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The conditional solution table

To model the host's behaviour opening door 2 or door 3, if he has the choice, completely we consider The solution shows the six possible equally likely arrangements of one car and two (distinguishable) goats behind three doors. and The table shows the result of staying or switching after initially picking door 1 in each case. In the first case the host opened door 2, in the fourth case door 3, according to the rules of the game show:

Now we have to pay attention to With respect to the final situation in the game show we consider the cases with door 3 opened, corresponding to the rows 2, 4 and 5. according to the final situation in the game A player who stays with the initial choice wins in only one out of three of these equally likely possibilities, while a player who switches wins in two out of three. The probability of winning by staying with the initial choice is therefore 1/3, while the probability of winning by switching is 2/3. We can see that the cases with door 2 opened are giving the same result and we are able to observe a symmetry between the non-chosen doors 2 and 3. --TotalClearance (talk) 09:03, 3 November 2012 (UTC)

So the goats are called "Goat 1" and "Goat 2" and the difference between the bottom and top halves of the table is the order of Goat 1 and Goat 2 behind the doors. And the host's rule is that he opens the door of "Goat 1" if he has a choice: the left hand goat in the first three rows, the right hand goat in the second three rows. The host can distinguish between the goats but the player can't (or rather: doesn't, probably because in advance he doesn't know the two goats personally, nor the rule being used by the host). Then indeed there are six equally likely possibilities, with Door 2 and Door 3 being opened by the host equally often. Given that Door 3 is opened, there are three equally likely possibilities. Switching wins the car in 2 our of 3 cases, both unconditionally and conditionally.

Note that TotalClearance says, dogmatically, we have to pay attention to the cases with door 3 opened according to the final situation in the game. Not everyone agrees with this (neither editors, nor readers, nor sources). The conclusion of the RfC is indeed that in the first part of the article we are not going to pay attention to this issue.

I have not seen this solution before. It's a cute and I think original "conditional solution". Richard Gill (talk) 09:20, 3 November 2012 (UTC)

Why do you need all that? This is all you need. We know the door numbers are unimportant so everything else is certainty.

Martin Hogbin (talk) 10:08, 3 November 2012 (UTC)

TotalClearance is proposing what he considers to be a "simple" conditional solution. It gives you a stronger conclusion than your table, Martin, but you have to do more work to get it. You could also add to your table, Martin, the comment that the door numbers are irrelevant by symmetry, and then you would also get the conclusion "given the host has opened Door 3, switching gives the car with probability 2/3". The more you put in, the more you get out. No free lunches.

But in view of the conclusion of the RfC, TotalClearance is going to have to try to get his preferred solution somewhere later in the article. And he's going to have to source it. Richard Gill (talk) 10:40, 3 November 2012 (UTC)

What exactly is stronger about the conclusion you reach from TotalClearance's table. We know door numbers and goat ID are irrelevant so we can conclude from my table that "given the player has originally chosen door 1 and the host has opened Door 3, switching gives the car with probability 2/3". The door numbers are discarded at the start.

One can *argue* that door numbers are irrelevant and then one can *conclude* what you just wrote. But you did neither. TotalClearance put in more assumptions and got out a stronger conclusion. You put in less assumptions and got a weaker conclusion. I think it's a matter of opinion which solution is better. TotalClearance on the other hand seems to think that you have to do it his way. His solution seems to me to be quite original. Richard Gill (talk) 15:31, 3 November 2012 (UTC)

TotalClearance's table chooses to distinguish between the two goats but does not include all the possibilities. For example, there is no cases where Goat 2 is revealed behind door 2. If we are going to distinguish between goats then we must have a complete table showing all the possibilities. THis will be quite complicated, bearing in mind the next problem with the table.

Yes there is. It's the third of the six possibilities (six permutations of three distinguishable objects: Car, Goat 1, Goat 2). If Goat 1 is behind door 1 and Goat 2 is behind door 2 then the host will be forced to open door 2 and reveal Goat 2, even though he prefers to reveal Goat 1. Richard Gill (talk) 15:31, 3 November 2012 (UTC)

Sorry, I should have said that there is no case where the player has originally chosen the car and Goat 2 is revealed behind door 2. Richard's version of the table can be found as the first tablehere for reference.

Yes. Of course. TotalClearance assumes (I think) that the host prefers to reveal Goat 1. If the car is behind door 1, the order of items behind the three doors is Car, Goat 1, Goat 2 in the first half of table (row 1) and Car, Goat 2, Goat 1 in the second half of the table (row 4). In the first case the host opens door 2, in the second case he opens door 3. In neither case does he reveal Goat 2. If the player knew about this, and knew the two goats in advance (goats are distinguishable, right!?) then the host's behaviour would give further information to the player, changing his probabilities that the other door hides the car, yet again. The player would not only condition on which door was opened but which goat he saw there, and get probability 1/2 in one case, 1 in the other case. Averaging out - when correctly weighted - to 2/3. Richard Gill (talk) 07:53, 4 November 2012 (UTC)

TotalClearance has also fallen for the Morgan conjuring trick. There is nothing special about the door number opened by the host. If we decide that door numbers are important (and goats are not) than we must show all the possible doors that the player might have initially chosen. That would result in this table.

You can argue there is nothing special about the door number opened by the host but you didn't - you took it for granted. This is nothing to do with some conjuring trick. TotalClearance wants to know what is the chance that the car is behind door 2 when you chose door 1 and the host opened door 3. (So he's not interested in two-thirds of the big table which you put below here: the player didn't choose door 2 or door 3). People like TotalClearance think that that is how the question has to be answered. Some editors think that (I don't, by the way), some writers think that, some readers will think that. The result of the RfC is that people like TotalClearance are going to have to be patient, and do their thing later in the article. Richard Gill (talk) 15:39, 3 November 2012 (UTC)

You still have not spotted Morgan's sleight of hand. The number of the door that the host opens is no more important than the door number that the player originally chooses. We know that neither makes any difference but the pedantic way of doing things is to show all the possible doors that the player might have chosen and all the doors that the host might have opened and then condition on the doors mentioned in the question. The intuitive way is to accept that door numbers are irrelevant, thus arriving at my table. Martin Hogbin (talk) 01:02, 4 November 2012 (UTC)

Just like the other table, this reduces to mine. If we want to distinguish between goats then the table gets more complicated still. How does this help our readers? Martin Hogbin (talk) 14:11, 3 November 2012 (UTC)

In case of vos Savant's solution, row 3 with the car behind door 3 doesn't fit to the standard problem conditions and in my experience many people argue consequently, that the odds must be the same for door 1 and 2 referring to the rows 1 and 2 being equally likely and row 3 being impossible. But we know that this reasoning is wrong...

