Talk:Monty Hall problem

My understanding

Opening a door with a goat after the contestant has made a choice effectively lets him look behind two doors, rather than one. It doesn't matter whether the host knew there was a goat there or not, every time a goat is revealed the benefit of switching is 2/3. Is this correct? Rumiton (talk) 02:24, 13 January 2013 (UTC)

Suppose the player chose door 1. Suppose the host will now choose at random door 2 or 3 and open it. Suppose he does this, and we learn that he happened to pick door 3 and that there happened to be a goat behind it. Before we got this information the odds on the car being behind door 1 or 2 are equal. If the car had been behind door 1, the chance the host would pick door 3 and there'ld be a goat there is 1/2. If the car had been behind door 2, the chance the host would pick door 3 and there'ld be a goat there is also 1/2. So seeing the host pick door 3 and seeing a goat behind it does not change the odds for the car behind behind door 1 or 2. Answer: the chance the car is behind door 2 is now 1/2. So it has changed because of the new information (it used to be 1/3) but it hasn't changed to 2/3. Richard Gill (talk) 09:54, 13 January 2013 (UTC)

We can also solve this variant of the problem ignoring door numbers. By symmetry it is also 1/2. If you switch (to the other closed door) whenever the host happens to open a goat door you'll win 50% of the times that he reveals a goat.

I imagine you would switch to the opened door if the host happened to reveal a car! Overall, you'll always switch (but to different doors depending on what the host reveals). This means that your overall chance of winning is now 2/3. Since just as in standard MHP, you'll get the car if and only if your initial choice of door hides a goat. Richard Gill (talk) 10:38, 13 January 2013 (UTC)

Yes, that is correct, the chances of winning by switching is 2/3, provided that the host always reveals a goat. The connection with the host's knowing what is behind the doors is that the host cannot guarantee to always reveal a goat without knowing what is behind the doors. Martin Hogbin (talk) 09:59, 13 January 2013 (UTC)

Yes, Marilyn vos Savant explicitly says "most significant is that the host always opens a losing door on purpose. (There’s no way he can always open a losing door by chance! Anything else is a different question."

Because if the host just randomly opens one of his two doors, then in 1/3 he shows the car, so just from the outset inevitably destroying the chance to win by switching in that 1/3. In that scenario the odds on the door chosen and the door offered to switch on remain 1:1. Richard is completely right in clearly saying that consistency would be given only if the host, after having shown the car in 1/3, would offer to switch to the "car". 11:37, 13 January 2013 (UTC)Gerhardvalentin (talk)

@Rumiton (edited "parallel"):Thank you for your clear and factual question which goes to the heart of the matter of the widespread MHP debate. The answer is No. A short but correct argumentation is: If the host is committed to open an unchosen door with a goat (knowing where it is) the probability to open his door is twice as large (p=1) if the chosen door has a goat as if it has the car (1/2 (assumption)) which leads to a 2/3 chance for the switch. If the host does not know where the car and the goats are, both considered probabilities are 1/2 leading to a fifty-fifty chance.

Amusingly, Rumiton is partly right. It depends *which* probability is claimed to be 2/3. Suppose the host always opens, completely at random, one of the two initially unchosen doors. So sometimes he'll reveal a goat, sometimes a car. Consider the strategy of "always switching" by which I mean the sensible strategy: if the host reveals a goat, switch to the other closed door; if the host reveals a car, switch to the open door (and certainly get the car). The player who follows this strategy wins 2/3 of the time and again one can say that this is because the host is essentially offering the player the choice to exchange his initially chosen door for the two unchosen doors.

You are completely right with the overall chance of 2/3 in this case. And if you have read may comments you will believe me that I considered this version already 20 years ago (all this is really not difficult). But the answer to the question of Rumiton is clearly No. For he asks exactly for the chance if the host has opened a door with a goat. And this chance is 1/2.--Albtal (talk) 16:39, 13 January 2013 (UTC)

All this shows that ordinary language and ordinary thinking is often inadequate for dealing with probability and uncertainty. Professionals have had to develop a more refined language in order to take care of important distinctions which ordinary people are usually not even aware or. Richard Gill (talk) 14:28, 13 January 2013 (UTC)

This remark is completely inappropriate here. And all essential statements concerning MHP can be made with normal language. But why do you defend a wikipedia article which starts putting up a huge smokescreen? Is this the professional refined language?--Albtal (talk) 16:39, 13 January 2013 (UTC)

This is quite the silliest argument I have seen here for some time. It is about a hypothetical game show in which the host may open any door, including one which hides the prize, and then give the player the option of switching to the prize that they have revealed. Rather than showing 'that ordinary language and ordinary thinking is often inadequate for dealing with probability and uncertainty' it shows that mathematicians can easily lose touch with reality. I have no argument against the mathematics being proposed but the idea that these events could happen on a game show is absurd. It is equally absurd to assume that this was the question that either Selvin or vos Savant wanted us to answer when they set their puzzle. Martin Hogbin (talk) 21:24, 13 January 2013 (UTC)

You not only ignore all arguments here, but you completely misunderstand the roles of task sets, proofs, examples, counterexamples, and extreme examples. So good luck, wikipedia and wikipedia readers, with the moral of the story (see below)!--Albtal (talk) 09:41, 14 January 2013 (UTC)

The following scheme which shows the probabilities for the "knowing" host (assuming that the contestant has chosen door 1) is also a good base for variants of the game (see Ein Auto und zwei Ziegen).

1. car behind door 1 - host opens door 2  : p = 1/6 (assumption)

2. car behind door 1 - host opens door 3  : p = 1/6 (assumption)

3. car behind door 2 - host opens door 3  : p = 1/3

4. car behind door 3 - host opens door 2  : p = 1/3

Probability for door 2 if the host opens door 3: p = (1/3)/(1/3 + 1/6) = 2/3

If the host does not know where the car and the goats are the probabilities 1. to 4. all are 1/6 leading to a chance of 1/2.

You may also easily analyse the different variants by playing the game with three playing cards. And maybe you see the result before starting the game, especially if analysing the variant with the knowing host.

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

We can see that the case that the host does not know the position of the car is excluded here. But even so, as I wrote above: Your question goes to the heart of the matter. For the problem set indeed requires the knowing host, but we can't see any reason why the host opens the door with a goat offering a switch. But this matters. Just above I cited Martin Gardner, Persi Diaconis, Marilyn vos Savant, and Monty Hall himself. Here an additional passage of the New York Times article: On the first, the contestant picked Door 1. "That's too bad," Mr. Hall said, opening Door 1. "You've won a goat." "But you didn't open another door yet or give me a chance to switch." "Where does it say I have to let you switch every time? I'm the master of the show. Here, try it again."

The problem set indeed does not say anything about the motivation of the host; and the articles which had been going around the world and claiming the 2/3 solution based their solution on the pure fact that the (knowing) host opens an unchosen door with a goat. And indeed: The widespread "proofs" by computer simulations or by other reenactments seem not to need the knowledge about the motivation of the host. But what they overlook: The assumption that the host is committed by the rules of the game has implicitly slipped in their "proof". So they seem to have a watertight proof for their solution, but really this is a joke. Clearly that overlooking this is the base for very much nonsense; and to present the "greatest paradox of the world". It is indeed an invincible paradox obstrusively to claim a solution which is wrong for the presented task set.

Now what do if the error is discovered? Simply say that the necessary assumption is obvious. So we have a "great paradoxon" which is deliberately worded wrong; posing the Parade question, waiting for the answer of the dialog partner, and afterwards now saying: In the usual interpretation of the question you are wrong. A truely great paradox!

Next step: Claimig that (most? all? the majority?) people interpret the question as if the crucial rule would hold. But this helpless attempt unfortunately is too late; for there are too many "proofs" which show the opposite: Why does the teacher in the film 21 say But the host may frame you? And why does a women in the internet answer I should switch, because the host is not framing me? And why does a disputant here write it matters not if he is forced or chooses to (see above)? And so on ...

A source which is often cited around wikipedia is Krauss and Wang, 2003. There is also a (German) paper Krauss and Atmaca, Unterrichtswissenschaft, 32(1), 2004 S. 38-57. There the problem set is exactly that of Parade, but the passage opens another door is supplemented to the remarkable passage saying "now I'll show you something" the host opens another door, which is incompatible with the "forced" host, and the reader as a contestant sees himself confronted with a situation which he couldn't expect - which lets the 2/3 solution crash like a house of cards. And after prompting the reader to answer the question before continuing reading they give the answer The candidate should switch, with a footnote in which the correct rules are listed, and with the ridiculous claim that these rules are implicitly assumed by almost all (nahezu allen) experimental subjects.

The temporary summit of the nonsense about MHP may be sources who say now, that of course the host does not play the game which leads to a 2/3 solution - but paradoxically claim the 2/3 solution:

Devlin, 2005:

As the game was actually played, some weeks Monty would simply let the contestant open their chosen door. The hypothetical version of the game described here, where Monty always opens the door and makes the "switch or stick" offer, is the one typically analyzed in statistics classes.

Drösser 2010:

But the chance to loose – assumed a nasty host– would be 100 percent. Therefore for the player there is only one solution: Staying hardheaded.

At the end a passage which in my opinion is the best summary of what had happened (better summaries are welcome, even from reputable sources):

Ein Auto und zwei Ziegen:

In my opinion many within the 2/3 fraction up to today don't see the difference between the blank fact that the host opens a not chosen door with a goat, and the enforcement by the rules of the game to do so, which is crucial for the 2/3 solution, especially too for "reenactments" and "computer proofs".--Albtal (talk) 12:38, 13 January 2013 (UTC)

I'm convinced people who furiously defend the point of view the the strategy of the host is unknown, and hence may lead to a solution with equal chances for the remaining closed doors, do this because they initially have given the answer of equal chances. However, their "mistake" was definitely based on other assumptions - two remaining doors, hence equal chances - than on considerations about the host's strtategy. Nijdam (talk) 13:33, 13 January 2013 (UTC)

Yes, I agree entirely. People get the answer wrong and then try to defend their wrong answer by inventing bizarre scenarios in which their answer is correct. In the section above I said to Albtal, 'If you do not mind, I would hazard a guess as to how you came to propose this option' thinking exactly the same.

Interestingly, Nijdam, I suspect that that is how the Morgan paper arose, at least one of the authors got the answer wrong and was determined to prove themselves right and did so by inventing the unrealistic scenario that the host does not choose uniformly. Unfortunately, Morgan et al were academics and published in a reliable source, so their craziness is still with us. Martin Hogbin (talk) 13:54, 13 January 2013 (UTC)

I don't think this is how the Morgan et al. paper arose. Their solution is the standard solution which any trained probabilist will give. It was already given by Selvin, in his second paper, 1975b. He was clearly criticised for his sloppy first try, 1975a. Morgan et al.'s only inovation was also to consider the rather unnatural biased host possibility. Richard Gill (talk) 14:33, 13 January 2013 (UTC)

Richard is right to call it a "rather unnatural biased host possibility". I call this "biased host possibility", attacking a one time show as a clear attempt to defraud, because they are unable to come to any sensual "solution", as such bias is forever unknown. And Ruma Falk confirms that it only makes sense to use any host's bias if the host IS BIASED indeed, and if he is KNOWN to be biased. So Morgan et al. are to be blamed to have presented a ridiculous dirty hoax with their so called "solution". For they forever are completely unable to give any so called "closer rate of probability to win by switching" that differs from 2/3. I miss the abundantly clear branding of their abundantly obvious deficiency of their argument in the literature. Gerhardvalentin (talk) 19:51, 13 January 2013 (UTC)

Unless one of the authors makes contact we shall never know. I think your confidence in probabilists is a little misplaced. As you know, after the (re)publication of the problem in 'Parade' (the original problem having been long forgotten) vos Savant got 10,000 letters telling her she was wrong, with 100 from PhDs and some from professional statisticians; not one of these mentioned the Morgan solution. Morgan's paper was published over a year later and in the intervening time no one suggested that there was any problem with the simple solutions, Morgan themselves say, 'That we have not seen the solution offered in this article,...motivate us to submit this note'. Hardly the words of people who think they are just giving the standard solution.

