and puts themselves forwards as a popularizer of science. It's clear to me that Devlin is not a probabilist. He hasn't thought MHP through carefully. Devlin tries to give the conditional solution (B) but doesn't give a correct argument for it!

On the other hand, arguments on the lines of "the host is offering you the choice of Door 1 versus Door 2+Door 3, therefore you should switch (2/3 versus 1/3)" are completely correct. The probability Door 1 hides a car doesn't change on revealing a goat behind one of the other doors, as yet unspecified. So it remains 1/3, and the other as yet unspecified door hides the car with probability 2/3. Richard Gill (talk) 08:20, 10 January 2011 (UTC)

I started a new section on Martin's talk page. This has nothing to do with Devlin, obviously. Do you have an answer to the question in this section's title? Or is there no way for some different strategy to improve on the very simple 'switch'? Glkanter (talk) 08:55, 10 January 2011 (UTC)

"Ask a stupid question, get a stupid answer". No I can't give an example. The title of this section is silly. It's a section about a non-topic. . Richard Gill (talk) 09:23, 10 January 2011 (UTC)

All that matters is that you, or anyone, can't describe a scenario where it's better to stay. That means that in every play of the game it's best for the contestant to switch. All that stuff about 'optimal' solutions and 'min/max' is of no incremental value. And, as Martin pointed out, each play of the game is, of course, 2/3 & 1/3. Glkanter (talk) 10:37, 10 January 2011 (UTC)

It is not only better always to switch than always to stay (2/3 versus 1/3 unconditional success-chance), it is also better to switch than to stay in each of the six possible situations you can find yourself in (door chosen, door opened). And this proves that 2/3 unconditional can't be improved on. And indeed (in the case of no-host-bias) it is also 2/3, in every one of the six recognisably different cases. The simple solutions *only* tell us that it is better always to switch than always to stay (2/3 versus 1/3 unconditional success-chance). The conditional solutions give us all that extra stuff. The incremental value is clear for all to see. I'm not saying that it's very important, but it is extra value. Is a middle-range Volkswagen enough for you or would you prefer a top-of-the-range Mercedes? You'll need more money to buy the second, but on the other hand it does have some extras... but maybe you're not interested in those extras... Richard Gill (talk) 17:25, 11 January 2011 (UTC)

Again, your analysis is absurd. There is no other choice that is 'best' other than to switch. Each and every time. This is evident from the simple solutions. And from the question I raise in this section that demonstrates the foolishness of any counter-arguments, even the hypothetical ones. Because they cannot exist. Please, Richard, just stop. Glkanter (talk) 17:34, 11 January 2011 (UTC)

Of course, it's nice to see you write unambiguously:

"On the other hand, arguments on the lines of "the host is offering you the choice of Door 1 versus Door 2+Door 3, therefore you should switch (2/3 versus 1/3)" are completely correct. The probability Door 1 hides a car doesn't change on revealing a goat behind one of the other doors, as yet unspecified. So it remains 1/3, and the other as yet unspecified door hides the car with probability 2/3. Richard Gill (talk) 08:20, 10 January 2011 (UTC)"

I think you properly described 'indifference', rather than 'before a door is opened'. Of course, that supports Devlin's solution as well. I'm anxious to hear what the other editors think of your statement. Glkanter (talk) 09:03, 10 January 2011 (UTC)

No @Glkanter, I am not talking about indifference, I am talking about before/after. Because Devlin is doing that himself, explicitly. Before Door 3 is opened the probability the car is there is 1/3. Afterwards it is 0. He says so himself. Nothing to do with indifference. He nowhere says anything explicit about his indifference as to whether Door 3 or Door 2 is opened; but he should have done. Because of his indifference (because of symmetry), the probability Door 1 hides a car is independent of whether Door 2 or Door 3 is opened, i.e., the conditional and unconditional probabilties that Door 1 hides a car are the same. See any first course in probability. If B has non-zero probability, then A is independent of B if and only if the conditional probability of A given B equals the unconditional probability of A. Richard Gill (talk) 09:19, 10 January 2011 (UTC)

Simple solution

What does the simple solution say? With the assumptions that the car is placed randomly and the player is completely unknown with the position of the car, the initial choice gets the car with probability 1/3, and hence switching will let you win the car with probability 2/3. So you should switch! Let's see what happens further. The player chooses door 1. The host opens door 3 showing a goat. And then, surprisingly, the host for a short moment opens door 1, the one you've chosen, and it appears the car is behind it. What does the simple solution say? Switch!!! That's where it leads if you do not condition on the information provided to you. Nijdam (talk) 22:53, 14 January 2011 (UTC)

If you are suggesting that the host revealing the car behind door 1 is a premise of the MHP that needs to be considered in any solution, I disagree. And if you were correct, the result would no longer be 2/3 & 1/3, and the famous Monty Pall Problem paradox would not exist. Glkanter (talk) 22:59, 14 January 2011 (UTC)

What if the information available to me is that the host has opened either door 2 or door 3 to reveal a goat? Martin Hogbin (talk) 23:01, 14 January 2011 (UTC)

What if you try to understand what I'm pointing at?? Nijdam (talk) 23:34, 14 January 2011 (UTC)

Are you pointing out that the opening of a door by the host can change the probability that the car is behind the originally chosen door? If not, perhaps you could explain further. Martin Hogbin (talk) 02:07, 15 January 2011 (UTC)

You almost got it: not "can change", but "does change". Nijdam (talk) 09:12, 15 January 2011 (UTC)

That is an odd statement. So in the K&W formulation the opening of door 3 by the host to reveal a goat changes the probability that the car is behind door 1 from 1/3 to 'a different' 1/3? Martin Hogbin (talk) 10:41, 15 January 2011 (UTC)

We've through this before, and I'm not in the mood explaining this to you for the so many'th time. Nijdam (talk) 11:31, 15 January 2011 (UTC)

You are obviously using 'probability' in a different sense from me. As far as I know it is a simple number, a scalar, it has no properties except a magnitude. In the sense that I use it, the probability of winning the car in the MHP if I swap is the same as the probability of initially picking a car in a game with 2 cars and one goat, it is 2/3. The questions may be completely different and the probabilities my refer to different concepts but they are both 2/3. They are both have the same probability. What exactly do you mean by 'probability'? Martin Hogbin (talk) 11:55, 15 January 2011 (UTC)

Well. yes, a probability is a number (etc.), but it is a number assigned to what is called an event, a function in fact. It is however not a property of the event, other than in the given circumstances. So, even when the event stays the same, its probability may change, vary. For instance a fair die, i.e. a nicely symmetric constructed die, does not necessarily have a uniform distribution. It is not only the physical properties of the die, but also the way it is thrown, that counts. And also what we already know of the outcome! If the die is thrown randomly in a cup (or how do you call it), and some one has seen the outcome and says it is an odd number, then the outcome is (conditional) uniform on the numbers 1, 3 and 5. Nijdam (talk) 12:36, 16 January 2011 (UTC)

A point less understood in the simple "solution" is the word switch. The argument is: switching will give you the car with chance 2/3. Is that true? If the player switches, she only has the possibility to switch to door 2. Although the car is with probability 1/3 behind door 1, it is also with probability 1/3 behind door 2. Hence unconditional switching will give you the car 1/3 of the times, just the same as if you stick to your original choice. Nijdam (talk) 00:08, 18 January 2011 (UTC)

No, W.Nijdam. –  Please no higgledy-piggledy pell-mell. You have to keep things apart. Everyone can distinguish those two different things you are disarranging. Please keep apart the fact that the host will open "one of his doors" from your "bare and unproven assumption" that, by his procedural method of choosing that door, he eventually might be giving or might have been giving some closer hint on the actual location of the car. In case you assume that he opens (resp. has opened) his more or less "avoided door", then you have to assume that the car is more likely to be behind his preferred door, so the door selected by the guest will be "less likely" to hide the car, the odds will no longer be 1/3 but will fall below 1/3, as a consequence. And vice-versa. That's just all of your secret matter. But all of your most extreme unproven assumptions will never provoke that, even in "your very special case", staying can ever be better than switching. So the actual conditional probability to win by switching will in any case be within the range of at least 1/2, but never less, to 1 !  On average even 2/3 !  The weakness of the "conditional solution" is that it just can "assume" that the odds on the door originally selected by the guest could change or could have changed (just as a consequence of your never guaranteed assumptions), but without any consequence to the decision asked for.