The conditional solution table can be considered as a "mental model representation" of the MHP as described in table 1, page 5, in Krauss' and Wang's paper. They say:"Most representations found in the literature consist of more than three single arrangements and specify Monty Hall’s behavior in each arrangement." Such an illustration often is used also if you have several arrangements which are not equally likely. In that case it is very helpful to expand the respective table to all arrangements being equally likely. So you are able to solve a problem by counting off the matching cases. This is an easy and intelligible way for ordinary people to understand conditional probability and meets didactic purposes.

IMO this solution table should be given to the reader directly after the formulation of the standard problem because it fits perfectly. Since the result of the RfC is not a law, with the help of a few editors here we can insert this table early in the article. --TotalClearance (talk) 22:12, 3 November 2012 (UTC)

The result of the RfC represents a consensus after years of arguing here. Content is determined by consensus. Martin Hogbin (talk) 01:06, 4 November 2012 (UTC)

IMO the result of the RfC is not a consensus but a decision of a small majority, which could be changed in certain cases. --TotalClearance (talk) 08:24, 4 November 2012 (UTC)

That may be your opinion but, as I have said, we have a conclusion to years of argument. Please do not start it again.

Anyway: TotalClearance's solution is completely original, as far as I know! Congratulations. Do you agree with my additions to your table? I put it also on my talk page at . TotalClearance also says, tellingly, "This is an easy and intelligible way for ordinary people to understand conditional probability and meets didactic purposes". I think that TotalClearance is interested in using MHP in a probability class, in order to explain conditional probability. But the wikipedia page on MHP is (I think) mainly about explaining MHP to a very wide audience, not about explaining conditional probability to students in a probability class. What the trained probabilists tend to forget, is that there are other ways to explain why staying with your initial choice is an unwise strategy if you want to get the car. Richard Gill (talk) 07:53, 4 November 2012 (UTC)

I am interested in the phenomenon that people are arguing wrong if they apply vos Savant's solution to the standard problem, with door 1 chosen and door 3 opened. Expanding her table for modelling the host's behaviour completely helps people to understand what happens in the game show. Furthermore this conditional solution table is the next logical step following vos Savant's solution, and it doesn't need any elaborated arguments like strategy, symmetry and "combined doors" to solve the problem. --TotalClearance (talk) 08:55, 4 November 2012 (UTC)

OK, now i understand your solution. It's the same as the decision tree in the article, but in which you split the two lower branches (each of probability 1/3) into two identical branches, each of probability 1/6. As if the host tosses a coin to aid his decision which door to open whether he has a choice or not. The top branch of each pair is heads, the bottom branch is tails. If he has a choice, then heads means open door 2 and tails means open door 3. Instead of having four distinct outcomes, with probabilities 1/6, 1/6, 1/3, 1/3, you have 6, all with probability 1/6. This is neat and a useful pedagogical device. Much better than (what seems like) pages of formulas, in the two Bayes sections (which in my opinion should both be deleted). Richard Gill (talk) 14:48, 4 November 2012 (UTC)

I do not see how that table is helpful. Lines 2 and 5, for example, are identical (as you are not distinguishing between goats). So this represents the host tossing a coin to decide which door to open. If it comes up heads he opens door 3 and if it comes up tails he opens door 3. Two distinct outcomes? Crazy! Maybe if I toss a coin and it comes up heads I become a different person. Martin Hogbin (talk) 21:31, 4 November 2012 (UTC)

Decision tree: 1/3 = 2/6, isn't it? --TotalClearance (talk) 22:14, 4 November 2012 (UTC)

Yes 1/3 = 2/6 = 7/14 = 243/486 So what? Where do the two distinct outcomes come from? Looks like double Dutch to me. Martin Hogbin (talk) 22:21, 4 November 2012 (UTC)

What is the problem? The aim is to show that it makes no difference at all which of those three door has been first selected by the guest, nor which one of "his two doors" has been opened by the host. Sources say that there is "no difference". The scenario is described and known: The guest first selected his door, and thereafter the host is to open one of his two unselected doors with a goat behind and to offer switching. Now it is a fact that there are many various weighty reasons to switch, nobody "must" only base on maths. The second goat and the car (or vice versa) are now behind the two still closed doors: the guest's door and the still closed host's door. As the modus operandi of the host, in case he has got two goats, forever will be completely unknown in this one time problem, it is completely useless to jump to conclusions based on "which one" of his two doors he opened. Doing this anyway, that may serve for purposes of maths only, but never to support the decision asked for. Using conditional probability in maths classes is out of position to ever be able to decide whether – in the actual problem – the probability to win by switching "could be" between , or "could be" between . Any of such assumption is completely worthless for the actual problem and for the actual decision asked for. Sources say therefore that the host's opening of his nearby door or of his distant door is "all the same". For the reader, it should clearly be shown that this makes "no difference". Use decision trees or various different tables, all of this will be helpful for the reader to "see" that it makes no difference "which" door the host has opened. And use the "combined door" visualization. Redundancy is most welcome to achieve this goal. Gerhardvalentin (talk) 00:05, 5 November 2012 (UTC)

The aim is to show that under K&W conditions the player should switch because the probability of winning by switching is 2/3. It is but a step from vos Savant's illustration, stated as the first simple solution, to the solution table which is solving "the full problem":

The first simple step is to expand the row 1 according to the host's behaviour, if he has a choice, similar to the decision tree:

The next simple step is to expand the remaining rows, too. Because of 1/3 = 1/6 + 1/6:

Therewith you have the correct solution table for the situation: door 1 chosen. Analoguous the tables for the other situations have the same appearance, with other doors chosen and other doors opened. But we don't need them, because of the car is initially equally likely to be behind each door, the player's initial choice doesn't change anything on principle. --TotalClearance (talk) 17:10, 6 November 2012 (UTC)

Neither does the host's door choice change anything, obviously.

What event exactly does the probability of 1/6 in line 3 refer to? Martin Hogbin (talk) 23:29, 6 November 2012 (UTC)

It doesn't refer to any event. As mentioned before, you can make this work by letting the host always flip a coin, and then add "coin flip result" column in the table.