It also seems too much of a coincidence to me that at one extreme of host behaviour the answer actually is 1/2. If there was a clear way of resolving this issue I would gladly place a small bet on at least one of the Morgan authors thinking the answer to the problem was 1/2. Martin Hogbin (talk) 15:13, 13 January 2013 (UTC)

When I read the first time about MHP in Gero von Randow's article in the German newspaper DIE ZEIT in July 1991, I confirmed the 2/3 solution of Marilyn vos Savant in a letter to DIE ZEIT with several arguments; one was the argument with 100 doors. A few months later a friend showed me an article in "Spektrum der Wissenschaft" about MHP. I was astonished to see that Marilyn's first argument had been exactly the same. I didn't see the original problem set for more than ten years; but I had introduced my 100 doors argument with the words The host now according to the rule of the game has to .... (Incomplete descriptions of my argument you can find here and here, a complete despription in Gero von Randow, Das Ziegenproblem, Reinbek bei Hamburg, März 2001, S.10). In a second letter I wrote that the 2/3 solution is wrong without this rule. Again I was surprised when more than ten years later I read in the here cited article in the New York Times, that Martin Gardner's argument matches exactly an argument of my second letter.--Albtal (talk) 17:17, 13 January 2013 (UTC)

Interpretation of the problem

Albtal, I agree that there is ambiguity in the Parade problem statement but in our discussion of that fact in the article we should take account of the following:

1 We know vos Savant intended that problem to be interpreted that the host would always open an unchosen door to reveal a goat and always offer the swap, because she says so.

2 We know that Selvin intended the same interpretation from his published solutions.

3 We know both vos Savant and Selvin intended us to take it that the host would choose uniformly according to the above rules when he had a choice because they have both said so.

4 In a Bayesian interpretation of the problem we must take the positioning of the car, the player's initial door choice, and the host's choice of legal door to be uniform.

5 We know that most people interpret the problem in the way stated above because this is stated in K&W.

6 We know that most solutions in the literature tacitly or explicitly (or belatedly in the case of Morgan) use the above problem interpretation

7 The whole point of the problem is to present a simple puzzle that most people get wrong. It was clearly never intended to be an exercise in imaginative interpretation.

I therefore think that any other interpretations of the problem must be considered as variants (or perversity) and that this must be clearly indicated in the article. Martin Hogbin (talk) 10:11, 14 January 2013 (UTC)

In a Bayesian interpretation of the problem we do *not* have to take the player's initial choice as uniform. The sensible approach is to take it fixed, and then take the other probabilities uniform, given this choice. In a Bayesian interpretation we are after the *player's* probabilities given everything he knows. He chooses a door and once chosen he knows it. I don't know where you get your dogmatic assertion from. In any case, it is simply not true that most people assume the player's choice uniform. (In fact the people who do are typically those who do not take a Bayesian interpretation at all, but a frequentist interpretation. They assume that the player's choice is uniform and the location of the car and action of the host are fixed. Because they're interested in optimal strategies and initial door choice is part of your strategy). The important Bayesian's assumption about the player is that the player doesn't know anything about where the car is and hence after choosing his door, the car is still initially equally likely to be behind any of the three doors.

I think your point about the player's initial choice is just one of words. We do not have to make use of any distribution if we can solve the problem without doing so. My point was that, if we choose to consider the distribution of the player's original door choice, then we must, as Bayesians, take it as uniform.

Otherwise I completely agree with Martin here, especially points 1, 2, 3, 5, 6, 7. Other interpretations should not be considered perversity. Other interpretations provide test cases for testing reasoning. We don't just want the right answer, we also want correct reasoning. Varying the conditions of a problem is a standard heuristic tool for learning how to solve it. (See e.g. Polya's "How to solve it"). Richard Gill (talk) 09:57, 15 January 2013 (UTC)

Again I gave no objection to considering variants of the problem as test cases but we should be consistent about this. I have often suggested considering the variant in which the host has a goat preference as a test case, only to be told that we are not interested in variations. Martin Hogbin (talk) 18:28, 15 January 2013 (UTC)

The moral of the story

What the above discussion shows is that the two things that are most important for our readers are that the answer is 2/3 and that it matters that the host knows where the car is. We still need to find ways to make these points clearer in the article. Martin Hogbin (talk) 16:04, 13 January 2013 (UTC)

You often say things like this. I do wonder, however, to which question in the problem statement the answer is 2/3?Nijdam (talk) 22:51, 14 January 2013 (UTC)

My unquestioned answer: as the hypothetical one-time host purposely opened a losing door in order to offer a switch to his second unselected door, the chance to win by switching to the door offered is 2/3. And basing on "unknowns like q" looks pretty smart and is mathematically correct, but actually is completely impotent and infertile to name any closer "value". Any hypothetical but wrong value differing from 2/3 is unquestioned in the given situation, because necessarily just plainly wrong if it differs from 2/3. Believe it or not, the question was not on pretty smart but actually wrong infertile hypothetical values. Gerhardvalentin (talk) 00:29, 15 January 2013 (UTC)

Gerhard: two thirds of what? Nijdam is completely right. It's like asking how tall is someone, and getting the answer 5. This is not complete. 5 what? Feet? Meters? ... Richard Gill (talk) 08:47, 18 January 2013 (UTC)

Martin, we should explicitly say what Marilyn vos Savant herself did say in "Parade" from 1990 to 1991:

• The host randomly hides the one and only price behind one of the three doors.
• The guest randomly selects one of the three doors.
• The host then purposely opens a losing door from his two unchosen doors, being determined in any case to offer a switch to his second unchosen door.
• If the prize is behind #2, the host opens #3, and if the prize is behind #3, the host opens #2. So if the prize is behind #2 or #3, you win by switching.

That's all we know, and we never will know more. MvS did not say that she met that hypothetical one-time host in person, and we don't know more about him than MvS did explain. We don't know more, and we never will know more. Anything different is a quite deviant scenario. Consequently, to base (to "condition") on additional (never to be given) "knowledge", and to claim instead to "know" that probability to win by switching actually clearly differs from 2/3 is a rather ridiculous hoax, in my opinion. This should be branded in the literature. You can discuss other scenarios, yes, but that should honestly be named as "other scenarios, differing from the scenario intended by MvS". Gerhardvalentin (talk) 20:11, 13 January 2013 (UTC)

Yes, I agree that this was the scenario intended by vos Savant and Selvin when they formulated their puzzles. I think the article does make that clear. Certainly neither of the two puzzle setters envisaged that the host might open a door to reveal the prize and then ask the player if they want to switch to that door! Thankfully the article does not consider that bizarre possibility.

What vos Savant and Selvin intended is not really relevant: the important thing is what the reliable sources of nowadays think is the Monty Hall Problem. What vos Sareally intended has historical interest. But fortunately it's clear that the main sources as well as these originators all agree that the problem is about the situation in which the host is certain to open a door different from the player's choice and reveal a goat, and that he can do this because he know where the car is hidden.

Sources do not agree what are the probability ingredients to a solution. Selvin, the originator, gave two completely different solutions with different assumptions. My personal opinion is that choosing and motivating sensible probability assumptions is part of a solution, not part of the problem statement.

Making assumptions which you do not use in your solution means that your solution can be improved. Because it means you can drop the assumptions which you have no use for. And that makes your solution stronger (more people will be able to buy it). If we assume that all doors are equally likely to have the car behind it we don't need to assume anything about how the player chooses his door initially. If we are interested in a "simple solution" we don't need assume anything about how the host chooses a door to open when he has a choice. Richard Gill (talk) 17:38, 14 January 2013 (UTC)

Fortunately the crazy game show in which the host reveals the prize and then asks the player if they want it is ruled out not only by common sense but by the problem statement, which does not allow the player to choose the door opened by the host. Martin Hogbin (talk)

Where the article still fails, in my opinion, is to answer the question that we get asked regularly on the talk page, 'Does it matter if the host has to open a door revealing a goat (and therefore has to know where the car is) and, if so, why?'. What can we do to improve that failing in our article? Martin Hogbin (talk) 21:33, 13 January 2013 (UTC)

Let's agree the host has to open a different door from the door chosen by the player. Does it make a difference whether he is obliged to open a goat-door or not? It doesn't make any difference at all to the unconditional probability of winning by switching. Because the host opening one of the other two doors is equivalent to the host offering the choice between your first door and the best that is behind the two other doors (if the host reveals a goat you go for the other closed door, if the host reveals a car you go for the car). It does make a difference to the conditional probability! Maybe the article can be improved by having a simple conditional solution among the simple solutions. Richard Gill (talk) 17:44, 14 January 2013 (UTC)

Using Bayes' rule, odds form, does explain everything, exactly. That's why many modern authorities like to use it to solve MHP both in its standard form as well as all kinds of variants. Richard Gill (talk) 18:02, 14 January 2013 (UTC)

I do not think your arguments that mathematicians have the monopoly on understanding of the MHP will convince many. What have you been drinking over the Christmas break Richard? Martin Hogbin (talk) 19:03, 14 January 2013 (UTC)

I am not saying that mathematicians have any monopoly on understanding MHP. I'm saying that Bayes' rule is simple, intuitive, and explains everything. To understand MHP one just needs to be able to do some clear thinking. The player chose door 1. Since we're told the host opens door 3 and reveals a goat there, the big question is going to be whether the car is behind door 1 or door 2. Initially both possibilities are equally likely. Odds car is behind door 1 : car is behind door 2 are 1:1. Now lets look at how the two possibilities influence the chance that the host will open door 3. If the car is behind door 1 the chance is 50/50 that the host will open door 3. If the car is behind door 2 the chance is 100% that the host will open door 3.

In short, the information we obtained (host opens door 3 and reveals a goat) is twice as likely to be received if the car is behind door 2, than if it's behind door 1. That's why the odds change from 1:1 to 1:2.

The reason therefore that one should switch is that it is much more likely that the host reveals a goat behind door 3 when the car is behind door 2 (because in that case the host had no choice which door to open), than when it is behind door 1 (because in that case the host did have a choice which door to open).

The numerical answer is 2/3 because this probability corresponds to odds of 1:2 (in favour of switching). The odds are 1:2 because probability 1/2 (when the host could open door 2 or door 3) is half of probability 1 (when the host has to open door 3).

I wouldn't call this mathematics. It is simple arithmetic and simple logic. As long as you have got the right concepts in your head.

I have nothing whatever against Bayes' rule and welcome it inclusion, in the appropriate, place in the article. I also agree that mathematics is not at the core of the MHP. You said above, ' Professionals have had to develop a more refined language in order to take care of important distinctions which ordinary people are usually not even aware of'. That may be so but when mathematicians try to add to the MHP by changing things and making them more complicated they display a degree of mathematical snobbery but they actually make the problem less interesting. The point is that, although the wording is ambiguous, nearly everybody, lay people and mathematicians alike, know what is meant yet most of them still get the answer wrong.

Read Jeff Rosenthal's book "Struck by lightening: the curious world of probabilities". Read Jason Rosenhouse's book "The Monty Hall Problem". Bayes' rule is for ordinary people, not for mathematicians. Lawyers and doctors and journalists need to know Bayes' rule. Also, never just say "2/3" but say "2/3 of what".

These are not my arguments but the arguments of people who have published major *popular* books on MHP. If you've really understood MHP then you will be able to instantly and correctly solve small variants of MHP. Bayes' rule is great for that. Why do people have this enormous reluctance to learn such a simple and fundamental tool for reasoning with uncertainty? Scared of the unfamiliar? Because everyone who first manages to get the answer 2/3 by whatever reasoning whatever, has their mind blocked to learning a different technique? (Different solutions are threatening - intuitively we feel that they might lead to a different answer and bring back uncertainty; we hate uncertainty. So we stick to our own first good solution and never try to learn other ways of looking at the problem.). Richard Gill (talk) 09:43, 15 January 2013 (UTC)

PS, it seems that variant problems are also threatening, if they show that some reasoning can't be entirely correct! We have to ban the variant where the host opens a different door at random and the player can choose to switch to either of the two other doors, because the simple "combined doors" solution also gives the correct (unconditional) 2/3 answer to this variant, yet it's clear that given the host happened to reveal a goat, the chance of winning by switching is 1/2. This is a variant where conditional and unconditional solutions are different, and everyone realizes we want the conditional solution. Very threatening. It's not devious to invent problem variants in order to test reasoning. It's how we debug computer programs, for instance. We don't just want the right answer, we also want the right reasoning to get that answer. Richard Gill (talk) 09:51, 15 January 2013 (UTC)

To get back to the main purpose of the talk pages, we have yet another reader who is puzzled by or interested in the fact that it matters (for the standard problem) that the host knows where the car is. This is indicates to me a failing in the article that we need to put right, in a way that everyone can understand. Martin Hogbin (talk) 19:00, 15 January 2013 (UTC)

That's precisely what I'm trying to help with.

Standard MHP: The player has chosen door 1 and the host knows where the car is. He proceeds to open door 2 or door 3 in order to reveal a goat. If he has a choice then it is 50-50 which door he opens.

• If the car is behind door 2, the host is 100% certain to open door 3 and reveal a goat. On the other hand, if the car is behind door 1, the host is only 50% certain to open door 3 and reveal a goat.

• Because the host is twice as likely to open door 3 and reveal a goat when the car is behind door 2 (100% is 2 times 50%) than when it is behind door 1, it it twice as likely, given the host opened door 3 and revealed a goat, that the car is behind door 2 rather than behind door 1.

