Talk:Monty Hall problem/Arguments/Archive 4

Winning by switching
May be this is a good moment to comment on the "event" 'winning by switching'. Not everyone who uses this term, uses it in the right way. The following reasoning, that some people use, is wrong: ''initially the player has probability 1/3 to get the car. Hence if she switches the probability increases to 2/3.'' In this way it is equivalent to the "simple solution". Swichting can only be judged after a door has been opened by the host, leading to the known calculation of the conditional probabilities. Nijdam (talk) 11:54, 28 May 2009 (UTC)


 * Please explain why your opinion is that this is wrong. Perhaps you could critique the 'Combining Doors' solution, which assigns the value of the selected door at the outset as 1/3, then shows the value is still 1/3 after a door is opened. It would be even more valuable if you could provide a reference in addition to your OR. Glkanter (talk) 12:15, 28 May 2009 (UTC)


 * I've explained this to you before, many, many times so I won't try again. Perhaps you'll believe Nijdam, who (like Boris Tsirelson and C S) claims to be a professor of mathematics (Nijdam: care to be more specific about your real life identity?  Glkanter at least claims to understand that math professors know more about math than he does).  BTW - Morgan et al. is the reference you're asking for.  As an unconditional solution, "combining doors" is in the general family of solutions they criticize.  It is equivalent to their F1: If, regardless of the host's action, the player's strategy is to never switch, she will obviously win the car 1/3 of the time.  Hence the probability that she wins if does does not switch is 2/3. -- Rick Block (talk) 13:44, 28 May 2009 (UTC)


 * Nijdam appears to have many skills. Ventriloquism among them. Glkanter (talk) 15:27, 28 May 2009 (UTC)
 * That's the least. Nijdam (talk) 18:50, 28 May 2009 (UTC)


 * Ah, yes, not to forget C S. Here's a complete paragraph he posted:


 * "Actually, I don't believe this material is over any reasonably intelligent adult's head. So let's give it a go. Let's address why people might think Marilyn's statement is conditional. By the way, I hope I haven't given the impression that I think the problem must be understood as conditional. As for Marilyn's exact statement, I think it's ambiguous enough on several fronts, and there's no reason to state with absolute confidence it should be conditional."


 * Of course, one is encouraged to read his entire response, and all his others, in archive 9 of the talk page. Glkanter (talk) 17:35, 28 May 2009 (UTC)
 * Here is another quote of C S. from your talk page: Since you like to misrepresent Boris' quotes, perhaps we could press him on whether the turbulence is caused by your refusal/inability to understand basic probability. --C S (talk) 05:12, 15 February 2009 (UTC)Nijdam (talk) 11:56, 29 May 2009 (UTC)


 * Yeah, that's been out there. Please provide where I actually mis-quoted Boris. Because it didn't happen. Maybe you could focus instead on the questions at hand. Glkanter (talk) 12:07, 29 May 2009 (UTC)

Morganian Fallacy
The argument for the continued emphasis on Morgan's paper (and those written by 2 or 3 others) has been primarily that they are published, reliable sources that are (allegedly) peer reviewed. Any logical counter arguments are therefore wasted effort.

This is a dangerous conceit, and should be rejected.

I could provide links to official US Government documents written by highly educated men and women that have been reviewed by the highest members of government. These documents would indicate that waterboarding suspected terrorists is not torture, and is therefore legal. There is great dispute about these papers and their conclusions. Fortunately, thinking, rational people are engaged in debating the merits of this proposition. Simply being written does not make it so.

I could provide a link to the US Constitution, in which slavery in certain of the 13 original states is allowed. Again, highly intelligent people wrote and reviewed this document. This same document denies women the right to vote.

Martin has pointed out an error in the Morgan paper. Once corrected, if I understand correctly, the formula determines that the likelihood of winning by switching is 2/3, rather than some other result. This is the same outcome as the unconditional solution, which is no coincidence.

I'd like to point out a logical fallacy in the Morgan paper. That is that the host may provide information as to the car's location based on which goat door he reveals. This contradicts the very notion of a game show. And in the US, is illegal. The producers of the show are obligated not to reveal any information. So, the probability of this happening is 0.

Morgan's paper was written by human beings, and as such, is subject to error. To blindly accept these errors as gospel only compounds these errors. Many papers have been written subsequent to Morgan's that solve the MHP using an unconditional solution. Devlin is listed in the article's references as one, there are countless more.

But, the Morganians discount these sources for various specious reasons. Essentially arguing of their sources, "Mine is bigger."

I'd still like to know a logical argument explaining how the original 1/3 likelihood of selecting a car is dependent on which goat door Monty opens. The strategy is to always switch. When you (the contestant) select a goat, this works 100% of the time. When you select the car, this works 0% of the time, regardless of which goat Monty reveals. As you will select a goat 2/3 of the time, the strategy works 2/3 of the time. Glkanter (talk) 08:11, 26 May 2009 (UTC)


 * As I have explained before, it is a conjuring trick. Just like the rabbit that the magician pulls out of a hat is a real rabbit, Morgan's solution is a real solution.  Just like it is no use checking out the rabbit to see how the magic trick is done, it is no use arguing about Morgan's solution to find out the problem.  Like many good magic tricks, it is accomplished before the main action starts.


 * The main bit of trickery is what I have been discussing above. In Whitaker's statement there is no information given about the way that the producer initially places the car, the way the player chooses a door, or the way that the host chooses a door.   To formulate the problem we need decide how we will deal with this lack of information.  What Morgan do, possibly unwittingly, is to take it that the car is placed randomly, the player chooses randomly, but the host chooses non-randomly.  There is no justification whatsoever in Whitaker's original question for this assumption although it is possible that others had previously made the first two assumptions and Morgan merely followed suit.  Once this decision has been made, all that is needed is a little sleight of hand and, hey presto, you have an 'elegant solution'.


 * I would be happy to explain why Morgan's solution is correct once you have set the trick up, as understanding this properly helps you see how limited their solution really is. Martin Hogbin (talk) 11:09, 26 May 2009 (UTC)


 * Oh, no, the canard occurs when Morgan says the unconditional solution is false. First, they offer a 'definitive' interpretation of what Whitaker is asking. As if somehow, they alone can divine this. Then they dismiss solutions like (but not exactly) the 'Combining Doors' solution by saying something like 'it answers the wrong question'. What hogwash! The MHP itself is a trick, as someone else posted a couple of months ago. By opening a door, the reader is lured into the 50/50 trap. Nothing has changed, (it is still 1/3 for the door the contestant selected, right?), but it 'appears' substantially different. That's why it's the world's most famous paradox.


 * That was the sleight of hand that I was referring to. Morgan do what I say above but also change the question asked in two stages to one that, given the non-random host door choice they have assumed, must be treated conditionally. Martin Hogbin (talk) 15:58, 26 May 2009 (UTC)


 * Good luck with your Morgan error effort. I'm doubtful that you'll get any 'acceptable' experts to help you out. My guess is that such experts studiously avoid the MHP (the Wikipedia Math Braintrust sure does!), and specifically Morgan. What is there to gain for someone with 'standing' to get involved in this quagmire? Heck, we all know we're wasting our time trying to fix this insignificant Wikipedia article. But, I've been at it for over 7 months now, and you, even longer. Of course, we're all essentially 'anonymous', and have no real 'standing' to lose. No, I can see no benefit to any Professor in either taking on Morgan, or proving his rantings invalid. To me, that would be like challenging some 'expert' who claims the sun rises in the west. Why even bother? Glkanter (talk) 13:34, 26 May 2009 (UTC)


 * Well at least no one seems to be arguing that Nijdam and I are wrong. Martin Hogbin (talk)

While we're out fishing, why don't we look at the MHP FAQs and the Bayes' Theorem article.
The FAQs are pretty much one person's personal POV.

The Bayes' Theorem page is interesting, in that it has a MHP example. http://en.wikipedia.org/wiki/Bayes%27_theorem#Example_3:_The_Monty_Hall_problem

I'm no expert, so could someone tell me what this means:


 * In the situation where the prize is behind the red door, the presenter is free to pick between the green or the blue door at random. Thus, P(B | Ar) = 1 / 2

Somehow, the doors are colored, rather than numbered. But what's that 1 / 2 stuff all about? I thought we've been told that it's not 1 / 2? And certainly not random!

Here's the conclusion:


 * Note how this depends on the value of P(B). Another way of looking at the apparent inconsistency is that, when you chose the first door, you had a 1/3 chance of being right. When the second door was removed from the list of possibilities, the conditional probability for the chosen door to be the winning door is also 1/3 and hence this left the last door with a 2/3 chance of being right.

And here's how it was before Nijdam's edit:


 * Note how this depends on the value of P(B). Another way of looking at the apparent inconsistency is that, when you chose the first door, you had a 1/3 chance of being right. When the second door was removed from the list of possibilities, this left the last door with a 2/3 chance of being right.

I think the word 'still' was intended rather than 'also' in Nijdam's edit. Is this correct?
 * [Nijdam]Definitely not. I mean 'also', because 'still' suggests it is the same probability, and it is not. Nijdam (talk) 18:16, 29 May 2009 (UTC)


 * At what point did it change from 1/3, and what did it change to? Then, when did it change back? Please provide a reference. Thank you. Glkanter (talk) 18:30, 29 May 2009 (UTC)


 * Apparently at a point far beyond your reach. I have explained this a dozen or more times, and won't repeat it any more. Nijdam (talk) 10:56, 31 May 2009 (UTC)


 * Still unwilling to provide a thoughtful response, Nijdam? Glkanter (talk) 09:31, 4 June 2009 (UTC)


 * Help a guy out. Can you provide a link to one of those diffs? Thanks. Glkanter (talk) 11:11, 31 May 2009 (UTC)

So, is the 'Combining Doors' (with a meaningless, superfluous 'conditional probability' thrown in there) solution good enough for the Bayes' Theorem article, but not the MHP article? Glkanter (talk) 12:35, 29 May 2009 (UTC)


 * The MHP article is a featured article, the Bayes' Theorem article is not. The line you quote


 * In the situation where the prize is behind the red door, the presenter is free to pick between the green or the blue door at random. Thus, P(B | Ar) = 1 / 2


 * is an assumption not present in this version of the problem statement (or the one presented in Parade) that many people think is reasonable but if it's not given in the problem statement it shouldn't be used in the computation of the conditional probability. Assuming instead that this is an unknown probability q leads to the 1/(1+q) solution.  -- Rick Block (talk) 14:06, 29 May 2009 (UTC)


 * You mean just like the assumption that the car was initially randomly placed or that the player picks randomly? Above all we should be consistent in our assumptions. Martin Hogbin (talk) 17:23, 29 May 2009 (UTC)


 * How about consistency between Wikipedia articles? It seems downright unfair that the MHP gets all the Morganian's attention, but the Bayes' Theorem readers remain unenlightened. But then, the Bayes' Theorem article wasn't personally 'shepherded' through the FAR process.
 * So, I guess it's valid to throw a 'condition probability' phrase in there, but not bother addressing any of the other obvious issues a Morganian would have with it. Glkanter (talk) 18:07, 29 May 2009 (UTC)

I can't exactly put my finger on it, but the Morganian's resistance to answering simple and relevant questions brings to mind this link, plus the 4 days that follow it. http://www.doonesbury.com/strip/dailydose/index.html?uc_full_date=20090427 Glkanter (talk) 23:58, 29 May 2009 (UTC)

What exactly is the real Monty Hall problem?
Before we start arguing about who is right, which I am sure that we will, can we just make sure what people actually think is the real MHP. Let me give some options. Please agree or disagree or add your own options to clarify your opinion. Normal game rules assumed.

1 The unambiguously worded problem treated unconditionally. In other words, what is the best player strategy, decided in advance. Although this should be a very simple probability problem, most people still get it wrong.

2 The unambiguously worded problem where the host chooses randomly, thus the issue of conditional probability is ignored or considered irrelevant. The player decides after a door has been opened but it is known that the door chosen by the host makes no difference to the result. Although this should be a very simple probability problem, most people still get it wrong.

2a The problem where the host door policy is not stated but the issue of conditional probability is ignored. The player decides after a door has been opened but it is assumed that the door chosen by the host makes no difference to the result. Although this should be a very simple probability problem, most people still get it wrong.

3 The problem described by Morgan et al, where the host is assumed to act non-randomly, thus the door opened by the host could affect the probability of winning by switching. The confusion arises because people do not realize that the host door opening policy is important. Were it not for this fact most people would get the answer correct. Martin Hogbin (talk) 19:27, 29 May 2009 (UTC)


 * 4 Same problem as your #2, but entirely different "thus": The unambiguously worded problem where the host chooses randomly (if the player initially picks the car), thus the conditional solution is forced to be the same as the unconditional solution.  The unconditional solution produces the correct answer, but it needs an argument for why it is correct involving the constraint that the host pick randomly.  Any solution that does not mention this constraint is incomplete, since the solution then applies to versions of the problem without this constraint (and is then incorrect).  Although this is a very simple probability problem, nearly all people (including most of those who think they understand it because they know one of the unconditional solutions) get it wrong. -- Rick Block (talk) 00:33, 30 May 2009 (UTC)


 * Thanks, I understand your position now. Martin Hogbin (talk) 10:26, 30 May 2009 (UTC)


 * Due to the popularity of the Parade version, there's a reasonable argument in favor of #3 as well. Most people get this wrong, and most explanations are incomplete (per above). -- Rick Block (talk) 00:38, 30 May 2009 (UTC)


 * As you know, I see no justification for the inconsistent application of the principle of indifference to the Parade problem. Martin Hogbin (talk) 10:26, 30 May 2009 (UTC)


 * Martin, regarding #3, how do you square it with the fact that Monty is prohibited from 'telegraphing' any bias to the contestant? Remember: "Suppose you're on a game show". #3 can't exist. Nor can any so-called 'variants', for the same reason. Glkanter (talk) 00:53, 30 May 2009 (UTC)
 * As I said above, I am trying initially just out find out what people think rather than to criticize or argue. I will answer your question in another section below. Martin Hogbin (talk) 10:26, 30 May 2009 (UTC)


 * I would go with #1 or either of the #2s. As there can be no new knowledge gained when Monty opens a door, there is no statistical (logical?) difference among them. You didn't list it, but I reject that Whitaker (or anyone who has ever been associated with the MHP) was interested in doors #1 and #3 to the exclusion of any other combination. This may be a nitpick, but I don't agree/understand the "most people" portion of choice #3. "Most people", including 1,000 PHDs got it wrong, hung up on the 50/50 trap. "Most people", to paraphrase Freud, think a goat is just a goat. They have never heard of Morgan, or the "equal goat door constraint", and there lives are no less rich for this gift. Glkanter (talk) 14:55, 30 May 2009 (UTC)

Some philosophical issues
Glkanter, this is to show how answer 3 in the section above can be right.

There are several definitions of probability. Below are some edited extracts from relevant WP articles to show the principles of each. (Note for Rick: yes, this is the same old host chooses randomly argument).

Classical definition: Initially the probability of an event to occur was defined as number of cases favorable for the event, over the number of total outcomes possible in an equiprobable sample space.


 * 1. Frequentists talk about probabilities only when dealing with experiments that are random and well-defined. The probability of a random event denotes the relative frequency of occurrence of an experiment's outcome, when repeating the experiment. Frequentists consider probability to be the relative frequency "in the long run" of outcomes.


 * 2. Probability, for a Bayesian, is a way to represent an individual's degree of belief in a statement, given the evidence.

Modern definition: The modern definition starts with a set called the sample space, which relates to the set of all possible outcomes in classical sense... It is then assumed that for each element [in the sample space] an intrinsic "probability" value, is attached...

I do not believe that any of the statements above are contentions. It is certainly not my intention that that should be the case.

In all cases there are two philosophical issues that must be addressed before we can come to any conclusion. The first is how to formulate the problem and the second is how to interpret the answer once we have done our calculations. The different definitions can treat these issues differently.

Consider this (badly posed) example:

A bag contains 20 red balls and 80 white ones. You, who know only that the bag contains a mixture of red and white balls, pick a ball at random from the bag, what is the probability that the ball picked is red?

a) Using the modern definition we define our sample space, based on the information given in the question, and arrive at the answer 1/5 as the ball is defined to be taken at random from this sample.

b) On the other hand, you do not know the ratio of red to white balls in the bag. From your state of knowledge (Bayesian perspective) the probability of picking a red ball is 1/2 as you have no more reason to believe that you will pick a red ball than a white ball. (Of course, from our perspective, the answer is the same as above)

So we have two answers to the same question. This paradox is resolved when we consider what the two answers mean. The first answer is to the question, 'given 20 red and 80 white balls ... what is the probability...'. The second answer is to the question, 'from your perspective, as you have no idea how many of each colour there is... what is the probability...'. Not that hard to understand, so what is the point of all this? The point is that it is no use thumping the table and saying answer a) or b) must be right; there is no right answer to the question until you have decided what it means. Martin Hogbin (talk) 16:49, 30 May 2009 (UTC)


 * Martin, I appreciate your efforts to educate me. As you may have already inferred, I am a little slow on the absorption of this new information. Help me narrow down your alternatives as it applies to the MvS recitation of the MHP, specifically, "Suppose you're on a game show". So, this excludes the knowledge of the producer, at one extreme, but does recognize that the contestant knows the 'history' of the two remaining doors.


 * Is this a narrow view point? Absolutely. I understand that the unconditional solution does not generalize to other permutations of the puzzle. Why would I expect it to? For purposes of improving the article, I'm only interested in the most efficient solution to the world's most famous paradox, not a comprehensive understanding of probability theory. Which I would be happy to engage in at the appropriate place and time. Glkanter (talk) 17:21, 30 May 2009 (UTC)


 * Don't forget, I agree with most of what you say. "Suppose you're on a game show" is an important part of the MH question and it does indeed imply that the question should be answered from the point of view (state of knowledge) of the contestant.  But I am trying to show that there is another way of looking at it.  What would be your answer to my question above?  Martin Hogbin (talk) 17:45, 30 May 2009 (UTC)


 * As far as the balls go, I guess I would have to say 'Hell, I don't know'. Or, 'something between 0% and 100% inclusive.' But the guy who placed the balls in the bag would know the precise answer. But the MHP starts: "Suppose you are on a game show'. And then I'm (the contestant) provided the 5 premises that you, me, Rick and others have previously agreed upon.


 * But, there is an assumption that you're making. And that is that those are the only techniques available to derive an answer. I'll be consistent and refer to the 'Combining Doors' solution. I think it's more "Logic' than 'Probability'. (Otherwise, please help me understand which definition the 'Combining Doors' falls under.) Equally valid approach. And it renders moot the entire discussion of host behaviour, etc. Maybe the first sentence of the article: "The Monty Hall problem is a probability puzzle..." itself points us in the wrong direction.