Both of you are saying "doesn't change anything" without specifying what that means. That's not useful. TotalClearance's argument can be made to work by saying something like "The proofs that switching wins with probability 2/3 given any other initial choice of door and choice of host's door are similar." at the end, or "Assume without loss of generality that the player initially picked door 1 and the host opened door 3." at the beginning. Either way, the meaning is that a proof template has been provided, and that template works for any door choices with trivial changes. I do not see any such concrete interpretation for your words. TotalClearance's table shows that the probability is 2/3 for one particular case, but your first table does not show that it is 2/3 for any of these cases. If you throw in mumbo-jumbo about the law of total probability, you do end up showing that the average of the probabilities for all possible combinations of doors chosen is 2/3, but that still doesn't show that it is 2/3 for any of them, only that their average is 2/3; to finally show that they are all 2/3 requires even more mumbo-jumbo. -- Coffee2theorems (talk) 19:06, 9 November 2012 (UTC)

Distinguishable goats?

If you really want to distinguish the goats then vos Savant's solution is incomplete respectively you have to determine which of the possible arrangements is occuring in her solution. For example:

behind door 1	behind door 2	behind door 3	result if staying at door #1	result if switching to the door offered
Car	Goat 1	Goat 2	Car	Goat 1 or Goat 2
Goat 1	Car	Goat 2	Goat 1	Car
Goat 1	Goat 2	Car	Goat 1	Car

If the player knew about this arrangement, and knew the two goats in advance, in case of the host showing Goat 2 he should switch, in case of the host showing Goat 1 the player should stay. Then his odds of winning the car are more than 2/3. --TotalClearance (talk) 09:53, 4 November 2012 (UTC)

Distinguishing between goats certainly complicates things. Martin Hogbin (talk) 10:46, 5 November 2012 (UTC)

Brilliant solution (6 equally likely cases)

I think TotalClearance's idea for making the conditional probability solutions simple and transparent is brilliant. Imagine the host has a preference to open door 2 or door 3, and his two preferences are equally likely. Then (given the player chose door 1) we have 6 equally likely cases:

• car is behind door 1, host prefers to open door 2, opens 2
• car is behind door 1, host prefers to open door 3, opens 3
• car is behind door 2, host prefers to open door 2, opens 3 (forced)
• car is behind door 2, host prefers to open door 3, opens 3
• car is behind door 3, host prefers to open door 2, opens 2
• car is behind door 3, host prefers to open door 3, opens 2 (forced)

As far as the player is concerned, nothing has changed. What the player gets to see, and the probabilities thereof are the same. In three of the six equally likely cases, the host actually opens door 3 (once forced, twice by choice). Of those three occasions, two times the car was behind door 2, once it was behind door 3. So the probability of winning by switching given the host opened door 3 is 2/3.

A brilliant solution, hopefully already known in the literature, and completely in line with Kraus and Wang's advice to do probability with frequencies, not relative frequencies. Better still, by counting equally likely cases out of admissible cases. Richard Gill (talk) 13:37, 18 November 2012 (UTC)

What preference is it the host has? I do not understand what it means the host has a preference to open door 2 or door 3. Seems to me something like: do you want Pepsi Cola or Coca Cola? Oh I prefer Pepsi or Coca. Nijdam (talk) 21:27, 18 November 2012 (UTC)

If the car happens to be behind door 1, the host will either open door 2 or 3. I call that his "preference". I don't know it, so for me either preference is true with probability 1/2. I imagine him having this preference (to open door 2 or door 3) independently of where the car actually his. He can follow his preference if the car is behind door 1. Otherwise, his choice is forced.

I'm trying to make the conditional solution easier for people who are not mathematicians, who didn't learn probability theory. So it doesn't matter if you Nijdam don't understand this. The question is whether someone like Martin or Gerhard understands it.

You as a mathematician understand that if I have a probability space in which some outcomes (little omega) have probability p and others have probability 2p, I am allowed to split each of the "double" outcomes into two outcomes of probability p. Now I have a bigger probability space (big Omega) on which all the old events and their probabilities are still defined, unaltered. But now I can compute probabilities by counting.

Use your imagination, man! Richard Gill (talk) 09:46, 19 November 2012 (UTC)

The six equally likely cases solution can be found in Jason Rosenhouse's (2009) book, page 54, and is attributed to Steven Krantz (1997), Techniques of Problem Solving, Publisher: American Mathematical Society, Providence. It's also in Kraus and Wang where it is attributed to Johnson-Laird, P. N., Legrenzi, P., Girotto, V., Legrenzi, M. S., & Caverni, J. P. (1999). Naive probability: A mental model theory of extensional reasoning. Psychological Review, 106, 62–88. So mathematicians recommend this little trick as a way of making probability problems more simple, and psychologists recommend this this little trick as a way of making probability problems more simple. Yet neither Martin (a simplist) nor Nijdam (a conditionalist) can make any sense of the solution at all. I can't see how to explain it any better. Sigh... Richard Gill (talk) 17:29, 22 November 2012 (UTC)

Conditional solution diagram

Should the diagram of Morgan's conditional solution have some introductory text, similar to that for the decision tree, for example: 'Diagram showing probability of every possible outcome if the player initially picks door 1'. Maybe we should also add, 'Similar diagrams exist for the cases where the player initially chooses door 2 or door 3'.

This is not Morgan's conditional solution! It is Selvin's (1975b) solution. (A few lines of completely elementary, direct calculation; absolutely routine for anyone who knows a tiny bit of probability theory).

Why consider the other cases? Keep it simple. Richard Gill (talk) 18:32, 15 November 2012 (UTC)

The diagram may be confusing to some because it contains no explicit renormalisation of the conditioned sample space. the reader sees two figures 1/6 and 1/3 and has to know how to obtain the required probability from these. This is shown but not explained in the text above. Might there be some advantage in clarity in showing the results of a large number (say 900) trials in this diagram? This would leave the more natural explanation that of the 450 trials in which the host has picked door 3, the player who switched wins 300 times. Maybe it would be better to consider just 6 trials or combinations? Martin Hogbin (talk) 16:36, 14 November 2012 (UTC)

Is there a case for another diagram showing what happens if the host does not choose evenly. Martin Hogbin (talk) 16:38, 14 November 2012 (UTC)

I don't see the problem. The present text has "Assuming the player picks door 1, the car is behind door 2 and the host opens door 3 with probability 1/3. The car is behind door 1 and the host opens door 3 with probability 1/6. These are the only possibilities given the player picks door 1 and the host opens door 3. Therefore, the conditional probability of winning by switching is (1/3)/(1/3 + 1/6), which is 2/3." Do you mean that this needs to be further explained?