Nonstandard (silly) MHP: The player has chosen door 1 and the host doesn't know where the car is. He proceeds to open door 2 or door 3 at random. There happens to be a goat behind it.

• If the car is behind door 2, the host is 50% certain to open door 3 and reveal a goat. Also, if the car is behind door 1, the host is 50% certain to open door 3 and reveal a goat.

• Because the host is equally likely to open door 3 and reveal a goat when the car is behind door 2 than when it is behind door 1, it equally likely, given the host opened door 3 and revealed a goat, that the car is behind door 2 as that it is behind door 1.

Isn't that something everyone can understand? Isn't this a simple solution? In fact, two simple solutions of two contrasting problems? Doesn't it give insight? Remember Whitaker's actual words in his question to vos Savant (quoted in Morgan et al 2011, response to Hogbin and Nijdam): he too wanted to understand why it makes a difference whether or not the host knows where the car is. What makes a difference is actually not that the host knows the location of the car, but that we know that he uses this knowledge in order to certainly reveal a goat. Richard Gill (talk) 09:33, 16 January 2013 (UTC)

No, I do not think everyone can understand what you say. I have no complaints but many people will be completely lost by your argument. I think we need to start by showing why the answer (to the standard problem) is not 1/2.

We know where most people go wrong, K&W tell us. This is how most (Bayesian)people think:

There are three doors and we have no information about which one hides the car therefore we must assume that it initially has 1/3 probability to be behind any given door. Correct

When the host has a choice of two doors we have no information about which one he will chose so we must assume that he chooses either of the two possible doors with probability 1/2. Correct

What there are two doors left which might hide the car we have no information about which door is more likely to hide it therefore we must assume that each door hides it with probability 1/2. Wrong!

K&W and other sources say that people make the mistake of assuming that probability is always equally distributed between the available possibilities. We need to explain to them, in words, why when the host knows where the car is (always reveals a goat) we do get information about the probability that the car is behind the unopened, unchosen, door but when the host reveals a goat by accident things are different. Martin Hogbin (talk) 11:14, 16 January 2013 (UTC)

I gave the explanation, in words. In the standard problem, the probability that the host will open door 3 and reveal a goat is different depending on whether the car is behind door 1 or 2. In the nonstandard problem, the probability that the host will open door 3 and reveal a goat is the same whether the car is behind door 1 or door 2. That's why in the standard problem we get information about whether the car is behind door 1 or 2, in the nonstandard problem we don't.

Standard problem: the host is more likely to open door 3 when the car is behind door 2 than when it is behind door 1. Given he actually opened door 3, the relative likelihoods that the car is behind door 2 or 1 have shifted in favour of door 2.

Nonstandard problem: the host is equally likely to open door 3 and reveal a goat when the car is behind door 2 as when it is behind door 1. Given he opened door 3 and revealed a goat, nothing has changed regarding the relative likelihoods that the car is behind door 2 or 1.

Of course, in both problems we get information about whether or not the car is behind door 3. In fact, we get to know for sure that it isn't.

I'm not aware of any other explanation than this. I would even say that, fundamentally, there cannot be any different explanation, precisely because Bayes' rule is the universal law telling us how probabilities are updated on getting information. (My explanation is a verbal and intuitive representation of what Bayes' rule tells us). Possibly my explanation can be expressed better, but I don't see how the content can be anything different.

You may or may not like Bayes' rule, you may or may not have something against it, but if your reasoning is correct it will lead to the same conclusion. Resistance is futile!

This is not original research and it is not a matter of personal opinion. It's the explicit message coming from the major sources. Richard Gill (talk) 14:50, 16 January 2013 (UTC)

Richard, you are misunderstanding me, I have nothing whatever against Bayes' rule (I am a strong Bayesian, remember) but the average person does not know Bayes' rule. You may think that they should but the fact is that they do not. Maybe it is just the way you say things but I do not think the average person will follow what you say. Martin Hogbin (talk) 14:43, 17 January 2013 (UTC)

Is there a better way to say this, or is there a different explanation altogether? I know the average person doesn't know Bayes' rule. My explanation does not assume that they already know Bayes' rule. It's a combination of pure common sense and easy to check facts. Richard Gill (talk) 08:44, 18 January 2013 (UTC)

I do not claim to have all the answers but is is clear what the questions are. We have just had two more users who have asked or argued about the same two things that everybody asks or argues about; that the answer is 2/3 and it matters that the host knows where the car is (and can therefore always reveal a goat). The fact that we still get users asking or arguing about these two things tells me that the article could be better, because it has failed to explain, in a way which the reader understands and believes, the answers to these questions.

We can should work together to improve the article but it helps if we know where its deficiencies lie. Martin Hogbin (talk) 10:21, 18 January 2013 (UTC)

I agree. That's exactly why I put forward here an answer to the question "why does it matter that the host knows where the car is?". I suggested that the answer is the following. Because the host knows where the car is and is not going to show us a car, the chance that he opens door 3 is different depending on whether the car is behind door 1 or door 2. He's much more likely to open door 3 when the car is behind door 2 than when it's behind door 1. So given he opened door 3, the car is much more likely behind door 2 than behind door 1.

Compare this with the silly alternative puzzle in which the host opens door 2 or 3 regardless and it just happens that he opens door 3 and it just happens that there's no goat behind it. In that scenario, what actually happened (door 3, goat) is equally likely, whether the car is behind door 1 or door 2, so doesn't give us any new information about whether the car is behind door 1 or door 2.

Do you agree with this explanation?

Are you aware of a different and better explanation?

It seems to me that if this is the only explanation going, then it's our task to find a good way of saying it! Richard Gill (talk) 12:22, 18 January 2013 (UTC)

I think that is getting better. This all shows why the problem is so unintuitive. Ther is no really simple way, that I can think of, of showing why the host always revealing a goat helps you guess where the car is. Can you, or anyone, think of a good analogy or example wher the above logic is clearer. Martin Hogbin (talk) 17:27, 18 January 2013 (UTC)

Here's a another variant. Suppose the host, who knows where the car is hidden, opens one of the two doors hiding a goat at random. Note: in my new variant he could open the door already chosen by the player. Suppose he happened to open door 3. (Always I'm supposing the player initially chose door 1 and that the car is equally likely anywhere).

Showing a goat at door 3 rules out door 3 for the car. We're interested in whether the car is behind door 1 or door 2. If the car is behind door 1 the chance is 50% that the host opens door 3. If the car is behind door 2 the chance is 50% that the host opens door 3. It's the same in both cases. So his opening door 3 does not give any new information about whether the car is behind door 1 or door 2.

Notice that in this variant the host is supposed to know where the car is and he is supposed to use his knowledge to only ever open a goat door. But the answer is 50%, not 2/3. So it's not just that the host knows where the car is hidden which is important, and it's not just that he uses this knowledge to reveal a goat, whatever. It's also that he is determined not to open the original door chosen by the player. Because of that, the number of choices he has differs according to whether the car is behind door 1 and door 2. Because of that, the chance he opens door 3 differs in those two cases. Because of that, his opening door 3 gives us information about whether the car is behind door 1 or door 2. Richard Gill (talk) 12:31, 18 January 2013 (UTC)

Again, I do not dispute your logic but I am not sure if that helps or not. Martin Hogbin (talk) 17:27, 18 January 2013 (UTC)

The original question again

Hi. Remember me? I am the non-mathematician whose question above started this section. After reviewing the 7300 words that have appeared in explanation, am I any clearer? Well, no. Let me put my question another way. Suppose that earlier that day the quizmaster has had an argument over pay with his employer, and he storms off the set just as the contestant makes his first choice of door 1. Nobody else is quite sure how to continue, except that the contestant should be offered the chance to change his choice. Then a funny thing happens: the contestant glimpses a tiny pair of hooves under door 3. The mind of the quizmaster has been eliminated; the rules of the game subverted, but he is now invited to change his choice. Should he switch to door 2? My guess is yes, by a factor of 2. Rumiton (talk) 14:20, 18 January 2013 (UTC)

Thanks for coming back with your question, it is always interesting and helpful to get feedback from readers.

Yes, you are correct (with some assumptions that you no doubt intend). The important thing is that the host must reveal a goat, in other words, he always reveals a goat. The host's mind only comes into it because the host would need to know where the car is so that he can guarantee to reveal a goat. Martin Hogbin (talk) 15:54, 18 January 2013 (UTC)

Just to explain a bit more. Your reformulation of the problem introduces all sorts of new possibilities which might complicate the problem. I am taking it that glimpse of the hooves is intended to be the same thing as saying that a goat is always revealed behind one of the unchosen doors but, to avoid someone complicating the argument, I think it is best to stick to the original problem formulation.

The crucial distinction in the normal formulations is between the case where the host always reveals a goat, in which case you double your chances of winning the car by swapping, and the case where the host just happens to reveal a goat in this particular game, in other words the host might have revealed the car. In the latter case if, for example, the host tosses a coin to decide which of the two unchosen doors to open and it just happens that doing this reveals a goat then there is no advantage in switching. Martin Hogbin (talk) 17:06, 18 January 2013 (UTC)

@Rumiton: Your question again is quite clear, and my answer is also as clear as my answer to your first question: No. So your guess is wrong.--Albtal (talk) 18:17, 18 January 2013 (UTC)

@all: Why should in the widespread Parade version anybody assume that the host is committed to open a not chosen door with a goat (and to offer a switch) if she believes that she doesn't need this assumption for the 2/3 solution? - And why should anyone assume this rule if she gets to the fifty-fifty solution? - So the proponent of the 2/3 solution for the Parade version produces a joke, whereas the fifty-fifty proponent gives a correct answer.--Albtal (talk) 21:09, 18 January 2013 (UTC)

Albtal: wikipedia merely summarizes what is written about the Monty Hall problem in reliable sources. Read the literature. For instance, the book by Rosenhouse (who cites hundreds of papers) or the book chapter by Rosenthal. Whatever Marilyn wrote, whatever she intended, the Monty Hall problem today is about the problem in which the host is guaranteed to open a different door to the door chosen by the player and reveal a goat (which he can do because he knows where the car is). Richard Gill (talk) 17:30, 19 January 2013 (UTC)

Others may score your arguments.--Albtal (talk) 18:33, 19 January 2013 (UTC)

Thanks both. Having been brought slowly and reluctantly to the 2/3 solution, I am reluctant to give it up. It seems that Martin applauds my understanding, and Albtal disagrees. I cannot see where I am wrong. If the contestant gains, by whatever possible means, information as to the presence of a goat behind one of his unchosen doors, then the entire 2/3 likelihood that there is a goat among the two, lands on the remaining door. I find this elegant and clever and entirely persuasive. How is it wrong? Rumiton (talk) 12:05, 19 January 2013 (UTC)

Albtal is pushing his own unique interpretation of the question in which the host may open the door originally chosen by the player. To avoid arguments can we take it that you are interested in an answer to the interpretation of the question in which the host always opens an unchosen door to reveal a goat and always offers the swap?

@Rumiton: In your second version you can easily see that your guess is wrong: If the contestant had chosen door 2, you would now get to a 2/3 chance for door 1. But I assume that the goat does not know the rules of the game... I think that your arguments belong to a class of arguments which we can represent by a single argument: If there are three cases, and I consider a group of two of these cases then the chance for the remaining case in this group is 2/3 if one of these two cases can be eliminated. But this argument is clearly wrong. And no one would have made this assertion before MHP appeared. The 2/3 chance only arises if someone who knows the right door is committed to open one of two doors with a blank after the contestant has named these two doors. So in the task set which really has a 2/3 solution the first choice is not really a choice but a request for the host to open one of the other doors with a goat. Without this (or an equivalent) rule the 2/3 solution is wrong. If you are interested you may read my other comments here which contain many arguments, examples, and counterexamples; especially for your questions. --Albtal (talk) 17:11, 19 January 2013 (UTC)

On the above basis I am not sure if you have fully understood the point about how the goat is revealed is important. Let me give two examples. In both cases the host knows where the car is.

A fuller explanation

We take it that the rules require that host always opens an unchosen door and always offers the swap.

1) The player chooses a door (1) then the host says, as he always does, 'I know where the car is so I am going to open one of the two doors that you have not chosen to reveal a goat'. He then reveals the goat behind door 3.

The player now has a 2:1 advantage by swapping.

2) The player chooses a door (1) then the host says, as he always does, 'I know where the car is but I am going to toss this coin and if it comes up heads I will open door 2 and if tails I will open door 3, regardless of what is behind the doors.' The coin shows tails and the host opens door 3, which happens to reveal a goat. (We presume that there is some rule to decide what to do had the host revealed the car, for example the game could be replayed from the start).