 * I don't agree with your comment that while this is a game show problem, there is another way of looking at it. For purposes of this puzzle, I don't think these ambiguities exist. And I'm still waiting for someone, anyone, to demonstrate any failings in the 'Combining Doors' solution, other than 'Morgan calls it false'. Glkanter (talk) 18:42, 30 May 2009 (UTC)


 * The essential point is whether the probability is the average probability across all initial player picks and doors the host happens to open (this is the "unconditional" probability) or the probability that applies when we know which door the player has picked and which door the host has opened (the "conditional" probability). From the Parade wording of the problem statement, the latter is what we should be talking about.  The use of door 1 and door 3 does NOT mean we're only interested in the case where the player picked door 1 and the host opened door 3, but that we're interested in a SPECIFIC case where we know what door the player has picked and what door the host has opened (focusing on door 1 and door 3 should end up with the same answer as any other door the player picked and door the host opened).


 * From Martin's description of a frequentist, the difference is whether we watch the show (of course, there never was a show quite like this) and count only 4 things (player switches and wins, player switches and loses, player stays and wins, player stays and loses) or we watch the show and keep track of these four things for each initial player pick and door the host opens (i.e. we keep track of 24 things). If we keep track of all 24 things, then if asked by  a player who has picked door 1 (or any other door) and has seen the host open door 3 (or any other door) what are her chances of winning by switching, we can look at our data samples and produce an answer specific to this player's condition (the "conditional" probability).


 * An unconditional solution (e.g. combining doors) answers the first question (average across all players and any door the host opens), not the question specific to any individual player's case. If the car is initially randomly located, and if the host chooses randomly when the player initially selects the car, the answer must be the same - but exactly why these conditions force this to be true is not exactly obvious.


 * Since the question seems to be asking about a specific case (door 1 and door 3), responding with the overall average doesn't really answer the question. It's as if someone asks what a baseball player's batting average is when playing against the Yankees and the answer provided is the player's overall batting average.  These might be the same, but they might not.


 * Back to MHP, what you're saying is because it's a game show the conditional probability must (well, at least should) be the same as the unconditional. I'm not going to disagree with this, but if you only solve the problem unconditionally you really can't tell whether the show is "honest" or not.  Placing the car randomly at the beginning (and having the host know where the car is, open a door showing a goat, and always make the offer to switch) is not sufficient to guarantee every player has the same odds.  You also have to make sure the host picks randomly if the player initially selects the car.  The unconditional solution doesn't make this clear, because it averages out the host's preference across all initial player picks and door the host opens.  If you worked for the FCC and wanted to make sure the show was honest, you'd have to record the results for each pair of player pick and door the host opens.  What it means for the show to be honest is that each of these will be the same (and they'll each be 2/3).


 * What Morgan et al. show is that each of the conditional probabilities (assuming the car is randomly placed) is of the form 1/(1+q) where q is the host's preference when the player initially selects the car. If you don't solve the problem as a conditional probability problem, you don't discover this.  If you're working for the FCC and you want to make sure the show is honest, you need to check 6 things: is the car randomly placed, is the initial player pick random (if the car is randomly placed, this doesn't actually matter), does the host always open a door showing a goat (and, thus, knows where the car is), does the host always make the offer to switch, does the player get to decide to switch after seeing which door the host opens, AND does the host open a random door when the player initially picks the car.  If this last point is not satisfied, different players can have different odds even though the overall odds will still be 2/3.  -- Rick Block (talk) 20:12, 30 May 2009 (UTC)


 * You just wrote that if the host doesn't pick randomly when the contestant picks the car, he would be in violation of the FCC. Hence, it need not be stated outright as a premise, but it is, indeed a premise of the MHP. Glkanter (talk) 21:15, 30 May 2009 (UTC)

You both seem to be determined to stick to your standard lines on the MHP, I am trying to establish some philosophical points that might help to reach some sort of agreement.

Glkanter, you keep saying that the player does not know about the host's policy. I know that. In my question, you say that there is no answer, but the question asks for 'the probability that the ball picked is red' and we know that 1/5 of the balls are red. Surely you must see the argument for saying that the answer to the question is 1/5.

Suppose the question told us that the balls are all white, could there be any answer other than 0?

Rick, can we please forget conditional/unconditional for the moment, I will come to that later. Do you agree that, from Bayesian perspective of someone who does not know the proportions of the balls, the probability is 1/2?

Martin Hogbin (talk) 21:46, 30 May 2009 (UTC)


 * Martin - In a strict sense I agree, but I don't see what your question has to do with the problem at hand. We're NOT asked what does the player think her probability is, but what IS her probability given the problem statement (in the context of your problem the answer would be 1/5 if we're given the distribution of balls).  If in your problem we're told there are red and white balls in the bag but not how many of each, 1/2 is only a weakly correct answer.  A better and much more useful answer is R/(R+W).  -- Rick Block (talk) 22:07, 30 May 2009 (UTC)


 * That R(R+W), where R and W are both completely unknown, is a better answer than 1/2 is debatable. In fact it can be argued that it is not an answer at all, having no defined numerical value. It is no better than saying 'I have no idea'.  Supposing that you wer pressed to give your best estimate numerical value, it could only be 1/2.


 * If you wanted to do things the hard way, you could take your formula and using a noninformative Bayesian prior calculate the probability to be 1/2. Martin Hogbin (talk) 08:53, 31 May 2009 (UTC)


 * Can we leave the MHP for the moment, it is all a matter of how you interpret the question. which I will move on to later.


 * Martin, I answered 'I don't know' on the understanding that I hadn't been told the distribution of the red and black ball in the bag. (Re-read your paragraph, it's tricky.) I'll stand by that. I'm a 'premises' guy. If you told me that it was 20/80, then I could confidently answer 1/5. So I focus on the premises of the MHP contestant only. I, for one, think the ambiguity of the 'random goat door constraint' has been eliminated. That the host picks randomly is a premise. But I'm more interested in your response to the questions I posed above. Where does the 'Combining Doors' solution fit in your narrative? Thanks. Glkanter (talk) 23:39, 30 May 2009 (UTC)


 * In my (deliberately ambiguous) question you are clearly told that there are 20 red and 80 white balls. Why is the actual probability of picking a red (rather then your estimate from a position of ignorance) not 1/5?   Martin Hogbin (talk) 08:53, 31 May 2009 (UTC)
 * Glkanter, do you see my point, the answer depends on who is being asked the question. Martin Hogbin (talk) 13:47, 1 June 2009 (UTC)


 * Yes, I do. I got confused by this portion of your question:


 * "You, who know only that the bag contains a mixture of red and white balls, pick a ball at random from the bag..."


 * So I answered, "I don't know". But my point is, in the MHP, it's the contestant being asked the question. Of this, there can be no doubt. As I said, ask the producer, and you'll know with 100% certainty where the car is.


 * By the way, I don't think I would ever answer 50%. This is just my opinion, but I can't see 'indifference' making that assumption for a bag with an unknown quantity of balls. (Maybe, 50% of the time I would guess each color, but that's not the question you asked). For contrast, with 100 doors and one car, I can see 1% from 'indifference'.


 * Please leave the MHP out of it for the moment. Both you and Rick seem to keep going back to it.  At the moment I am trying to make some general points and will then go back to the MHP when we have agreed them. Let me rephrase my question.  A person picks a ball from a bag containing 20 red and 80 white balls.  What is the probability that they will pick a red ball? Martin Hogbin (talk) 18:04, 1 June 2009 (UTC)


 * As I have been informed of the distribution of the balls in the bag, I am comfortable saying there is a 1 in 5 chance that the first ball chosen randomly from that bag will be red. Glkanter (talk) 18:30, 1 June 2009 (UTC)


 * What if the person picking the ball does not know the distribution of balls in the bag? Martin Hogbin (talk) 19:49, 1 June 2009 (UTC)


 * As I don't believe the picking person's knowledge state affects his random selection of balls from a bag, I'll stick with 1/5. But you're scaring me Martin. I can't tell yet where this is all going. Glkanter (talk) 20:31, 1 June 2009 (UTC)


 * Will you answer my question? Where does the 'Combining Doors' solution fit in your Probability narrative? Glkanter (talk) 14:28, 1 June 2009 (UTC)


 * I will answer that in a new section below. Martin Hogbin (talk) 18:04, 1 June 2009 (UTC)


 * About Martin's problem. If I'm just shown a bag with 100 balls, red and white, without any further info, and I'm asked to tell the probability a random drawn ball is red, I really wouldn't know. Also a Bayesian will not say 1/2. He would model the situation with a parameter R, the number of red balls. However instead of merely saying: "I don't know the value of R", he would assume a prior distribution of R, to be used to improve his knowledge of R by calculating the posterior distribution after one or more drawings have been done. Of course he may compute the prior expected value of R if you want to know. That's just calculus. Nijdam (talk) 18:32, 31 May 2009 (UTC)


 * Nijdam, are you saying that you would never estimate the probability as 1/2 in such circumstances? Martin Hogbin (talk) 13:47, 1 June 2009 (UTC)
 * Well, I wouldn't speak of estimating. No info from a sample is available. You are merely asking about a good guess. I put some euro coins in a bag. How many did I put in? Nijdam (talk) 17:42, 1 June 2009 (UTC)


 * As I am sure you must know that is not the same type of question as there is no upper limit on the number. A bag contains only red and white balls, you pick one, what is the probability that it will be red? Do you assert that the only answer that can be given to this question is 'unknown'  (or maybe a formula such as R/(R+W) where R and W are both unknown)? Martin Hogbin (talk) 18:04, 1 June 2009 (UTC)


 * You have gone quiet, Nijdam, perhaps you can see where I am going. Martin Hogbin (talk) 08:20, 2 June 2009 (UTC)


 * For the relevant issue it is the same type. The answer 'unknown' or R/100 or R/(R+W), would be in my question: 'unknown' or N (=the number I put in). In the end all answers mean: 'unknown'. Nijdam (talk) 08:52, 2 June 2009 (UTC)

[Outdent] How about this question then. A car is placed behind one of three doors numbered 1 to 3. What is the probability that it is behind door 1? Martin Hogbin (talk) 11:37, 2 June 2009 (UTC)


 * Strictly speaking, and in line with the above: I don't know the probability. But similar as when you say: I throw a die, instead of: I throw a fair die, often randomness is implicitly assumed. Nijdam (talk) 17:15, 2 June 2009 (UTC)


 * Nijdam, is randomness of the host's behaviour implicitly assumed when you are talking about a game show problem, from the contestant's point of view, in which it would be illegal and illogical for the host to provide any information to the contestant as to the location of a car when faced with a choice of 2 doors? If no, why not? Glkanter (talk) 23:52, 2 June 2009 (UTC)


 * It was hard to make you assume randomness when I was talking about the balls. The one thing that is needed is consistency of approach. Martin Hogbin (talk) 22:22, 2 June 2009 (UTC)


 * There is quite a difference between the two situations. To compare them: with the car the experiment may be considered as the choice of 1 out of 3 doors. With the balls, the experiment is the choice of 1 out of 100 balls, but with unknown distribution of the colors. Even if you wouldn't explicitly say "draw random", mostly this is assumed, but the big difference is the unknown number of red balls. Nijdam (talk) 13:38, 4 June 2009 (UTC)

Nijdam, do you see my point now? There are three door choices made in the MHP, the producer's choice of door to place the car behind, the player's choice of door, and the host's choice of door. None of these is defined in the question to be random. There are only two consistent ways to treat the problem, treat them all as random or treat them all as non-random. What is your justification for treating the host choice differently from the others? Martin Hogbin (talk) 09:48, 3 June 2009 (UTC)


 * I wouldn't speak of "consistent way" to treat the problem. The placement of the car to be random seems quite natural. The distribution of the initial choice of the player is unimportant for the problem and is not specifically considered to be random. The strategy of the host is considered to be random in the MHP and doesn't take away the necessity of the conditional probability to be considered. Only for the sake of showing the dependency of the host's strategy it is varied, as any mathematician would have done. So, I'm sorry, I do not see your point. Nijdam (talk) 13:38, 4 June 2009 (UTC)


 * Saying, 'The placement of the car seems natural' is no way to deal with a probability problem. We must answer the question based on the information given in the question. What may seem natural to you may not seem natural to anyone else.  If we are to apply the principle of indifference to the producer's door choice we must apply it to the host's door choice. I agree that, as it turns out, the player's choice is unimportant if the car is placed randomly (but not otherwise).  Why would a mathematician not show how the producer's choice is important also?  Only because this is obvious and does not lead to an 'elegant solution'.


 * I really do not understand you Nijdam, you seem like someone who likes to do things properly yet you are content to approach this problem in a thoroughly sloppy manner. This is a highly contentions probability problem and we therfore must approach its formulation in a consistent manner.  Saying 'this seems natural' but 'that does not' is simply not good enough, especially as it has no basis in the real world. Martin Hogbin (talk) 19:35, 5 June 2009 (UTC)

Why the combining doors solution fails if the host chooses non-randomly
This section is to answer a question put by Glkanter in the section above.

Note that, in my opinion at least, the host must choose non-randomly for this argument to be valid. In other words when the host has a choice of which door to open, for example when the player chooses door 1 and the car is behind door 1, the host has a preference for opening one particular unchosen door. An extreme case would be that the host always choses door 2 when possible.

The questions of whether in the real MHP the host chooses randomly or not, and whether we are assumed to know that, and what difference that might make, are being addressed in the section above. This section is strictly concerned with the case where the host does in fact have a preference for one door.

Before I go any further, Glkanter, do you accept what I have said so far? Martin Hogbin (talk) 19:46, 1 June 2009 (UTC)


 * Thanks for asking, as I do not agree with the implied conclusion, to wit: By showing other puzzles have different results (actually, I'll even add in 'the same results'), the 'Combining Doors' solution is disproved. I believe the only appropriate method to disprove the 'Combining Doors' solution would be to show that:


 * with the 5 previously agreed premises
 * + what I have demonstrated is an unstated, but existing premise: any host behaviour cannot impart information to the contestant as to the location of the car
 * + the Combining Doors solution


 * there can be results other than: always switching results in winning 2/3 of the time. Glkanter (talk) 20:25, 1 June 2009 (UTC)


 * As discussed in the section above, depending on how you read the question, it may not matter what the contestant knows. Martin Hogbin (talk) 22:30, 1 June 2009 (UTC)


 * By the way, the question was 'Where does the 'Combining Door' solution fit into the Probability narrative?' It's multiple choice! Glkanter (talk) 20:35, 1 June 2009 (UTC)


 * That is what I am trying to answer. But as I have said, the combining doors solution only fails if you accept that the host acts non-randomly.  Whether that is a reasonable assumption is another matter. Martin Hogbin (talk) 22:30, 1 June 2009 (UTC)


 * If you're asked what a player's batting average is against the Yankees and you say "divide the player's total hits by total at bats" is this a legitimate solution? Or does it answer a different question?  Does it answer the question if you then say "well, only in cases where it's right"?  -- Rick Block (talk) 23:56, 1 June 2009 (UTC)
 * Rick, I understand what you are saying but I am trying to address one issue at a time. For the moment let me retract the word 'only' from my statement above. In this section I am now making only this assertion:


 * If the host chooses non-randomly, the combining doors solution fails.


 * I am not, in this section, making either of the following assertions:


 * That the host does choose non-randomly.
 * The combining doors solution does not fail if the host chooses randomly.


 * Of course, both of the above statements do need to be addressed, but can we agree one thing at a time. Does everyone agree with my first statement? Martin Hogbin (talk) 08:18, 2 June 2009 (UTC)

Martin Hogbin (talk) 08:18, 2 June 2009 (UTC)


 * The 'combining doors idea' always fails (for the given problem). When the host acts 'randomly' it produces the number 2/3, but so does the division 4/6 or the addition 1/3+1/3. The flaw, as Rick and I have stipulated over and over, lies in the arguments. Nijdam (talk) 08:57, 2 June 2009 (UTC)


 * With the host acting 'randomly', will it ever produce a result other than 2/3? Glkanter (talk) 09:08, 2 June 2009 (UTC)
 * No. Martin Hogbin (talk) 17:45, 2 June 2009 (UTC)
 * Glkanter, as you are not prepared to even consider the possibility that the question might be interpreted in a way that the host acts non-randomly, I will not continue with this argument. Martin Hogbin (talk) 18:29, 2 June 2009 (UTC)


 * G - the problem is "combining doors" says 2/3 whether the host acts "randomly" or not. You're asking if something that always says 2/3 is a correct solution if the problem is constrained to have the answer 2/3.  It is producing the correct number, but so does anything else that results in the number 2/3.  Consider "Rick's stupid approach" - with your first pick your probability is 1/3 and if you switch it's your second chance so the numerator changes from 1 to 2 (i.e. your chance of winning by switching is your guess number divided by 3).  This produces the correct numeric answer, so it must be correct reasoning, right?  Does it ever produce a result other than 2/3?  Is this not the correct numeric answer?  What's confusing you about "combining doors" is that there is a question that it correctly answers and the answer to this question is 2/3 - but it's not the question that we're asking.  The way to see it is not the question we're asking is to change the assumptions slightly so the answer to the question we're asking is different from the answer to the question "combining doors" solves.  This is exactly like using an overall batting average when asked about a team specific one.  If the player's overall average happens to be the same as his average against a specific team the error in how we're computing the average is not readily apparent.  We can see the error by looking at players for whom these two averages are different. -- Rick Block (talk) 18:43, 2 June 2009 (UTC)

Martin, I think that's the best course. I thought I read a few days ago where a Morganian agreed that the FCC would require Monty to act randomly. Because it's in the definition of a game show. And it's US law. How is it still a debatable point?

Rick, how come when I disagree with a professor, I'm described as ignorant and arrogant, but when you disagree with Devlin and dozens of other published authors, you claim to have a valid point? As always, you did not address a single facet of the 'Combining Doors' solution in order to refute it. You just go off on that gibberish, whether it's changing the premises, or cards, or now, batting averages.