Yes that is exactly what I mean. You know about these things and have automatically normalised the probabilities. The diagram is surely not aimed at the expert or even the student of probability. For them, the explanation in proper notation that you have given below is better. The diagram is aimed at the interested general reader, they will see 1/3 and 1/6 and from those figures must calculate a figure of 2/3. Surely it is easier in natural language to say something like, 'in 2 out of the 3 cases that met the conditions specified in the question the player who swaps will win'. I am not going to fight over this, I was rather hoping that those who were so adamant that the conditional solutions were presented in the article would comment. Martin Hogbin (talk) 20:00, 15 November 2012 (UTC)

Here is some such explanation: The calculation is done by using the fundamental rule P(A|B)=P(A&B)/P(B), in words: "probability of A given B is the probability that A and B occur together, divided by the probability that B occurs". In this case, A="car behind door 2" and B="host opens door 3"; and the probabilities (P) are all computed given the player's initial choice (door 1). Therefore, P(B) = P(host opens door 3) = P(host opens door 3 and car is behind door 2) + P(host opens door 3 and car is behind door 1) = 1/3 + 1/6 = 1/2 because "host opens door 3 and car is behind door 2" and "host opens door 3 car is behind door 1" are mutually exclusive events which together make up "host opens door 3".

Filling in yet more details: of course, P(host opens door 3 and car is behind door 2) = P(host opens door 3 | car is behind door 2) x P(car is behind door 2) = 1 x 1/3 = 1/3, while P(host opens door 3 and car is behind door 1) = P(host opens door 3| car is behind door 1) x P(car is behind door 1) = 1/2 x 1/3 = 1/6. Two more instances of the same fundamental rule as before, now in the form: P(A&B) = P(A) x P(B|A). Richard Gill (talk) 18:22, 15 November 2012 (UTC)

PS at the end of the section is the rather odd reference to Behrends. I read this source and expanded this sentence as follows: "Behrends (2008) concludes that 'One must consider the matter with care to see that both analyses are correct'; which is not to say that they are the same. One analysis for one question, another analysis for the the other question".

Behrends does not say that the two analyses are somehow the same. He is at pains to emphasize that they have different aims. Each achieves its own aim in an appropriate (correct) manner (according to Behrends). Richard Gill (talk) 18:37, 15 November 2012 (UTC)

PPS. I still don't see the point of later in the article rewriting this little calculation in huge scary mathematical formulas, after introducing a heap of notation. That just doesn't belong in an article on Monty Hall Problem. In an elementary text book on probability theory for university mathematics courses, after introducing notation, giving Bayes' theorem etc, you might like to show how it all works out in a simple example. The present article is an article about Monty Hall Problem. Not an article about formal probability theory. Just refer to the article on Bayes Theorem, in which MHP is used as a cute example! Richard Gill (talk) 18:49, 15 November 2012 (UTC)

I amplified the text next to the decision tree and the caption to the decision tree. Is that clear enough, now? Richard Gill (talk) 19:27, 15 November 2012 (UTC)

I would not use the word 'renormalise' it is too technical. Martin Hogbin (talk) 09:32, 16 November 2012 (UTC)

Please suggest an alternative expression. To get conditional probabilities, one declares all probability zero, for outcomes not satisfying the condition. The probabilities of the remaining outcomes no longer add to one (they add to the probability of the condition). We divide by this sum to make them add to one again. The technical word for this is renormalization. The text implicitly explains exactly what it is - multiplying everything by the same number, 1/sum, so that they remain in the same proportion, but now adds to one. (This is why odds are convenient - no need to renormalize). Richard Gill (talk) 12:27, 16 November 2012 (UTC)

It is hard to think of anything simple with probabilities, that is why I changed the pictorial diagram to show plays of the game. We can then intuitively normalise by saying that the player wins by switching 2 times out of 3.

TotalClearance had a very clever suggestion, precisely for this purpose. Let me present it in my own words. Imagine that the host always tosses a fair coin. If he needs to choose a door to open, he consults his coin (eg: heads and tails correspond to lower numbered, higher numbered door). Given the player initially chooses door 1 there are 6 = 3 x 2 equally likely cases: 3 locations for the car, times 2 outcomes for the coin toss. Make the table. Or if you like, the decision tree, but with 6, not 4, endponts. In 3 of those 6 cases, the host opens door 3. Of those 3 cases, the cars is behind door 2 twice and behind door 1 once. So 2 out of 3 times that door 3 is opened, switching to door 2 wins the car.

People don't gain any understanding from doing arithmetic with fractions. But they do understand counting. Kraus and Wang emphasize this as a succesful strategy for solving this kind of problem: from fractions to whole numbers. And counting equally likely cases is, I think, more simple and direct and intuitive than imaging 600 repetitions, say, in about 300 of which.... Small numbers are better than bug numbers. But talking about 6 repetions is not helpful. Talking about 6 equally Ikely cases is. Richard Gill (talk) 14:16, 17 November 2012 (UTC)

In the decision tree we would have to say something like considering only the half 1/2 (1/6 + 1/3) of the total cases where the host has opened door 3, of those 2/3 the player wins by switching. Not very good English or very clear. Martin Hogbin (talk) 10:00, 17 November 2012 (UTC)

I have boldly added a counting argument to the discussion of the decision tree, inspired by TotalClearance's table of six equally likely cases, and following Kraus and Wang's advice to calculate chances by plain counting, not by arithmetic with fractions. References also now added (I chose two out of the at least four citations possible). Richard Gill (talk) 15:00, 19 November 2012 (UTC)

I think the diagram is awful now, but I will leave the decisions on the conditional diagrams to those who think such things are necessary in the first place. Martin Hogbin (talk) 08:45, 23 November 2012 (UTC)

Since I put in the "six equally likely cases" explanation as a separate subsection (proof by expansion of Vos Savant's table) we could just as well remove the splitting of one third into two times one sixth from the conditional probability diagram, again. But I agree: those who think this diagram is useful are the ones who should be doing such modifications (or not). I tried to make the diagram more useful by leading it to six equally likely cases. But if this doesn't work, we shouldn't do it.