Conditional probability is correct and clear, and therefore it should at least remain honest. Some mathematicians say they can give a closer result for any actual game. But they should also say that they only can do that just based on flat assumptions they make, assumptions that never can be guaranteed to be true. Just flat assumptions.

The simple solution says that probability to win by switching is 2/3 on average, so you should switch in any case. And the conditional solution (forever only based on mere assumptions) can proof that nothing can ever be better than switching. That's good, maybe better, but doesn't dispraise the simple solution. So no necessity to distinguish which of his two doors the host will open resp. has opened. Because all of your subsequent conclusions can only be and will only be based on your very flat assumptions concerning your secret "additional hint". It was fine if we could praise conditional probability to be as clear as honest. Gerhardvalentin (talk) 01:57, 18 January 2011 (UTC)

They Are *Shortcuts* That Are Only Utilized For A Specific Problem Statement

The simple solutions will not derive the correct result to the door 1 and door 3 problem if certain (any?) premises are changed, omitted, or added. In the case of the MHP, with Selvin's and K & W's premises, those who recognize the simple solutions as a valid and accessible shortcut to the conditional solutions, have analyzed the problem definition correctly. Further, there are no conclusions regarding the switch vs stay question put to the contestant that may be drawn by using the conditional solution that cannot also be derived from the simple solutions. No statement as to the suitability of these shortcuts (simple solutions) to a problem with different premises is given or implied. This use of a shortcut, or *simpler tool*, in lieu of a more complicated tool is a proper and often used technique in many disciplines, including mathematics. Glkanter (talk) 23:36, 14 January 2011 (UTC)

Door Numbers Are Not *Conditions*

There's only one condition in the MHP: That the host will reveal a goat behind another door each time the game is played. And that is a 100% condition, which with all the other premises, eliminates the need for a tool that can solve a conditional problem. The tool used only needs to solve the door 1 and door 3 problem statement. The specific case of door 1 and door 3 is merely one of the finite, unique, but equally likely door selections and car & goat placement permutations. Glkanter (talk) 00:12, 15 January 2011 (UTC)

You *Could* Present The Simple Solutions As Simple, Elegant, But Only Useful In Certain Cases

The reliable sources that point out that the simple solutions don't generalize when certain assumptions are changed do not have to be considered critics. Rather, the Wikipedia article could introduce the simple solutions as wonderfully accessible devices to explain the 2/3 & 1/3 vs 50/50 paradox of door 1 and door 3, *because* of the unique premises of the problem. It would be very useful to point out that if certain premises were changed, one might need a tool that can handle the complexities of conditional problems (unlike the 100% likelihood of revealing a goat behind another door). Then refer the reader to a section which gives voice to the reliable sources who offer up a host bias or the random Monty, or whatever, to explain to the interested reader what a conditional problem looks like. Glkanter (talk) 00:37, 15 January 2011 (UTC)

The conditional decision tree solution and Bayes Theorem solution could likewise be introduced as applicable in a great many more situations than the simple solutions that solve the MHP, but are more complex, and generally require some amount of formal education to understand how to apply them. The same reference to the other section would be provided for the interested reader. The conditional solutions *only* would be presented, all talk about variants, host bias, reliably sourced criticisms of the simple solutions, etc. would exist elsewhere in the article. Glkanter (talk) 00:59, 15 January 2011 (UTC)

The Simple Solutions Do Not Rely On "Intuition"

The simple solutions properly solve the door 1 and door 3 because of 'symmetry'. Not intuition. That the 6 different door choice & opening combinations all have equal switch or stay probabilities is readily evident from Selvin's & K & W's premises that the car is placed at random, the contestant does not know where the car is, and the host chooses randomly when faced with 2 goats. This is consistent with the TV game show watching experience and expectations of a puzzle solver hearing the MHP for the first time.

Otherwise, most simple solutions rely on rote listings of equally likely outcomes, or the initial 2/3 & 1/3 probabilities for each door, the lack of any new information as to the car's likelihood of being behind the contestant's door when the door revealing a goat is opened as promised, and that the total probability must equal 1.

That some sources didn't feel it necessary to include the statement that all door combinations are equally likely, or whatever your criticisms are, was each source's own judgement. No Wikipedia editor can state that the source was 'required' to mention a particular item. I've heard enough about each source's flaws in how they present their simple solutions. The facts are the simple solutions are capable of solving the door 1 and door 3 problem, they come from reliable sources, and there is no single 'correct' way to present these solutions. Glkanter (talk) 03:12, 15 January 2011 (UTC)

Why Morgan's Criticisms Are Not A Significant Minority Viewpoint

Morgan criticizes 6 solutions as false:

1. Regardless of what the host does, and the player never switches, the player wins 1/3 of the time, hence switching wins 2/3

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

2. vos Savants 3 row listing

Also solves the unconditional problem as door 2 being opened is listed as an outcome

3. vos Savant's card game simulation

Also solves the unconditional problem

4. A 50/50 solution

Wrong

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

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

6. Mosteller's formula version of the conditional decision tree

Relies on the unstated 50/50 host bias

They conclude their paper saying that the above solutions are all false, and that their solution that it's always 50% or better to switch even without knowing the host's bias is the 'correct resolution'.

In their rejoinder to vos Savant, they state regarding vos Savant's card game simulation:

"Certainly the condition p = q = 1/2 should have been put on via a randomization device at this point. It could also have been mentioned that this means that which of the unchosen doors is shown is irrelevant, which is the basis for solving the unconditional problem as a response to the conditional one. But then, it may be that only an academician, and no one connected with a game show, would ever consider p <> q."

So, Morgan claims:

1. The 50/50 host is not a premise of the MHP

No other sources say this

2. The 4 simple solutions are false because they really solve a different unconditional problem that they created. However, with the 50/50 host premise (see #1, immediately above), they acknowledge that the simple solutions do indeed, solve the conditional problem.

Because of #1, the simple solutions are not 'false' at all.

3. The conditional solution is false because it relies on the 50/50 host premise

Because of #1, the conditional solution is not 'false' at all.

4. Their 'correct resolution' is that winning by switching is always 50% or better, even without a host bias premise.

Every other source says the answer is 2/3 & 1/3.

Morgan acknowledged something about 2/3 being the answer in their 2010 response to Martin and Nijdam

Conclusion: Morgan's paper does not represent a significant minority viewpoint. Glkanter (talk) 04:54, 15 January 2011 (UTC)

And I don't need Morgan's paper to know that it's never worse than 50/50 if I switch. There's 2 doors, 1 car and 1 goat. And the car placement was initially randomized. That little green man who only knows 2 doors, 1 car and 1 goat has a 50/50 chance. Morgan's 'correct resolution' adds nothing. Glkanter (talk) 18:08, 15 January 2011 (UTC)

And, Really...

OK, without the 50/50 host premise, symmetry does *not* enable the simple solutions to solve the conditional door 1 and door 3 MHP, returning the result of 2/3 & 1/3. So what? Without the 50/50 premise, the conditional solutions *don't* solve the MHP returning the result of 2/3 & 1/3 *either*. In fact, the conditional tools are useless, unless you have a Bayesian Prior, which the problem statement does not provide.

And, really, is Morgan's "Although a specific numerical result cannot be determined, it's always greater or equal to 50% when you switch" any better than "Because that specific probability can't be given a numerical answer, I can still calculate the average of all door combinations and car placements as 2/3 & 1/3, so I'll switch"? Both answer the question, "Should the contestant switch?". Good luck winning your bar bet, or benefitting a Wikipedia reader with Morgan's solution. Glkanter (talk) 10:40, 15 January 2011 (UTC)

A Starting Point For Peaceful Co-existance.