Martin, could you indulge me? Into which category does the 'Combining Doors' solution belong in your Probability narrative? Glkanter (talk) 20:04, 2 June 2009 (UTC)


 * I do not understand what you mean. It is an unconditional solution, thus it has all the problems of unconditional solutions, if you believe that there are any. Martin Hogbin (talk) 22:16, 2 June 2009 (UTC)


 * Martin, which category does the 'Combining Doors' solution belong to:


 * Classical definition - Frequentists
 * Classical definition - Bayesian
 * Modern definition
 * Other - 'Logical Notation'
 * Glkanter (talk) 00:33, 3 June 2009 (UTC)


 * I'd say
 * Other - Informal explanations meant to provide an intuitive understanding for why the solution is not 1/2.
 * Although I sincerely believe such explanations are offered with the best of intentions, I think they have interfered with the general public developing an actual understanding of this problem. Approaching it as a conditional probability problem works if the host is constrained (or assumed) to pick randomly if the player initially picks the car, or if the host picks non-randomly in this case, or if the host forgets where the car is and happens to open a door showing a goat, or many other versions.  Teaching people an unconditional approach gives them a "solution" that works in only one of these cases. -- Rick Block (talk) 01:29, 3 June 2009 (UTC)


 * Which case does the 'Combining Doors' solution work for? Glkanter (talk) 02:04, 3 June 2009 (UTC)


 * I assume this is a rhetorical question, but I'll answer it anyway. It "works" in any case where the answer to a) "what is the chance of winning by switching for a player deciding to switch when standing in front of two closed doors and one open door showing a goat" (given the other assumptions we all agree on) is the same as the answer to b) "what is the chance of winning by switching for a player who must decide whether to switch before the host opens a door".  In my opinion, this means it doesn't ever actually "work" if you want the answer to a) because it's always telling you the answer to b).  If you constrain the problem so a) and b) have the same answer it seems to work.  But this is like saying your stopped clock that says "2:15" works as long you only ask what time it is at 2:15. -- Rick Block (talk) 02:50, 3 June 2009 (UTC)

Not rhetorical at all. You wrote:


 * "Approaching it as a conditional probability problem works if the host is constrained (or assumed) to pick randomly if the player initially picks the car, or if the host picks non-randomly in this case, or if the host forgets where the car is and happens to open a door showing a goat, or many other versions. Teaching people an unconditional approach gives them a "solution" that works in only one of these cases."

So, which of the above cases does the 'Combining Doors' solution work for? It's the first one, isn't it? Glkanter (talk) 03:00, 3 June 2009 (UTC)


 * Note the quotes around "solution". What about my explanation above don't you understand?  It never answers a) since it actually answers b).  It turns out they have the same numeric answer if the host picks randomly in the case the player initially selects the car, but unless you can give a reason for this you're not answering the question that's asked - which is a) not b).  -- Rick Block (talk) 03:28, 3 June 2009 (UTC)


 * You're not arguing with me. You're arguing that Devlin and many other published sources are not as reliable as Morgan. How can you be so certain that they're all wrong? Glkanter (talk) 03:43, 3 June 2009 (UTC)


 * You also wrote this:


 * "Although I sincerely believe such explanations are offered with the best of intentions, I think they have interfered with the general public developing an actual understanding of this problem."


 * The way I read that is you feel that while they may provide a valid solution, your personal opinion is that they're not properly addressing general probability theory. This strikes me as a POV contrary to those published sources. I don't believe that published sources are to be discounted based on an editor's personal POV. Glkanter (talk) 04:01, 3 June 2009 (UTC)


 * I've emailed Devlin and asked him to comment. Maybe he will, maybe he won't.  I've read and understand the Morgan et al. paper (and Gillman, and Grinstead and Snell, and Falk).  What they all say is that the MHP is a conditional probability problem.  The Morgan et al. paper is specifically about this exact issue - and it's in a peer reviewed math journal which means it is more reliable than Devlin's column (and BTW, I was the one who found Devlin's column and added it as a reference for the "combining doors" section of the article).  Searching for references for the article I've read probably 50 papers on this problem.  I really do understand it.  I understand POV as well and have never argued that reliable published sources should be stricken from the article.  -- Rick Block (talk) 04:40, 3 June 2009 (UTC)

I believe your personal POV as described above, plus your strong affinity for the Morgan paper, have been the primary barrier to the unqualified recognition of the validity of the various published unconditional solutions. Such unqualified recognition would, of course, diminish Morgan's claim as the only 'true' solution. Noteworthy, perhaps, but not due the emphasis it currently enjoys. There are numerous citations in the article of the unconditional solutions being false, etc., and these would be removed. Extensive discussion of 'variants' would also be less important, as 'host behaviour' would no longer be a necessary component in understanding the solution. I believe these changes would result in a more concise article, leading the reader to a better comprehension of the Monty Hall problem.Glkanter (talk) 05:57, 3 June 2009 (UTC)

Mutually Exclusive Published Reliable Sources

 * Are you trying to understand the problem, or suggesting a change to the article, or what? Suggestions for changes should be at talk:Monty Hall problem.  If you're seeking to understand, this is the appropriate place.  -- Rick Block (talk) 12:23, 3 June 2009 (UTC)

I created a new section with the same name on the regular talk page. Glkanter (talk) 12:35, 3 June 2009 (UTC)

Analysis
I model the problem as follows:

C1 is the event that the car is behind door 1. Similar C2 and C3. We have:


 * $$P(C_1)=P(C_2)=P(C_3)=\tfrac 13.$$

K1 is the event that the player chooses door 1. Similar K2 and K3.

The choice of the player is independent of the position of the car, hence:


 * $$P(C_i\ and\ K_j)=P(C_i)P(K_j)=\tfrac 13P(K_j).$$

M1 is the event that the host opens door 1. Similar M2 and M3.

According to the rules:


 * $$P(M_i|C_i\ and\ K_j)=0.$$ (The door with the car is never opened)
 * $$P(M_j|C_i\ and\ K_j)=0.$$ (The chosen door is never opened)

Thus:
 * $$P(M_3|C_2\ and\ K_1)=P(M_2|C_3\ and\ K_1)=1.$$


 * $$P(M_3|C_1\ and\ K_2)=P(M_1|C_3\ and\ K_2)=1.$$


 * $$P(M_2|C_1\ and\ K_3)=P(M_1|C_2\ and\ K_3)=1.$$

According to the rules we also have:
 * $$P(M_2|C_1\ and\ K_1)=P(M_3|C_1\ and\ K_1)=\tfrac 12.$$


 * $$P(M_1|C_2\ and\ K_2)=P(M_3|C_2\ and\ K_2)=\tfrac 12.$$


 * $$P(M_1|C_3\ and\ K_3)=P(M_2|C_3\ and\ K_3)=\tfrac 12.$$

The probability the car is behind door 3 is:
 * $$P(C_3)=\tfrac 13.$$

What happens when the host opens door 3? What is the probability the car is behind door 3? As above:
 * $$P(C_3)=\tfrac 13.$$

But there is a goat behind door 3? There is no chance there is a car? Right, but the car is placed randomly, hence in 1/3 of the cases the car is behind door 3. Also is the probability the car is behind door 3, 1/3. A new situation is originated, but that doesn't change the probabilities. In the new situation we have also new probabilities. And the new probability the car is behind door 3 is clearly 0! To distinguish them from the old ones, the new probabilities are called 'conditional' probabilities. They differ from the original, unconditional probabilities. They refer to the new situation. The conditional probability the car is behind door 3 is 0.


 * $$\!\,P(C_3|M_3)=0.$$

After the choice of door 1 by the player all probabilities are conditional probabilities, given the event K1. We have i.e.:
 * $$PC_1|K_1)=P(C_2|K_1)=P(C_3|K_1)=\tfrac 13.$$

Then the host opens door 3. All probabilities are from now on conditional probabilities, given the events K1 and M3. What will be the probability the car is behind door 3? As always:
 * $$P(C_3)=\tfrac 13.$$

But!! the conditional probability the car is behind door 3, given K1 and M3 is:
 * $$P(C_3|K_1\ and\ M_3)=0.$$

And what will be the probability the car is behind door 1? No different:
 * $$P(C_1)=\tfrac 13.$$

And what will be the probability the car is behind door 2, the unopened door? Well, also:
 * $$P(C_2)=\tfrac 13.$$

We have:
 * $$\!\,P(C_1)+P(C_2)+P(C_3)=1.$$

But in the new situation the probability the car is behind door 2, i.e. the conditional probability the car is behind door 2, is:
 * $$P(C_2|K_1\ and\ M_3).$$

We don't know it's value, we have to calculate it.

For the the conditional probabilities also holds:
 * $$P(C_1|K_1\ and\ M_3)+P(C_2|K_1\ and\ M_3)+P(C_3|K_1\ and\ M_3)=1.$$

And we already know:
 * $$P(C_3|K_1\ and\ M_3)=0.$$

hence:
 * $$P(C_1|K_1\ and\ M_3)+P(C_2|K_1\ and\ M_3)=1.$$

It is thus sufficient to calculate one of these probabilities.

With the use of Bayes' law:



P(C_2|K_1\ and\ M_3)= $$

= \frac{P(M_3|C_2\ and\ K_1)P(C_2|K_1)} {P(M_3|C_1\ and\ K_1)P(C_1|K_1)+P(M_3|C_2\ and\ K_1)P(C_2|K_1)} $$

= \frac{1\times \tfrac 13}{\tfrac 12\times \tfrac 13 + 1\times \tfrac 13} = \tfrac 23.$$

As the choice by the player is independent of the position of the car and door 1 is chosen, we may only consider the situation K1 and leave the mentioning of K1 as a condition.

Another possibility is the use of the symmetrie in the problem to proof that the conditional probability the car is behind door 1, given K1 and M3 is also 1/3, just like the unconditional.

We have:
 * $$P(C_1|K_1)=P(C_1|K_1\ and\ M_3)P(M_3|K_1)+P(C_1|K_1\ and\ M_2)P(M_2|K_1)$$

Because of the symmetry:
 * $$\!\,P(C_1|K_1\ and\ M_3)=P(C_1|K_1\ and\ M_2)$$

Hence:
 * $$\tfrac 13=P(C_1|K_1)=

$$

=P(C_1|K_1\ and\ M_3)P(M_3|K_1)+P(C_1|K_1\ and\ M_2)P(M_2|K_1)= $$

=P(C_1|K_1\ and\ M_3)\left[P(M_3|K_1)+P(M_2|K_1)\right]= $$

=P(C_1|K_1\ and\ M_3).$$

If someone thinks any solution may be formulated without conditional probabilities, please use the above notation to proof it.

Nijdam (talk) 23:35, 3 June 2009 (UTC)

[Please do not put comment in the text of the analysis]
 * Yes indeed, symmetry is a very good, simple and obvious option. Why not use it. Martin Hogbin (talk) 17:23, 4 June 2009 (UTC)
 * Okay, but so is Bayes' law. Nijdam (talk) 21:50, 4 June 2009 (UTC)


 * The main problem with your analysis is with the problem formulation. What is the question that you are answering?  Is it Whitaker's question in Parade?  How are you interpreting the problem?  Are you giving an answer to what you believe Whitaker wanted to know, in which case you should make clear what that is, or are you treating the question as a formal probability problem, to be interpreted from what is actually written, in which case you must make clear your treatment of all points that are not made clear in the question and your rational behind each decision.


 * You cannot have a definitive answer without a definitive question. Martin Hogbin (talk) 09:39, 5 June 2009 (UTC)


 * Quit right of course. My question is clear. What question do you want to consider?Nijdam (talk) 07:52, 7 June 2009 (UTC)


 * Suppose we take the question to ask whether is it better to switch or stick and by how much, for a player the show Whitaker describes. This is, in my opinion, what Whitaker actually wanted to know.  The door numbers etc were just to make clear what happened in the show. Martin Hogbin (talk) 16:57, 7 June 2009 (UTC)

When?
Consistent with "Suppose you're on a game show", these premises seem warranted:
 * The puzzle is to be solved from the contestant's POV/knowledge-state
 * The host acts randomly when faced with 2 goat doors

Could someone please explain to me at what point the contestant's selected door does not have a probability of 1/3 of being the car, and what is that probability?

It seems to me the only value it could change to is 1/2. I don't think anyone would argue that it goes to 0, 2/3, or 1.

And the only point at which it could change is when Monty reveals a goat. Otherwise, nothing changes.

I look forward to your responses. Glkanter (talk) 13:39, 5 June 2009 (UTC)


 * As Nijdam and I have said, more times than I care to count:


 * Initially, the chances are 1/3 + 1/3 + 1/3 = 1


 * After the host opens (say door 3), the chances are a + b + 0 = c


 * Since the host doesn't always open door 3, c != 1 (c is the probability the host opens door 3).


 * Since the host always opens door 3 if the car is behind door 2, b=1/3.


 * Since the host doesn't always open door 3 if the car is behind door 1, a != 1/3.


 * If the host acts randomly when faced with 2 goats, this happens 1/3 of the time (the times when the car is behind door 1), in which case c=1/3 + (1/3)*(1/2), which totals 1/2, and a=(1/3)*(1/2) = 1/6. So, under your assumptions, the chances after the host opens door 3 are:


 * 1/6 + 1/3 + 0 = 1/2


 * Under the "normal" rules, the host opening door 3 doesn't affect the chances that the car is behind door 2, but halves the chances that the car is behind door 1. So switching doubles the player's chances of getting the car. -- Rick Block (talk) 15:28, 5 June 2009 (UTC)


 * Thank you, Rick, for the thorough response. As you know, I am not an expert in probability. I'm only familiar with the branch of probability theory where the sum of the possible outcomes must = 1. Would I be following proper form in adjusting the "1/6 + 1/3 + 0 = 1/2" by multiplying it by 2 on both sides of the '=' sign? Glkanter (talk) 16:35, 5 June 2009 (UTC)


 * When you do this (multiply by 2) you're changing what you're talking about from the original (unconditional) probabilities to the conditional probabilities that are in effect in the case where the host opens door 3. This makes the statement "the host opening door 3 doesn't change the player's initial 1/3 chance of picking the car" sort of true, but entirely misleading since it is simultaneously referring to the initial (unconditional) 1/3 probability and the resultant (conditional) 1/3 probability.  Looking at it this way, the host opening door 3 certainly does change the player's initial 1/3 probability - it changes into a conditional probability that is also 1/3.  These are both 1/3, but they are completely different probabilities.


 * A less confusing way to look at it is to keep the unconditional perspective throughout. The other 1/2 is the case where the host opens door 2, so the whole (unconditional) equation is:


 * Initially 1/3 + 1/3 + 1/3 = 1


 * We know the host will open door 2 or door 3


 * (1/6 + 1/3 + 0)host opens door 3 + (1/6 + 0 + 1/3)host opens door 2 = 1


 * By keeping the unconditional perspective we avoid the switch to conditional probabilities and the host opening either door doesn't change any of the initial probabilities (player's door is 1/6+1/6, and the other doors are 1/3+0) and you can see switching doubles your chances of winning the car (from 1/6 to 1/3). -- Rick Block (talk) 03:11, 6 June 2009 (UTC)
 * Clear as mud. Martin Hogbin (talk) 09:52, 6 June 2009 (UTC)


 * What about this do you find hard to follow? -- Rick Block (talk) 15:34, 6 June 2009 (UTC)

The question asked is "Could someone please explain to me at what point the contestant's selected door does not have a probability of 1/3 of being the car, and what is that probability?"

You wrote:
 * "1/6 + 1/3 + 0 = 1/2", which, despite your response, is equivalent to 1/3 + 2/3 + 0 = 1
 * and
 * "(1/6 + 1/3 + 0)host opens door 3 + (1/6 + 0 + 1/3)host opens door 2 = 1", which is equivalent to 1/3 + 1/3 + 1/3, which is where we obviously started
 * and
 * "These are both 1/3, but they are completely different probabilities."

So, you've agreed that it is 1/3. But you haven't shown where it ever was not 1/3, much less what that value might have been. Which, of course, were the only questions asked. Glkanter (talk) 16:40, 6 June 2009 (UTC)


 * I think I've been perfectly clear that there are two different probabilities that are both 1/3. You don't seem to think there's a difference between 1/3 of something and half of 1/3 (which is 1/3 of something else).  Please answer the following questions (with numbers, not fractions):


 * 1) If 300 players all pick door 1, about how many of these would you expect to have picked the car?


 * 2) Of the 300 players in #1, about how many would you expect to see the host open door 3?


 * 3) Of the players who see the host open door 3 (your answer to #2), about how many have picked the car?


 * Are your answers to #1 and #3 the same? If you say "the host opening door 3 doesn't change the player's probability of having selected the car" how many players does this statement refer to, or does "probability" in this statement mean two different things?  I think a better way to say this is "the player's initial probability of having selected the car is the same as the conditional probability in the case where the host opens door 3".  These are different probabilities (100 out of 300 vs. 50 out of 150), and (yes) they are both 1/3.  The constraints on the host force these both to be 1/3, but they are distinctly not the same thing. -- Rick Block (talk) 17:41, 6 June 2009 (UTC)


 * You know, Rick, absent the canard that Monty may 'tip off' the contestant in some way by having a bias, your arguments just doesn't have much heft. Like a cigar, sometimes 1/3 is just 1/3. Glkanter (talk) 04:02, 7 June 2009 (UTC)
 * I would answer: the contestant's door (always) has probability 1/3 to hold the car. But it is not obvous that the conditional probability to hold the car is 1/3. So better to formulate your question and ideas in the terminology I proposed. And yes: 1/3 is always 1/3, but not everything with the value 1/3 is the same as something else with that value. I.e. my wife is called Petra, just like one of her friends. I better be careful not to think they are the same. Nijdam (talk) 08:00, 7 June 2009 (UTC)

Well, I understand (and agree with) this:
 * "I would answer: the contestant's door (always) has probability 1/3 to hold the car. "

Which differs from this, just a week ago:
 * "At what point did it change from 1/3, and what did it change to? Then, when did it change back? Please provide a reference. Thank you. Glkanter (talk) 18:30, 29 May 2009 (UTC)"


 * "Apparently at a point far beyond your reach. I have explained this a dozen or more times, and won't repeat it any more. Nijdam (talk) 10:56, 31 May 2009 (UTC)"

It appears you changed your mind on this question. But I would not want to put words in your mouth. Can you explain this apparent dichotomy?

As you probably expected, I don't understand the relevance of rest of your response. In what situation would 'the 1/3 probability of the contestant's door holding the car' not exactly equal 'the 1/3 probability of the contestant's door holding the car'?

If I didn't know better, I would think you're just playing word games with me. But a professor of Mathematics wouldn't do that, would he? Glkanter (talk) 10:36, 7 June 2009 (UTC)


 * Glkanter - are you unable to answer my questions above? -- Rick Block (talk) 17:35, 7 June 2009 (UTC)


 * Rick, given that Nijdam just 2 hours ago proposed an entirely new "Probabilistic solution" section which is entirely devoid of any 'host behaviour', or references to the unconditional solutions being 'false' or 'incomplete', I see no need to continue these topics on this 'arguments' page. I hope we can focus our attention on attaining a consensus that will propagate these long over due enhancements throughout the entire article and the FAQs. Glkanter (talk) 18:15, 7 June 2009 (UTC)


 * You've missed the point of Nijdam's suggestion - he's suggesting starting the Solution section with this, not replacing the "Probabilistic solution" section. I'd still like your answers to my questions above. -- Rick Block (talk) 20:00, 7 June 2009 (UTC)


 * Not much interested. And just so no random reader thinks I'm insane, here's what you wrote 3 hours ago on the talk page:
 * "Just to be clear, Nijdam is suggesting this new text come before (or perhaps even instead of) the current "Popular solution" section. Are you saying you'd be OK with this? -- Rick Block (talk) 17:57, 7 June 2009 (UTC)"
 * Why not just drop it, OK? Glkanter (talk) 21:01, 7 June 2009 (UTC)

Why?
Why is it that in the 18 years since Morgan and at least 3 others published their papers, other reliable sources continue to publish solutions to the MHP using the unconditional solutions? And, as Rick points out, no reliable source has ever published a paper directly challenging Morgan and the others? Although, "The American Statistician" felt it appropriate to include a 'comment' from Professor Richard G. Seymann immediately following Morgan's paper. I presume they had Morgan's approval to include this.