The player now gains no advantage by swapping. Martin Hogbin (talk) 14:25, 19 January 2013 (UTC)

Again @Rumiton: In the beginning, the chances of the three doors are 1:1:1. Your example of the contestant glimpsing a tiny pair of hooves under door 3 is rather interesting. My view is the following. As I said in the last year's RfC we should point out just in the beginning of the article that it really matters that the host INTENTIONALLY shows a goat and not just "happens" to show a goat in randomly opening one of his two doors, accordingly ruining the chance to win by switching in 1/3 by showing the CAR instead of his goat (game over).

In the original scenario given by MvS, with a host who KNOWS the location of the car and INTENTIONALLY shows a goat, the chance staying:switching goes up to 1:2.

But in the strange variant of a host who just "happens" as though by chance to show a goat and not the car, chance staying:switching remains 1:1.

That can be explained by using maths, yes.

Maths? I gave a common sense explanation in plain words. In scenario (1) we gain information concerning the relative likelihood of door 1 or door 2, in scenario (2) we don't, because in scenario (1) the chance the host will open door 3 and reveal a goat depends on whether the car is behind door 1 or 2, while in scenario (2) it doesn't. Richard Gill (talk) 17:46, 19 January 2013 (UTC)

Splendid, yes. By common sense explanations in plain words. Redundancy welcome, imho. Very clearly shown and imaginable for grandma and grandson also, to easily comprehend that crucial difference. See the literature of psychologists on exactly that unintelligible matter Gerhardvalentin (talk) 18:17, 19 January 2013 (UTC)

But it should also be said verbally that the latter strange variant does require sth. to match the original version of MvS.

The bizarre version of a host, in 1/3 disposing of ONE GOAT AND THE CAR, but in showing the CAR instead of his goat and leaving his second door with the goat CLOSED, would achieve the success rate of MvS's original version only if he offers then the switch to his open door SHOWING THE CAR. Only then the demand of equality would be fulfilled.

And as to the tiny pair of hooves: that may be possible or impossible in the original version of MvS as well as in the strange variant described. If the host, who knows the location of the car, does not INTENTIONALLY show a goat, the chances REMAIN 1:1 under all other circumstances whatsoever. --Gerhardvalentin (talk) 17:29, 19 January 2013 (UTC)

Gerhard, to clarify a bit, it is not really the intention of the host or his knowledge that matters but the fact that he must always reveal a goat or, to put it another way, he can never reveal the car. Of course, knowledge and intention are practical prerequisites to the host always revealing a goat but it is the fact that he always reveals a goat that counts. If there is a possibility that the host might reveal the car then the problem, and answer, are different. Martin Hogbin (talk) 17:51, 19 January 2013 (UTC)

Martin, I guess you ignore that 1:1 versus 1:2 very very very obviously is difficult to understand for newcomers. Don't forget the vast literature written by psychologists on exactly that matter. So what do you mean? This diference should be explained as colourful as necessary, just in the beginning. Gerhardvalentin (talk) 18:17, 19 January 2013 (UTC)

How you put the problem depends on what you want to achieve. You can make the problem easier for most people to get right by, stating the problem very clearly and unambiguously, and not mentioning door numbers, at least this is my reading of K&W. I think it would also help most people get the answer right and understand the problem better if it said that host must always reveal the goat and never the car.

On the other hand, if you want to show the problem at its most unintuitive best it probably helps to leave a little ambiguity in the problem, distract people with door numbers, and talk about what the host knows rather than what he must do. Martin Hogbin (talk) 18:29, 19 January 2013 (UTC)

An intuitive explanation of why the answer to version 2 above is different from the answer to version 1

In version 2 there is a possibility that the host will reveal the car (1/3 in fact). If the host reveals the car, the result does not count because the game cannot continue as per the original description. The player cannot sensibly be asked to choose between his original door, which how knows hides a goat, and the unopened door, which he also knows hides a goat. If the car is revealed the game is void and has to be abandoned or replayed.

If the player has originally chosen the car the host cannot reveal the car so the only time that a car is revealed is when the player has originally chosen a goat, and will therefore win by swapping. What this means is that all the games in which the player wins by sticking count but some of the games where the player wins by swapping are eliminated, thus swapping is no longer advantageous. Martin Hogbin (talk) 18:12, 19 January 2013 (UTC)

You can say it in words for grandma, also: The aberrant version of a host, in 1/3 disposing of ONE GOAT AND THE CAR but showing the CAR instead of his goat and leaving his second door with the goat CLOSED, would achieve the success rate of MvS's original version only if he offers then the switch to his open door SHOWING THE CAR. Only then the demand of equality would be fulfilled. The well-known false common 50:50 intuition is correct only if the host just randomly opens one of his two doors, careless what's behind it, and if he happens to show the car: game over. Only then the common false 50:50 intuition would apply. Help grandma and grandson to easily finally grasp that crucial difference. I don't understand your intent, do you? Gerhardvalentin (talk) 18:26, 19 January 2013 (UTC)

Questions to Rumiton

Having been brought slowly and reluctantly to the 2/3 solution, I am reluctant to give it up. So there seem to be three steps: 1. You didn't believe the 2/3 solution. 2. You were confident of the 2/3 solution. 3. Now it looks as if your explanation of the 2/3 solution has been wrong.

Which task set had been presented to you? Why didn't you believe the 2/3 solution if somebody said: Suppose the contestant chooses door 1. The host now has to open door 2 or 3 with a goat. If he opens door 2, you finally choose door 3, and if he opens door 3, you finally choose door 2. So you will win the car if it is behind door 2 or door 3. (Or another simple explanation)? And why did you change your opinion? I think it would be helpful for the discussion here and for the article, if you could explain us your way in MHP. --Albtal (talk) 22:05, 19 January 2013 (UTC)

Short comment: the chances staying:switching are 1:2 in the scenario presented by Marilyn vos Savant only, where the "paradox" (not 1:1 but 1:2) arises: The host knows the location of the car / objects, and he intentionally opens a door hiding a goat in order to offer a switch to his second still closed door. Only in that scenario the "clean paradox" arises. Gerhardvalentin (talk) 23:37, 19 January 2013 (UTC)

Important: intentionally is not enough. Only if the host is committed by the rules of the game to open another door hiding a goat the solution is 2/3. In the counterexamples of Martin Gardner and Monty Hall (see above), and in real game shows (see Devlin and Drösser above) the host indeed acts intentionally, but the solution is not 2/3. But this section was intended by me for Rumiton to answer.--Albtal (talk) 10:37, 20 January 2013 (UTC)

OK, ignoring the pain I am feeling in my brain and in the hope our efforts might improve the article, here goes. Initially I was like most other educated but not specifically mathematically inclined people; it seemed to me that discovering that door 3 hid a goat did not tell me anything about door 2. So it was just as likely to conceal a car as was door 1. I have presented this problem to two high school math teacher friends, and they both emphatically agreed that the chances are equal. All three of us were wrong, but why? My new understanding, which I tried to illustrate above, is that the intentions or knowledge of the host are irrelevant: as soon as one of the unchosen doors is eliminated, either through being deliberately opened, or through an accidental discovery, the 2/3 probability that the car was in that pair devolves onto the one remaining door...door 2. Over to you. Rumiton (talk) 13:06, 20 January 2013 (UTC)

??? Intentions or knowledge of the host are irrelevant??? – The host, in only 1 out of three disposing of two goats, but in 2 out of 3 having the CAR and only one goat, inevitably in 1 out of 3 will open a door with the CAR behind. If this happens, the chance to win the price by switching to his second still closed door is WHAT? So intentions and knowledge of the host are irrelevant? You have to consider the consequences. Gerhardvalentin (talk) 13:42, 20 January 2013 (UTC)

Rumiton, have you missed the section above in which I explain exactly what matters when the goat is revealed. What matters is whether the car could ever be revealed. Whether the revealing is intentional or accidental does not matter. What matters in whether thecar might have been revealed.

@Rumiton: Thank you for your answer. All three of us were wrong, but why?. But if you were wrong or not depends on the problem which you tried to solve. And whether your actual opinion about the solution is right or wrong, also depends on the question, which you answer. See my comments above. I think the best method to avoid headache with MHP is to formulate a clear question (without any not formulated additional assumptions). Then every one can solve the problem with three playing cards. But the authors of the actual wikipedia article seem to be friends of headache. If you have further questions or comments, I shall try to answer you. But for the rest I say goodby.--Albtal (talk) 14:41, 20 January 2013 (UTC)

Another example for Rumiton

Case 1

The player chooses a door and the host stands in front in such a way that it cannot open. Before the host can do any more, one of the two unchosen doors accidentally swings open (because one of the stage hands had not secured it properly) revealing a goat.

The host, trying to keep the show going, asks the player if they would like to switch.

Here there is no advantage in switching.

Case 2

The player chooses a door and the host stands in front in such a way that it cannot open. Before the host can do any more, one of the goats nudges a door open revealing himself (we assume that this cannot happen to the door hiding the car).

The host, trying to keep the show going, asks the player if they would like to switch.

Here there a 2:1 advantage in switching. Martin Hogbin (talk) 14:15, 20 January 2013 (UTC)

Rumiton, take things easy! Because both goats don't know the rules of the game, the chances for switching are 1:1 in this version, too.--Albtal (talk) 15:09, 20 January 2013 (UTC)

Martin, I feel that Albtal's conclusion is incorrect and obfuscating. As to me, I guess your "Case 1" pictures the scenario of the forgetful host, true?
And I guess your "Case 2" pictures exactly the scenario of the standard version of the MHP:

The door selected by the guest (1/3 chance on the car) remains closed, as it is to remain closed anyway.
And from the group of the two unselected doors (together 2/3 chance on the car), the car will never ever be shown.
So, even if there should reside the car in 2/3, then the only goat will appear.
And if there should reside both two goats in 1/3, then one of them will appear just at random. Is that the genuine scenario that the standard version of the MHP is based on?
Chance of the door selected by the guest (chance on the car 1/3) versus the only other still closed door (chance on the car 2/3) is 1:2.

Am I right? -- Gerhardvalentin (talk) 19:40, 20 January 2013 (UTC)

I think you are right, but I would only say so in a quiet, unassertive way, almost a whisper. I will try to reply to Albtal's comments later today. Rumiton (talk) 01:37, 21 January 2013 (UTC)

By all means reply if you like but I agree with Gerhard. Albtal is pushing his own unique interpretation of the question in which the host may open the door initially chosen by the player. No source treats the problem this way and sources tells us that practically nobody interprets it that way.

To avoid further confusion it might be useful if you would confirm that, in the interpretation that you are interested in, the host always opens an unchosen door to reveal a goat (if that is indeed the case).— Preceding unsigned comment added by Martin Hogbin (talk • contribs) 11:06, 21 January 2013 (CEST) (UTC)

Solutions.

Regarding Case 2, I think you are all wrong.

I suppose in each case the player had selected door 1 and it was door 3 which opened revealing a goat (and nothing else happened!). Introducing the door numbers makes the problem easer to analyse.

Case 1. Whether the car is behind door 1 or door 2, the chance that door 3 accidentally swings open and reveals a goat(and nothing else happened) is the same (at least, this seems to me to be a reasonable assumption). Hence the information that this event took place doesn't alter the relative likelihood the car is behind door 1 or behind door 2. Everyone seems to agree on this one.

Case 2. Whether the car is behind door 1 or 2 can very reasonably affect the chance a goat bumps into door 3 and opens it (and nothing else happened), because what else might have happened - which competes with what actually did happen - is different. Suppose for instance any goat behind any door has a 1 in 10 chance to nudge a door open within the length of time in question. Suppose this is true, independent of what happens at other doors. Then, the chance that door 3 is nudged open (and only door 3) is 1/10 when the car is behind door 2, and 1/10 times 9/10 when the car is behind door 1 (9/10 for the chance the other goat didn't nudge door 2 open, "times" for my assumption of independence). So what we actually see (and don't see) is slightly more likely if the car is behind door 2 than if it is behind door 1. The posterior odds on the car being behind door 2 are 10 to 9.

And if another contestant (or any other person) had first chosen door 2? Which chances does she now have with door 1?--Albtal (talk) 22:47, 21 January 2013 (UTC)

Yes Richard, it is always possible to contrive complicating factors and your application of Bayes rule is most instructive in showing how to deal with them but it is out of place here. I was trying to help Rumiton understand exactly what matters and what does not. At the moment, this page gives the impression that there is some doubt or disagreement about the essential point in question, which is that it matters how the host reveals the goat. I am always happy to discuss interesting variants of the problem (probably better in user space) but Rumiton's question shows that we need to improve in the article. Perhaps that you could confirm for the benefit of Rumiton and other interested readers, that, barring complications like a goat sensing a car in the room next door or the stage hand leaving the handbrake of the car off, my explanation is correct. Martin Hogbin (talk) 10:58, 21 January 2013 (UTC)

Martin, I dd not contrive complicating factors. I tried to solve your case 2 under a perfectly reasonable interpretation of that case, and I discover that your answer was very wrong!