The following is copied verbatim from a discussion on a mediation sub-page. I added a subsection called 'This is where it gets good. You can skip to it if you like.

Most editors may want to standardise on the K&W formulation but this does not make it what Whitaker actually wanted to know. The most well known statement of the MHP was a question from a member of the public to a column in a general interest magazine. Before we attempt to answer it we must ask ourselves exactly what he actually wanted to know. Martin Hogbin (talk) 19:48, 9 January 2011 (UTC)

Martin, what you describe above is not required, outside the scope of our roles as Wikipedia editors, and insures a continued stalemate in this mediation. Otherwise, I support it fully. Glkanter (talk) 20:11, 9 January 2011 (UTC)

Well, many sources give solutions that, according to some, only apply to the unconditional formulation. Do we assume the authors were all idiots or might we assume that some intentionally answered the unconditional formulation because that is how they interpreted Whitaker's question. Also, Seymann raises this point, so it should have a mention. Martin Hogbin (talk) 00:03, 10 January 2011 (UTC)

Look at this very carefully. The critics say 2 things, one factual, one an opinion. And, these critics are a minority. They say that the solution offered is an unconditional solution. That is a true statement. Then they say the author can only be solving an unconditional problem with the simple solution, which is a different problem than the door 1 and door 3 100% conditional MHP. That statement is not directly attributable to the sources (who included no such delineation), rather it is the critic's own conclusion. And, based on our conversation below, that the unconditional solutions, thanks to symmetry, do solve the conditional problem, it is incorrect. Glkanter (talk) 13:46, 15 January 2011 (UTC)

Not at all. Each solution from a reliable source should be considered for inclusion in the Solution section. That the reliable sources saw no need to artificially bifurcate the problem statement and choose a side only means to me that we would be overstepping our editorial roles if we did so. The controversies would be properly addressed elsewhere. Glkanter (talk) 04:06, 10 January 2011 (UTC)

I strongly disagree with the 'cut-and-paste' approach to sources; we should attempt to do better than that. WP requires that what we say be supported by reliable sources. Martin Hogbin (talk) 12:51, 15 January 2011 (UTC)

Martin, please, don't Aunt Sally me. Glkanter (talk) 12:56, 15 January 2011 (UTC)

I have written extensively today/yesterday, and previously, that reliable sources, Wikipedia editors who are math professionals, and mathematics itself, all say that the simple solutions do indeed solve the 100% conditional door 1 and door 3 symmetrical MHP. (Not to mention the countless reliable sources who simply offer the simple solutions when presented with the 100% conditional door 1 and door 3 symmetrical MHP without further comment that they are solving a different problem.) Do you disagree with any part of this? Glkanter (talk) 13:04, 15 January 2011 (UTC)

Good to see you using the British term ;-). It is what you appeared to me be saying, but evidently not what you meant. Martin Hogbin (talk) 13:06, 15 January 2011 (UTC)

I keep telling people that I respond favorably to coaching, yet I have many doubters. More importantly, do you agree with the above paragraph? Glkanter (talk) 13:10, 15 January 2011 (UTC)

Yes. Martin Hogbin (talk) 13:20, 15 January 2011 (UTC)

Then I think you would agree:

1. While there have been many different ways of saying it, all sources are describing the same conditional problem where door 3 has been opened, and they give their solution to that problem.

2. Therefore, there are no sources mistakenly solving an unconditional Monty Hall *problem*, but there are many unconditional Monty Hall *solutions* which due to symmetry solve the 100% conditional Monty Hall problem.

Do you still agree? Glkanter (talk) 13:35, 15 January 2011 (UTC)

1 Many sources describe the conditional formulation, some may not, others give Whitaker's ambiguous question and then provide an answer without saying exactly what question they are answering.

2 Based on Whitaker's question, I do not think it is a mistake to solve the unconditional formulation.

I would add that I do not know of any sources which clearly say that the simple solutions to the symmetrical problem are wrong. That seems to me to be a view that exists only in this forum. Martin Hogbin (talk) 15:27, 15 January 2011 (UTC)

1a. Have you ever, in all your MHP readings, come across a problem statement that is obviously the unconditional problem as described by Morgan, which is presented as the MHP, and solved using simple solutions? That is, without trying to make a point about rigor, or generalizing, or as a launching pad to conditional probability, etc? If so, I would suggest it is not a significant minority viewpoint and can be disregarded for Wikipedia purposes.

1b. I would say any problem statement that says door 3 has been opened must be defined as the door 1 and door 3 100% conditional problem. Have you seen any that omit this?

2. I'm sorry to hear that. If, as you say above, no reliable sources criticize the simple solutions as not solving the conditional problem (what about Morgan?), and every reliable source has door 3 being opened, and no reliable source states they are solving the unconditional problem, by what basis is it appropriate to prominently include an unconditional problem statement in Wikipedia? Where does the unconditional problem description exist in the reliable sources as being referred to as the MHP? Again, if you find 1 example, I would say it did not represent a significant minority viewpoint, and could be treated accordingly. Further, is that conclusion consistent with Selvin and K & W? Honestly, it sounds like OR to me.

Bonus. Then why can't we remove or otherwise reduce the OR, NPOV, and Undue Weight violating criticisms from the article? Glkanter (talk) 16:08, 15 January 2011 (UTC)

This is where it gets good

Now, if you wanted to say that the same symmetry rule that makes the simple solutions valid for the conditional problem lets you reduce the conditional problem to an unconditional problem, I would say you are mathematically correct. And that's what goes on in bars every night. But I would also say that is OR, and not appropriate for the article. And if you said it was not OR, but a proper application of mathematics, I would say it was not appropriate for the solutions section. Glkanter (talk) 16:18, 15 January 2011 (UTC)

Upon further review, I think some simple solution sources actually *do* reduce the problem from conditional to unconditional, then offer a simple solution. But they are not describing a different problem. They are applying symmetry, along with all the other premises in order to use the various shortcuts, or simple solutions. I don't think there are two different problems. If you read my paragraph on your talk page about praising the simple solutions, rather than criticizing them, I could see including this topic in the intro. In fact, saying that symmetry and the other premises lets you restate/reduce the problem as unconditional, then using a simpler tool, is really outstanding as far as reader understanding goes. But not delineating it as a separate problem. Glkanter (talk) 16:31, 15 January 2011 (UTC)

For example, here's Carlton:

"5. Comments and variations on the Prisoner’s Paradox & the Monty Hall Problem

"Before presenting a formal solution to the Monty Hall Problem to my students, I find that it helps to give an intuitive explanation for the 1/3 - 2/3 solution. Imagine you plan to play Let’s Make a Deal and employ the “switching strategy.” As long as you initially pick a goat prize, you can’t lose: Monty Hall must reveal the location of the other goat, and you switch to the remaining door - the car. In fact, the only way you can lose is if you guessed the car’s location correctly in the first place and then switched away. Hence, whether the strategy works just depends on whether you initially picked a goat (2 chances out of 3) or the car (1 chance out of 3)."

Ok. He calls it 'intuitive', which is his prerogative. What he's done, of course, is reduce the problem using the symmetry, then used a simple solution. So here's a guy who actually gives the door 1 and door 3 problem statement, tells you he's reducing it, then gives a simple solution. There's the reason I always liked this paper. Glkanter (talk) 17:02, 15 January 2011 (UTC)

I agree that the sharp distinction between conditional and unconditional is somewhat artificial and very unhelpful. Martin Hogbin (talk) 11:09, 16 January 2011 (UTC)

For anyone with some formal (mathematical) training in probability theory, if only at very elementary level, there is a very very very sharp distinction between an unconditional probability and a conditional probability. Conditioning in probability theory is not the same as listing conditions (premises, assumptions) to a problem. But I think there has been some confusion between possible meanings of the word "conditional" in the past. Richard Gill (talk) 12:25, 16 January 2011 (UTC)

Let's cut the crap, shall we? This has nothing to do with my level of understanding of probability theory. If I die tomorrow, nothing changes.