Since Morgan's POV is that the unconditional solution is 'false', they can't both be right, can they? Glkanter (talk) 14:05, 5 June 2009 (UTC)


 * I suggest that you write a paper yourself if so inclined, or blog about it, or take it privately with the editors of "The American Statistician". However, I fail to see how the above is a "mathematical argument concerning the MHP", which is the subject of this page (or any argument at all, for this matter).glopk (talk) 22:14, 5 June 2009 (UTC)


 * Thank you for the thoughtful suggestions. I will certainly take them under advisement.
 * In the context that on these pages Morgan is often trotted out as the definitive word on the MHP because no subsequent publisher has challenged their analysis directly, and that being peer-reviewed makes it the 'uber'-reference, I saw fit to comment that even in the absence of direct conflict, the fact that many reliable sources continue to use unconditional solutions indicates that many professionals do not accept Morgan's argument. I think it's worth re-examining many previous arguments on these pages that have occurred over 4 years and 9 archives, in light of two recent developments: A mathematical error has been found in Morgan's work, and that the premise that the host must act randomly when faced with two goats may indeed be part of MvS's problem statement.
 * I didn't edit the article. I didn't even put this on the talk page. I think this is the appropriate venue for this discussion. Glkanter (talk) 02:31, 6 June 2009 (UTC)


 * Do you live anywhere near a major city that might have a university library that has copies of The American Statistician? If you've purchased the online copy of the Morgan et al. paper, it has a truncated version of Seymann's response and is missing Morgan's rejoinder.  If it's at all possible, please go to a university library and look up these references.  While you're there you might look up papers that reference Morgan's paper (the librarian will be happy to help you).  Contrary to what you apparently think, Morgan et al. is not some perverse oddity that mathematicians dismiss as the ravings of some lunatic.  It is instead the first published rigorous treatment of the MHP (deserving a prominent place in Wikipedia's article on this problem).  Do some subsequent sources not just ignore but conflict with this paper?  Absolutely.  Shame on them.  Perhaps they might read the Wikipedia article and learn something. -- Rick Block (talk) 05:03, 6 June 2009 (UTC)


 * Do you continue to claim you have a NPOV on this article? Glkanter (talk) 06:30, 6 June 2009 (UTC)


 * WP:NPOV applies to articles, not to editors. Of course I have a POV.  You do, too.  Everyone does.  The point is for articles to fairly represent what is published about the topic.  -- Rick Block (talk) 15:31, 6 June 2009 (UTC)


 * Rick, your claim that Morgan's paper is it all rigorous does not stand up to scrutiny. It has a misquotation and several inconsistencies, it does not make clear the premises on which it reaches its conclusions, and most amazingly it contains a error in the application of conditional probability!  These are matters of fact that I will be pleased to point out to you. Martin Hogbin (talk) 10:23, 6 June 2009 (UTC)


 * Continuing to harp on the alleged error in this forum is relatively pointless. Morgan and his co-authors have (as far as I know) never commented on this article.  As I've said, I've previously contacted Morgan by email and he has politely refused to comment here.  If you want to pursue this, please contact the journal.  Regardless of what you think of the paper, it is the first peer reviewed paper published in a math journal about the MHP.  As such it should be a primary reference for the article here.  -- Rick Block (talk) 15:31, 6 June 2009 (UTC)


 * I will be happy to stop pointing out the deficiencies of the Morgan paper if you stop promoting at as the last word on the subject. I accept that it is a reliable source but there are others, such as Krauss and Wang, that are much more balanced in their approach and more relevant to the article in that they discuss the basic problem and why people get it wrong.  I also see no reason for us to duplicate the unscholarly tone of the Morgan paper in the article, with descriptions of the good    work of others as 'false' and insistence on 'one correct solution'.  We have an equally reliable source, which no one has challenged, in the form of Seymann's comment that, 'Without a clear understanding of the precise intent of the questioner, there can be no single correct solution to any problem'. Morgan's paper does indeed present a correct solution to their particular formulation of the problem but there are clearly other ways to interpret the problem and these require different solutions.


 * Regarding two of the errors in the paper, the misquotation of Whitaker's question is there for all to see and the error in the application of conditional probability is being pursued via the journal. Martin Hogbin (talk) 12:54, 7 June 2009 (UTC)

What was Whitaker's question?
I think it is universally agreed that to answer a question we must understand what the questioner is actually asking. In the case of Whitaker's question I see two ways of doing this. We can either take an open view of the question and ask ourselves, regardless of what he wrote, what do we think Whitaker actually wanted to know. Although this, in my opinion, is the better way to answer the problem it has the difficulty that the intended question becomes purely a matter of opinion that can never be settled. If we are to take this approach the least that we can do is make clear the exact formulation that we are providing a solution to.

The other way is to answer the problem as written. This has the advantage that we have a clear problem statement to start with although, personally, I doubt that this is the right approach, bearing in mind that the question was a letter from a reader of a popular general interest magazine, rather than one in a statistics exam. If we do take this approach, where we have a clear problem statement, there are many issues to be decided before we have a precise enough formulation of the problem to get a single correct solution. Here is Whitaker's question:

''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?''

Firstly the game rules. Certain aspects of these are not clear from the problem statement but there is general agreement that the standard game rules should state that the host always offers the swap and always opens an unchosen door to reveal a goat, even though this may not have always been the case on the assumed show.

Starting with the stated question, first we have, 'Suppose you're on a game show...'. What do we take this to mean? To many it might be taken to mean that we should answer the question from the perspective or state of knowledge of a contestant on a game show, that is to say someone who has no idea of the host door choice strategy. Many might argue that this is the only way to take it. For some reason Morgan ignore this part of the question.

Next we come to, You pick a door, say No. 1. Here we could have two contradictory possible meanings depending on which part of the phrase we give priority to. 'You pick a door', means that you could pick any door but, say No 1 strongly suggests that we do, in fact, pick door 1. To most English speakers this phrase means that we pick any door, an example of such a door being door 1. In other words Whitaker does not want to ask specifically what the best action is only if you happen to open door 1 but he is asking a general question.

Then we come to, the host, who knows what's behind the doors, opens another door, say No. 3. Once again we could take this to mean either that the host opens another (unspecified) door or that the host does, in fact, open door 3. Which of these two meanings we give to this phrase is of vital importance for it determines whether the questioner considers the door opened by the host to have any significance and therefore be identified in the question. Curiously, Morgan misquote this section as the host, who knows what's behind the doors, opens No. 3, giving the strong impression that Whitaker actually intended the door opened to be identified.

It may remain debatable which of these interpretation we should assume but we should at least be consistent in our choice. Either we take it that Whitaker meant only for the specific case that the player chooses door 1 and the host opens door 3 to be considered, which seems unlikely to me, or we must take it that the question means that the player chooses any one of the three doors and the host then opens either one of the two unchosen doors.

To sum up, there are two issues to be decided before we can provide a solution. Should we answer the question from the state of knowledge of a player on the game show? Does the questioner intend the door opened by the host to be identified? Martin Hogbin (talk) 22:18, 8 June 2009 (UTC)


 * Why are you so focused on Whitaker's version of the problem statement? As I've mentioned before (the article says this) the Monty Hall problem predates Whitaker's question to Parade (by 15 years!), and even Selvin's original version is clearly a reformulation of the Three Prisoners problem, which itself did not even originate in Gardner's 1959 Scientific American column.  The problem is a classic problem in elementary conditional probability.  It has gained popular notoriety precisely because people are really bad at correctly reasoning in the realm of conditional probability.  -- Rick Block (talk) 04:19, 9 June 2009 (UTC)


 * I have focussed on Whitaker's question because it is the question that Morgan quote in their paper and it is by far the most well known problem statement. What can Morgan be referring to when they talk of the 'problem at hand'?  They insist that the problem is one of conditional probability, yet it is easy to formulate the problem such that this is not the case.  What Morgan do is start with Whitaker's problem, reformulate it so that it is a conditional question, add in bogus 'information' that makes the condition important, them criticize everyone else for not treating the problem the way that they do. Martin Hogbin (talk) 08:51, 9 June 2009 (UTC)


 * I'm sorry Martin, but the MHP is always a question for a conditional probability. Already the choice of the player is in principle a condition, although one may omit it, but the opening of the door certainly is. That's why they're critical, and so am I. Any effort to answer the question without referring to a conditional probability is wrong.Nijdam (talk) 18:01, 9 June 2009 (UTC)
 * Even if the numbers of the chosen and opened doors were not mentioned in the problem statement, the solution would be the conditional probability as a function of all the possible combination of doors. That's why it is made more specific and one of the combination - doors 1 and 3 - are taken as an example.Nijdam (talk) 18:06, 9 June 2009 (UTC)


 * So this is yet another attack on the Morgan et al. paper? We all know you don't like this paper, why not just drop it? -- Rick Block (talk) 12:57, 9 June 2009 (UTC)


 * It is not specifically an attack on the Morgan paper, it is an attack on the concepts, that Morgan started, that the question identifies the door opened by the host and that the question should not be answered from the state of knowledge of the player. How about you address the content of what I have written, which is that the question in general, and Whitaker's statement in particular, is open to other interpretations.Martin Hogbin (talk) 16:24, 9 June 2009 (UTC)

I guess it's my turn. For purposes of editing the article, it doesn't matter which is right. They both go in. The question is how much credence/emphasis do we, as thinking, analytical editors, give to each solution? Specifically, since the conditional solution does not rely on them, should 'host behaviour' or 'unconditional is false' be part of the conditional solution? Or where do they best belong, if at all? Nijdam has presented a conditional solution which does not include either of them. I heartily support that approach. Glkanter (talk) 18:20, 9 June 2009 (UTC)

@Martin - we've been through your questions about the interpretation of the problem innumerable times. Is Whitaker's question open to interpretation? Of course it is. Do you realize yet that Whitaker was paraphrasing an already well-known probability puzzle he'd no doubt heard elsewhere? Regarding your questions:

Should we answer the question from the state of knowledge of a player on the game show? If the player knows everything given in the problem statement, yes. We should answer the question from the state of knowledge of someone reading the question (which may or may not match the player's knowledge).


 * Here you have neatly sidestepped answering the my question, which was whether we should answer the question from the state of knowledge of the player. You seem to have given two answers.Martin Hogbin (talk) 20:29, 10 June 2009 (UTC)


 * No, I've given one answer. What is not clear about it?  If you think the player knows everything stated in the problem statement, then the answer is yes.  Otherwise, the answer is no.  If you insist on a one word answer, then it would have to be no.  -- Rick Block (talk) 02:09, 11 June 2009 (UTC)

Does the questioner intend the door opened by the host to be identified? In the sense you mean, yes, absolutely. There is simply no other reasonable way to read the question. And, if you think you can read it some other way Whitaker's paraphrasing was lousy. The classical question is constructed to contrast the initial situation in which the player clearly has a 1/3 chance of having selected the car, with a conditional case in which the player is standing in front of two closed doors and an open door, not knowing which door the car is behind, and deciding whether to switch. If you're going to take the player's viewpoint the player can clearly see which door the host opened (whether you say which one it is in the problem statement or not). Anyone reading the problem statement can infer the player can see which door the host has opened (again, whether or not it's specifically mentioned in the problem statement). Your continued insistence that not identifying the door the host opens means it is not a conditional probability problem is simply bizarre. Perhaps this is not so clear to you since probability is apparently not your field - but this would be like someone arguing a problem that starts with "in a situation where your velocity is close to the speed of light ..." is not a relativity problem. Of course it is. -- Rick Block (talk) 04:17, 10 June 2009 (UTC)


 * What classical question is, 'constructed to contrast the initial situation ... with a conditional case...'? How do you know this? Not identifying the door that the host opens does make the problem unconditional. Nijdam has already agreed this. If the question merely states that the host opens one of the other two doors to reveal a goat, what is the condition? Martin Hogbin (talk) 20:29, 10 June 2009 (UTC)
 * Let's be specific. What I agreed on in not identifying doors, is that another type of problem arises. One in which the player doesn't know which door is opened. The MHP definitely is about identified doors. And as I wrote somwhere above: the MHP always concerns conditional prob's. Nijdam (talk) 10:35, 11 June 2009 (UTC)
 * On what basis do you make the statement, 'the MHP always concerns conditional prob's'? You seem to be resorting to a circular argument that the doors must be identified because that makes the problem conditional and the problem is conditional because the doors are identified.
 * This is a little unfair: I never said the doors must be identified to make the problem conditional.Nijdam (talk) 14:02, 11 June 2009 (UTC)
 * If you ask yourself what Whitaker (or any one else) actually wants to know. It could quite easily be that all he wants answered is the question, 'You will be offered the choice of three doors, and after you choose the host will open a different door, revealing a goat.  What is the probability that you win if your strategy is to switch?'. Martin Hogbin (talk) 13:06, 11 June 2009 (UTC)
 * Well we discussed this before and I repeat: the decision would be based on the conditional probabilities as a function of the numbers of the chosen and the opened door. Nijdam (talk) 14:02, 11 June 2009 (UTC)
 * I cannot understand why you are so fixated on conditional probability. The formulation I have given above is that given by Morgan as a statement of the unconditional problem. Martin Hogbin (talk) 20:21, 12 June 2009 (UTC)
 * @Martin: let us be practical, the problem we want to address is the formulation of Krauss & Wang. This is generally considered the MHP. At least it has all the aspects needed. May be we will never be sure what Whitaker really wanted to ask, although I'm pretty sure it was what K&W formulated. May be even Whitaker himself didn't exactly know what he wanted. The article already accounts for the ambiguities in his formulation. Happily this gives the opportunity to discuss the alternative host strategy. Please what exactly do you want to discuss in the article other than these aspects. Nijdam (talk) 11:32, 10 June 2009 (UTC)
 * I agree that K & W do give a formulation that they consider is the way that most people see the problem. From their formulation I see two directions to go in, both leading to valid solutions.  Firstly, the academic direction in which we consider that the host door choice may not be random (although car placement remains so).  This is what Morgan do and it has a place in the article.


 * Alternatively we can notice that in the K & W formulation there are several very simple arguments to tell us that the 'condition' that the host opens a specific door is irrelevant and thus this factor can be completely ignored in our solution. In other words the problem can be treated unconditionally. Martin Hogbin (talk) 13:06, 11 June 2009 (UTC)


 * No, the arguments that tell us that the 'condition' that the host opens a specific door is irrelevant, lead to the conclusion that the conditional probability, that needs to be calculated, equals the unconditional, and hence is also 1/3. That's the only simplification in what may be referred to as "unconditional solution". There is no way the decision is based on the unconditional probability. Nijdam (talk) 13:37, 11 June 2009 (UTC)


 * I think you are just arguing about words here. The simple solution is correct in principle and practice. Martin Hogbin (talk) 20:21, 12 June 2009 (UTC)

Host acts randomly?
Whitaker begins, "Assume you are on a game show..."

Rick, given that Whitaker's question is a story problem about an American game show, please expand on your response on the talk page that the host doesn't necessarily have to act randomly when faced with two goats. Thank you. Glkanter (talk) 02:14, 11 June 2009 (UTC)


 * This either is simply setting some informal context for what we should take as a word problem, or is suggesting the answer should consider a real life scenario. The predominant interpretation is by far the first.  If instead we're to treat this as a real life scenario, US laws about game shows could be considered to be in effect (which would mean the car should be placed randomly and the host should pick randomly if faced with two goats).  On the other hand, in a real life scenario the motives of the host distinctly come into play, and since we're not told the host makes this offer to every player we should consider the possibility that the host is only (or more often) making this offer when the player has initially selected the car.  I mean, what kind of show is going to create a situation where with optimal play 2 out of 3 players can win a car?  If we allow the host to make the offer more often (or exclusively) when the player has initially picked the car, the player's chances of winning by switching are anything from 0 to 2/3.


 * So, clearly, it's a word problem. As many, many sources point out, as a word problem it's underspecified.  Any solution that clarifies it's assumptions and solves the problem according to those assumptions is therefore valid.  Whitaker's version doesn't say anything about the host preference when faced with two goats, so assuming the host might have a preference is at least as valid as assuming the host will pick randomly in this case.  But, whatever you assume, the problem is a conditional problem.


 * It seems to me you're wanting to have it both ways. The answer is 2/3 only if the host always makes the offer to switch (which wouldn't be the case on a real game show) AND the host chooses randomly if picking between two goats (which might be considered to be required on a real game show).  -- Rick Block (talk) 04:02, 11 June 2009 (UTC)


 * Relative to the question I asked, I haven't assumed anything. It's in the dna of a game show. I'm talking dictionary, not judicial statutes.


 * Always offers the switch? I've never commented on that. From what I've read, MvS agrees she should have stated it, believes her answer implies it, and everyone agrees its a premise to the MHP. Just more stalling. Glkanter (talk) 04:23, 11 June 2009 (UTC)
 * Of course it is. Surely we can take it as read now that the host always opens and unchosen door to reveal a goat and always offers the switch. Martin Hogbin (talk) 12:46, 11 June 2009 (UTC)


 * You both seem to be missing the point. Because we're assuming the host always makes the offer to switch we're obviously not considering a real-life game show scenario, but rather a word problem.  The argument that the host must choose randomly when faced with two goats is based on the premise that a real-life game show would have to work that way in the US - but since it's not a real-life scenario, this constraint (or any other based on how real-life game shows operate) doesn't necessarily apply.  -- Rick Block (talk) 14:11, 11 June 2009 (UTC)


 * "Assume you're on a game show, well, not exactly a game show, I'll specifically use the words 'game show', but anyways, don't feel compelled to even understand what a game show is, just make up any exceptions that don't fit your narrow paradigm..." Glkanter (talk) 14:53, 11 June 2009 (UTC)


 * Actually Rick, you are confusing the definition of a game show with the rules of the two goats and one car game on Lets Make A Deal. Whitaker says, 'Assume you are on a game show..', then spells out the unique rules of his Lets Make A Deal contest, specifically, that the host will always offer the switch. This does nothing to negate the obvious understanding (the guys who didn't understand this went to jail) that the host will not actually tell the contestant where the car is. Glkanter (talk) 15:02, 11 June 2009 (UTC)


 * Whitaker spells out that the host will always offer the switch? Here's the quote from Parade:
 * 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?
 * Where does this say the host always makes the offer to switch? In the typical analysis we assume this to be true even though if you were on a game show I think we both know the host wouldn't always be giving you a 2/3 chance to win a car.  In any event, you asked me to expand on my response from the talk page.  I have.  It's clear you think "Suppose you're on a game show" means the host must pick randomly in the case in question.  I've said why I disagree.  Can we drop this now?  -- Rick Block (talk) 18:46, 11 June 2009 (UTC)


 * You're right, as I posted earlier, MvS agrees she should have stated "the host always offers the switch", believes her answer infers it, and it is universally accepted as a premise of the MHP. Of course, you ignore the point, which is by the definition of a game show, the host will not tell the contestant where the car is. "Offering the switch" was brought up by you in an attempt to negate that point.