Whether or not you like conditional solutions, the question is, does the diagram help you to understand them? Does the diagram help people like Martin and Gerhard at least to understand what the conditionalists are talking about? Richard Gill (talk) 12:15, 23 November 2012 (UTC)

I presume that you are talking about Gerhard and myself when we first encountered the problem. The strange thing is that all this conditional stuff is much easier to understand that the problem itself, and far less interesting. The question that you should be asking yourself is, 'Do the conditionalists know what I (and Gerhard) are talking about?'. Speaking for myself, I found that once I had got my head round the basic puzzle, and things like why it matters that the host knows where the car is, the conditional solution was quite simply explained by considering the case where the host does not choose evenly. You can then, very pedantically (and inconsistently), insist on treating standard case in the same way.

To come back to the question of what pictorial diagrams will help explain the conditional solutions best I think you still need to ask who exactly is the audience for them. I think it is vanishingly small. At one end of the scale we have those who have just come to terms with the basic puzzle (or maybe not even done that). They are unlikely to be interested in making it any more complicated at all. At the other end of the scale we will have experts, who surely will not need the pictures. Then we have students of elementary probability. I would have thought that some standard structure with which they might be familiar (like the tree diagram) would be best for them. So, my honest opinion is that they are unnecessary, but I am aware that some people insist on having them, and that this was agreed to, so have them we must. Martin Hogbin (talk) 15:07, 23 November 2012 (UTC)

Yes that's what I meant. And yes, I agree. My guess is that the tree and the table are enough, the diagram is superfluous. And yes, to explain the solution, pedagogically it is smart to start with a biased host. Maybe we can try a rewrite on these lines in a few weeks. Richard Gill (talk) 08:23, 26 November 2012 (UTC)

I think we should wait and see what supporters of the 'conditional' solutions think before we make any major changes. As you know, my personal preference would be to treat the 'conditional' solutions in a slightly more 'academic' way with tables and formulae etc but some may consider this to reduce their prominence in the article, which we agreed not to do. Martin Hogbin (talk) 09:33, 26 November 2012 (UTC)

Changes made

I have been bold and made changes to the 'conditional' diagram to explain exactly what I mean. Comments, improvements and reversions welcome, especially from the 'conditionalists'. Martin Hogbin (talk) 10:01, 16 November 2012 (UTC)

I've been bold also, and reverted your changes. The diagram does not show six plays. Nijdam (talk) 11:31, 16 November 2012 (UTC)

By all means revert, but you have only reverted half of my changes. I changed the table to show the average result of 6 plays.

Do you see my point though? In the original, the reader sees two figures at the bottom of the unshaded part of the diagram showing 1/6 and 1/3, and from these they must deduce that the probability of winning by switching is 2/3. No doubt, normalising the conditioned sample space is second nature to people like yourself and Richard but I think the general reader will find it more natural to use whole numbers. Martin Hogbin (talk) 15:02, 16 November 2012 (UTC)

Okay, I'll have a look at it. Nijdam (talk) 20:20, 16 November 2012 (UTC)

How about adding one row in the conditional probability table just after the row with four cells giving the probabilities 1/3, 1/6, 1/6, 1/3.

Let the next row have six cells all with probability 1/6. In other words, split both of the two 1/3's.

Now we have six equally likely cases and we can just count. In 3 of the 6, door 3 is opened. Of those 3 cases, in 2 the car is behind door 2, in 1 the car is behind door 1. Richard Gill (talk) 15:18, 17 November 2012 (UTC)

I boldly made these changes. OR? Or simple arithmetic. In any case, following Kraus and Wang's advice (do probability by counting, not by arithmetic on fractions). Richard Gill (talk) 15:51, 17 November 2012 (UTC)

Not OR! The six equally likely cases solution can be found in Rosenhouse's book, page 54, and is attributed to Steven Krantz (1997), Techniques of Problem Solving, Publisher: American Mathematical Society, Providence. It's also in Krauss and Wang, 2003, and they have yet another reference to this solution. Richard Gill (talk) 13:27, 19 November 2012 (UTC)

The first sentence of the "conditional probability" section had a frequentist slant: "assuming the host chooses evenly" means, is think, that we imagine many repetitions and an "unbiased" (non-signalling) host. I added the Bayesian version: because we know nothing of how the host chooses, either choice is equally likely for us.

it's my opinion that a lot of the confusion around MHP is due to the fact that different authors have a different preferred interpretation, though often only implicit, not explicit. EG Kraus and Wang, and Morgan et al., always use a frequentist story in explaining probabilities. For subjectivists a different story has to be told. And for a subjectivist, the irrelevance of door numbers can be argued in advance, before looking at probabilities. Hence so many people's mystification about the need for a conditional solution or difference with a simple solution.

Well: this is my personal opinion; I think there is a gap in the literature on this point. Nobody (except yours truly, who therefore has a conflict of interest) has written explicitly about the matter. Anyway, I hope the subjectivist line, which I added, makes the conditional solution more easy to understand for everyone. Richard Gill (talk) 08:56, 19 November 2012 (UTC)

I agree with adding the Bayesian perspective but I think the 'conditional' solution would best be explained from a frequentist perspective showing the average result of 6 plays of the game in which the player has chosen door 1.

"The average result of 6 plays of the game" makes no sense to me. 6 plays can have all kinds of different outcomes including 6 times the same outcome and the player losing by switching. Or are you supporting the "six equally likely cases" solution? Or do you mean the approximately expected result of 600 plays?

I mean the average of many sets of six plays, but you know the words better than I do. My point is simple, that whole numbers are easier to understand than fractions. The reader has to get a 2 out of 3 chance from the diagram somehow. You know how to do that and so do I but it may not be so obvious to everyone. Martin Hogbin (talk) 17:57, 19 November 2012 (UTC)

The diagram shows that there are two ways to arrive at the situation with door 1 chosen and door 3 opened: one, having probability one sixth, with the car behind door 1, the other, having probability one third, with the car behind door 2. One third is twice one sixth: given door 1 chosen and door 3 opened it is twice as likely that the car is behind door 2 than door 1.

If therefore you think of many, many repetitions of the game, then on average out of every six repeats there will be three with door 2 opened and three with door 3 opened. Of the latter three, the car will be behind door 1 once and behind door 2 twice.

This is exactly what the "six equally likely cases" table introduced by TotalClearance is saying. And giving the argument with simple whole numbers (indeed, just counting) instead of arithmetic with fractions. Which according to Kraus and Wang is the best way to build the right visual picture of the problem (the right one, is the one which makes the right answer obvious).