1. Given that the host has a 50/50 bias, because of symmetry, are the conditional and unconditional solutions solving the same interpretation of the conditional Monty Hall Problem?
2. If no, are they both *capable* of doing so?
3. If 'yes', is this response based on the reliable sources of Morgan and G & S, Wikipedia editors Boris and Richard, and the discipline of Mathematics?
4. Is this excerpt from Richard's 2011 paper on page 8 relevant to this discussion?:

Prop. 3: If all doors are equally likely to hide the car then by independence of the initial choice of the player and the location of the car, the probability that the initial choice is correct is 1=3. Hence the unconditional probability that switching gives the car is 2=3. If the player's initial choice is uniform and the two probability distributions involved in the host's choices are uniform, the problem is symmetric with respect to the numbering of the three doors. Hence the conditional probabilities we are after in Proposition 3 are all the same, hence by the law of total probability equal to the unconditional probability that switching gives the car, 2/3.

Please, if any answer is no, *try* to respond in a manner that I might comprehend. It would be much appreciated. Glkanter (talk) 15:03, 16 January 2011 (UTC)

Richard, perhaps you could explain to me exactly what this sharp distinction is as I obviously have not had the appropriate training. Martin Hogbin (talk) 17:22, 16 January 2011 (UTC)

New Edits commence

Added bonus: Carlton explains why the simple solutions should precede the conditional solutions in the article. Glkanter (talk) 17:47, 15 January 2011 (UTC)

That's right. It motivates his students to get down to the nitty-gritty tricking conditional probability solution, the only way you can be guaranteed to find the optimal solution without thought or luck - all you need is perseverance. Richard Gill (talk) 12:27, 16 January 2011 (UTC)

Once again (as always), the non-value-added tangent about 'optimal', when the grownups are discussing 'should you switch?'. Glkanter (talk) 14:09, 16 January 2011 (UTC)

Whatta you guys think of this OR simple solution?

How's this solution to Selvin's and Whitaker/vos Savant's question 'should you switch, given that the host has opened door #3'? It's simple, answers the question asked, *and* does not require the 50/50 host premise.

With the initial random car placement the contestant knows that his chance of selecting the car was 1/3. With only doors 1 and 2 remaining, 1 car and 1 goat, he knows door 2 can't be any worse than 50/50. Heck, his door might still be only 1/3 for all he knows, he's no rocket scientist, and it sure can't be any better than the compliment of door 2's 50/50 worst case. Well then, he summarizes, the odds for my door are between 1/3 & 1/2 inclusive, and door 2's odds are between 1/2 & 2/3 inclusive. So, if its 50/50 I'm no worse off, otherwise, I've gained *some* advantage, by switching. Glkanter (talk) 01:05, 16 January 2011 (UTC)

You say, '...he knows door 2 can't be any worse than 50/50'. How does he know this? Martin Hogbin (talk)

10:13, 16 January 2011 (UTC)

Good question. Morgan et al prove that that the posterior odds on Door 2 are between 50/50 and 100/0 (this is also Gerhard Valentin's Leitmotiv). I have previously given really short proofs using the odds form of Bayes rule. Indeed, it can't be worse than 50/50, but it can even be a whole lot better than 2/3 ! Richard Gill (talk) 12:31, 16 January 2011 (UTC)

Richard, how can it 'be a whole lot better than 2/3 !' without changing premises? Do you understand the problem we are discussing on these pages? Glkanter (talk) 20:44, 16 January 2011 (UTC)

Martin, I tossed and turned on that extensively. And the answers I came up with were:

• Well, no one has *ever* suggested that staying was a *better* option
• The same way Morgan knows
• For an observer who only sees the 2 doors and is told of 1 goat and 1 car, the SoK would be 1/2 & 1/2
• Common Sense
• There is no counter-argument

I wonder if the hurdle of 'proof' is different if one is to make a decision vs calculating a probability.

Thanks, for nothing, Richard. Your response, as usual, goes off on a tangent far afield from Selvin & Whtitaker/vos Savant's question, is OR that cannot be used for Wikipedia purpose, and, in my opinion obscures this entire wonderful discovery, without adding any value whatsoever. For a guy who hangs his hat on the question being, 'should the contestant switch' rather than, 'what are the probabilities', the answer to 'what is optimal' sure comes up an awful lot. And how hard is it to figure out that given an infinite number of discreet game plays in which the premises never change, the best decision never varies? Glkanter (talk) 13:34, 16 January 2011 (UTC)

You told me so!

Well actually you didn't. If any one claims to have already given me this answer I ask them to show me exactly where.

I have asked several times on what basis an event must be considered as a condition in a probability question and have never received a complete and satisfactory answer. However I have found a reason why, on being given the event that he host has opened door 3 to reveal a goat, we should take this as a condition.

According to some formulations of the MHP we are given that the player has chosen door 1 and the host has opened door 3. As these are events are not part of the rules of the game, and therefore do not happen with certainty, we might reasonable start with a sample space based on the rules of the game and condition it based on the two events given.

There is a difference in principle though between these two events and that is this: we could, it turns out, make the event that the player picks door 1 part of the game rules without changing the nature of the problem. Provided the producer initially places the car uniformly at random and the host opens an unchosen goat-door uniformly at random, the game rules could, rather bizarrely, specify that the player must choose door 1. The paradox would be exactly the same.

In contrast, we cannot do this with the host's choice of door. If we make it part of the game rules that the host must open door 3, the whole problem falls apart. The car cannot be initially placed uniformly and the player cannot choose uniformly. Even if we restrict these choices it is very difficult to retain the original paradox.

I do not think this fact changes many of the arguments but it may explain why some editors are so insistent that the opening of door 3 by the host must be taken as a condition.

I do not think that this difference has been noticed before. Anyone who thinks that they have noticed it, please show me where. Martin Hogbin (talk) 12:54, 16 January 2011 (UTC)

Well, I hit on at big parts of it, anyways. I've said repeatedly that the only condition, based on K & W's premises (or any problem statement, really), is that the host always reveals a goat behind one of the other doors, and always offers the switch. Since doors 1 and 3 are not premises, they cannot be required conditions. I've also stressed, on numerous occasions to you and Richard, think about the decision tree I derived from Carlton's solution, that there *is* importance to the 100% condition. You and Richard both blew me off on that, and I never agreed with you guys. By the way, one of those Morgan false solutions is the same as Carlton's, and specifically mentions multiplying by 1. Their only criticism is incomprehensible, "Solution F5, like F1, is a true statement that answers a different problem. F5 is incorrect because it does not use the information in the number of the door shown." By my way of thinking, this supports the table as *not* being OR. And gives a clearly understandable visual explanation of why it's 2/3 & 1/3, that any bar patron or Wikipedia reader could easily understand. I've also said that this proves that '1/3 = 1/3' in the MHP, which I believe some editors are still discussing, without signs of progress.

So, do you agree both solutions are solving the identical MHP, and that there is, indeed, only 1 problem statement? So we need not define the problem, etc...? Have we broken the stalemate? Glkanter (talk) 13:46, 16 January 2011 (UTC)

No. I am just showing that the 'conditionalists' have a point if the question is construed to be specifying door numbers. Martin Hogbin (talk) 14:08, 16 January 2011 (UTC)

I'm certain you are wrong. Glkanter (talk) 14:49, 16 January 2011 (UTC)

I say a lot in that post. Maybe you could tell me which parts you disagree with. Glkanter (talk) 15:04, 16 January 2011 (UTC)

Monty Hall and Centrifugal Force

Martin, it seems that you have strong views both on centrifugal force and Monty Hall. And it seems that you may actually agree with me when it comes to Monty Hall. So since we can't agree on centrifugal force, I'd like to examine your reasoning on the Monty Hall issue. I'll tell you my reasoning.

On being asked would he change his mind once the door is opened, the simple solution applies correctly before the door is opened. Rick Block claims that when the door is actually opened, we need to do a more rigorous calculation in order to establish the same answer. I counter argue that we know in advance that the mere opening of the door (neglecting host bias) is not going to change the probability, hence the simple solution is adequate.