 * No, we can't drop it. Morgan says it. NPOV says we have to include it. It has to be addressed. It's not needed for the conditional solution, so my point is it shouldn't be in the solutions section. It is there now. And Nijdam, seemingly, proposed a conditional solution without it. And without 'host behaviour', Morgan's only remaining point is that the conditional solution provides a solution to the MHP, as does the unconditional solution. So, the question will ultimately be, 'how extensively should the 'dispute' be covered in the article'. Since we won't try to prove one side or the other wrong or right in the article, my proposed statement "'Morgan says the unconditional solution is false. This viewpoint is not shared universally in the professional community." may be all that is required in the article. Outside of this Wikipedia article, where does the so-called 'controversy' exist, anyway? The professional community certainly don't address it. How does the reader benefit, when both solutions are deemed valid by reliable published sources? Glkanter (talk) 19:17, 11 June 2009 (UTC)

Self-analysis tool
The following is a method designed to help people who are uncomfortable with the puzzle's resolution to pin down which specific features lead their intuition astray, without requiring them to delve into any but the most superficial aspects of probability maths.

After reading some of the preceding discussion, I should clarify right away that it's based on the only version of the puzzle I was familiar with before reading the article - namely one in which one considers the three doors to be a priori indistinguishable. This is what justifies using an urn problem analogy by removing most of the ambiguities that otherwise need to be addressed.

Incidentally, the reason for my not having come across the version the article focusses on before may have to do with my being German. While the name "Monty Hall problem" is in use here, the names "Let's Make a Deal", "Parade magazine" and "Marilyn vos Savant" were completely unknown to me. If there is a German standard phrasing of the problem at all, it may be based on an earlier version than the Whitaker one, or on a very free translation that didn't preserve all the details.

No matter, the aspect that this is meant to address is one that comes up in any version.

Thought experiment
We can construct an equivalent problem by using two urns and three balls, one white and two blacks. The host puts one (random) ball into the first urn and the other two into the second. The player knows the number of balls in either urn, but not the colour distribution. The game consists in choosing one urn or the other, and the aim is to end up with the white ball.


 * In version A, the player simply picks an urn.


 * In version B, the host first looks into the second urn, takes out a black ball, shows it to the player, and drops it back in. Then, the player picks an urn.

For version A, it is intuitively obvious that it is favourable to pick the second urn, and that the odds are 2:1.

However, it is also intuitively obvious that the difference between the versions is immaterial - all the host does is confirm that there is at least one black in the second urn, which was already a certainty.

What the experimental subject has to decide is what their intuition tells them about the favourable choice and the odds for version B. If it is the same as for version A, one now transforms version B step by step into the MHP (with indistinguishable doors), which is trivial, and marks the point at which their intuition flips. This may, for example, be to discard the black ball instead of putting it back, or the replacement of a single urn by two separate doors, or ...

If it is different from version A without any transformation, the subject is confronted with a simple and objectively contradictory situation consisting of the two versions and their equivalency, which can only be resolved by recognizing and localizing the flaw in their intutitive understanding.

Unfortunately, none of the regulars here will make good test subjects themselves, I imagine. :)

62.180.36.2 (talk) 04:23, 28 June 2009 (UTC)


 * On the German Wikipedia a user, Wilbert, proposed the same, in his opinion, similar experiment. I already answered there that this experiment is not similar to the MHP, because of the unindistinguishability of the black balls. Nijdam (talk) 11:43, 28 June 2009 (UTC)

I thought I addressed that above, sorry if I wasn't being clear enough - this is meant to mirror a version of the MHP in which the doors (and, if you like, the goats, though unless I'm mistaken this is not strictly required) are indeed indistinguishable. Whether such a version is too different from the standard one to still be called "the Monty Hall Problem" is a matter for debate, I suppose.

Either way, the purpose of the thought experiment isn't to solve the problem, it's to help anyone who's at the "I have to accept the numerical solution but I still don't really believe it" stage to figure out where the disbelief stems from. As such, I find it helpful even with regards to the standard version, as the reluctance it evokes is qualitatively the same, I should think.

(dynIP) 62.180.36.10 (talk) 12:26, 28 June 2009 (UTC)

Thanks for pointing me to the German discussion page, I found the passage you were referring to. However, my version is quite a bit more powerful as a tool for understanding, IMO, as it reduces the complexity by several more steps. This makes the intuitively contradictory situation much tighter and thus easier to resolve - the equivalency between versions A and B has to be overwhelmingly obvious.

62.180.36.10 (talk) 12:53, 28 June 2009 (UTC)


 * To be equivalent, I'd suggest using balls of three different colors, perhaps white, red, and green, with one ball in one urn and two in the second urn. Then, in version B, the host withdraws a non-white ball, say the red one, and doesn't drop it back in but discards it leaving both urns with only one ball.  Note that if this is your own idea it can't be incorporated into the article as it would be original research, which is prohibited (all material added must be from reliable sources).  This is actually fairly close to one of the experiments reported by Fox and Levav (referenced in the English version of the article).  In their experiment they dealt 5 cards including only one ace face down, 2 to the subject and 3 to the "host" and then the host either looked at the host's cards and put them back face down, or looked and turned a non-ace card face up, or looked and turned two non-ace cards face up, and then asked the subject to estimate the probability that the "host" had the ace and whether the subject wanted to trade their two cards for the host's remaining face down cards (3, 2, or 1).  In about 35 trials each way, the mean estimate of the probability that one of the host's cards was the ace was .6, .5, and .33 depending on whether the host was left with 3, 2, or 1 face down cards (with a total of 5, 4, or 3 face down cards between the host and the subject) and the percentage of subjects willing to trade their two cards for the host's remaining face down cards was .58, .23, and .15 respectively.  Assuming the host was randomly selecting non-ace cards to turn face up, the actual probability was always .6 and to maximize their chances of winning (winners were actually paid a small sum of money) all subjects should have switched.  Like it says in the article here, people apparently strongly believe that if there is one winner among N unknowns, the probability of each is 1/N whether this is true or not.  The Fox and Levav experiment would suggest in your urn version that it's important to actually remove and discard one of the balls rather than put it back in the urn.


 * BTW - in German the problem is apparently often called "Das Ziegenproblem". There's a book about it by van Randow called Das Ziegenproblem: Denken in Wahrscheinlichkeiten  (I don't read German so haven't read this book, but it is referenced in the German Wikipedia article about the problem). -- Rick Block (talk) 16:22, 28 June 2009 (UTC)

It's entirely OR, which is why I posted it here (looking for feedback) rather than on the main talk page.

I agree that changes along the lines you suggest would make the thought experiment more compatible with the standard version, but any added complexity runs contrary to the experiment's purpose, which relies on statements being what I call "intuitively obvious".

Basically, I'm asserting that the reluctance of many people to accept the puzzle's solution is based on features other than the distinguishability of the doors and/or goats. Anyone unhappy with the idea that there is an objective advantage in switching in the standard MHP will be just as unhappy with the idea that there is an objective advantage in switching in a related urn experiment (one urn, three balls, player picks one without looking, host removes a non-white ball, stay or switch), and vice versa. If this is so, any insight gained about the intuitive thought processes leading to the disbelief should transfer from one to the other.

I'm fairly certain that the explanation that "people apparently strongly believe that if there is one winner among N unknowns, the probability of each is 1/N" is at best incomplete. People are familiar with non-equiprobable distributions (you seem to be using the term "non-random" on these talk pages, btw, which seems imprecise to me), and would be perfectly happy with accepting and using reliable information along the lines of e.g. "most of the time, the stage hands put the car behind the middle door". My impressions is that it has more to do with the notion that information revealed about one door should not affect the probabilities assigned to another door.

Your impression that many people will see a crucial difference between the host's replacing or discarding a black ball after showing it is likely correct, and is precisely part of the reason for the design of the experiment as it stands. These people should agree that the odds in version B are still 2:1, and will disagree with the step along the subsequent back-transformation to an MHP-like form which changes this aspect. All that remains is to convince them (or wait for them to convince themselves) that the transformation step is valid, without being encumbered by any other aspect of the puzzle.

ps: As a sidenode, I'd just like to say that after reading a significant portion of the talk page archives, I'm very impressed with the amount of patience you have shown with even the most stubborn and quarrelsome parties.

62.180.36.29 (talk) 17:38, 28 June 2009 (UTC)

Why is the chance 50/50 when the host doesn't know?
Can someone explain why it doesn't matter if you stay or switch when the host doesn't remember which door has the goat and which the car? Assuming he still (randomly, by accident) chooses a door with a goat, wouldn't it still be a 2/3 chance of winning if you switch, simply because you had a 2/3 chance of initially choosing a goat? --JohnJSal (talk) 15:46, 29 June 2009 (UTC)
 * This confusion is exactly the problem with the so-called "unconditional" solutions. At the point there are two doors and a goat, however you get there, you're in a different situation than you were originally.  To figure out the probabilities in this (new) situation you have to use concepts from conditional probability.  In the "standard" version, the host opens Door 3 whenever the car is behind Door 2 (which is 1/3 of the time) and half the time the car is behind Door 1 (i.e. 1/2 * 1/3, which is 1/6), so the car is behind Door 2 twice as often as it is behind Door 1.  The overall probability of being in this case (where the host has opened Door 3) is 1/3 + 1/6 = 1/2, and the conditional probability of the car being behind Door 1 is (1/6)/(1/2) = 1/3, while the conditional probability of the car being behind Door 2 is (1/3)(1/2) = 2/3.  In the "host forgets" version, the host opens Door 3 half the time regardless of where the car is.  So the chances for each door are 1/6, 1/6, and 1/6.  But in addition, we know the host didn't reveal the car so the 1/6 case where the car was behind Door 3 simply hasn't happened.  We're left with a 1/6 chance of the car being behind Door 1 and a 1/6 chance the car is behind Door 2, so switch or not doesn't matter.  Note that these only add up to 1/3, which means if the host opens Door 2 or Door 3 completely randomly, the case where the host opens Door 3 and happens to not reveal the car occurs only 1/3 of the time.  In the "host forgets" version this is the only case we're interested in, and so the conditional probability the car is behind Door 1 (or Door 2) is the same, i.e. (1/6)/(1/3) = 1/2. -- Rick Block (talk) 16:33, 29 June 2009 (UTC)
 * Yikes! I appreciate the in-depth response...just a little beyond my league I think. :) But it definitely helps me to see the overall shape of the problem, at least. --38.108.205.180 (talk) 17:45, 29 June 2009 (UTC)
 * Conceptually, you can think about it like this: If player initially picked car, there is no chance of host revealing car. If player initially picked goat, there is a chance of host revealing car. Hence, the observed fact that host didn't reveal car makes it more likely that player initially picked car.
 * 62.180.36.26 (talk) 12:04, 1 July 2009 (UTC)


 * The basics of conditional probability are not that difficult. The probability of this given that is written P(this|that) and is the joint probability of this and that together divided by the probability of that, i.e. P(this and that)/P(that).


 * In the normal version of the problem, assuming we're asking about the probabilities when the player has picked Door 1 and the host has opened Door 3, the probabilities of interest are the conditional probability that Door 1 hides the car given the host has opened Door 3 and the conditional probability that Door 2 hides the car given the host has opened Door 3. P(Door 2|host opens Door 3) is slightly easier to figure out, and is P(Door 2 and host opens Door 3) / P(host opens Door 3).  P(Door 2 and host opens Door 3) is the probability the car is behind Door 2 and the host opens Door 3.  We're assuming the player picked Door 1, so since the host always opens Door 3 if the car is behind Door 2 and the player picked Door 1, this is the same as P(Door 2), which is 1/3.  P(host opens Door 3) is the probability the host opens Door 3.  We might assume the problem is symmetric, meaning we're assuming the answer is the same whether the host opens Door 2 or Door 3 - another way to approach this is to specifically figure out the probability the host opens Door 3 as opposed to Door 2 (perhaps assuming the host picks randomly if the car is behind Door 1).  Either of these approaches results in the probability of the host opening Door 3 being 1/2.  Putting these together, we get P(Door 2|host opens Door 3) = (1/3) / (1/2) = 2/3.  Using similar reasoning P(Door 1|host opens Door 3) works out to 1/3.  The above is the "show your work" step missing from all of the "unconditional" solutions that say something like "when the host opens Door 3 the initial 1/3 chance of the car being behind Door 1 doesn't change" (or "your initial chance of picking a goat is 2/3, so ...").


 * In the host forgets version, the probabilities of interest are a little bit different and are the conditional probability that Door 1 (or Door 2) hides the car given the host has randomly opened Door 3 and has not revealed the car. Lets figure out P(Door 1|host opens Door 3 and car is not behind Door 3).  The joint probability, P(the car is behind Door 1 and the host opens Door 3 and the car is not behind Door 3), is the same as P(the car is behind Door 1 and the host opens Door 3) because if the car is behind Door 1 it is not behind Door 3 and this is (1/3)*(1/2) = 1/6.  The probability the host opens Door 3 (by itself, this is 1/2) and the car is not behind Door 3 (i.e. the probability Door 3 hides a goat) is (1/2) * (2/3).  So, the conditional probability Door 1 hides the car given the host has randomly opened Door 3 showing a goat is (1/6) / (1/3) = 1/2.


 * If you approach this problem (or any of its variants) as a conditional probability problem you shouldn't get confused. The approach works for other "paradoxes" as well, like the Boy or Girl paradox (I know a family has two children, I see one and the one I see is a boy, what is the probability both are boys?).  The instinctive answer is 1/2, but the question here is actually P(both are boys|at least one is a boy) which if you work it out you'll find is not 1/2.  -- Rick Block (talk) 14:26, 1 July 2009 (UTC)


 * "I see one and the one I see is a boy" is usually not taken to be equivalent to "at least one is a boy", but to "one which was chosen at random is a boy", which makes all the difference, no?
 * 62.180.36.6 (talk) 00:22, 2 July 2009 (UTC)


 * Ah yes, that is the problematic wording. How about "You run into the mother of a family you know has two children and since you're involved with the local Boy Scouts organization ask her if at least one is a boy.  She says yes.  What is the probability both are boys?" (the intent is "at least one is a boy"). -- Rick Block (talk) 12:23, 2 July 2009 (UTC)


 * The common feature is the method in which partial information (location of goat, minimum number of children of given sex) is obtained: Deliberately, by a knowledgable agent (Monty-with-good-memory/mother), as opposed to random sampling (Monty-with-bad-memory/observer). The solutions are different because the methods affect the sample space differently.
 * 62.180.36.7 (talk) 14:36, 2 July 2009 (UTC)

This is Marilyn Vos Savant's answer:


 * "Here’s one way to look at it. A third of the time, the clueless host will choose the door with the prize, and the game will be over immediately. In our puzzle, that didn’t occur. So we’re considering the two-thirds of the time when either: 1) You have chosen the door with the prize; or 2) The prize is behind the unopened door. Each of these two events will occur one-third of the time, so you don’t gain by switching."

Please note there are 3 different premises to this puzzle from the MHP:
 * Host does not always show a car
 * Host does not always offer the opportunity to switch
 * Contestant doesn't always makes a decision

Since it's random (as per vos Savant), anyone could pick the 2nd door. Even the Contestant. And now you have Deal Or No Deal with just 3 suitcases. Glkanter (talk) 19:13, 29 June 2009 (UTC)

(from talk) Knowing the host will open a door makes the probability 1/2
Moved from Talk page, 28 July 2009 per Martin's suggestion.

Given the statement of the problem, that the host knows what is behind all 3 doors, and he is going to reveal a door with a goat (or whatever) behind it, the original choice made by the contestant had a 1 in 2 chance of winning from the beginning. Not after the reveal, but before it. Because the host was ALWAYS going to reveal a booby prize behind one of the unchosen doors. Although the contestant might not know he has a 1 in 2 chance from the beginning, that's what he has. And that probability remains the same whether he switches or not. I think this is a misunderstanding in the statement of the problem. I agree with the logic leading to the 2 in 3 chance of winning if the a priori condition is left out. However, that's a different problem, not the one that was in Parade or discussed in this article. PatPM (talk) 04:24, 27 July 2009 (UTC)


 * If you want a response to this, I suggest that we move to the /Arguments page. Martin Hogbin (talk) 08:21, 27 July 2009 (UTC)

PatPM (talk) 00:47, 29 July 2009 (UTC)


 * So, is it your claim that all the references about the problem including probability textbooks are incorrect and all the simulations showing a 2/3 chance of winning by switching are flawed? Are you actually interested in understanding why 2/3 is the correct answer even though you don't apparently believe it, or have you completely made up your mind that you're right and the rest of the world is wrong? -- Rick Block (talk) 01:40, 29 July 2009 (UTC)
 * Rick, this is another example of where the current article fails to do its job. By obfuscating the central problem with talk of conditional probability we fail to address the real problem of why people get the answer wrong. Martin Hogbin (talk) 13:34, 29 July 2009 (UTC)
 * I am of the exactly opposite opinion on this. By highlighting that the problem is a conditional probability problem we address the precise reason people get the answer wrong, which is that they fail to realize the problem is about conditional probabilities.  The structure of the problem puts the reader in a conditional case (thinking about the player in front of 2 closed doors and one open door).  The "simple" approach says to just ignore that and think instead about all possible outcomes rather than the specific case the problem asks about (with the implicit assumption either that the answer in the specific case matches the answer considering all outcomes or, more dubiously, that the probability of interest is actually the one considering all outcomes).  My claim is that most people read the problem as asking about the conditional case but then fail to correctly analyze the conditional probability, and the article should therefore show them how to do this correctly.  The "simple" approach (ignore what's actually asked, and answer something else that is easier to figure out) doesn't generalize to any other conditional probability problem and doesn't even work for all variants of the MHP.  It is by obfuscating the conditional nature of the problem that we fail to address the real problem of why people get the answer wrong. -- Rick Block (talk) 02:39, 30 July 2009 (UTC)
 * So would you claim that, if the question were asked in a completely unconditional format (as in the example given in Morgan) most people would get it right? Martin Hogbin (talk) 15:44, 30 July 2009 (UTC)


 * The completely unconditional format is, for example, "the host will open a door showing a goat, what is your chance of winning if your strategy is to switch" (i.e. deciding whether to switch before the host opens a door). If you don't switch, this problem is logically equivalent to "what is your chance of correctly choosing the car given it is behind one of 3 closed doors", which is a completely uninteresting problem - one that nearly all people would get right.  On the other hand I suspect most people would probably get the "unconditional" form of the MHP wrong, because they'll actually treat it like the conditional problem (not realizing it is simply the completely uninteresting problem with a bunch of extra words around it) - meaning they'll imagine the situation after the host has opened one of the two unpicked doors and (mis)judge that the probabilities are 50/50 because there are only 2 doors left.  I know you don't agree with this, but IMO the unconditional form completely misses the entire paradoxical nature of the MHP which is that you're deciding whether to switch when you're looking at two doors (not three) and don't know where the car is, but the chances are not 50/50.  Understanding that picking one out of 3 doors has a 1/3 chance is trivial.  Understanding how two unknowns can end up with different probabilities is not so trivial, but not so hard that it's beyond the grasp of, say, a reasonably bright high school student. -- Rick Block (talk) 00:03, 31 July 2009 (UTC)


 * Pat, is what you are saying essentially this? If the host opens a door to reveal a goat right at the start, the contestant clearly then has a 1/2 chance of initially choosing a car and the same chance of getting the car if he later switches.  Because we know that the host is going to open a door later on, the odds are exactly the same in the Monty Hall problem. Martin Hogbin (talk) 13:34, 29 July 2009 (UTC)


 * What I interpret Pat to be saying is that the knowledge of the game rules, in particular that the host will with certainty open a door revealing a goat, makes the odds of picking a car or a goat the same (each 1/2) before the host opens a door. You know you will be left with two alternatives, so your chance is 50/50 of initially picking either one of them.  Then, when the host follows through and does open the door, the odds remain 1/2.