Hopefully anyone who is interested in the conditional solution will find an explanation which they can easily grasp. Richard Gill (talk) 17:43, 20 November 2012 (UTC)

I think the article has been improved significantly. Will it be possible to find somewhere Marilyn's first "answer" after 9 Sept. 1990, and before 17 Feb. 1991? Gerhardvalentin (talk) 17:56, 20 November 2012 (UTC)

I agree, huge improvement! I'm interested in how accessible you find the conditional part! Does it help people to understand the issues? And to understand what the solution *means*?

Jason Rosenhouse says in his book that MHP was discussed by MvS on September 9, 1990; December 2, 1990; February 17, 1991; and on a fourth date (unspecified). He got much of his information from her book "The Power of Logical Thinking". Maybe you can find what you are looking for in that book?

The article itself cites: Parade magazine p. 16 of issue of 9 September 1990, p. 25 of issue of 2 December 1990, p. 12 of issue of 17 February 1991, and p. 6 of issue of 26 November 2006.

In any case, if the conditionalists want to convince the simplists of the value of their solution, they had better bear in mind that most simplists are probably using probability in a very intuitive way, and therefore most likely Bayesian way. Richard Gill (talk) 14:29, 19 November 2012 (UTC)

As you know, I do not like these solutions anyway and wonder who the audience is for these diagrams. I have made some constructive suggestions but I am happy to leave the decisions on the best way to explain the 'conditional' solutions to those who consider them important. Martin Hogbin (talk) 13:32, 19 November 2012 (UTC)

In my opinion, the audience for these diagrams are the people who were perfectly happy with simple solutions but who want to find out what the fuss is all about, about conditional solutions. The aim should be that such intelligent and inquisitive readers can find out what the conditional solutions mean, so that finally they can draw their own conclusions as to which solutions they actually prefer. Richard Gill (talk) 15:04, 19 November 2012 (UTC)

My first question is, 'What fuss?', most of the fuss has been in these pages; there is much less of a dispute in the literature. If you want to explain to interested amateurs why the conditional solutions are considered necessary or desirable by some people then I would do what most sources do and consider the (variant) possibility in which the host does not choose evenly. There are very few sources that talk about the need for a conditional solution if the host is taken to choose evenly. If we start from the Bayesian (or host chooses evenly) perspective the requirement for a conditional solution is somewhat pedantic, and not even that rigorous (as discussed). Having considered the variant case you can then go in to say that mathematicians?/some people? consider the conditional solution is required even in the standard case, although there are not many references for that. Martin Hogbin (talk) 17:57, 19 November 2012 (UTC)

Rosenhouse's book seems to me to be the definitive resource on MHP at the moment and he pays a lot of attention to the matter.

Fuss or no fuss, most people trained in probability theory will "automatically" produce the conditional solution. Because it is follows automatically from following basic principles. On the other hand, most popular sources automatically produce simple solutions. Misplaced Pages pages on MHP will be visited just as much by students learning probability (where they will automatically meet MHP as a popular pedagogical example) as by ordinary people who heard about the puzzle from an aquaintance. The *fuss* is not important, but it is important that there are two common and differing approaches to the problem and the article has to cover both. If the article is good, it will also show that both have their merits, that the conditional solution is also easy to understand, and hopefully even that the two approaches are linked in numerous ways. Richard Gill (talk) 09:43, 21 November 2012 (UTC)

Speculation about 'The Economist' solution.

Richared, you gave your understanding of what 'The Economist' were intending in their solution:

We think of the location of the car and the choice of the host (if he has one) as being fixed, unknown. All randomness is in the player's initial choice of door. If he initially picks the door with the car (probability 1/3), then by switching he'll get a goat. If he initially picks a door with a goat (probability 2/3), then by switching he'll get the car.

My own opinion is that they were following an approach that I have proposed, which to ignore the doors completely and concentrate only on the objects behind them: car, goat A, and goat B. There is no need to make any assumptions as to which door was chosen.

As neither of us can tell what was really in the minds of the writers I suggest we both leave out our personal interpretations. Martin Hogbin (talk) 16:07, 21 November 2012 (UTC)

The illustration shows a fixed picture of car, goat, goat. What varies (with equal probabilities 1/3) is the initial choice of door of the player. I have read the original and it tells, as I recall, the same picture. All randomness in the story lies in the choice of door by the player. The location of the car is fixed. The choice of the player is random. (And we don't even talk about the choice of the host). Richard Gill (talk) 18:18, 21 November 2012 (UTC)

That may be the way that you see it but it is not the way that I do. The pictures show the car and the two goats (A and B) being in fixed positions relative to the page but I see this a diagrammatic only. The description makes no mention of doors: it refers only to the objects behind the doors. I see this solution as a vindication of my assertion that the doors are completely irrelevant and that it is only what is behind them that matters. The problem can be perfectly well described without the use of doors or any form of numbering of the objects. The player picks one object, the doors only being present to ensure that the player does not know what object they have picked, the host then removes an object from the remaining two, and the player has to choose between his original choice and the object remaining. No doors, numbers, or positions are necessary. For some reason, the doors seem to have taken on a life of their own in the minds of many solvers of this puzzle. Martin Hogbin (talk) 21:44, 21 November 2012 (UTC)

"For some reason, the doors seem to have taken on a life of their own in the minds of many solvers of this puzzle". For two very good reasons. One, because vos Savant named them. Two, because if you want to solve MHP through principled use of probability theory in which you carefully go step by step through the history leading up to the moment of the player's choice, you are forced to. People will this background will automatically solve the problem in this way. And they'll use it in the courses they teach and the books and articles they write. It is not some kind of mass-delusion, started by evil Morgan. No. It's the careful solution given by Selvin 1975b, in response to the just criticism which he got from numerous correspondents on his rather sloppy first attempt Selvin 1975a. Have you read these papers???? Richard Gill (talk) 15:47, 23 November 2012 (UTC)

I have no objection to being thorough and careful I only object to solutions which claim to be this but are not. I thought you had agreed to this on my talk page. I agree that most solutions in sources use doors but there really is no need to. There are sourced solutions that do not mention doors at all. Martin Hogbin (talk) 10:47, 24 November 2012 (UTC)

Martin, now you are imposing your personal interpretation on the article in the Economist. Read it! Bear in mind that the author comes from economics, and writes for readers from economics. MHP was placed in the mathematical economics literature (which is all about decision theory, game theory) by Nalebuff in 1987. It's a trivial exercise in game theory, which anyone from that background will understand immediately. Already Nalebuff mentioned the game theory approach to MHP (strategies of player, strategies of host). The central result is von Neumann's minimax theorem which solves two-party finite zero sum games if the parties may use randomized strategies. The contestant's optimal strategy if he knows nothing of the game organizers plus host's (and the contestant wants to get the car, the game organizers plus host want to prevent that) is to pick an initial door completely at random and then switch.