Rick Block argues that the simple solution, after a door is opened, is like a crude way of getting the correct result, and he makes the analogy of using an average distance and time, as opposed to an instantaneous distance and time, for a uniform speed calculation. It's an interesting analogy but it doesn't, in my view, detract from the fact that in Monty Hall, the simple solution is adequate, even when a door has been opened.

Have I missed anything? David Tombe (talk) 16:27, 16 January 2011 (UTC)

Yes, you have missed a lot. There are actually two forms of what may be considered the simple solution. (1) As the car is placed randomly, it is with probability 1/3 behind the chosen door 1. As the opened door 3 shows a goat, the remaining closed door 2 must hide the car with probability 2/3. (2) As the car is placed randomly, it is with probability 1/3 behind the chosen door 1. Hence switching wins the car with probability 2/3. (1) has a logical error in it, by not distinguishing the probabilities before anything happened and after the host opened a door. (2) is a very simple reasoning, but does not calculate the required (conditional) probability. It solves a different problem, in which the player is asked if she will switcht even before she has made her first choice. It should not be difficult to understand these things, yet some participants in the discussion haven't coped, despite more than 2 years of explaining. Nijdam (talk) 23:17, 16 January 2011 (UTC)

Your disagreement concerning centrifugal force is not so much with me as with physicists in general. I would be happy to explain it all to you if you will agree to forget about conspiracy theories amongst physicists to deceive the world.

Regarding the MHP you will have to make up your own mind and contribute to the discussion as you wish. Martin Hogbin (talk) 19:34, 16 January 2011 (UTC)

Martin, In that case, we'll have to forget about discussing centrifugal force. As regards MHP, is the key point that we know in advance what the probability will be even before the host opens the door, hence the mere fact of him actually opening it is not going to make any difference to the probablility? But while Rick Block agrees that it will not change the probability, he argues that it does change the manner in which we should calculate that probability? My counter argument to Rick Block would be that we have already calculated the probablity in advance and so the simple solution is adequate. David Tombe (talk) 22:44, 16 January 2011 (UTC)

No, Rick does not argue that the probability is not changed. On the contrary: the (?) probability itself changes from unconditional to conditional. The confusing fact for some participants in the discussion is that both probabilities have the same value. It has nothing what so ever to do with the way of calculation. Your counter argument is incomprehensible, as you (i.e the simple solution) have calculated nothing in advance, Nijdam (talk) 23:25, 16 January 2011 (UTC)

I've been on this cruise ship from Hell alongside Martin for over 2 years, David. There's no point in looking for any logic in Rick's, or any sympathetic editor's postings. They use every method of gamesmanship imaginable. It's horrible. Intellectual dishonesty at it's worst. And as I think we've discussed before, I have no clue what motivates them. Glkanter (talk) 22:57, 16 January 2011 (UTC)

Glkanter, OK. You can go to my talk page and take up where Martin left off. See if you are still on Martin's side when it comes to centrifugal force. I would certainly appear to be on Martin's side on MHP. But I think I know what motivates my opponents at centrifugal force. David Tombe (talk) 23:27, 16 January 2011 (UTC)

No thanks. I'm a one-trick pony on Wikipedia. Anything more complex than 1/3 * 100% = 1/3 is beyond my job classification. Glkanter (talk) 23:35, 16 January 2011 (UTC)

ps Don't do it, man. Martin has gone right done that rat hole you're standing a little too close to. Look at any of his or Nijdam's talk pages. There is *no* payoff. You will not prevail within Wikipedia's rules. Glkanter (talk) 23:35, 16 January 2011 (UTC)

Glkanter, That's OK.David Tombe (talk) 23:48, 16 January 2011 (UTC)

:Nijdam, If we know in advance what the probability will be, why should the probability be altered when a door is actually opened? You of course agree that it is not altered. But you seem to think that it needs to be recalculated nevertheless. I don't see why. David Tombe (talk) 23:29, 16 January 2011 (UTC)

Nijdam, I didn't see your second posting. You say that we don't know the probability in advance. But we do. We know, even before the host has opened the door, that if he does open the door and offer the choice to switch, then it will be 2/3 to our advantage to switch. Hence nothing is going to change when he does actually open the door. The probability will still be 2/3. David Tombe (talk) 23:51, 16 January 2011 (UTC)

You definitely have some problems with logic. Neither you nor me said anything about knowing a probability in advance. You said in your first posting: I counter argue that we know in advance that the mere opening of the door (neglecting host bias) is not going to change the probability, hence the simple solution is adequate. and in your second: My counter argument to Rick Block would be that we have already calculated the probablity in advance and so the simple solution is adequate. See? In your reply however you mention that we know before the door is opened, it will be 2/3 to our advantage to switch. I would not call this knowing in advance, it has been calculated without taking notice of the opening of the door, if that is what you mean. This is true as long as you don't know which door is opened. Once a door is opened different probabilities govern the situation. Although it turns out it is also 2/3 to our advantage to switch, it is a different 2/3 than before. Compare it with 2/3 of all the British people and 2/3 of the inhabitants of London. Study probability theory! Nijdam (talk) 10:02, 17 January 2011 (UTC)

Let's take the usual assumptions car initially equally likely to be behind any door, host equally likely to open either door when he has a choice. Player chooses Door 1: odds the car is behind there are 1:2 against. Host opens a different door revealing a goat: no change to the odds, since he was equally likely to do that under either hypothesis (it was certain, in each case). He tells you the identity of the opened door: Door 3. Again, the odds for and against Door 1 do not change since the "information" you have gained is not informative for the issue at stake ... It comes to you with probability 50% in either case. If the car *is* behind Door 1, the host is equally likely to opened Door 2 or 3, thus it's Door 3 which is opened with probability 50%. If the car isn't behind Door 1 it's equally likely to be behind Door 2 or 3, hence equally likely the host will open Door 3 or Door 2. Thus it's Door 3 which is opened with probability 50%. So knowing the identity of the opened door doesn't change the odds for and against Door 1. Of course the Odds on Door 3 do change at this moment, and hence also on Door 2.

The difference between simple and full solutions is merely whether you bother about the possible informativeness of the identity of the opened door, regarding your strength of belief that the car is behind Door 1. The answer is that it turns out you needn't have bothered. It would have been possible to argue that in advance. But either way, both solutions are right, as far as the man in the street is concerned, but the conditional solution is more complete. On the other hand the man in the street has difficulty in appreciating the subtleties involved. Richard Gill (talk) 09:49, 17 January 2011 (UTC)

Maybe the moment has come to discuss the problem with the simple solution, the one just concerning the unconditional probability. It is not a solution to what you call the full MHP, i.e. the formulation in which the player is offered to switch after the door with the goat is opened by the host. You said so more or less yourself. The answer is that it turns out you needn't have bothered. It would have been possible to argue that in advance. but the simple solution neither says nor argues this way. A (simple) solution that calculates the conditional probability by using the symmetry, or reasons that its value must be 2/3 in some way, would be correct. But alas, that's not what the simple solution does. Nijdam (talk) 10:17, 17 January 2011 (UTC)

Richard and Nijdam, I think that Richard has stated the issue correctly when he says that both solutions are correct but that the conditional solution is more complete. But I counter argue that if the simple solution is correct, is the more complete solution actually necessary? David Tombe (talk) 11:57, 17 January 2011 (UTC)

I guess, David, you're neither an expert on probability theory nor even on mathematics. Also you're not a party in the present mediation process. I don't intend to explain much more as I have done. Study it, or study the lengthy discussion of the last two years. And if you want to really understand, then study what I explained to you above and comment on it. Nijdam (talk) 17:37, 17 January 2011 (UTC)

+1 You're so right, David. – The more complete solution is a clear solution, yes, but of no direct benefit for answering the MHP-question "is it of advantage to switch", but it obviously is a welcome side-arena for teachers of conditional probability theory in showing their students what the conditional probability would be after a goat appears behind door #3, under the condition that the host eg just never uses to open door #3, if any possible, – but, contrary to expectations, actually did so. Although not impairing the "paradox" of the MHP, it is a famous and very welcome example in teaching cond.prob.theory, indeed. Its clarity to be shown in the article it is absolutely necessary to show its real aspect, without confusing wishy-washy ("Before&After") statements that it was "indispensable" for correctly answering the simple question of the Monty Hall Paradox:  "Is it of the contestant's advantage to switch?"   Gerhardvalentin (talk) 13:37, 17 January 2011 (UTC)

Outstanding David. You may want to rephrase your question to Richard as:

Which questions *do* the simple solutions answer equally as well and completely as the conditional solutions?