 * Pat - the reason this is not what happens is because there are three different initial conditions (assuming the player picks door 1), each equally likely, as follows:


 * CASE1) Car is behind door 1. In this case the host can open either door 2 or door 3.


 * CASE2) Car is behind door 2. In this case the host must open door 3.


 * CASE3) Car is behind door 3. In this case the host must open door 2.


 * If we see the host open door 3 we must be in CASE1 or CASE2 and cannot be in CASE3. If we're in CASE1, and if we assume the host picks which door to open randomly in this case, and we see the host open door 3, we're in this case 1/2 of 1/3 of the time (i.e. 1/6 of the time).  We're in CASE2 1/3 of the time.  These are the only possibilities if we've seen the host open door 3, so we have a 1/6 chance of the car being behind door 1 and a 1/3 chance of the car being behind door 2 when combined with the fact that we've seen the host open door 3.  The crux of the problem is that if we pick door 1 the host always opens door 3 if the car is behind door 2 but only opens door 3 1/2 the time if the car is behind the door 1 (the other 1/2 the time the car is behind door 1 the host opens door 2, but since we've seen the host open door 3 this didn't happen).  So if we pick door 1 and see the host open door 3, the car is twice as likely to be behind door 2 as behind door 1.


 * The same argument applies if we pick door 1 and see the host open door 2, in which case the car is twice as likely to be behind door 3 as behind door 1.


 * Make sense? -- Rick Block (talk) 02:39, 30 July 2009 (UTC)

New book as mentioned on Talk Page: search for: Morgan
http://www.amazon.com/Monty-Hall-Problem-Remarkable-Contentious/dp/0195367898/ref=sr_1_1?ie=UTF8&qid=1249163746&sr=8-1#reader —Preceding unsigned comment added by Glkanter (talk • contribs) 22:10, 1 August 2009 (UTC)
 * Is there some point here you're wanting to discuss? This book was already mentioned and it's account of the exchange between Morgan et al. and vos Savant was discussed on the talk page at talk:Monty Hall problem.  -- Rick Block (talk) 14:54, 2 August 2009 (UTC)

JeffJor's comments to your redirection here from the MHP discussion page.
Martin, you are wrong about Morgan's so-called "surprising error." The integral on the bottom of p286 expresses an a posteriori probability in terms of the related a priori probabilities according to the definition of conditional probability. Correctly. You erroneously translated it to a formula based on a posteriori probabilities: your q' is literally "the probability the host opens Door #3 given that the host opens Door #3." And whatever your "symmetry" argument is, it seems to be incorrect. That same argument should apply to any value of q, and different values of q can (with a catch) lead to answers other than 2/3.

The "catch" I referred to is subtle, and often confuses people about probability. In fact, it is ultimately the cause of the MHP's controversy, even among PhDs. I'll explain by using an example: Suppose I have a bag with 101 coins. Each coin has an integer N printed on it, ranging from 0 to 100, and has the property that it lands on heads N% of the time. I draw a coin out at random, and flip it. What is the probability that it lands heads? There are actually two answers, based on whether I tell you what N is. P(heads|know N) = N/100, and P(heads|don't know N)=1/2. The first probability can be thought of as representing the fraction of the time I would get heads if I flip that same coin over and over; while the second probability is when I flip different coins, replacing each and drawing another one at random each time.

The subtle point is that probability is a property of the process, not of the coin, per se. The problem statement I gave you only described a single example of the process I intended, and I left whether the process required replacement ambiguous. So the two kinds of knowledge represent different processes. In Morgan, if q=0 is a known property of the process that is repeated every time the show airs, then the probability P(Ws|D3)=1; and if q=1 the same way, then P(Ws|D3)=1/2. But it is 2/3 without that knowledge, since you can't attribute any specific value to it. The ln(2) answer represents an average where the host picks a q at random each show, and tells it to the contestant; the 2/3 answer represents results over many shows where we do not know q. (NOTE: This seeming paradox is not Bayesian Inference, as some people will claim. It is a result of including parameters that are not specified in the problem as though they are both included and known.)


 * It is an odd way of putting it, to say, 'the host picks a q at random each show, and tells it to the contestant', this never actually happens, but I understand what you mean so let us consider that to be that case. At the start of the show let us say that the host randomly chooses a value of q, this the a priori distribution of q is uniform (over the interval 0 - 1).  The quesion that Morgan are answering is 'what is the average probability of winning by switching, given that the host has opened door 3 ?'  Although the distribution of q was uniform at the start of the game, we are only considering cases where door 3 has been opened.  this is not a random selection from the original sample set thus the posterior (meaning in this case after the host has opened a door) distribution of q is no longer uniform.
 * Rick, you either didn't read, or didn't understand, what I wrote. (1) The concept of the uniform q is not consistent with the problem statement, so yes I am saying it is absurd, and (2) the integral equation on p286 is a correct expression for the a posteriori probability expressed in terms of the a priori probabilities. There was no error there; and if you think there was, you need to revisit your elementary course in probability. For a similar example, the conditional probability that a playing card is a spade, given that it is black, is 1/2. P(spade|black)= [a priori probability that a card is both black and a spade]/[a priori probability that a card is black]=(1/4)/(1/2)=1/2. Changing q to q', as was done to correct this non-error, is equivalent to changing that card formula to P(spade|black)= [a priori probability that a card is both black and a spade]/[a posteriori probability that a card is black]=(1/4)/(1)=1/4. That "correction" is wrong. JeffJor (talk) 21:42, 28 October 2009 (UTC)
 * It was Martin who wrote the above comment, not Rick. Let us go through this more slowly.  You say that a uiniform distribution of q is inconsistent with the problem statement.  Why do you say this.  The problem statement (Whitaker's) says nothing whatever about the host's strategy thus the prior distribution of q could be anything.  Martin Hogbin (talk) 17:12, 29 October 2009 (UTC)

However, Morgan is still a mixture of inconsistent pedantry that is both correct, and incorrect, at points. Ignoring the misquote of the problem (which invalidates much of Morgan's argument, but does not invalidate the correct results if you adjust them properly), it treats "Player chooses Door #1" as a fixed part of the probability experiment (i.e., a priori information) while treating "Host opens Door #3" as the condition which separates a priori from a posteriori. Since most of Morgan's argument is based on "Solution FX is wrong because it did not prove why an a posteriori probability is equivalent to an a priori one," that was an inconsistency - Morgan ignored the same proof w.r.t. Door #1.


 * I agree completely that Morgan is a mixture of inconsistent pedantry. This is the main error in their paper.  Above, you say that 'the host picks a q at random each show, and tells it to the contestant'.  Of course we know that the host does not, in fact, tell the contestant of his door choice policy, this the parameter q should be fixed at 1/2, just as Morgan tacitly fix the initial distribution of the car at 1/3 per door.

I also notice the result at the top of page 286 is wrong: P(Ws|D3)=(p23+p33)/(p13+p23+p33). This allows for the possibility that p33>0, which is the probability the host opens Door #3 when the car is there. It also assumes the contestant will always pick the car when she can deduce where it is, something Morgan's pedantry at this point does not allow us to assume. But p33>0 is what I mean by Morgan's pedantry, since including p33 as a parameter requries us to allow that p33>0. Morgan also ignored the possibility the host would not open a door, which is more justifiable (Monty Hall did not always open a door) than p33>0.

Assuming the contestant does not know the host's strategy (remember my catch?), Morgan's approach can be treated as correct if you make one simple adjustment: The player ignores any numbers printed on the doors, and mentally labels the door she choose Door #1, the unopeneed door #2, and the opened door #3. This can be done with no loss of generality, since each door is clearly specified in this system and the contestant does not know how they relate to the host's choices in the process I am assuming. But we can also deduce p=q=1/2 by the principle of indifference. Morgan's formula on the top of page 286 is correct, and strictly speaking was derived while many of the so-called incorrect student answers were not. But the rest of the discussion, including that integral, really have no bearing on the problem as asked, because the problem does not say we know what q is. We can't assume it is a constant from show to show. The only possible answer is 2/3, by the formula on the top of page 286. JeffJor (talk) 14:58, 28 October 2009 (UTC)


 * I agree that the only possible answer is 2/3. The article gives gives the K&W statement as the problem to be answered, where the host chooses randomly.  If this is the question then the answer is 2/3 plain and simple and Morgan is an irrelevance. Martin Hogbin (talk) 18:07, 28 October 2009 (UTC)


 * I've had this discussion with Martin before, but I'll note that the context given for the problem ("Suppose you're on a game show") means this problem is not an abstract urn problem with indistinguishable doors.
 * Yes, it is an abstract "urn problem." Even if the host distinguishes the doors, the player does not now how he is doing that. Whatever the host does differently to distinct doors is not revealed to the player. In her context, they are indistinguishable. So the rest of your comment, which I indented for consistency, is irrelevant. If you need to re-think this, look at what I said about the player using her own numbers for the doors, not the numbers the host uses. JeffJor (talk) 21:54, 28 October 2009 (UTC)


 * I guess we'll have to agree to disagree. The doors on a game show are clearly distinguishable since they have (as I say below) distinct, persistent, physical locations and, since we're later told (in the Parade version) to imagine the player picking "say, No. 1" the doors in addition are almost certainly labeled (consistent with the photo of the Let's Make a Deal stage).  The Parade version doesn't say the host chooses randomly (if given the choice) - analyzing the effect of a preference (even if the contestant is unaware of it) seems perfectly reasonable. -- Rick Block (talk) 23:16, 28 October 2009 (UTC)


 * (A) Whether or not a game show typically labels doors, this problem didn't say they were labeled. Stop projecting your experience into the problem. (B) The mention of labels in the problem we are dealing with were examples, not meant to be taken literally in the answer. Morgan MISQUOTED the problem, (C) Even if they are labeled, and the problem were about Doors #1 and #3, the probabilities you calculate based on the labeling only apply if the player knows how the host uses the labels. Without that knowledge, the probability is 2/3 for switching. (D) No, it is not a contradiction that two people can attribute different probabilities to the exact same action in the exact same game. Say I draw a card, and note to myself that it is the King of Spades. I then tell Alice it is black, Bob it is a spade, Cindy it is a face card, and Doug I tell nothing. I ask them the probability it is the King of Spades. To me, it is 100%. To Alice, it is 1/26. Bob says 1/13, Cindy says 1/20, and Doug says 1/52. All are different, all are correct, and there is no contradiction. (E) It is an abstract urn problem to the player, based on the information the player has according to the problem statement. JeffJor (talk) 11:02, 29 October 2009 (UTC)


 * Oh, and finally: (F) The Principle of Indifference (used in Baysian Perspective) does not mean "assume a uniform distribution for anything you need a distribution for." It means you can assume that the same distribution values can be assumed for elements that are indistinguishable. It is only if all elements are indistinguishable that you can assume a full uniform distribution. Applied to this problem, it would mean you assume p=q=1/2, not that q is uniformly distributed between 0 and one. Because each q is definitiely distinguishable from another. The ln(2) calcualtion is correct when you do assume that distribution, but there is no justification for assuming it. JeffJor (talk) 16:47, 29 October 2009 (UTC)


 * This context provides ample justification for assuming:


 * 1) the doors are distinguishable (physical objects on a stage with at least persistent physical identity if not big numbers 1, 2, 3 on them like the photo at http://www.letsmakeadeal.com/), and because of this we can mentally label the player's door Door #1, the unopened door Door #2, and the opened door Door #3


 * 2) the player has the opportunity (by having viewed past airings of the show) to observe p relating to her specific initially chosen door - if players initially pick randomly, roughly 1/3 will have picked the same door our player has picked. Of these, roughly 1/3 will have initially picked the car.  By observing how many of these players win by switching, p can be estimated.


 * Perhaps this is debatable, but I think the problem is obviously intended to be a straight conditional probability problem where the probability of interest is exactly what Morgan et al. analyze, i.e. the player has picked a specific, identified, door and the host has in response opened a different, also identified, door. Removing these aspects of the problem essentially turns it into a very wordy version of "what's the probability of correctly picking a car from behind 3 doors".  Perhaps the wordiness obfuscates the problem sufficiently that people get it wrong as well.  However, the conditional probability problem where your initial 1/3 pick is now one of only two possible choices is (IMO) FAR more paradoxical. -- Rick Block (talk) 20:15, 28 October 2009 (UTC)


 * But note, Rick, that the article now gives the K & W unambiguous description as a statement of the problem. In this the host is stated to choose randomly.  The problem that Morgan answer thus clearly becomes not just an obfuscation  but an extension of the MH problem in which the host is allowed to chose non-randomly.  Martin Hogbin (talk) 21:22, 28 October 2009 (UTC)


 * But note, Martin, that even the K&W version is clearly a conditional probability problem. Morgan et al. address both the "usual" case where the host must choose randomly as well as variants (like the Parade version) where this aspect of the host behavior is not specified. -- Rick Block (talk) 23:16, 28 October 2009 (UTC)


 * This is pretty much where we started. To a degree it is a matter of opinion.  Let me state some things that I think we agree on in the unambiguous K & W case.


 * The doors are distinguishable.


 * The sample set is reduced when the host opens a door and the conditioned set is dependent on the door opened.


 * On the other hand I think that you must agree (and other have noted) that there is an obvious symmetry between the two conditioned sample sets. In other words the answer clearly must be the same whichever door the host opens.


 * As well as the symmetry argument I am sure that you agree that it is generally accepted that no information is revealed by a random choice thus the probability that the player has originally chosen a car must be unchanged by the host's action.


 * I accept that there is no reliable source that treats the problem this way. (If there were I would be busy editing the article).


 * My opinion is that, in the light of the above, a conditional treatment of the problem is an unnecessary complication and an obfuscation of the real problem. You think that the problem must still be treated conditionally.  On that point we must agree to differ. Martin Hogbin (talk) 17:00, 29 October 2009 (UTC)


 * (1) The doors are indistinguishable, and the problem is not supposed to be about Doors #1 and #2. This is clear from reading K&W, since thier solutions actually use other arrangements. THE DOOR NUMBERS ARE USED ONLY AS EXAMPLES.


 * (2) Even if they are distinguished, it does not matter UNLESS THE PLAYER KNOWS HOW THE HOST USES THE METHOD OF DISTINGUISHING THEM. This is how "the answer clearly must be the same whichever door the host opens" is achieved.


 * (3) There is a more convienent way to distingusih the doors: Chosen, Unopened, Opened. Once again, these names are not associated to any bias the Host has, so they can (and must, by the principle of indifference) be treated as having equal a priori probabilities for anything that needs a probability. The (a priori) chances are 1/3 that the car is behind each, and 1/2 that the two unchosen doors would be opened by the host. By the principle of indifference.


 * (4) No problem "needs" to be treated conditional, it just helps express the answer sometimes. Bayes' Theorem works to solve this particular problem. P(Unopened has car|Host can find a door with a goat to open)=P(Unopened has car AND Host can find a door with a goat to open)/P(Host can find a door with a goat to open)=(2/3)/(1)=2/3. The a priori probability that "Unopened has car AND Host can find a door with a goat to open" is 2/3 because the Host can always find such a door, and in 2/3 of the them the door he declined to open has the car. No, this isn't a strict proof of the answer, but a strict proof can be made of it. 23:55, 29 October 2009 (UTC) —Preceding unsigned comment added by JeffJor (talk • contribs)

The so-called “conditional solution” is invalid.
(1) The problem statement Morgan uses is a misquote. The source Morgan attributes actually says “You pick a door, say #1, and the host … opens another door, say #3.” Not “You pick Door #1, and the host … opens Door #3.” The door numbers are meant to be examples, not conditions of the problem, so that the solutions are easier to describe. It is just shorter to say “Door #1” than “The door that the contestant originally chose,” “Door #2” rather than “The door the contestant didn’t choose and the host didn’t open,” and “Door #3” rather than “the door the host opened.”

(2) While this problem sounds similar to many games used on Let’s Make a Deal, no game like it was ever played there. Specifically, Monty Hall never offered a player the chance to swap to a prize that player had already turned down (see teh letter at http://www.letsmakeadeal.com/problem.htm). He did offer players the chance to trade in a prize for new options, but never old ones. And the only game that used door numbers was the “Big Deal of the Day,” and it was played with two contestants, and all prizes had significant monetary value. The doors were revealed in the reverse order of value, regardless of whether they were chosen. So don't refer me to pictures of the show's set. Since the problem sttement didn't say the doors had numbers printed on them, or say they were distinguishable in any way, THEY AREN'T. That doesn't mean that a contestant can't create names for them, like #1, #2, and #3; it just means that they can't be related a priori to anybody else's names.

(3) The 1/N discrete probability distribution is required by the Principle of Indifference when the random elements in a discrete random process are indistinguishable, except for any names given to them. Door numbers are names. So in the prior distribution, this principle applies even when doors are numbered. They need to be treated identically.

(4) The Principle of Indifference is very hard to apply to a continuous random element, since what makes the different values “indistinguishable” is all but impossible to define. For an example, see Bertrand’s Paradox. The three answers there try to use the same technique that Morgan used with q.