I did this because I wanted to help Rumiton; and you *need* help too. I'm telling you and Rumiton that there is a simple, principled, systematic, intuitively understandable way to solve all conceivable variants. We initialy chose door 1, and now we get information that the car is not behind door 3. Answer the two questions: what is the chance that I get this information when the car is behind door 1? And what is the chance when the car is behind door 2? If those two chances are the same, then the odds on door 1 versus door 2 have not changed; the information was not biased. If the chances are different, then the odds on door 1 versus door 2 have changed by the same proportion; the information was biased. To modify the words of Marshal McLuhan: the medium and the messenger are part of the message. It's not just the face-value definite facts which you have learnt which are important, it's also important why/how those facts came to your knowledge.

So at a very general level, I'm confirming that it indeed matters how the host comes to reveal the goat. I'm moreover saying exactly *how* it matters: in a very simple way, intuitively highly understandable, easy to remember, easy to apply. Richard Gill (talk) 14:37, 21 January 2013 (UTC)

Sorry Richard I misread what you said so my criticism was unjustified. Also I accept that I did not properly describe my intended scenario properly. I was essentially considering the case where the host waits for a goat to nudge a door and then immediately offers the swap. There is no possibility that both doors will be opened or that no door will be nudged, regardless of the position of the car. So, yes, it was a bad example. Martin Hogbin (talk) 15:52, 21 January 2013 (UTC)

The answer in Case 2 depends on a lot of things which were not specified in advance. However, using Bayes' rule tells us what questions we have to ask, even if it might not determine the answer unequivocally. Moreover: Bayes's rule is simple, it's intuitive, and it comes with a gold-plated guarantee. Richard Gill (talk) 09:48, 21 January 2013 (UTC)

I really do not have anything against Bayes' rule. Martin Hogbin (talk) 10:59, 21 January 2013 (UTC)

@All. I have read all your arguments carefully, and also Albtal's zwei Ziegen link, though I found it quite repetitive and rather obscure in its aguments. Frankly I am still missing something. The concepts revolve around each other, biting each other's butts. Maybe my layman's notion of what probability actually means needs overhauling. And I am not at all singling any editor out, honestly, and my sincere thanks go to you all for your time and effort, but when I see, The host, in only 1 out of three disposing of two goats, but in 2 out of 3 having the CAR and only one goat, inevitably in 1 out of 3 will open a door with the CAR behind. If this happens, the chance to win the price by switching to his second still closed door is WHAT? my eyes start to water and I wish I was somewhere else. I think it is best if I pull back and watch this page. I really have enjoyed the mental stimulation this subject supplies, and if current or future editors can identify where the confusion is coming from and formulate some changes to the article that will make it clearer to the non-mathematician, I will be very happy to offer my help with the wording. I apologise for my obtuseness, which was not willful. Rumiton (talk) 13:16, 21 January 2013 (UTC)

Thank you for reading my comments. But this is not the discussion of arguments which I intended.--Albtal (talk) 05:33, 22 January 2013 (UTC)

Dear Rumiton, I told you a simple, principled, systematic, intuitively understandable, universal method to solve all conceivable variants. It's presented in numerous reliable sources. It's supported by mathematics but you don't need any mathematics to appreciate it.

Let me try again, in the hope some layman will understand this, and say it again in better layman's language.

The MHP variants we are talking about all have in common: we initially chose door 1, and now we get information that the car is not behind door 3.

Here's the method. Just answer the two questions: what is the chance that I get this information when the car is behind door 1? And what is the chance when the car is behind door 2? If those two chances are the same, then the odds on door 1 versus door 2 have not changed; the information was not biased. If the chances are different, then the odds on door 1 versus door 2 have changed by the same proportion; the information was biased.

To modify the words of Marshal McLuhan: the medium and the messenger are part of the message. It's not just the face-value definite facts which you have learnt which are important, it's also important why/how those facts came to your knowledge. What is so difficult to understand? Richard Gill (talk) 14:33, 21 January 2013 (UTC)

I'm sorry, Richard, I wish I could say Eureka! I see it! But I don't. And when you write, what is the chance that I get this information when the car is behind door 1? I am not even clear what you are talking about. Rumiton (talk) 15:05, 21 January 2013 (UTC)

Thanks! The "information" I'm talking about is the set of new facts which comes to the attention of the player after he's chosen door 1 (up to the moment when he may reconsider his initial choice). In the standard Monty Hall problem, it's "the host opens door 3 and reveals a goat" (door 3 rather than door 2). If you don't like the word "information" substitute it with a different word. Like "new facts" for instance? Or "message"?

The reason the odds on door 1 versus door 2 change on learning these new facts is because the chance of getting precisely those facts (receiving exactly that message and not a different one; getting exactly that news) is 50% when the car is behind door 1, but 100% when it is behind door 2.

Sure, MHP is a brain-teaser, and thinking about several dfferent variants is certain to cause head-aches. Fortunately there's an intuitive, systematic way to solve all known varants as well as any new ones anyone might have nightmares about. Richard Gill (talk) 15:16, 21 January 2013 (UTC)

OK, I get that! I really do. The host has truthfully said to the contestant, "There is no car behind door 3." If there is a car behind door 1, he might still have provided the same message, but he might equally have said, "There is no car behind door 2." So the chance of the contestant receiving that precise message is 50%. Hallelujah. Let us proceed! Rumiton (talk) 10:24, 22 January 2013 (UTC)

Hallelujah! Richard Gill (talk) 09:28, 24 January 2013 (UTC)

It really matters that the host knows the location of the objects
and that he purposely opens a goat-hiding door, intending
to offer a switch to his other still closed door

Rumiton, I see you really are helping to uncover the weak points / entrapments of the article. Please stay helping.

"If this happens, then the the chance to win the price by switching to his second still closed door is WHAT?" – That was just my provocative edit addressing the statement

"... the intentions or knowledge of the host are irrelevant: as soon as one of the unchosen doors is eliminated, either through being deliberately opened, or through an accidental discovery, the 2/3 probability that the car was in that pair devolves onto the one remaining door...door 2."

Because in the MHP the history of development is most significant:

• The host randomly hides the one and only prize behind one of the three doors.
• The guest randomly selects one of the three doors.
• The host then purposely opens a losing door from his two unchosen doors, being absolutely determined in any case to offer a switch to his second unchosen door (we know that he already decided to do so, beyond repeal).

Please regard the "core of the paradox":

In this very special situation,  resulting from a special history,  there remain two still closed doors. It is well known that one of those two doors does hide the prize for sure, while the other door must be hiding the second goat for sure. The location of the prize and the location of the second goat are unknown / in dispute.

So the great majority of people forget about that special history of development and are just apt to say that the chances are 1:1, while in effect, in that given special situation, the chances are 1:2 indeed. So this veridical paradox can cause head-aches. The key for "not 1:1 but 1:2" is the history of development in a given special scenario.

In other scenarios or if you forget about this special history of development, the answer "1:1" could even be correct with a host who does not matter to him which one of his doors he opens:

Anyhow the host has got two unselected doors     A: [goat  goat]     or B: [goat  car] or     C: [car  goat]   (together 3 possible locations of the car, each one with two unselected doors that the host has to chose from).

If he just opens one of his two doors, no matter what it hides, goat or car, then

in "A:" he will show (1) the left goat offering a switch to the second goat,   or he will show (2) the right goat offering a switch to the second goat
in "B:" he will show (3) the goat and offer the switch to the car, or he will show (4) the car (game over as switching to his goat makes no sense)
in "C:" he will show (5) the car (game over as switching to his goat makes no sense) or he will show (6) the goat and offer the switch to the car.

The door first selected by the guest has 1/3 chance to hide the prize. In the scenario of a host who does not care which one of his two doors he opens, be it goat or car, a switch will give the car in only 2 out of 6 (1/3): in (3) and in (6). – It would only equal the outcome of a host who never shows the car, when it is given that in (4) and in (5) he would offer a switch to the shown car.

Only then the result would equal the scenario of a host who intentionally shows a goat and never the car, giving a chance to win by switching in 4 out of 6 (2/3).

So it really matters that the host knows the location of the objects and that he purposely opens a goat-hiding door, intending to offer a switch to his other still closed door. In this scenario, the clean "paradox" arises: 1:2 and not 1:1. Regards, Gerhardvalentin (talk) 17:22, 21 January 2013 (UTC)

Thank you, Gerhard. I still have difficulties with your argument. You are using some common English words in an idiosyncratic way and it is throwing me. Mainly, these words are "disposes", determined", "indifferent", and "matter" (with a host who does not matter...) If I already understood the argument you present I am sure I could catch your meaning, but as it is, I am defeated. Can you rephrase? Rumiton (talk) 11:55, 22 January 2013 (UTC)

"he disposes of": I tried to say that he has got; - "he is determined to do so": I tried to say that he has irrevocably decided to do so; - "indifferent": I tried to say that it does not matter to him; – Pardon Rumiton, I'm no native speaker, as you can see. I wanted to help to point out the following:

The door first selected by the guest has 1/3 chance to hide the car, so by staying he wins in 1/3.
And it matters whether the host offers a switch to the prize (!) in 4 out of 6, as in the original scenario of a host who is absolutely determined to ... / who irrepealably decided to show a goat and never the car in order to offer a switch to his second still closed door in 6 out of 6, or if he offers a switch to the prize only in 2 out of 6, by ignoring what the door opened did hide.  Regards,  Gerhardvalentin (talk) 15:47, 22 January 2013 (UTC)

Thank you, Gerhard. I have worked as a German-English translator so I am very sympathetic to people experiencing the treacherous nature of the English language. And here we (or all of you) are translating from the pure language of mathematics to that same perfidious language, which makes it so much more challenging. I am still thinking about your statements. Thank you. Rumiton (talk) 11:49, 23 January 2013 (UTC)

Thank you Rumiton. And I just saw what you said below to Martin: "And as you say, it is the declared intention of the ..."

In my opinion this is the "key to the correct different answers"

1:1 (correct common sense answer, based on the assumption that the host did not care what the door he opened did hide, whether car or goat) – versus

1:2 (correct answer based on the declared intention of the host to show a goat in any case, but never the car).

Imo this should be pointed in the article as an "eye opener", as soon as possible. Regards, Gerhardvalentin (talk) 14:08, 23 January 2013 (UTC)

 

Rumiton, I am glad to see that you are still here, I hope that you brain has cooled down a bit and you are not put off by what might seem like continual argument here. I said above that Richard had complicated things regarding my example where the goat kicks a door open. It was in fact me who complicated things by giving a bad example. I was trying to create a scenario where an unchosen door was always opened to reveal a goat and never the car but I got it wrong as Richard correctly pointed out. As I worded the example it was possible that a goat would not open a door at all. This makes a difference.

I hope you will not mind a simple example to show that how information is revealed forms part of the information itself and this can affect probabilities.

A fair coin is tossed and, depending on the result, either 9 black balls and 1 white ball, or 9 white balls and 1 black ball are secretly placed in a box. The question to be answered is 'What is the probability that 9 black balls and 1 white ball were originally placed in the box?'. Simple, as it depends on the fall of the coin, the answer is 0.5 (50/50).

Now a black ball is removed from the box. What now is the probability that 9 black balls and 1 white ball were originally placed in the box? It depends on the circumstances surrounding the ball's removal.

Case 1. A person says, 'I will look in the box and remove a black ball'. This tells us absolutely nothing about what was originally in the box. Would you agree?

Yes, I would. Yes. So the chance is still 50:50 on whether the coin came up heads or tails. Rumiton (talk) 11:49, 23 January 2013 (UTC)

Case 2. A person says, 'I will take a ball at random from the box, without looking inside'. The ball taken proves to be black. I would bet on there originally being 9 black balls and 1 white ball in the box. Would you? Martin Hogbin (talk) 19:30, 22 January 2013 (UTC)

Yes. Absolutely. It seems much more likely. And as you say, it is the declared intention of the ball remover that changes the probability. This smells like progress. Rumiton (talk) 11:49, 23 January 2013 (UTC)

Excellent. I guess you would say the same if, instead of tossing a coin to decide how to start, a fair die is rolled and if it come up with 1 or 2 then 9 black balls and 1 white ball are put in the box, otherwise 9 white balls and 1 black ball are placed in box.

Now in case 1 the probability of 9B 1W is 1/3 but in case 2 it is more than 1/3. Agreed? Martin Hogbin (talk) 19:17, 24 January 2013 (UTC)

Yes, I am sure that would be true. Rumiton (talk) 02:12, 25 January 2013 (UTC)

You may be able to guess what is coming next. Suppose we have either two goats or one goat and one car behind the two unchosen doors. The probability of the a latter is 2/3 as this occurs if the player has originally chosen a goat. Exactly as above, if the host always reveals a goat, the we get no information about whether there were originally two goats are two goats and a car because the host always reveals a goat regardless of whether there were two goats or a goat and a car, so the probability that there were a car and a goat behind the two unchosen doors remains 2/3. We do, of course get information about what is behind that the door that has been opened, we know for sure that the car is not there and, as we also know that there was a 2/3 chance that there was a car and a goat, we now know that there is a car behind the unopened door with probability 2/3.