You are speculating that the writer of the Economist article thinks like you. A way of thinking which is not even discussed in published literature. Pure speculation and highly biased. I am arguing from knowledge of the context in which the article was written, and knowledge of the preceding literature. Do I have a personal bias? Sure, I have personal favourite approaches, and I have a personal bias to value diversity. Apart from that, I study the literature and use "good faith" (assumptions of intelligence, thoughtfulness) on the part of the various writers out there, in order to obtain a fair (undogmatic) panoramic view of diversity. Richard Gill (talk) 08:43, 22 November 2012 (UTC)

Richard I am not suggesting that we add my interpretation of this solution to the article. Just that we leave our own personal interpretations out of it.

I have to admit that I have based what I have said on what was shown in the article where there is no mention of doors. Do you have a copy of the article that I can read. If it clearly supports your view that the door chosen by the player is to be taken as fixed then we can, of course, say that in the article. Martin Hogbin (talk) 09:17, 22 November 2012 (UTC)

The *picture* which we have in our article shows the location of the car as fixed, and the door chosen by the player as variable. I have always supposed that our picture was derived from the diagram which was originally present in the published Economist article. And this was also the reason why I asked if we could have this proof back in the article: as representative of lots of sources which take the choice of the player as random, the actual location of the car as fixed. We have also frequently seen editors on this talk page spontaneously offer the same solution as their own original idea. It seems to be a common way ordinary people like to solve the problem. Richard Gill (talk) 15:39, 23 November 2012 (UTC)

Hm. Our Misplaced Pages article links to a web page of The Economist which contains the text of the original article but not the accompanying illustration. So I cannot check this now. I have always imagined that the author of the diagram in our article had copied it from the original article. Richard Gill (talk) 12:37, 22 November 2012 (UTC)

Yes, I have just had a look too. It makes me wonder of we even have the right reference. The words in the Economist section in the article bear little relationship to those in 'The Economist', and there is no diagram there either. Martin Hogbin (talk) 18:50, 22 November 2012 (UTC)

There is a reference to a diagram - a probability tree. So the actual published (paper) version of the article contains much more information than what is on internet. Can you get a photocopy from a UK library? Richard Gill (talk) 12:07, 23 November 2012 (UTC)

I will give that a go. Martin Hogbin (talk) 10:36, 24 November 2012 (UTC)

Intro to Conditional Solution: biased host

I added to the introductory text to the conditional solutions the derivation of the conditional probabilities of 1/2 and 1 corresponding to known, maximal (in either direction), host bias. So the section now begins with an elementary derivation of conditional 2/3 by consideration of the expected outcomes of 600 repetitions in the standard situation, as well as illustration that if we have certain knowledge of host bias, we get different answers. The same motivation as many writers, e.g. Rosenthal, give for the conditional solution. Richard Gill (talk) 19:49, 26 November 2012 (UTC)

The section goes on to give three more derivations of conditional 2/3: by a conditional probability diagram, by a decision tree, and by extension of Vos Savant's table to 6 equally likely cases. I propose that the conditional probability diagram is deleted - it seems superfluous among all the other graphical and tabular methods, and it takes a huge amount of space. Any opinions on this? Richard Gill (talk) 19:54, 26 November 2012 (UTC)

By all means! Delete all traces of any accessible conditional solution - it is of course merely an academic contrivance! We find your recent work here to be most excellent, although it is easily seen that the authors should perhaps read Misplaced Pages:Manual of Style/Mathematics.

In case there's anyone clueless enough to miss the sarcasm, I'm saying no, the diagram should not be deleted. -- Rick Block (talk) 05:08, 27 November 2012 (UTC)

Rick, you think the conditional probability diagram is more accessible than anything else? I really do wonder what "first time readers" find the most enlightening. Under the motto "less is more" the conditional probability part of the article would be more accessible (I think) if it was trimmed of superfluous fat, started with the a really good motivation of *why* a conditional solution, and clearly showcased the most accessible/appealing solution. Alternative derivations can then be relegated to a collection of "alternative approaches". (PS thanks for the tip regarding manual of style and thanks for the compliment, which I take as being seriously meant!)

The article was reorganized by Martin a few weeks ago and this left the conditional probability parts rather disjointed. There was the big diagram alongside some other solutions but no explanation of where you "see" that the answer is 2/3 in that diagram. I have tried hard to remedy this, but I don't know if I succeeded (I added a row in the table, splitting 1/3 into two times 1/6).

At the moment my feeling is that the discussion of the expected numbers of different outcomes in 600 repetitions, and the table of six equally likely cases extending vos Savant's table of three, are the most accessible derivations for people who never saw conditional probability before, can't read maths formulas, and are scared of arithmetic with fractions. If you have any familiarity with probability and conditional probability at all, then the decision tree is useful visual aid. Maybe I'm wrong. Maybe the big conditional probability diagram is a wonderful pedagogical tool for most readers. Richard Gill (talk) 15:55, 27 November 2012 (UTC)

Richard, I did not (intentionally) change the organisation of the 'conditional' section, I moved it as it was.