1. Should the contestant switch?
2. What are the probabilities?
3. Are the probabilities 2/3 & 1/3?
4. What is the optimal strategy for the contestant?
5. How does knowing the optimal strategy further benefit the contestant who already knows the probabilities are 2/3 & 1/3, and therefore he should switch?

And then, simply for clarity, you might ask Richard:

Does:

• Knowing the uniformly at random distribution of the car is 1/3, 1/3, & 1/3

• Knowing the host will select uniformly at random (1/2 & 1/2) between doors when the contestant has selected the car

• Allow the contestant using a simple solution to correctly conclude that the probability of a goat being behind door 2 or door 3 are 2/3 & 2/3 after he has selected door 1?

• Yes Richard Gill (talk) 18:36, 17 January 2011 (UTC)

• Allow the contestant using a simple solution to correctly conclude that the odds the car is behind his door is 1/3 after door 3 (or door 2, for that matter) is opened?

• The answer given here is correct but it takes more than "the simple solution" to draw this conclusion. A few words of explanation are needed. Richard Gill (talk) 18:36, 17 January 2011 (UTC)

Might Richard be willing to share those 'few words of explanation" required to draw this conclusion? Glkanter (talk) 00:45, 18 January 2011 (UTC)

Richard, are you referring to the need for the host to choose equally between goats when the contestant has chosen the car? Glkanter (talk) 00:45, 18 January 2011 (UTC)

• Allow the contestant using a simple solution like Monty Hall's to correctly conclude: "So knowing the identity of the opened door doesn't change the odds for and against Door 1." as you wrote above?

• Again, no. Glkanter jumps to a conclusion but does not give any argument for it. So we cannot tell if he is brilliant but fast and careless, or dumb because he doesn't even realise that he has missed a logical step. Richard Gill (talk) 18:36, 17 January 2011 (UTC)

Richard, if Glkanter had prefaced his questions with "based on the premises set out in Selvin's 1st and 2nd letters and/or K & W's formulation", would he still be jumping to conclusions? Glkanter (talk) 00:45, 18 January 2011 (UTC)

And Richard, your esp skills are failing badly. Glkanter is *not* jumping to conclusions. He would have to making statements in order to do that. Instead, Glkanter is asking questions. He prepared 2 well designed, though not perfect, I'm afraid, checklists in order to understand where people like me, and Martin, and Gerhard, and David are going off the tracks. Instead of making what I did better, you sling insults, keep secret the answers, and pontificate on shit like 'optimal strategy'. I do this in an orderly fashion, not unlike one might troubleshoot a problem. Or apply heuristics to increase the skill levels. Or maybe like a binary search technique. If I can narrow down where I'm not 'getting it', then I can increase the granularity on the smaller area. But after 2 years, I'm not jumping to conclusions. Common sense, and my advice to David should make that pretty goddam obvious. And I'm not claiming to be brilliant (except for my new OR solution to 'should you switch') or any other uncalled for, derisive, horseshit from you. This one was especially nice, professor. "So we cannot tell if he is brilliant but fast and careless, or dumb because he doesn't even realise that he has missed a logical step." Glkanter (talk) 06:41, 18 January 2011 (UTC)

Good luck, David! Glkanter (talk) 13:19, 17 January 2011 (UTC)

Glkanter and Gerhardvalentin, Yes. But we need to move on now to the issue of how to present all of this in the final MHP article. I still hold to the view that both the simple and the conditional solutions need to be presented in the article. My compromise would be that the simple solution must come first, and after that the conditionalists should be given total freedom to have their say, with the proviso that it must finally be emphasized that the simplists, who may well agree with the conditionalist solution as such, nevertheless still believe that the simple solution is adequate and that the conditionalist solution is unnecessary. David Tombe (talk) 14:40, 17 January 2011 (UTC)

+1  ! Yes, David, that's exactly the point. It should be undisputed and clear that such beautiful and elegant addition is nice and helpful, but not essential or even indispensable to understand the famous glaring Monty Hall Paradox. Gerhardvalentin (talk) 15:21, 17 January 2011 (UTC)

David, you're getting *way* ahead of yourself. Trust me. You've not introduced anything that hasn't been argued at length for over 2 years, including 1 year in mediation. I suggest you ask Richard to address the questions I asked. I've found that *very* often what I think Richard has said is *not* what Richard has actually said. And Nijdam is not in agreement with anything you think about the simple solutions. You can take my suggestion, and help me & Martin out, or you're on your own, insofar as any other editors care to re-hash 2 years of stuff with you. I am not one of those editors. Glkanter (talk) 15:24, 17 January 2011 (UTC)

@David, I have written what I think about MHP in a peer-reviewed and published paper, you can read it on internet at Gill (2011): html version on my homepage. So I'm not editing the wikipedia page any more; at least, not for quite a while. I've tried to explain various subtle points of logic and probability theory to various editors here till I'm blue in the face - the point of diminishing returns was reached several months ago. I'm presently devoting myself to write a more accessible and more neutral version on Citizendium at the invitation of Boris Tsirelson. Richard Gill (talk) 18:23, 17 January 2011 (UTC)

"Of course I taught them! Can *I* help it if *none* of them learned?" Glkanter (talk) 20:39, 17 January 2011 (UTC)

Richard and Nijdam, I'm not trained in advanced probability, and I cannot grasp why you think that the simple solution is not adequate. I believe that I can identify the probability as 2/3 without recourse to anything beyond knowing that there is a 2/3 chance that the car will be behind one of the other two doors, and that this 2/3 probability will concentrate onto one door in particular after the host opens a door revealing a goat. I have seen cases in physics where mathematics has become derailed from physical reality, and I suspect that we might be dealing with something similar here. There is a school of physicists who believe that they can use mathematics to blur the distinction between two distinct physical scenarios. If we are on a roundabout, we may witness an object sliding outwards due to physical inertia. We may also witness an object, not on the roundabout, that is sitting stationary on the ground. It will not be seen to be moving radially outwards, but it will appear to be tracing out a circle from the perspective of the man on the roundabout. Mathematicians use maths to argue that both of these objects equally have a fictitious centrifugal force of the same magnitude acting upon them, from the perspective of the rotating roundabout. I would argue that only the object that is being dragged by the roundabout, and is hence sliding outwards, is being acted upon by a centrifugal force, which is of course an inertial force. I was once locked in a prolonged argument with mathematicians over this issue, and as a result, I don't always have full confidence in expert mathematicians when it comes to them applying their mathematical expertise to real situations. And it strikes me here at MHP, that a perfectly simple problem, which makes an interesting article, has become an unnecessary battleground because some advanced probabilitists may have perhaps lost the plot, just as some advanced mathematicians seem to have lost the plot at centrifugal force.

I really have got nothing more to say on MHP other than that, since I'm not a deletionist, I do recognize the rights of the conditionalists to have their say, but providing it comes second, and providing it is made clear that the simplists are adamant that the simple solution is adequate. No matter how hard I try, I cannot see how the conditional solution is necessary. It may be correct. But it's not necessary. David Tombe (talk) 19:40, 17 January 2011 (UTC)

You said it yourself, you're not trained in advanced probability, and in my opinion even not in basic probability. Nijdam (talk) 23:47, 17 January 2011 (UTC)

Criticism of Richard's paper

Richard, maybe it is too late now but I have some criticism of your paper. You show a difference between the subjectivist and frequentist approaches to the problem but, in fact, this difference comes not from the difference in approach but from the way that you choose to treat the problem differently in the frequentist case. You say:

For a frequentist, MHP is harder – unless the problem has already been mathematized, and the frequentist has been told that the car is hidden completely at random and the host chooses completely at random (when he has a choice) too. Personally, I don’t find this a very realistic scenario.