Now let us consider the case where the host chooses randomly between the two unchosen doors. If he reveals a goat it makes it more likely that there was originally two goats, thus reducing the benefit of swapping.

To make Richard happy we can calculate how much more likely it is that there were originally two goats behind the two unchosen doors if the host chooses randomly. Before a door is opened, it is twice as likely that we had a car and a goat than that we had two cars. When the host opens a door, he is twice as likely to reveal a goat if there were two goats compared with the case where there was a car and a goat. As we only count times where the host reveals a goat (because the question tells us that is what happened) we can double the odds that there was originally two goats. This now makes it evens between two goats or a goat and a car, so swapping gains us nothing. Bayes rules OK! Martin Hogbin (talk) 09:53, 25 January 2013 (UTC)

Martin, I have taken the extreme liberty of rewriting your post in a form I can understand. Do I have it right?

You may be able to guess what is coming next.

Suppose we have either two goats or one goat and one car behind the two unchosen doors. The probability of the latter is 2/3, as this occurs if the player has originally chosen a goat.

Exactly as above, if the host always reveals a goat, then we get no information about whether there were originally two goats, or one goat and one car, because the host always reveals a goat, regardless. So the probability that there were a car and a goat behind the two unchosen doors remains 2/3. We do, of course get information about what is behind the door that has been opened. We know for sure that the car is not there, and as we also know that there was a 2/3 chance that there was a car and a goat, we know now that there is a car behind the unopened door with probability 2/3.

Now let us consider the case where the host chooses randomly between the two unchosen doors. If he reveals a goat it makes it more likely that there were originally two goats, thus reducing the benefit of swapping.

To make Richard happy, we can calculate how much more likely it is that there were originally two goats behind the two unchosen doors if the host chooses randomly. Before a door is opened, it is twice as likely that we had a car and a goat than that we had two goats. When the host opens a door, he is twice as likely to reveal a goat if there were two goats compared with the case where there was a car and a goat. As we only count times where the host reveals a goat (because the question tells us that is what happened) we can double the odds that there were originally two goats. This now makes it evens between two goats or a goat and a car, so swapping gains us nothing. Bayes rules OK!' Martin via Rumiton (talk) 11:31, 25 January 2013 (UTC)?

Yes, thanks for clarifying what I wrote (I have taken the liberty changing the bold to italic as it looked a bit aggressive). Does it make sense now? Martin Hogbin (talk) 16:19, 25 January 2013 (UTC)

Actually, there is still a problem which I missed. Would you mind reading carefully over the whole post? We have it is twice as likely that we had a car and a goat than that we had two cars. Clearly there is only one car in the game. Cheers. Rumiton (talk) 12:52, 27 January 2013 (UTC)

Sorry Rumiton that is just carelessness on my part, it should, of course, have read, it is twice as likely that we had a car and a goat than that we had two goats. I have amended the section above for the benefit of anyone else who may be trying to follow this. Martin Hogbin (talk) 21:51, 27 January 2013 (UTC)

Still: The host never shows the car

How about saying it with words that can easily be understood? Valid for the host who has the declared intention to show a goat in order to offer a switch to his other still closed door, according to the scenario of Marilyn vos Savant. Redundancy is most welcome imo, so I would say it also in that way for grandma and grandson:

The door first selected by the contestant has a 1 in 3 chance to hide the car, and the group of two unselected host's doors may contain with equal chance of 1/3:

1. goat + goat
2. goat + car
3. car  + goat

Whether the unknown one-time host now did open his leftmost door or his rightmost door to intentionally show a goat and to offer a switch, gives us no additional hint whatsoever on the contents of the door first selected by the contestant.

After having opened one of his two doors showing a goat behind, his second and still closed door now will contain with equal chance of 1/3:

1. one goat
2. one car
3. one car

So the door offered to switch on has a 2 in 3 chance to hide the car. Am I correct? Too easy?  Gerhardvalentin (talk) 17:29, 25 January 2013 (UTC)

I was trying to show how we can use Bayes' rule to calculate the odds that there were originally two goats behind the unchosen doors, given that the host has revealed a goat.

The prior odds of two-goats:goat-and-car are 1:2. If the host chooses randomly then the probability of revealing a goat in the two-goats case is 1 and in the car-goat case it is 1/2. To get the odds of two-goats:goat-and-car given a goat has been revealed we multiply the prior odds by the probability of a goat being revealed in the two-goat case divided by the probability that a goat would be revealed in the goat-and-car case, namely 2. Thus the odds become 2:2 or evens.

If the host always reveals a goat then the probability of a goat being revealed in the two-goat and probability that a goat would be revealed in the goat-and-car case, are both 1 so the original odds remain unchanged at 1:2 after the host has revealed a goat and the player gains by swapping. Martin Hogbin (talk) 18:43, 28 January 2013 (UTC)

And not helping the reader to decode the paradox? Gerhardvalentin (talk) 20:15, 28 January 2013 (UTC)

Editor note

Moved from article, see this edit

• • • • Editor note: This edit I suggest should replace the next two paragraphs of the current page as neither are helpful nor reliable.

It is Vos Savant's response that is inadequate. She simply states as though it were self-evident that switching to the other unopened door doubles one's chances. It can however be demonstrated that staying with the first choice, whether it be Door 1, 2 or 3, increases the probability that it holds the prize to 50%, equally with switching. The already-opened door does not form part of the probabilities of the next event. The chosen door and the unopened door form 100% of the probabilities, 50% and 50%. The probability of the already-opened door holding the prize is zero %. You can't have zero % of an happening.

Vaszonyi's article, and it's only a chatty article in a column of a University magazine, does not state that distinguished mathematician Paul Erdos agreed with Vaszonyi that switching increased one's chances. In fact Vaszonyi ran a computer simulation in 1995 (How good was the simulation? How many times was it run? How good was the randomization process?) that favoured switching and Vaszonyi himself did not understand why. Vaszonyi states that later on Erdos talked to mathematician Paul Graham about it but that Vaszonyi could not understand what Erdos was saying (Erdos would have been 92 at the time)(Vazsonyi 1999).

• • • • • • delete

--Tito Dutta (talk) 08:05, 23 January 2013 (UTC)

Thank you Tito. This is another of the long stream of users who refuse to believe that switching increases your chances of winning. That is why this problem is so famous, the correct answer is remarkably unintuitive. Martin Hogbin (talk) 09:29, 23 January 2013 (UTC)

"It can however be demonstrated that staying with the first choice, whether it be Door 1, 2 or 3, increases the probability that it holds the prize to 50%". I would like to see the demonstration, Titodutta! Are you talking about the same MHP as we are? Richard Gill (talk) 09:31, 24 January 2013 (UTC)

It was User:203.3.64.20 who said that on in the article. Tito Dutta deleted those paragraphs of IP 203.3.64.20 from the article page to put them here on the discussion page. It's what IP 203.3.64.20 did say. Gerhardvalentin (talk) 09:53, 24 January 2013 (UTC)

Yes, it is not very uncommon that readers leave feedback in article. (I) generally remove/revert useless feedback (like "I like it", "Thank you for information", "Hi, add me in Facebook" etc) and move helpful/detailed/(comments which I can not understand whether helpful or unhelpful) to talk page. Found another one today. --Tito Dutta (talk) 17:10, 24 January 2013 (UTC)

Edit to article by 203.3.64.20

I have removed this edit from the article as the explanation given is already in the article.

Perhaps the reason why so many maintain that switching makes no difference is that the reason for switching has never been explained clearly. The reasoning is that the initial choice of a door is likely, by two to one, or 66%, to have a goat behind it. If there is no switch then the contestant is 66% likely to be left with a goat. If a contestant switches from the door that is 66% likely to have a goat behind it to the other unopened door, it is 66% likely to contain the car.

Yet again this shows the need to make the correct solution clearer and more convincing. Martin Hogbin (talk) 09:22, 25 January 2013 (UTC)

Agreed. The intro to the article has "Vos Savant's response was that the contestant should switch to the other door. If the car is initially equally likely to be behind each door, a player who picks door 1 and does not switch has a 1 in 3 chance of winning the car while a player who picks door 1 and does switch has a 2 in 3 chance, because the host has removed an incorrect option from the unchosen doors, so contestants who switch double their chances of winning the car." The long second sentence with Vos Savant's alleged explanation is extremely clumsy. How about replacing it with A player who sticks with his original door has a 1 in 3 chance of getting the car, since he gets the car if his initial choice was correct, he doesn't get it if his initial choice was incorrect. On the other hand, a player who switches has a 2 in 3 chance of getting the car, since he gets the car if his initial choice was incorrect, he doesn't get it if his initial choice was correct.

I would prefer this statement, You initially have a 1/3 chance of picking the car. If you pick the car and swap you get a goat, if you pick a goat and swap you get the car. Or maybe this longer version, You initially have a 1/3 chance of picking the car and a 2/3 chance of picking the goat. If you pick the car and swap you get a goat, if you pick a goat and swap you get the car. If you swap, you therefore have a 2/3 chance of ending up with the car and a 1/3 chance of ending up with the goat . Martin Hogbin (talk) 10:09, 26 January 2013 (UTC)

Both of these are good, and much better than what we have now! Richard Gill (talk) 13:27, 28 January 2013 (UTC)

Also, it is an 'unconditional' solution but it refers to (an assumed) 'unconditional' problem statement so there should be no arguments. The statements are true regardless of the door opened by the host or the goat revealed by him. Martin Hogbin (talk) 19:01, 28 January 2013 (UTC)

One could also add in the introduction to the article an executive summary of a conditional solution. Another way to see that switching is favourable is as follows. Given that the player initially chose door 1, if the car is behind door 1 the chance is only 50% that the host will open door 3 (he could also have chosen door 2). If however the car is behind door 2 the chance is 100% that the host will open door 3 (he has no other choice). So the host is twice as likely to open door 3 when the car is behind door 2 than when it is behind door 1. Given therefore that door 3 was opened, the car is now twice as likely behind door 2 than behind door 1.

As 'conditional' solutions go I quite like that but I still think it is too complicated for the introduction, and possibly against the long-argued consensus. I think it would make an excellent introduction to the 'conditional' solutions though. Martin Hogbin (talk) 10:09, 26 January 2013 (UTC)

The consensus was to drop the fight and to structure the article by first presenting simple solutions (without criticism), then conditional solutions, then criticism in general. So I don't think that neutral inclusion of a conditional solution in the introduction, which after all should give the reader some idea of the whole subject, is against the consensus, as long as it is written in simple words which everyone can understand. Maybe someone else can rephrase my proposal so as to be even more generally accessible? It seems to me that it could be useful for a very wide range of readers. It's the informal way (as informal and accessible as possible) to present the widely favoured solution of the academic/scientific world (the books of Rosenthal and Rosenhouse and numerous journal articles). Many good reasons for it to be there in the introduction to the article along with one or two other one- or two-liners. Different readers find different approaches useful. Different one- or two-line solutions give different insights or intuitions, all of them can be useful. Richard Gill (talk) 13:26, 28 January 2013 (UTC)

By 'introduction' you meant the lead. I was thinking of the first part of the body of the article. Although I have no objection in principle to including a 'conditional' solution in the lead but I think it will be almost impossible to get something short and snappy enough (you suggested a one or two-liner). Also, the lead should be a summary of the article as a whole. My proposal for a simple solution is in the body of the article already. Martin Hogbin (talk) 19:01, 28 January 2013 (UTC)

The solution I offered you is also in the body of the article already. I think it's short and snappy. If you don't think it is short and snappy, please help me rephrase it. Given that the player initially chose door 1, if the car is behind door 1 the chance is only 50% that the host will open door 3 (he could also have chosen door 2). If however the car is behind door 2 the chance is 100% that the host will open door 3 (he has no other choice). So the host is twice as likely to open door 3 when the car is behind door 2 than when it is behind door 1. That makes the car twice as likely behind door 2 than behind door 1, after the host has opened door 3. Richard Gill (talk) 19:02, 29 January 2013 (UTC)

How about this for short and snappy: Suppose the player chose door 1 and the host opened door 3 to reveal a goat. The car is now twice as likely behind door 2 as behind door 1 because the host's action was 100% certain in the first case (it was forced) but only 50% likely in the second (he had two choices). Richard Gill (talk) 19:46, 29 January 2013 (UTC)

To you, that explanation may be simple but, to most people, it is not at all clear why, 'The car is now twice as likely behind door 2 as behind door 1 because the host's action was 100% certain' true though that may be. Martin Hogbin (talk) 11:47, 31 January 2013 (UTC)

It seems to me that in order for the introduction to the article to be a good introduction to the actual content, it should mention that there are many ways to show that switching is beneficial. It would be helpful to all those readers who come to the article, don't believe that the answer is "2/3, switch", to emphasize explicitly that they're wrong and that there are many ways to see this. It would be helpful to write out a couple of the "one- or two-liner" solutions in the introduction, as clearly as possible,. This would also benefit those readers who do think the answer is 2/3 and don't understand why the article is so long and complicated. Richard Gill (talk) 13:09, 25 January 2013 (UTC)

'Yes I agree. I think some earlier versions of the article had more along those lines; I will have a look. Martin Hogbin (talk) 10:09, 26 January 2013 (UTC)

making progress

I appreciate these proposals to write the introduction in simple words which everyone can understand. And I like especially what IP:209.49.60.66 said here and here, I would say it in similar words, accentuating the key of decryption: In the first the contestant is free, in the second the host is "2/3 slave" to the act.