(5) If we are going to assign random probabilities to strategies, based on door numbers, we also need to consider the possibility that the car’s location is determined with a similar biased strategy. (Math removed - I shouldn't try to do math and figure out how to display it, all while hurried, as I did yesterday JeffJor (talk) 14:39, 6 November 2009 (UTC).) I'm not sure what that would do to the "conditional solution," but I doubt it leaves it unaffected. The point is that the Principle of Indifference tells us what probabilities to use for the strategies, not how to vary them. JeffJor (talk) 20:47, 5 November 2009 (UTC)


 * [Quoted forom the Monty Hall discussion page, as stated by Martin Hogbin on 11:32, 17 November 2009.]
 * To start with, is there anyone who can tell me whether Morgan are answering the problem from:
 * The point of view (state of knowledge) of the player.
 * The point of view (state of knowledge) of the audience.
 * Only on the information given in Whitaker's question.
 * Only on the information given in Morgan's restatement of the question.
 * Some other basis - please specify.


 * This is not a minor detail but something that is universally accepted as an essential requirement in statistics - that the question being asked is clear.

Exactly right, Martin. But I prefer to use this list, of States of Knowledge (SoK), for the general problem. I'll try to divorce them from anybody's problem statement for now, and define them a little better than you did:
 * The contestant's state of knowledge, which does not include any way to assign a value to p.
 * The audience's state of knowledge, which we can assume represents a weighted average of the many different values p could take on based on any number of factors. But those factors are not seen by the audience, which is why they see a weighted average.
 * The producer's state of knowledge. She does know the value of p Monty Hall will use today, but not where the car is nor the final choice Monty makes based upon that value of p.
 * The mysterious Morgan Solver, who doesn't know p, but suspects it may not be 1/2 and assigns a uniform distribution to it without giving a valid justification.
 * Monty Hall's state of knowledge. Yes, the Host's himself. He knows where the car is, and so knows that P(win|switch) is either 0% or 100%.

To answer your question, nobody explicitly says a contestant has an SoK other than #1. This includes those who mis-quote, or misinterpret, the problem statement as "the conditional problem." Now, some solve for #2 or #3 after stating it is an alteration of the actual problem. Even Morgan starts that way. They show that the solution still has the property that, at worst, switching does not reduce the probability. Which is true. They don't always make it as clear as they should that such a solution is to a DIFFERENT PROBLEM (and I don't mean "the conditional problem," I mean "the problem from a different SoK") than what is stated, but it usually is presented along the lines of "If the SoK included a p, then the probability would be 1/(1+p)". Where they lack accuracy is that, even if the question is the conditional problem, the SoK being used has to be associated with every answer. Because from SoK #1, even the conditional probability is 2/3. And I think some people, like Rick Block, feel that is a contradiction. Is that true, Rick? It really isn't a contradiction, you know.

The way to see that it isn't a contradiction is to consider SoK #5. It is just as valid an SoK to evaluate the probability within, as any of the others except #4. But it allows only two values for P(win|switch), either 0% or 100%, and is independent of p. The contestant, stuck in SoK #1, just can't know which of those values applies to him. Just like the contestant has no way to know what p is. Another way to say that, is that it is just as valid to say "When Monty Hall knows the car is actually behind Door #1, then P(win|switch)=0 and the contestant should not switch" as it is to say "when the producer knows what p is, P(win|switch)=1/(1+p)>=1/2 and the contestant should switch. Both statements are true. They give different answers for "the probability." Neither is directly useful to a contestant in SoK #1. There is not such thing as the one-and-only value of the probability. The latter answer does bound it, and is helpful, so it makes an interesting problem extension. What it doesn't do is tell the contestant what "the probability" is.

However, SoK #4 is wrong. Morgan tries to calcualte "the probability" from any SoK. There is no law, theorem, or proposition in the field of Probability that allows Morgan, et al, to say "the noninformative prior in the vos Savant scenario makes..." That portion of their analysis is wrong.

Now Rick, I agree that the article has to be based on citable references. But that does not mean that everybody is interpreting those references correctly, or even that all are correct. Not every reference has to be used, especailly if the valid point they are trying to make is contained in opthers. And as Martin points out, whenever you say "the probability is..." you have to make it clear what SoK it applies to, and that is where references are being misinterpreted (or if SoK cannot be made clesr from them, they can't be used at all because interpretation is impossible).

We can apply some educated intelligence to clearing that mess up. In particular, any statement like "The probability is 1/(1+p) has to be qualified: "The the probability from an SoK where p is known is 1/(1+p). If p is not known in the SoK, the probability is 2/3." This is said, with varying degrees of clarity, in most of your references. But the article fails completely to make that distinction in the "Probabilistic solution" section. In particular, it is not true in general that "The conditional probability may differ from the overall probability depending on the exact formulation of the problem" as the statement is intended. What is true is "The conditional probability from an SoK that includes p may differ from the overall probability depending on the exact formulation of the problem." And it needs to be clear that no version of the problem allows for the contestant to have that SoK. We can say that an extension problem using SoK #3 can be helpful to the contestant because they all say "switch" regardless of p. JeffJor (talk) 20:36, 17 November 2009 (UTC)

[Rick Block's reply begins here, indented once. My counter-comments are interspersed, indented twice and ID'ed individually] JeffJor (talk) 14:49, 19 November 2009 (UTC)
 * 1) Yes, the door numbers are examples. I don't think anyone (but perhaps Martin) is reading Morgan et al. to mean anything different.  The problem is seemingly about the conditional probability a player faces after having first choosing a door and then seeing which door the host opens.  The entire point of the problem is the difference between the situation before the host has opened a door and after the host has opened a door.  Talking about this in the general case, i.e. the overall success of a pre-selected "switch" strategy vs. a pre-selected "stay" strategy destroys the essence of the problem.  The analysis of a specific case, using door 1 as the player pick and door 3 as the door the host opens is representative (by renumbering the doors) of all cases.  This doesn't mean the analysis is of an unconditional problem, merely that the same analysis of the door 1/door 3 combination applies to any of the 6 possible combinations of player pick and door the host opens.  Similarly, an analysis of a problem about roulette using the number 17 as an example pertains to any specific number - not just the number 17.


 * If you allow the doors to be renumbered, it can only be about the general (unconditional) problem. That is what renumbering them accomplishes, after all. If you don't allow them to be renumbered, but the problem does not state what q is, or that a bias is even possible, you have to treat them as unbiased. I am not saying that the conditional problem, even using a specific value for q, is not useful. It shows that the host can't be trying to trick the contestant into making a choice that reduces her expected value. It just has no other relationship to the problem that was asked. JeffJor (talk) 14:49, 19 November 2009 (UTC)


 * Also, both "versions" of this problem represent a conditional probability. The difference between the "unconditional" and "conditional" versions, is whether a specific door or a gerneral door is opened and/or chosen. JeffJor (talk) 14:55, 19 November 2009 (UTC)

Later editing

 * Yes, I have tried to explain this before. The problem is that some people seem to treat the Morgan paper as a religious text and give it meaning to suit their own POV.  Morgan do actually state the question that they are answering, quite clearly, they say (with my emphasis), " To avoid any confusion here is the situation: The player has chosen door 1, the host has then revealed a goat behind door 3, and the player is now offered the option to switch"  The problem is that it is hard to imagine that Morgan were only considering this specific case but that is what they actually say.  Despite this, some people like to try to interpret Morgan's clear statement to mean that the door numbers are examples only.  The problem with this is that, as Jeff says, it blows the conditional answer out of the water.  Again sometimes the Morgan faithful try to explain that Morgan really meant to say that that the player chooses any door and the host then opens a numbered and identified door and then offers the swap.  It really is a great pity that Morgan did not say exactly what they meant.  Martin Hogbin (talk) 18:01, 23 December 2009 (UTC)


 * 2) Yes, no game show was ever played by these rules. However, the context of "game show" certainly implies the doors are real physical objects on the set of a TV studio.  If nothing else they have positional identities (left, middle, and right).  I don't know what universe you live in, but in the one I live in any two doors that are not the same door ARE DISTINGUISHABLE (and, if two doors are not distinguishable they are the same door).  This means even if they aren't already numbered they can be referred to by their role in the problem as the door the player initially picks, the door the host opens, and the other door.  Calling them #1 and #3 is a notational convenience.


 * And I don't know what mathematical universe you live in, but "able to be distinguished from each other" is not what allows the solver to treat them differently. I keep linking you to the Principle of Indifference, but you refuse to look at it. "The principle of indifference states that if the n possibilities are indistinguishable except for their names, then each possibility should be assigned a probability equal to 1/n." It is only when the problem statement allows you to distinguish how the random factors apply to the named doors, that you can assign a probability other than 1/n. "They could apply differently" is not enough, you need to know how. The problem statement does not do this. Having numbers, or having defined physical placements that allow you to create names but not to divine how random factors are applied, do not alter the problem for a simialr one where teh doors are completely indistinguishable.


 * It is the exact same reason EVERYBODY IN THE KNOWN UNIVERSE assumes that the car has a 1/3 probability of being behind any of the distinguishable-only-by-name doors at the start of the game, even though the problem doesn't say that, either. If we assume a bias can exist, by door number, for the host's chocie of a door to open, then we also must assume a bias is possible, by door number, for the stagehands when they drove the car into the studio. If the probability that livestock can be placed behind door #2 (the furthest from the stage hands who might have to deal with accidents) is zero, q is irrelevant and the probability the contestant wins by switching is 100%. JeffJor (talk) 14:49, 19 November 2009 (UTC)


 * 3) I don't think anyone (again, excepts perhaps Martin) is challenging the assumption of the prior 1/3:1/3:1/3 distribution of the car. Actually, Morgan et al. note that this assumption could be relaxed as well, although they suggest this is likely of less interest.


 * It is included essentially as an afterthought. What they say is that possibility is "unlikely to correspond to a real playing of this particular game show situation." The same argument applies to q. Sponsers are very particular about being able to estimate the expected payouts, and demonstrating a bias that can be observed doesn't allow that. But the point is, that unlikely possibility has just as much relevance to the problem statement whether or not it is interesting. If you ignore it for lack of relevance, you are admitting that using q is also not relevant to the problem as asked. JeffJor (talk) 14:49, 19 November 2009 (UTC)


 * 4) I think Morgan et al. would agree that applying the principle of indifference to a continuous random variable is difficult - SO THEY DON'T. I don't think anyone disagrees that the formula for the conditional probability of winning by switching (i.e. the probability in the specific example case of player picks door 1 and host opens door 3) is


 * P23 / (P13 + P23)


 * where Pij is the probability the host opens door j given the car is behind door i. By the standard rules, P23 is 1 (host must open door 3 if the car is behind door 2).  The Parade description (and vos Savant's clarifications) never address P13, SO THEY LEAVE IT AS A VARIABLE.  This makes the conditional probability 1/(1+p) where p is the probability the host opens door 3 given the car is behind door 1.  If the host is constrained to act equally randomly in this case, p is 1/2 and the resulting probability is definitely 2/3 - but if the host is not so constrained Morgan et al. don't insist on picking a specific value.  They do suggest one way to arrive at a definite value is to use a non-informative prior for the distribution of p resulting in the seemingly nonsensical answer of ln(2).  I think this is more like the equivalent of picking one of the Bertrand Paradox methods - but Morgan et al. clearly are not expressing any sort of preference for this answer.


 * Yes, they do apply it. They don't say so, but few people do when they use it. They just give no reason for assuming "the noninformative prior" means that q is uniformaly distributed, but that principle is the only possible justification, especially when it is just presented without an attempt at justification. They assume such a distribution when they decide P(WS|D3)=ln(2). Do you want references for why "noninformative prior" does not mean a uniform distribution?


 * And the point is not that they use it, but that they think there is such a prior (other than q=1/2) at all. There isn't - a basis for such a prior has to be described in the problem statement. Assuming any one prior means that you have to give equal weight to the exact opposite prior (because you can only distinguish the doors by name, not intent), and "equal weight" is what makes it difficult to apply distributions to priors. That's why the only possible prior to assume is q=1/2, since it is the only prior that needs no opposite. JeffJor (talk) 14:49, 19 November 2009 (UTC)


 * 5) Morgan et al. suggest analyzing the effect of non-random initial placement of the car is possible, but likely of less interest. WP:OR: I think if the probabilities are C1, C2, C3, and the player has selected door 1 and the host opens door 3, the probability of winning by switching is


 * C2P23 / (C1P13 + C2P23)


 * which, because P23 is 1, simplifies to C2 / (C1P13 + C2). In this version, one might further conjecture the host and player both know the non-random C1, C2, and C3 values and figure out the optimal "play" on both the player's and host's part.  I'm sure I've seen this published somewhere.


 * Again, the point is that any formulation of a solution with a q in it must (yes, MUST) also include C1, C2, and C3, and is incomplete without it. Isn't Morgan's entire thesis based on others' solutions to the problem are incomplete? JeffJor (talk) 14:49, 19 November 2009 (UTC)


 * Regarding the points about the probability being a function of the SoK, I'm fine with different SoK's resulting in different probabilities. I don't think there's any controversy about this and that everyone basically agrees the relevant SoK when solving a problem is that of someone who knows what is given in the problem statement.  In the case of the MHP this can be taken to be that of the player, or the audience.  However, (per my comments just above) the structure of the problem seems to be asking about the probability of winning by switching in a specific example case, leading to a straightforward conditional probability treatment.  Looking at it this way makes the dependency of the answer on P13 (or, more generally, the probability that the host opens the door he's opened in the case the player has initially selected the door hiding the car) quite obvious.  Unless you know this probability you can't compute an exact answer for any example case.  The question is literally "should you switch".  The answer, even not knowing P13 (or Pij for whatever door you've picked and door the host has opened) is YES - your specific probability of winning if you switch is between 1/2 and 1 so you're no worse off switching and you might be far better off if you do.  Will players in your specific situation who have initially picked door i and have seen the host open door j win 2/3 of the time?  If you say yes, you're asserting Pij (for your case) is 1/2. -- Rick Block (talk) 06:49, 19 November 2009 (UTC)


 * And here the point is that there is one, and only one, SoK that can be applied to the question "Should the contestant switch?" The contestant's, AS CONTAINED IN THE PROBLEM STATEMENT. And it is impossible for "the structure of the problem" to ask "about the probability of winning by switching in a specific example case" without stating what the random factors that affect that alternate case are. Branching out beyond the problem statement this way is "unlikely to correspond to a real playing of this particular game show situation."


 * Morgan, et al, address only the SoK of someone who knows what q is, calling their solution "the probability." They do it parametrically, but this is not the contestant's SoK, and the result is not "the probability" for anything to do with the problem as stated. Essentially, it represents an SoK from after the host chooses a q to use today, but before he decides between #2 and #3 based on that q. And it includes knowledge of what q is, which may not be observable from past history. To answer Martin, and contradict what you claim in that other discussion page about Morgan's question, it is the answer to "If you were watching the show with this fabricated SoK, how would you assess the contestant's options to switch?" It is only one possible answer among many possibilities once you start introducing unspecificed SoKs. Relying on it is no more valid than those who insist on using the SoK of Marilyn vos Savant's little green woman. It is indeed far more interesting, and has application, but not directly to the problem as asked. JeffJor (talk) 14:49, 19 November 2009 (UTC)


 * Oh, I just noticed I didn't address your last point, about "you're asserting [that] Pij (for your case) is 1/2." I guess I am, sorta. Because that value represents the typical host as represented in the problem statement. And yes, if the host is atypical, the calculated probability will not match experimental evidence over many games. THIS IS NOT A CONTRADICTION, because the actual game does not match the game implied by the problem statement. 2/3 is the correct probability for the game represented, but the actual game was misrepresented. Try it a different way: say the car never gets put behind door #2 on Tuesdays, but this Tuesday's contestant does not know that and is not told. The host tells her that he picks randomly if able, and opens #3 when she picks #1. To the best of that contestant's knowledge, should she switch? Yes. (If you say "no," what basis do you have for that decision, how does the contestant know when it applies, and why does it not apply to Morgan's problem?) To the best of her knowledge, her probability after switching is 2/3. The supposed "contradiction" here is that the game was not correctly represented to her, not that she miscalculated anything. JeffJor (talk) 19:25, 19 November 2009 (UTC)

What exactly makes a probability problem conditional.
The WP article on the subject states that, 'Conditional probability is the probability of some event A, given the occurrence of some other event B'. This means that we must make some kind of judgement as to whether event B can possible affect the probability of event A, otherwise we would have to take every event into account for every probability problem. Martin Hogbin (talk) 21:44, 20 December 2009 (UTC)


 * Your searching in vain for an argument to avoid conditional probabilities. The event "opening door 3" is part ot the "experiment" and hence we have to condition on it. Nijdam (talk) 13:42, 21 December 2009 (UTC)


 * Exactly why is this so and not me clapping my hands, for example. Martin Hogbin (talk) 21:34, 21 December 2009 (UTC)


 * As I said: clapping your hands is not an event in the problem (read the problem statement). Nijdam (talk) 16:30, 22 December 2009 (UTC)


 * So you re saying that any event mentioned in the problem statement is a condition? Martin Hogbin (talk) 17:14, 22 December 2009 (UTC)

So far I have.

1) Conditional probability is the probability of some event A, given the occurrence of some other event B.

2) The event must reduce the sample space.

3) The event must occur before the question is asked.

4) The event must be 'part of the experiment'.

5) The event must be mentioned in the problem statement.

For an event to be a condition must it be one of these, some of these, all of these?

Martin Hogbin (talk) 20:31, 22 December 2009 (UTC)


 * Do not make to much fuss about it. If you perform some experiment, the possible events are known to you. If you know one of these events has occurred, you just have to account for the outcomes that are still possible. In theory this is achieved by reducing the sample space to the given (occurred) event. Draw a card from a complete deck. If you see the card is a spade card, the outcome can only be a card of spade. That's all. Nijdam (talk) 10:03, 23 December 2009 (UTC)
 * But I am not the one making the fuss. All I want is a simple explanation of a simple problem but people keep telling me that the problem must be treated conditionally, yet no one can tell me precisely why.  Martin Hogbin (talk) 10:29, 23 December 2009 (UTC)
 * You're the one asking about what conditional prob. means. And you have been told more then once precisely how things are. Please study them. Nijdam (talk) 14:20, 23 December 2009 (UTC)
 * I only ask because you and others insist it is mentioned in every solution to the MHP, if you stop insisting that, I am happy to drop the subject. I have studied what you have said, plus the WP article on the subject but unfortunately no answer is provided.  Why not just answer my question?  What are the exact criteria for a given event to be considered a condition in a probability problem?  Martin Hogbin (talk) 15:10, 23 December 2009 (UTC)
 * I explained some lines above completely what cond. prob. is about. And: the only MHP that will be of interest to, I guess, all interested readers, is the one where a door has been opened and then the player is offered the choice to switch. If you have another version in mind, please tell me. Nijdam (talk) 13:09, 24 December 2009 (UTC)
 * BTW: I laid my cards completely open to anyone. But I have no idea what your opinion is of what version or versions to consider the MHP. You mentioned somewhere: what sources say. So please tell me what sources say. And formulate a solution. This is not to try you, I will be glad in assisting you doing so. Nijdam (talk) 13:14, 24 December 2009 (UTC)

Is the problem to be answered from the player's point of view?
We are to see the problem through the eyes of the player. The host offers her the opportunity to change her choice. And, as far as I know, almost everyone (except you and ...) put themselves in the position of the player and consider the door opened. Actually we also have the same information: door 1 (or another) chosen and door 3 (or another) opened. Nijdam (talk) 13:56, 21 December 2009 (UTC)


 * This depends on the basis on which we are addressing the problem. As I understand it there are three possible bases on which probability can be dealt with: frequentism, Bayesian probability, and the more formal modern probability theory.