Why not? In a modern quiz show it would undoubtedly be a regulatory requirement. Opportunities for presenters to influence the outcome are strongly discouraged. The car would undoubtedly be placed uniformly at random, much as the prizes are stated to be placed randomly in Deal or No Deal. The host would also be undoubtedly required to choose at random when he had the choice. The closest we have to this in a modern quiz show is Who wants to be a millionaire?. When the player asks for 50/50 the computer removes two random incorrect answers. Why random? Because it is not considered acceptable for the host to be able to influence the probability that the player will win by, for example, intentionally removing a wrong but plausible answer. There is no doubt that, in a modern show, both distributions would be uniform at random (from the available legal choices).

What about a historical quiz show like Let's make a deal? Who is to say? The Monty Hall problem scenario never actually happened on the show but my guess would be that, if it had have done, the producer (or stage hand) would have placed the car roughly randomly and the host would have chosen roughly randomly. You go on to say:

I can think of one semi-realistic scenario, and that is the scenario proposed by Morgan et al. (1991a). Suppose we have inside information that every week, the car is hidden uniformly at random, in order to make its location totally unpredictable to all players. However Monty’s choice of door to open, when he has a choice, is something which goes on in his head at the spur of the moment.

Semi-realistic is, in my opinion, a somewhat generous description of this scenario. Why is one choice random and the other not? The question tells us nothing about either choice and our real-world knowledge tells us that the method of choice is likely to be the same. The only reason for doing this is to artificially construct a slightly different problem so that we can get a different answer. Finally, you say:

In this situation we may as well let our initial choice of door be determined by our lucky number, e.g., Door 1. Proposition 2 tells us that not only is always switching a wise strategy, it tells us that we cannot do better. No need to worry our heads about what is the conditional probability. It is never against switching.

In this situation I would ensure that I chose a door uniformly at random, as you have often said, to guarantee a 2/3 win by swapping. If done consistently, the frequentist approach gives exactly the same answers and conclusions as the subjectivist approach. Martin Hogbin (talk) 19:44, 17 January 2011 (UTC)

Martin, Richard is substituting his personal opinions in place of Morgan's comment regarding symmetry, G & S's comment on symmetry, Boris' work, Richard's own agreement with Boris, and per Richard's own research, recognised mathematics principals. And I'm pretty sure he is contradicting his own 2011 paper. (See my talk page for more on that item.) And he is substituting his own premises for the standard ones. I hope Richard will be good enough to address the first group of questions I prepared for David. That will help all of us know at what point I am jumping to conclusions. Glkanter (talk) 20:15, 17 January 2011 (UTC)

A Little Help, Please

Richard posted this on one of Martin's talk pages, and Nijdam copied it to one of his talk pages a few days ago (emphasis mine):

"@Nijdam, you are absolutely right. "Hence" is wrong. What is said is a nonsequitur. The fact of specifically Door 3 being opened could change the likelihood that the car was originally hidden behind Door 1. So far we only used "all doors initially equally likely" and that isn't enough to get a conditional result, while this argument is about a conditional result, since it talks about how probabilities change (or don't change) on getting the "Door 3" information. There is a missing step where "no host bias" has to be explicitly used. Eg, appeal to independence: by symmetry, the number of the door opened doesn't change the probability the car is behind door 1. Or if you prefer, use Bayes' rule and show by explicit computation that the odds on Door 1 hiding the car isn't changed by the information *which* specific door is opened. Next task, find a reliable source which explicitly makes this point, because I'm afraid that some editors could consider this observation of yours "OR". (Or write it yourself: why not submit a small note to Statistica Neerlandica?) Richard Gill (talk) 07:24, 8 January 2011 (UTC)"

In my questions above, I include these statements:

•Knowing the uniformly at random distribution of the car is 1/3, 1/3, & 1/3

•Knowing the host will select uniformly at random (1/2 & 1/2) between doors when the contestant has selected the car

Richard said I still have it wrong, thusly:

"Again, no. Glkanter jumps to a conclusion but does not give any argument for it. So we cannot tell if he is brilliant but fast and careless, or dumb because he doesn't even realise that he has missed a logical step. Richard Gill (talk) 18:36, 17 January 2011 (UTC)"

Can *anyone* help this person (me) who readily acknowledges, 'I don't know', what he is doing wrong, or differently than what Richard has described? Glkanter (talk) 00:57, 18 January 2011 (UTC)

Maybe he's looking for these exact words?:

1. Because the initial distribution of the car is uniform at random, when the contestant has selected a goat, the car is equally likely (1/2 and 1/2) to be behind door 2 or door 3, as they both have an initial probability of 1/3 of having the car
2. When the contestant has selected the car, the host is equally likely to open either door 2 or door 3 because of the 50/50 host bias premise
3. Whether the contestant has selected the car or a goat, the opening of door 2 or door 3 is equally likely
4. There is no new information as to the likelihood of the contestant's door having the car whether door 2 or door 3 is opened
5. Once door 3 is opened to reveal a goat, the probability it has the car is 0
6. The contestant's door again has a 1/3 probability of being the car, as there is no new information
7. Door 2 has the complement, 2/3
8. The symmetry principal applies
9. [Some verbiage about 'the simple solutions properly solve the conditional problem, but I'm *sure* I would phrase it wrong]
10. The contestant will double his chances of winning the car by switching from door 1 to door 3

Assuming I'm close, does anyone know of reliable sources comprising a significant minority (at least) that insist on this level of detail? Anybody care to comment? Glkanter (talk) 01:21, 18 January 2011 (UTC)

The contestant will double his chance by switching only under the firm condition that, by the host's opening of one of his two doors, we have learned absolutely nothing to allow us to revise the odds on the door originally selected by the guest. You know that, but you have to say it explicitly.  Gerhardvalentin (talk) 02:05, 18 January 2011 (UTC)

It is a matter of opinion exactly what needs to be made explicit in the solution of any problem. Must we say exactly why the player's initial door choice is not important? Must we say that we assume that the words spoken by the host do not give any clue as to the location of the car. These things are both essential for understanding the solution but they are normally omitted because they are considered self-eveident.

So it is with the host's door choice. I have always agreed that, given a conditional problem in which the door chosen by the host is clearly given there is a step where we have to make the decision that the probability that the car is behind the initially chosen door is not affected by the number of the door opened by the host. The question is, bearing in mind that most people naturally make this assumption, how explicit do we need to make it in our solution.

I believe that the answer to this problem is simple. Use a bit of licence in the simple solutions section and ignore the requirement for this step then, later on and for those interested, explain all about the many possible assumptions and approaches relevant to this puzzle. Martin Hogbin (talk) 09:48, 18 January 2011 (UTC)

Yes Martin, you are right, but "self-evident" only for some, for others not...  –  No matter where you mention it, but you have to say it explicitly. You have to say that the host, by showing a goat, of course never can be assumed to be giving away any additional hint on the actual location of the car. You can say that later, but you got to say it: That the simple solution does not need to adopt any such additional "assumption" that never might be given.   Gerhardvalentin (talk) 11:58, 18 January 2011 (UTC)

Do you think the I have given the answer that Nijdam & Richard have been ever so patiently nurturing us to come to? Are you in agreement that they are relying on OR, and that the effect on the article is improper? Is your 'bit of license' resolution less 'improper', or perhaps, even 'proper' as per Wikipedia policies? Glkanter (talk) 10:51, 18 January 2011 (UTC)

Enough Of The OR Objections

We have *way* over-indulged the Wikipedia editors critical of the simples solutions. I know Gill says the simple colutions *can* solve the conditional problem, as do G & S, and Morgan and Selvin, and vos Savant, and Carlton, and Rosen(house/thal). Maybe Nijdam, too. So all Nijdam & Richard are going on about is their viewpoints of 'explanatory completeness of the simple solution presented', not the 'mathematical correctness of the simple solutions presented to solve the conditional problem'. If they have reliable sources that support this, I think it's way beyond time they showed them.