Even though door #2 and #3 have equal overall chance to be opened, it never is a coincidence which one of those two host's doors actually is opened by the host. The overall chance of 1/2 is valid only before the contestant has made his first selection. Afterwards however, the two unselected host's doors are no more 1:1 to be opened. As soon as the contestant made his first selection, in any specific case, the host never is completely free to choose, he definitely has a choice which door to open only in 1 out of 3, but in the remaining 2 out of 3 the host is slave to the act and never has a choice. In that 2 out of 3 he definitely is forced to open a specific door only, it's the one that he actually opens. So focusing on door numbers is misleading if we try to decode the paradox, and we better should ask WHY a specific host's door has been opened, and why not the other one?

In 2 out of 3 the host is definitely enforced to open that specific door that he opens, because his second still closed door (that he offers to switch on) actually is definitely hiding the prize. In any case, the door that has been opened was 2/3 forced to be opened, because in that 2 out of 3 the other door definitely hides the prize. As soon as the contestant did make his first selection, independent of door numbers, the door actually opened, in 2 out of 3 had been forced to be opened. You ask which one? Quite easy, it's the one that has bee opened that in 2 out of 3 has been forced to be opened. --Gerhardvalentin (talk) 18:33, 28 January 2013 (UTC)

Crazy idea of the day but would there be any benefit in having an article, 'Intuitive solutions to the MHP'? We would sill require solutions to be sourced, or else be 'routine calculations' but perhaps there could be more leeway for different types of solution. We might be able to categorise solutions. Martin Hogbin (talk) 19:10, 28 January 2013 (UTC)

You suppose that the host has limited choice in choosing which door to open is unsourced OR? You suppose that in 2 out of 3 being forced to avoid to show the car is unsourced OR? For user IP:209.49.60.66 it was the key to decode the paradox and to understand. You call it a crazy idea to help the reader to decode the paradox. Maybe. -- Gerhardvalentin (talk) 20:15, 28 January 2013 (UTC)

All simple solutions are intuitive solutions. The conditional solutions can be made intuitive too, but first of all you have to understand them; understand them so well that you even see their charm; and then you have to find ways to express them in a few short sentences - ideas rather than calculations. I have been trying and trying but no one wants to know. The people who support simple solutions don't want to find ways to make the conditional solutions simple too; the people who support conditional solutions seem to want to keep them difficult (specialistic, technical). It's very saddening. Richard Gill (talk) 18:58, 29 January 2013 (UTC)

Richard, it is sad that you say "making no progress at all, very sad." – As to me, it means progress

1. to stop baffling the reader with cryptic nonsense,
2. and stop to simultaneously refer to diverging scenarios, page per page, without saying what idiosyncratic scenario the actual sentence tries to refers to,
3. stop fooling the reader with reportedly given exact actual "knowledge about the host's given bias, its direction and its degree"
   (but yes, show that invincible range of "1/2 to 1" in the variants section, as an actually nonsensical tool that cannot give any result – but scrap and junk only if it uses any actual one-sided bias, as we never have nor will have any knowledge about),
4. stop to present a textbook in conditional probability and stop to offer private lessens in conditional probability theory page per page, but never addressing the core of that brilliant paradox, nor showing how to detect nor to decode it.

Can you help to find sources with succinct wordings like

• As soon as the contestant has chosen his door, the two non selected host's doors, as they inevitably hide at least one goat, have a risk to hide one goat each of either "1/3:1" or "1:1/3" and never any other value,  and
• It is most significant that the host in any case opens a losing door on purpose and never may show the prize, and
• therefore in 2 out of 3 the host is slave to the door first selected by the contestant,
• and to plausibly explain the difference of 1:1 to 1:2, concurrently say that - if the host could show the price in 1/3 – he would inevitably be destroying the perfect chance to win by switching in that 1/3, if he was not forced in such case to offer a switch to his open door showing the car.

And so on. It is most important to also verbally prepare the ground for decoding and understanding the paradox. And just in the beginning show conditional probability, but not for the polemical purpose meant by MCDD, but just to show that fair and proper authoritative tool (Bayes rule e.g.), because it is convincing and easy enough to be understood by the reader interested in. I still hope we are making progress. Am I wrong?  Gerhardvalentin (talk) 00:04, 3 February 2013 (UTC)

Asymmetry - no "jumping of probability" in the chronological order

To think of "jumping" is inaccurate. The paradox is hardly to decode, the article should help to understand the paradox. What I said for years has been discarded, now IP 199.227.123.6 yesterday said a similar thing in the article. It is about the chronological order, and it is about asymmetry.

Originally any three doors have equal chance of "1:1:1" to hide the car (or "1/3:1/3:1/3") and they have equal risk of "1:1:1" to hide a goat (or "2/3:2/3:2/3"). Three doors, but only one car. As soon as (just mentally) one door of three should be singled out, no matter which one, then any group of the remaining other two doors, no matter which ones, will hide one goat at least. Nijdam stereotypical used to object "but you don't know which one", although no one needs to know which one. It is just evident that asymmetry did emerge. The risk of any two remaining doors to hide a goat may still be "said" to be "2/3:2/3" and nothing else, but this is not the full truth. As a result of asymmetry that emerged, this even is bluntly inaccurate. You should consider that it can only mean "1:1/3" or "1/3:1", with equal probability. This is true before any door will be opened, and the reader should be aware of that fact. And after opening of a door with a goat behind, it will become evident that this door effectively had a risk of "1" to hide a goat, just from the start. No more "jumping of probability".

And as to the combined doors: To "assume an unbiased host" is never necessary, it's enough that we never can nor will know anything about such one-sided bias. Basing on any such forever unknown one-sided bias though, albeit mathematically fully correct, as a post-hoc fallacy will result in mandatory scrap and junk values for the effective situation actually given, never apt in any way to affecting the answer / decision actually requested. Gerhardvalentin (talk) 17:47, 1 February 2013 (UTC)

Misleading pictures

I notice the misleading pictures in the 'Simple solutions' section, trying to explain the combined doors argument, are still there. I suggest to remove them. Nijdam (talk) 00:43, 29 January 2013 (UTC)

The combined door argument is correct for the one-time show presented, with an unknown host who will show a goat but never the car. See the actual literature. He might be biased as much as he wants to be, but regrettably we cannot profit from such knowledge, as have no evidence whatsoever. While just "assuming" (?) some bias and its direction, without evidence, although mathematically fully correct, is a blunt illusion and will inevitably lead to a wrong actual answer. Value within the range of 1/2 to 1, yes. But any illusory "closer value" that differs from 2/3 is actually inapt and therefore (actually) just bluntly wrong.   Gerhardvalentin (talk) 02:30, 29 January 2013 (UTC)

Nijdam, good to see you back taking an interest in this page. I know these things have been discussed at length before but could you tell us exactly what you think is misleading about the pictures an captions as they are now. Martin Hogbin (talk) 11:04, 29 January 2013 (UTC)

This has been discussed before, and as you may recall, Richard agrees that a fixing step is needed. The pictures, without this step, are misleading. The upper one combines the doors 2 and 3, and indicates a combined probability of 2/3. This is correct, because each door has probability 1/3. The lower picture however, out of the blue, shows contradictory probabilities 2/3 and 0 for respectively doors 2 and 3. How come? Nijdam (talk) 12:00, 29 January 2013 (UTC)

There is nothing contradictory about assigning probabilities 2/3 and 0 to doors 2 and 3. But an explanation is required. Because we don't know anything about how the host chooses a door to open when he has a choice, either is for us equally likely. This creates a symmetry between doors 2 and 3 and because of that, whether the host opens door 2 or door 3 is independent of whether the car is behind the initially chosen door or the remaining closed door.

But anyway, so what? Reliable sources give this solution, and some of those sources give a proper explanation too, some don't. Editors of wikipedia are not entitled to stick labels "correct" and "incorrect" on different solutions. We can only report what is out there in the literature and attempt to give a coherent overview of all the different ways people like to look at MHP.

If you want to promote a particular (alternative) kind of solution, then you can better devote your energies to finding a way to make it as simple and as accessible as possible to a really broad audience. I tried to find ways to express what is actually the conditional probability solution in a completely non-technical way, in just a couple of sentences which everyone can understand. Still Martin thinks my latest attempt was too difficult for ordinary people and no-one else seems interested to find something better. So the conditional solutions are doomed to remain an academic side-show. I find that very saddening. Richard Gill (talk) 18:51, 29 January 2013 (UTC)

Whom you're preaching to? I'm only suggesting to remove the misleading pictures. Nijdam (talk) 21:00, 29 January 2013 (UTC)

I am not preaching, I am pleading. If you want to "sell" conditional solutions you must find ways to make the conditional solutions attractive and intuitive. Being negative about the simple solutions only makes ordinary people even more convinced that they don't want to learn about the conditional solutions.

Personally I find the conditional solutions very attractive and important, and I'd like to have one of them in the introduction to the article; I'd like to have some "conditional solutions" included among the so-called simple solutions. Why can't a conditional solution be simple? I think the proper distinction is not between unconditional and conditional, but between short verbal arguments which everyone can (think they) understand, and mathematical arguments which only very few people are going to look at.

Another distinction is between arguments which are essentially correct and complete, and arguments which are clearly wrong ... with a grey area in between. Whether or not the picture in question is misleading seems to be a matter of taste. Wikipedia reports general knowledge as reflected by published sources, independent of whether that "general knowledge" is actually correct or not.

MHP is a popular brain-teaser. Most people who come to wikipedia wanting to find out about it, want a popular solution. Richard Gill (talk) 12:34, 31 January 2013 (UTC)

What gives you the idea want to sell conditional solutions? Just read what I write. Nijdam (talk) 09:28, 1 February 2013 (UTC)

Nijdam, I am still not completely sure exactly where your objection lies. Do you object to the combined probability of the two door remaining at 2/3 after the host has opened a door? Do you therefore object to the probability of the initially chosen door remaining at 1/3 after the host has opened a door? We do say in the caption (my emphasis), ' With the usual assumptions player's pick remains a 1/3 chance'. Is that not enough? Martin Hogbin (talk) 11:55, 31 January 2013 (UTC)

I agree Martin. "With the usual assumptions" the information *which* door was opened doesn't change the probability the car was behind the *originally* chosen door. Since its 1/3 remains unaltered, and the opened door gets probability 0, the other door must get the remaining 2/3. Completely OK.

The picture is not misleading. The argument (i.e. the main text) is arguably incomplete. Devlin even believed it was wrong, and withdrew it. Richard Gill (talk) 12:34, 31 January 2013 (UTC)

Explain to me what a reader has to understand from the pictures. Nijdam (talk) 09:28, 1 February 2013 (UTC)

Would you accept it if we improve the pictures in such a way that the first just shows | 1/3| 1/3 | 1/3 | and the second | 1/3 | 2/3 | 0 | ?Nijdam (talk) 09:32, 1 February 2013 (UTC)

Nijdam, it would really help if you would explain exactly what your objection is. Martin Hogbin (talk) 17:32, 1 February 2013 (UTC)

Martin, better is you follow my suggestion and explain how the two pictures should lead to understanding. Don't hesitate to let me join your thoughts. If you think the pictures are useful, you certainly can tell me why. Nijdam (talk) 14:31, 2 February 2013 (UTC)

Yes, we say 'With the usual assumptions player's pick remains a 1/3 chance'. It could be argued that we should explain why this is so but this is what most people naturally assume and it also turns out to be correct, so in a simple explanation I would say that this is all we need.

What really puzzles me is why people make such a fuss about the unimportant and extremely obvious fact that it makes no difference which of the two unchosen doors the host opens but they do not mention the key issue that player's pick does not remain a 1/3 chance if the host does not always reveal a goat. This, if anything, is the deficiency of the combining doors solution. Has anyone else mentioned this?