 * The Morgan paper seems to be promoting the modern probability theory, in which the problem must be answered strictly from information given in the question, they make this point several times in their article. What we assume that a person involved may or may not know is unimportant, we must solve the problem only on the information given in the question, which is used to assign intrinsic probabilities to events in the sample space.  Information not given cannot be used to do this.  If the problem statement does not say which door the host has opened then this data cannot be used in answering the question.  Rick and others have made the point that according to Morgan it does not matter whether the player knows about the hosts door opening policy.  This only applies if they are using modern probability theory.


 * If on the other hands we are Bayesians we can decide on whose state of knowledge we are to answer the question from. I agree the players POV is the most obvious to take.  We might also choose to make some reasonable assumptions about reality.  How a host might act, what a player might see etc.  If that is to be the case I would have to rephrase my question to make clear that the player does not see which door was opened.  Or do you insist that, even in this case the problem is conditional?  Martin Hogbin (talk) 21:54, 21 December 2009 (UTC)


 * In its simple form it may be addresed in a frequentistic or formal way, both essentially equivalent. And both asking for conditional probs. The Bayesian approach is a theoretical extension, meant for experts, in my opinion not to be mentioned in the article. But it also asks for conditional probs. Where are you heading at? Nijdam (talk) 16:38, 22 December 2009 (UTC)


 * In my problem statement I did not say which door the host opens, thus, according to your propose rule above, the door opened cannot be taken as a condition. My problem statement is exactly equivalent to Rick's first problem, the door opened is not identified in the problem statement just as the we have no information on which ball the host reveals in Rick's problem.  Of course we are free to speculate on whether the host might actually be able to distinguish between the two red balls just as we may speculate on what the player might see but neither is part of the problem as stated.   Martin Hogbin (talk) 17:21, 22 December 2009 (UTC)

What urn problem is the appropriate model
There are basically two ways to turn the MHP into an urn problem.

1) The host puts three marbles, indistinguishable except two are black (representing goats) and one is white (representing the car), in a bag. The player withdraws one without looking at it.  The host looks in the bag and withdraws a black marble and shows it to the player and the audience.  The player can now switch the marble she's withdrawn for the remaining marble in the bag.  The choice is between two marbles, one that is white and one that is black.  Should the player switch?

2) The host puts three marbles, indistinguishable except one is red (representing a goat), one is black (representing a goat), and one is white (representing the car), in a bag. The player withdraws one without looking at it.  The host looks in the bag and withdraws a losing marble, and shows it to the player and the audience.  The player can now switch the marble she's withdrawn for the remaining marble in the bag.  Let's say the host withdraws the red one meaning the choice is between two marbles, one that is white and one that is black.  Should the player switch?

I think those who are favoring the unconditional solution are thinking the first version is the most appropriate analogy. For this problem there are two scenarios


 * 1a) Player withdraws a white marble, leaving two black marbles in the bag. Host now withdraws a black marble leaving a black marble in the bag. (probability 1/3)
 * 1b) Player withdraws a black marble, leaving a white and black marble in the bag.  Host now withdraws the black marble, leaving the white one. (probability 2/3)

The second version is more complicated and seems to have the same definitional issues as the MHP. Is the question specifically about the case where the host withdraws a red marble or not? Whatever the question is, it's clear that there are two actions (initial player pick and then host withdrawal) whose probabilities combine in ways that are not immediately obvious. Enumerating the possible cases as above is not that helpful


 * 2a) Player withdraws the white marble, host withdraws the black one.
 * 2b) Player withdraws the white marble, host withdraws the red one.
 * 2c) Player withdraws the black marble, host withdraws the red one.
 * 2d) Player withdraws the red marble, host withdraws the black one.

It's perhaps easy to see that the first two of these are subcases of a case with probability 1/3 where the player withdraws the white marble and the other two cases each have probability 1/3. However, if the question really is asking about the probability given the host withdraws the red marble we clearly have a conditional probability problem.

The sources presenting unconditional solutions to the MHP are treating it as if it is equivalent to the first of these urn problems. The sources that say it should be treated as a conditional probability problem are saying it's equivalent to the second of these urn problems. This is a POV issue. Neither one is "more right" than the other. It's simply different POVs. -- Rick Block (talk) 01:04, 22 December 2009 (UTC)


 * Only Bosons are truly indistinguishable so I suspect that Nijdam might insist that even this case [the first version] is strictly one of conditional probability as one or other of the black marbles must in fact have been shown.


 * There really is no fundamental difference between these two cases, in both cases there are two goats represented by two different marbles. My point is that there is no absolute clear distinction between a conditional problem and an unconditional one, your problems above demonstrate this clearly.  This it is, to some degree a matter of choice as to how we treat a given problem.  The question that we must ask is whether the potential condition could affect the probability that we are trying to calculate. Martin Hogbin (talk) 12:55, 22 December 2009 (UTC)


 * Some sources say that the MHP should be treated as a conditional probability problem, some do not, so you are quite right it is a POV issue but there are clearly two valid POVs, both of which should be represented in the article. All I have ever asked for is that a convincing explanation the simple problem, or what I will call the non-conditional problem (meaning that the problem is either stated in an unconditional form, or that it is treated unconditionally by virtue of its inherent symmetry) is shown at the start of the article, together with  a discussion of reasons for confusion for the non-conditional case.  After that the issue of conditional probability can be discussed for those interested.  Martin Hogbin (talk) 12:43, 22 December 2009 (UTC)


 * To me the first representation is not equivalent to the MHP. I know some sources say it is. May be there even are "sources" advocating the 50/50 "solution". What ever. Anyone (almost) accepts a door has been opened. Only because one wants the simple solution to work, one tries to argue the opened door is not identifiable. Nijdam (talk) 17:05, 22 December 2009 (UTC)


 * You are quite right, I do want to use a simple model to start with, but as you can see from the two problems above there really is no sharp division between a conditional and an unconditional problem. If you disagree you must tell me exactly what makes the distinction.  Martin Hogbin (talk)


 * Martin - Nijdam has answered this repeatedly. There is a sharp difference, and it is whether the condition reduces the size of the event space.  For example in the second problem above, withdrawing the red marble means we're only talking about cases 2b and 2c.  The only way the host opening a door in the MHP does not reduce the size of the event space is if the doors are indistinguishable.  You can certainly change the problem description to make this the case (the contestant is blind and the host says he's opened a door revealing a goat but doesn't say which door, or the two unselected doors are hidden behind a single curtain and a goat is produced from behind the curtain, or the player must decide before the host opens a door, or ...).  Do you agree any of these are changes?


 * Rick - Reduction of the sample space is not an answer because it then depends on how you set up your sample space. It is also not supported by Conditional probability which says just, 'Conditional probability is the probability of some event A, given the occurrence of some other event B'.   In my example in which I clap my hands before choosing a door I can set up my sample space to include the options of clapping and not clapping.  This makes it a condition.  Martin Hogbin (talk) 19:36, 22 December 2009 (UTC)


 * Nijdam - Do you agree the first problem is NOT "conditional"?


 * Assuming we get to mediation, we'll rehash all of this there so there's really not much point in continuing this conversation here. I offered these two urn problems as a way to illustrate the difference between the approaches.  As Boris says The coexistence of the conditional and the unconditional can be more peaceful.  Ultimately, these are two sides of the same coin.  The coin is not fully described unless both sides are considered.  NPOV says we must present both, and can't favor either one. -- Rick Block (talk) 18:52, 22 December 2009 (UTC)


 * Rick, would you describe the current article as adhering closely to "we must present both, and can't favor either one."? Glkanter (talk) 19:06, 22 December 2009 (UTC)


 * You're talking about this version, and I wouldn't claim it's perfect but I think it is fairly close - I would strongly prefer a single solution section. Note that "neutral" here means that it says what the sources say fairly, in a disinterested tone.  -- Rick Block (talk) 19:51, 22 December 2009 (UTC)


 * Rick, what about the text and graphic volume of the content? Or all the time spent on so-called 'variants'? Or the talk page FAQs, which are barely sourced, more like riffed? These are all part of the 'Article', not just the Solutions section. If they were allegedly NPOV when you insisted Morgan was dominant, how can they still be NPOV when you now argue both sides should be equal? Glkanter (talk) 20:19, 22 December 2009 (UTC)
 * But that must be all the sources, without giving one source the right to veto the others. Martin Hogbin (talk) 19:56, 22 December 2009 (UTC)


 * I agree that, 'The coexistence of the conditional and the unconditional can be more peaceful', this is what I have always wanted. One solution might be to treat the sources in rough chronological order, with earlier solutions first.  As far as I know no sources before Morgan state the problem must be conditional. Martin Hogbin (talk) 19:59, 22 December 2009 (UTC)

MHP?
1. The player is informed about the rules, but still in the dressing room, before being on stage, she is asked whether she will switch. Simple solution, no need of conditional probs.


 * I consider this version as the standard MHP: (Insert your name here)


 * Nijdam - Let us assume that we know that the host will always choose the highest available door, within the rules, to open. Can you show me how you would calculate the probability of winning by switching in this case using correct notation. Martin Hogbin (talk) 19:42, 22 December 2009 (UTC)

2. The player is informed about the rules, and after she made her first choice, the host tells her he will shortly open one of the remaining doors. Before doing so, he asks her whether she likes to switch to the door he will leave closed. Simple solution possible, although one formally has to condition on the player's choice.


 * I consider this version as the standard MHP: (Insert your name here)

3. The player is informed about the rules, and after she made her first choice, the host opens one of the remaining doors. He then asks her whether she likes to switch. The solution needs to consider the conditional probability. (And I add: the player knows which door she has chosen and sees which door has been opened, hence the solution is a function of these door numbers.Nijdam (talk) 10:06, 23 December 2009 (UTC))


 * I consider this version as the standard MHP: (Insert your name here)
 * Nijdam

4. Any other scenario.


 * I consider another version as the standard MHP: (Insert your name here)

For me only the 3rd form is of interest. Nijdam (talk) 17:39, 22 December 2009 (UTC)

Just out of curiosity and to know where everyone stands, I invite all the participants in the discussion to just mention there name.Nijdam (talk) 10:17, 23 December 2009 (UTC)
 * Nijdam, I do not think that you will get a good response to your request. As I am sure Rick will agree, it is not up to us to determine what the 'standard MHP' is, or even if there is such a thing, we must rely on sources.  On the other hand I do not know of any source that says 'This is the standard MHP', we therefore have some freedom to decide what the problem is and how it should be answered.  My suggestion is that we should use this freedom to start with a formulation that allows us to provide a simple, non-conditional solution.  We can then discuss other formulations and their solutions later.  So, to continue with a question you asked elsewhere: yes, I do want to change the question to suit the answer but I believe we can do this and still maintain the quality of the article. Martin Hogbin (talk) 10:43, 23 December 2009 (UTC)
 * I know it is not up to us to determine, yet I'm interested in were you and others stand. Nijdam (talk) 14:12, 23 December 2009 (UTC)
 * I cannot answer the question as asked as there clearly is no 'standard' formulation. I would say we should start with any non-conditional formulation (certainly this is the notable case) and then proceed to a conditional one.  This is in line with the sources, some of which treat it one way and some the other. Martin Hogbin (talk) 17:44, 23 December 2009 (UTC)


 * Nijdam - You now seem to be using time to determine what is conditional and what is not, can you show any reference that states that this is what makes an event a condition. In the first two examples we know the host will have to open a door and that it might matter which door he opens.  Why can we ignore this just because it will happen in the future? Martin Hogbin (talk) 19:42, 22 December 2009 (UTC)

Here's An Opportunity:
Convince me that Huckleberry could have done better if someone had explained the 'equal goat door constraint must equal 1/2' to him. Or, that 'maybe it didn't equal 1/2'.

I'll become open to compromise, and I'll stop believing that some editors are being obstructionist.

On the other hand, if no one comes forward with a convincing argument, my views will remain unchanged. Glkanter (talk) 13:15, 23 December 2009 (UTC)
 * Okay then, an oppoprtunity for you: how do you know Huckleberry wins 2/3 of the times?Nijdam (talk) 14:16, 23 December 2009 (UTC)


 * Various reliably published sources say so, some using the Combining Doors solution, like Devlin. Glkanter (talk) 14:46, 23 December 2009 (UTC)


 * Nijdam, please don't treat this as you giving me a test I must pass. I think prominent Wikipedia editors have answered all your questions and challenges about symmetry, and whatever else, and I'm not aware that the 2/3 is even being debated. I don't think it's necessary for each editor to 'prove' to your satisfaction his or her master of Probability. This is the MHP paradox article, only. Please respond only about the Huckleberry section as I originally started this section, thank you. Glkanter (talk) 15:02, 23 December 2009 (UTC)


 * Lets say Huckleberry is not a contestant but works for the FCC. His job is to make sure the show is fair.  He doesn't know much about probability so looks up the problem on Simplepedia.  The article has the "combining doors" solution and various other "simple" explanations that convincingly say that 1/3 of the time you should win by staying and 2/3 of the time you should win by switching.  He watches the show.  He keeps track of how many times players who switch win and how many times players who stay with the first choice win.  He sees 3000 shows in which 2700 players stay with their original choice and only 300 switch (people think it doesn't matter so most stay with their original choice).  Of the 2700 "stickers" 900 win, and of the 300 switchers 200 win.  1/3 of the stickers win and 2/3 of switchers win.  Perfect.  He sleeps soundly, convinced everything is OK.


 * But, it's a game show, so it's run by humans. It turns out what's really going on is they perfectly randomize the initial car placement but the host is told only "open one of the other doors if the player initially picks the car".  The host figures it doesn't matter which one he picks in this case, so doesn't pay any particular attention to it.


 * Huckleberry's boss also doesn't know much about probability and looks up the problem on Wikipedia which has a reference to the Morgan et al. paper and says that how the host picks between two goats matters. The boss thinks about it and decides to see if Huckleberry is really doing his job.  The boss looks at tapes of the show and keeps track of the 6 individual pairs of player pick and door the host opens separately, i.e. player picks door 1 host opens door 2, player picks door 1 host opens door 3, etc.  The boss determines that even though 2/3 of the players who pick door 1 win by switching (and 1/3 win by staying) players who pick door 1 and see the host open door 2 would win by switching with probability about .625 (rather than .66) and players who pick door 1 and see the host open door 3 would win by switching with probability about .714 (rather than .66).  Using the Morgan et al. formula he found on Wikipedia, the boss works backwards, and solves .625=1/(1+p) and .714=1/(1+q) and sees that instead of p=q=1/2 the host is exhibiting a p=.6 preference for the leftmost door and a q=.4 preference for the rightmost door.  That's odd, he thinks.  Exactly 2/3 of the players who switch win and exactly 1/3 of the players who stay win, but yet the host is exhibiting a slight preference for one goat door over another.


 * Looking into it some more, the boss notices that the host seems to open door 2 more often on Mondays, Wednesdays, and Fridays, and less often on Tuesdays and Thursdays. The boss talks to the host and the host says to make it "random" he always picks the leftmost door (if he has a choice) on Mondays, Wednesdays and Fridays, and the rightmost door on Tuesdays and Thursdays.


 * Huckleberry is fired.


 * All because Simplepedia never said anything about how much it matters how the host chooses between two goats when he gets the chance. -- Rick Block (talk) 20:21, 23 December 2009 (UTC)


 * Let's limit this to only those variants consistent with 'Suppose you're on a game show', and answers that are intended for the typical Wikipedia reader. The FCC has plenty of books they can rely on. Thanks. Glkanter (talk) 20:51, 23 December 2009 (UTC)


 * In your opinion Rick, are the explanation and the risk you described above your reasons for advocating for 'equal treatment' in the article of the simple solution and Morgan's solution? Glkanter (talk) 23:31, 23 December 2009 (UTC)


 * No. My reason for advocating for 'equal treatment' is because the simple solution and the "conditional" solution are both significant POVs published in reliable sources.  How about if we just leave this discussion for mediation? -- Rick Block (talk) 05:24, 24 December 2009 (UTC)


 * Rick forgot to mention the show where the car is placed behind door 1 on Mondays, door 2 on Tuesdays and door 3 on Wednesdays, the only days that the show is on. Huckleberry, who knows all about the effect of host preference from reading the paper by Morgan is still puzzled to discover that the chances of winning by switching are not what he expects.  Finally he wakes up from his dream to watch the real show where the contestant has no idea of either the host preference or the car placement so he just chooses randomly and wins 2/3 of the time by switching.  Martin Hogbin (talk) 00:03, 24 December 2009 (UTC)

In formulas

 * 2/3=P(Winning by switching)="average over the possible scenario's"=
 * =SUM P(winning by switching|scenario)P(scenario)

If we know P(winning by switching|scenario) is the same for every scenario, we may conclude that for every scenario:
 * 2/3=P(winning by switching|scenario)

And note: we decide on the basis of the conditional probability.Nijdam (talk) 18:08, 26 December 2009 (UTC)

For better understanding
Much of the confusion lies in the formulation. So, for a better understanding, I give you the following consideration:

What is the probability the car is behind the chosen door? Most people will not hesitate to answer: 1/3, assuming of course the car is placed randomly. And they are right, numerically, but do they understand the situation? Let us put the car half the times behind door 1, and 1/4 of the times behind each of the other doors. Again: what is the probability the car is behind the chosen door? Answer? Nijdam (talk) 10:34, 27 December 2009 (UTC)


 * If the choice is random without knowledge of which door is which, then from the player's state of knowledge 1/3. This is (of course!) a conditional probability problem.  The player's random choice has a 1/3 chance of being each door, so the player's chance is (1/3)(1/2) + (1/3)(1/4) + (1/3)(1/4) = (1/3)(1/2 + 1/4 + 1/4) = 1/3.  Uniform distribution means the choice might as well be random.  Random choice means the distribution might as well be uniform.  -- Rick Block (talk) 18:19, 27 December 2009 (UTC)


 * I never had any doubt you wouldn't know the answer. The point is of course, we don't know the distribution of the choice. Nijdam (talk) 23:22, 27 December 2009 (UTC)