Further, the Wikipedia editors critical of the simple solutions have not shown that there is a significant minority (at minimum) of reliable sources that support the criticism that the symmetrical simple solutions are inadequate in *any* way. (I think 'optimal strategy' is unique to Richard.) This is a Wikipedia requirement for the prominent inclusion in the article the criticisms enjoy now.

I think we should encourage AGK and Sunray to resume the mediation, and insist that the critics support their position with a significant minority of reliable sources, rather than just shooting off their own OR endlessly. Glkanter (talk) 02:17, 18 January 2011 (UTC)

Doesn't it become obvious at some point that Nijdam's constant harping that not a single MHP reliable source knows what they're talking about, and his inability or disinterest in finding even one solitary source that agrees with him should bring an end to his Reign Of Error? Pretty much the same with Gill, with his never-ending flip-flopping, and being pretty close to Nijdam's level of criticisms of the reliable sources, and his diarrheac, endless, pointless, lengthier & lengthier postings about game theory, min max, optimal solution, etc. Doesn't it at some point become obvious that whatever it is he's saying is OR, and again, no other reliable source can be found to support this far out, non-germain, irrelavent masturbation? Glkanter (talk) 07:20, 18 January 2011 (UTC)

Glkanter: You Wikipedia:NPA have to apologize. For your personal attacks you are to be blocked indeed. Sorry, Gerhardvalentin (talk) 08:32, 18 January 2011 (UTC)

Gerhardvalentin, you do whatever you think is right, as long as you do the same thing for the editor who thoughtlessly, needlessly, recklessly, and incorrectly posted this BS about me prior to my comments:

"So we cannot tell if he [Glkanter] is brilliant but fast and careless, or dumb because he doesn't even realise that he has missed a logical step."

Do we have a deal? Glkanter (talk) 11:44, 18 January 2011 (UTC)

Not attacks. True statements. Glkanter (talk) 08:45, 18 January 2011 (UTC)

It's clearly a deadlocked issue

Having not been too much involved in this controversy, I can stand back and view it all from a bit of a distance, and it is obvious to me that the situation has long passed the deadlock stage, and that nothing more will be accomplished from debate. The deadlock issue is that the conditionalists seem to be insisting that the conditional solution is necessary, but they have failed to convince the simplists including myself.

It has degenerated to the stage where I have essentially been accused of being incompetent to pass judgement on the matter because I am not trained in advanced probablity. I have counter argued that advanced mathematicians are prone to lose sight of the higher picture regarding the application of their maths in the real world, and that I can see the problem at hand quite clearly without having to know about advanced probability.

So we are dealing with a situation in which the conditionalists are pushing the idea that their advanced knowledge is necessary, and then using their claimed expertise as the basis upon which we are supposed to believe them. Personally, I don't believe them, but I notice that other simplists have fallen into the trap of endlessly trying to reason with the conditionalists. And as a point of clarification, I don't doubt the accauracy of the conditionalist solutions as such. I merely doubt that they are necessary.

Such a deadlocked situation requires a binding decision to be made by the mediation committee. In the absence of any binding decision by the mediation committee, any further debate is a total waste of time. David Tombe (talk) 12:27, 18 January 2011 (UTC)

A very reasoned analysis, but you omit one item: The simplists need nothing more than reliable sources, common sense and Wikipedia policy to make their case. The other guys have *no* legitimate way to make their case. That's why they rely on the gamesmanship, etc. that you just described. I counseled you on this a couple of days ago. Was I reasonably accurate? Glkanter (talk) 12:42, 18 January 2011 (UTC)

Glkanter, I believe that the conditionalists have sources too, perhaps even claiming that the conditional solution is necessary. If that is the case, then you can't win here. You can only draw. Normally you would be at the mercy of a judge as to how favourable the draw would be to you. But in this situation, we are all still waiting for a ruling. If I am wrong, and the conditionalists don't have any such sources claiming that the conditional solutions are necessary, then it is as you have said, and the mediation committee should be making a binding decision in your favour. David Tombe (talk) 12:51, 18 January 2011 (UTC)

I won't judge your qualifications to make that statement. I'll tell you that I've been at this over 2 years, and have (with Martin's leadership) made most of the 'discoveries' that allow us to define the debate as simply and succinctly as we are now doing. There is no 'significant minority viewpoint of the reliable sources' as required by Wikipedia. For the BS they're arguing above, there are *no* reliable sources, period. And the article continues to suffer from UNDUE WEIGHT, NPOV violations and OR supporting their POV. Have I steered you wrong, before? Glkanter (talk) 13:01, 18 January 2011 (UTC)

And while I can't say 'why', I *can* say it's interesting a long time admin/editor has contributed little to these discussions on Martin's talk pages. Especially the last week or so, when I started posting with the "They're Shortcuts" section as above. For 3 or 4 years, he was like the tides, very reliable in responding. But, like I said, while it's interesting, I cannot say why it is occurring. Glkanter (talk) 13:19, 18 January 2011 (UTC)

Glkanter, OK, I haven't examined the issue of sources in this matter in detail, so I'll take your word for it. I was just operating on the basis of rational argument, and I can see now that you are being taken on a merry-go-round that will go on forever. You are up against an argument which runs something along the lines of 'our expertise in the matter is crucial and we know this because we are experts'. In other words, you need to be an expert to know that your expertise is required. As it is, I am not an expert in conditional probablity, but I don't believe that expertise is necessary in order to know that expertise is not necessary for the MHP problem. I would say that you have adequately and correctly made your point, and that it is now time for the mediation committee to make a ruling. The trouble is that in this day and age, people in authority often defer to warring parties and urge them to reach agreement, when most people with common sense can see that in the circumstances, agreement will never ever be forthcoming. (After all, who is ever going to agree to have their goals quashed when they realize that their position will be sustained so long as they don't reach agreement?) Hence decisions are never made, and problems fester like a boil that never comes to a head. Just don't fall into the trap of thinking that you will ever persuade these conditionalists. It will never happen, no matter how you try to dress your arguments. All too often, I have watched situations like this only to see people changing sides out of desperation in order to impress their opponents. And doing that hardly leads to victory. David Tombe (talk) 15:20, 18 January 2011 (UTC)

Agree with, and have written every word repeatedly. Change sides? Read 'Conventional Wisdom' on my talk page. It's from February, 2009. Glkanter (talk) 15:28, 18 January 2011 (UTC)

The technical terms I use for this paradox are the "High Priests" and the "Emperor's New Clothes". Glkanter (talk) 15:53, 18 January 2011 (UTC)

Lost In The Wilderness, Or Just Lost?

I have been requested, badgered, cajoled, harangued, hassled, challenged, humiliated, etc. by Richard into reading Richard's MHP papers.

"Be careful of what you ask for, you may receive it." - Mother Theresa and others

Here's how Richard passionately concludes his 2010 paper:

"This little article contains nothing new, and only almost trivial mathematics. It is a plea for future generations to preserve the life of The True Monty Hall paradox, and not let themselves be misled by probability purists who say “you must compute a conditional probability”. - The Three Doors Problem...-s Richard D. Gill Mathematical Institute, University Leiden, Netherlands http://www.math.leidenuniv.nl/!gill February 15, 2010

Beautiful, isn't it?

What a shame that dumbasses like Glkanter have complicated and polluted the discussions and the article on Wikipedia so horribly contrary to Richard's dream, that he must retreat to some 'dumbass' friendlier encyclopedia to share his thoughts on game theory, min max, optimal solutions, the proper presentation of the symmetry principal, etc.

I give a dumbass's eye-view of Richard's accomplishments a grade of 'FAIL'. Glkanter (talk) 16:06, 18 January 2011 (UTC)