I have no objection to discussing both these points later on on the article. Bayes' rule (see my discussion above) can be used to intuitively show why it matters that the host always reveals a goat. Martin Hogbin (talk) 14:39, 31 January 2013 (UTC)

Suggestions for the article

Thanks to all here who have patiently helped me understand this problem, and to understand what is meant by probability in general. This now seems to me the essential thing that is lacking in the article. Tell me if I am wrong, but: There is really no such thing as probability; the car is definitely behind one of the doors and not behind any of the other doors. What we call probability is just a measure of our cleverness in extracting information from the host, who knows which is which. If we extract no information, our odds are 1/3. How am I going so far? Rumiton (talk) 12:25, 29 January 2013 (UTC)

I agree that the essential thing that is missing in the article is what we mean by probability. Unfortunately most of the sources are completely silent on this. On the other hand, there are different concepts of probability "out there" and (in my opinion) the different solutions which different sources seem to prefer, tells us a lot about what they mean by probability, even though they never say so explicitly.

For most laypersons in the context of problems like this, probability indeed represents our knowledge. There is no randomness involved in the problem. The car is already placed behind one of the doors; we chose door 1 because it's our favourite number; the host proceeded to reveal a goat behind door 3 but we have no idea how he chose between doors 2 and 3 if he had a choice. Because we have no information whatsoever about how the car was initially hidden, initially it is *for us* equally likely behind any of the three doors. Because we have no idea whatsoever about how the host opens a goat-door when he has a choice, either of his choices would be equally likely *for us* in the case that the car happened to be behind door 1.

Now we may as well imagine the host always chooses between door 2 and 3, but in the case that his choice would reveal a car, he opens the other door. Given we initially chose door 1, there are six equally likely possibilities corresponding to all combinations of the initial location of the car and the preference of the host (which he's not always able to exercise!). Door 3 actually gets opened in 3 of the six cases (in the other 3 cases, door 2 gets opened). So given the host opened door 3, there are just 3 equally likely possibilities, and in 2 of them the car is behind door 2, in 1 of them the car is behind door 1. So given door 3 was opened, the car is *for us* twice as likely behind door 2 as behind door 1. Richard Gill (talk) 18:34, 29 January 2013 (UTC)

I essentially agree with Richard that nobody really knows what probability is. There are different interpretations and I agree with Richard that the appropriate interpretation for the MHP is the Bayesian one, in which probability is a state of knowledge. A classic example of this is that I take a card from a pack. I tell one person that it is an ace, another that it is a spade, and another just that it is a card. I ask them all what the probability is that I have picked the ace of spades. From their perspectives they each have different answers (1/4,1/13,1/52). For my perspective the probability is 0 because I know I have picked the four of clubs.

Richard says, 'For most laypersons in the context of problems like this, probability indeed represents our knowledge'. I am am not sure why Richard refers to 'laypersons' (which includes me) as it seems to imply that experts have some better, or more appropriate, interpretation of probability (I do not know if he means to imply this). They simply do not, but various commentators here have, from time to time, confused the issue with other interpretations. Martin Hogbin (talk) 11:43, 31 January 2013 (UTC)

My remark was not meant imply criticism of anyone. It's a matter of fact that many (most?) people think about probability in the Bayesian sense, some think about it in the frequentist sense. Most people are not aware of the difference. Among people trained in probability and statistics the frequentist interpretation is more common than among laypersons. That does not mean it is better. In different contexts, different probability interpretations can be more or less appropriate. In some contexts it is very important to be aware of the difference.

Eskimos distinguish between many more kinds of snow and ice than most people. That does not imply that most people's understanding of snow and ice is inadequate for their purposes. Richard Gill (talk) 12:21, 31 January 2013 (UTC)

Richard, I was not suggesting that you were criticising anyone, just that when you say things like, 'For most laypersons...' and 'people's understanding ... is inadequate for their purposes', it can give the impression that there is some better way to approach the problem that is used by experts but it is too complicated to use here.

Right from the start, I and many others have said that the Bayesian approach to probability is the way to tackle the MHP and that discussions about probability interpretations do nothing but obfuscate the simple but unintuitive puzzle that the MHP was clearly intended to be. Martin Hogbin (talk) 14:28, 31 January 2013 (UTC)

Combining doors

Nijdam asked for an explanation of how the diagrams in the 'Adams and Devlin' section help our readers to understand that answer to the MHP.

The first picture shows three doors and has the caption, Player's pick has a 1/3 chance while the other two doors have 1/3 chance each, for a combined 2/3 chance. The diagram shows the first door having a probability of 1/3 (of hiding the car) and the two unchosen doors having a combined probability (the probability that the car is behind door 2 or the car is behind door 2) of 2/3.

The second picture shows the same probabilities after door 2 has been opened to reveal a goat, with the caption With the usual assumptions player's pick remains a 1/3 chance, while the other two doors have a combined 2/3 chance: 2/3 for the still unopened one and 0 for the one the host opened.

The 'usual assumptions' referred to are that the host always opens a door to reveal a goat and that when he has a choice of two doors he chooses uniformly between them (or that from a Bayesian perspective the player has no information requiring him not to take the probabilities to be uniform).

The caption then goes on to say that, with these assumptions, 'player's pick remains a 1/3 chance [after the player has revealed a goat behind door 3]'. This statement is intuitively obvious since, under the stated assumptions, no information about whether the car is behind door 1 or not is revealed when the host shows a goat behind door 3.

This intuitive observation turns out to be correct under the stated assumptions when analysed in more detail. A more detailed analysis shows that player's pick may not remain a 1/3 chance if the host has a known preference for one of the doors or does not always reveal a goat. Put more generally, the posterior probability that the car is behind door 1 is unchanged if the probability that the host will reveal a goat behind door 3 given the car is behind door 1 is the same as that the probability that the host will reveal a goat behind door 3 given the car is not behind door 1.

Th second picture also shows the probability that the car is behind door 2 or door 3 as 2/3 (the complement of its probability of being behind door 1) and also that the probability that the car is behind door 2 is 2/3 (as we now know it is not behind door 3). Martin Hogbin (talk) 16:16, 2 February 2013 (UTC)

What has combining to do with this explanation? Answer: nothing! Nijdam (talk) 22:58, 2 February 2013 (UTC)

Please say what you think is wrong with the pictures. Martin Hogbin (talk) 23:31, 2 February 2013 (UTC)

Not wrong, misleading. The combining of doors 2 and 3 bears no meaning in the explanation and suggest a wrong way of arguing. Nijdam (talk) 15:04, 3 February 2013 (UTC)

Why? If we know the probability that the car was originally behind door 2 or door 3, and now we know it is not behind door 3, we can calculate the probability that it is behind door 2. What is wrong with that? Martin Hogbin (talk) 16:33, 3 February 2013 (UTC)

Read!! Again: not wrong, but misleading. Why should the combined probability be mentioned. It serves no purpose. Nijdam (talk) 10:29, 4 February 2013 (UTC)

You seem to be the only person mislead by this approach. Please tell us in what way you are mislead.

You make a logical error: A mislead person does not know they're mislead!!

The statement leads most people to this way of thinking:

1) The car is behind door 2 or door 3 at the start of the game with probability 2/3.

2) The car was behind door 2 or door 3 with probability 2/3 at the start of the game given that the host has revealed a goat behind door 3. Most people intuitively assume this, and it turns out to be correct under the standard assumptions.

3) The car is behind door 2 with probability 2/3 after the goat has been revealed behind door 3. Martin Hogbin (talk) 10:50, 4 February 2013 (UTC)

Conclusion: you're also mislead. BTW: I do not understand point 2: at the start of the game no goat is shown. Also I do not see the relevance of point 1, unless, as most people do, you reason that doors 2 and 3 have 2/3 chance on the car at the start, and HENCE, as door 3 shows a goat door 2 must have this 2/3 chance. Nijdam (talk) 12:56, 4 February 2013 (UTC)

Nijdam, it would really help if you were not so cryptic in your responses. We might get to the point quicker if you explained what the problems are.

I do reason that doors 2 and 3 have a 2/3 chance of hiding the car at the start. Is this wrong in some way?

The next step is to reason that the probability that the car was behind door 2 or door 3 (if you like, the probability that the producer placed the car behind door 2 or door 3) is still 2/3 after the goat has been revealed. Is there a problem with this step? Martin Hogbin (talk) 14:26, 4 February 2013 (UTC)

No, your first statement is true; so is 1+1=2 or a die has 6 sides. Why is this of any help in understanding.

I didn't and don't understand your formulation The car was behind door 2 or door 3 with probability 2/3 at the start of the game given that the host has revealed a goat behind door 3. The words At the start refers to the unconditional probability; but given that the host has revealed a goat behind door 3 refers to a conditional probability. This should also puzzle you.

Did you notice the capitalised HENCE? And understood it? Nijdam (talk) 23:00, 4 February 2013 (UTC)

I see no problem in revising an earlier probability in the light of new information. I am talking about a single event, that a car and a goat are placed (in any order) behind doors 2 and 3 at the start of the game.

First I consider, at the start of the game, the probability that this event has occurred. We shall call this an unconditional probability and agree that it is 2/3.

Then I consider, after the host has revealed a goat behind door 3, the conditional probability that this same event occurred. In the standard case it turns out that this conditional probability is the same as the unconditional one. But, had the host tossed a coin and just happened to reveal a goat, the conditional probability, of the same event would have changed to 1/2.

After the host has revealed a goat we know that the conditional probability (given that a goat has been revealed behind door 3) that a car and a goat were originally placed behind doors 2 and 3 is the same as the unconditional probability, namely 2/3, 'HENCE' we can deduce that the probability that the car is behind door 2 is 2/3. Martin Hogbin (talk) 23:28, 4 February 2013 (UTC)

It is somehow tiresome: the point is not whether you can explain a solution, but how the PICTURES CONTRIBUTE to the explanation. Let me tell you: the first picture is correct, but misleadingly suggest that the combined probability will always be 2/3. What else could it possibly tell us? Nijdam (talk) 11:19, 5 February 2013 (UTC)

The second picture has an explanatory caption though. It starts, 'With the usual assumptions...' and, with the usual assumptions, the combined probability is always 2/3. In other cases it may be something different.

Surely the way to address your concern is to give a more detailed explanation later on in the article. There we can show how things would be different if the host chooses between doors 2 and 3 randomly or if the host has a known door preference. These are things that ought to be explained, I agree but, with the usual assumptions, the pictures are correct. Martin Hogbin (talk) 14:08, 5 February 2013 (UTC)

Misleading? "Always"? Just note: this is not about aberrant variants that never allow the paradox to arise: not about "host forgets" and not about "host is known to show a certain well-known bias when he got two goats to choose from". The combining doors approach is a correct argument for what it delivers, namely that switching gives the car 2/3 of the time. It helps to think in terms of "my door" and "Monty's doors." Somehow, the combination of grouping the doors and personalizing them in a possessive sense prompts many people to think carefully about the implications of selective evidence, and this often leads to correct and logically sound answers. Selective evidence can be misleading, and bringing it up makes people's eyes glaze over. For many newcomers to MHP, it's *the* eye-opener. It's used by many sources, both popular and academic. Some academic mathematical sources even use it as a step towards a conditional probability statement. Gerhardvalentin (talk) 13:20, 5 February 2013 (UTC)

The longest-running content dispute on Wikipedia

It has now been five months since I posted the following RfC: Talk:Monty Hall problem/Archive 33#Conditional or Simple solutions for the Monty Hall problem?. Since then I have been monitoring this talk page but (mostly) not posting. If you recall, at the start of the RfC there had been over 1,300,000 words written on this talk page (all seven Harry Potter books together have 1,084,170 words) plus multiple trips to various dispute resolution venues, all without any basic agreement about what should be on the page.

The result I was hoping for was, after a few months, for the discussion to die down and the page be stable. Of course there will always be new editors with new ideas which should be discussed, but those who were involved in the RfC should have arrived at a consensus by now.

I did a count of Dec 01 - Jan 31 edits for the last 3 years. The numbers are:
December2012 + January 2013: 466 talk page edits, 34 article edits
December2011 + January 2012: 35 talk page edits, 35 article edits
December2010 + January 2011: 7 talk page edits, 22 article edits

The above counts and the nature of the current threads tells me that we have no reached a consensus on what the content of this page should be, and that another million words written to this talk page are unlikely to change that.

Clearly, somebody or several somebodies needs to drop the stick and back slowly away from the horse carcass. I am going to ask all of you who have been at this for over a year to take a good look at the big picture and ask yourself "am I willing to walk away and let the other editors have their way with this article?" Another possibility is for two or more of you to negotiate an agreement to leave if the other fellow agrees to leave as well.

I am going to wait another month and then evaluate the situation. At that point I will have some other suggestions for resolving this, the longest-running content dispute on Wikipedia. And, I might add, the one with the smartest and friendliest participants. :) --Guy Macon (talk) 19:44, 5 February 2013 (UTC)