Rick, as you know, I do believe that the 'conditional' solutions are 'merely an academic contrivance', but the proposal was to give them equal prominence so, as far as I am concerned you can edit them in whatever way you think showcases them well and makes them accessible to your intended audience. This is what Richard was trying to do. I asked Richard to think about the intended audience for this section and would make the same suggestion to you. Martin Hogbin (talk) 23:45, 28 November 2012 (UTC)

The example of 600 repetitions explains the "conditional approach" that is viable with a biased host, but never does address the direct approach. Rick's mapping is pointless and incorrect. The article must stop to be kept blurry and unclear. Gerhardvalentin (talk) 21:50, 29 November 2012 (UTC)

What do you mean: the example with 600 repetitions never does address the direct approach? It tries to explain what conditional 2/3 means. The 600 repetitions are imaginary, not real. Later there is an equivalent description using 6 equally likely possibilities. Do you understand it? Richard Gill (talk) 06:42, 30 November 2012 (UTC)

What do you mean by "the direct approach"? The simple solution? 600 repetitions ilustrates the simple solution too. Richard Gill (talk) 06:44, 30 November 2012 (UTC)

The "600 repetitions", explaining all possible outcome, never belong to the section "Criticism of the simple solutions", where Rick did rearrange it. It belongs where it was, before Rick disarranged it. Gerhardvalentin (talk) 07:41, 30 November 2012 (UTC)

Rick made some big changes. The "criticism" section is now a mess. He had better revise it, too. Richard Gill (talk)

Is now cleaned up again. Richard Gill (talk) 08:11, 1 December 2012 (UTC)

The problematic choice of goats

I believe much of the confusion this problem has generated has been caused by an incorrect choice of animal. If the goats are replaced by Schrödinger's cats, everything becomes much clearer. Rumiton (talk) 07:14, 28 November 2012 (UTC)

Your suggestion is too late, a quantum version of the puzzle has already been proposed. Martin Hogbin (talk) 09:36, 28 November 2012 (UTC)

Gosh, what are the chances of that? Rumiton (talk) 10:32, 28 November 2012 (UTC)

Several different quantum versions. Ours (me and my friends') is the best (d'Ariano et al). Richard Gill (talk) 08:16, 1 December 2012 (UTC)

Source of confusion

The Sources of confusion section has this confusing sentence: "However, if a player believes that sticking and switching are equally successful and therefore randomizes their strategy, they should, in fact, win 50% of the time, reinforcing their original belief." It could be just due to English not being my native language, but I read that as going against the entire point of the paradox: that the player somehow suddenly has the intuitive 50/50 odds of winning at their last choice, should they simply choose to believe it so. I suggest that this be mended or expanded to better express whatever it is that it's supposed to mean. -Uusijani (talk) 10:56, 28 November 2012 (UTC)

You are right, Uusijani. The article is a mess, never clearly saying what it purports to say. It explicitly should say that (50% x 1/3) + (50% x 2/3) gives "1/2". And in presenting the paradox it does not clearly say what Marilyn vos Savant herself did say concerning the correct "scenario" where the paradox arises. Gerhardvalentin (talk) 12:16, 28 November 2012 (UTC)

It seems clear to me. If a player believes that there is no advantage in switching, and therefore randomises their strategy (in other words sticks half the time and switches half the time) they will win the car half the time. This will (illogically) reinforce their belief that it makes no difference whether you stick or switch. In this respect pigeons do better than humans.Martin Hogbin (talk) 23:31, 28 November 2012 (UTC)

Intro to conditional solution

Some days ago, I wrote an intro to the conditional solution, motivating it (as many authorities do) by showing that if the host has a preference, then the conditional probability answer is different from 2/3. This intro has now got spit over two distant sections, I think by Rick. Please Rick do things properly or not at all. The "criticism" section is now a mess. Richard Gill (talk) 19:15, 30 November 2012 (UTC)

The previous intro brought up the issue of biased host behavior - which is distinctly unnecessary to motivate the conditional solution. Most sources presenting conditional solutions do not (directly) bring this up. IMO, the biased host scenario is directly related to the criticism of the simple solutions - since these solutions produce the same "2/3 chance of winning by switching" result regardless of any host bias. The sources that are critical use the biased host variant to justify their criticism. -- Rick Block (talk) 01:10, 1 December 2012 (UTC)

Many sources would disagree with you, Rick. Rosenhouse and Rosenthal use the host bias variant to explain and motivate the conditional solution. This forces people to realise that by ignoring door numbers they are implicitly assuming that the host's choice gives no information to the player.

Traditional probability textbooks don't bother to motivate their approach, since in that context, one automatically figures out the conditional solution, because one works from first principles. First step: write down a probability model for the history of events leading up to the moment the player is asked whether he wants to switch. Second step: write down the probabilities of possible locations of the car given what the player has seen so far, using the definition of conditional probability. No need to be clever or creative. Just follow the rules and blindly (but carefully) calculate.

Why start half way through the problem? Martin Hogbin (talk) 23:32, 1 December 2012 (UTC)

Writers of probability texts start where they find it convenient and natural and sensible to start. It seems to me a very natural choice to start with (1) the car is hidden, (2) the player picks a door, (3) the host opens a door, (4) the player picks. Anyway, that's what everyone does, and it works fine. Next, since we will assume the player's choice is independent of the location of the car and later we want to condition on it anyway, we might as well consider the player's choice fixed. So we now have: (1) the car is hidden, (2) player picks door 1, (3) host opens a door, (4) player decides what to do. There is no point in making the story more complicated. We can't make it less complicated.

Having decided the sequence of steps leading up to the player's choice and specifying the probability of each outcome of each step given the preceding history, we are in business to routinely calculate what we want to know: the probability the car is behind any particular door given everything that has happened up to the moment the player must decide. Richard Gill (talk) 08:06, 2 December 2012 (UTC)

This may be what some sources do but it is incomplete. We are given a game in which the player can pick any of three doors. As we are given no information on how this is done we should take is as uniform at random. Does it matter how this initial choice is made? Of course it does. As you know, if the choice is made uniformly then the player's door choice and the location of the car are irrelevant. If the choice were not uniform, we would have to take account of these other two distributions (which, as it happens, we also take as uniform). So for the game, in general, the unconditional game, we must consider how the player initially chooses.

This is what many sources do, it is completely standard and seems to me completely reasonable too. If you are using probability in the Bayesian sense to measure the beliefs of the player at the moment when he must make his decision, then you are interested in the probability that the car is behind door 2 given the player chose door 1 and the host opened door 3. How you, the player, chose your initial door is irrelevant. Richard Gill (talk) 14:26, 2 December 2012 (UTC)

Some people (somewhat perversely) consider there to be two conditions implied by the question. These two conditions are: that the player initially chooses door 1; and that the host then opens door 3. The routine way to tackle this problem is to start with the sample space of the unconditional game and then to condition it according to the conditions that we wish to apply. To start half way through the problem with a hand-waving argument about starting where convenient is not acceptable if we want to be thorough. Martin Hogbin (talk) 10:14, 2 December 2012 (UTC)

This is not perverse, it is eminently sensible. If you want to decide rationally on a course of action you need to know the probabilities of the different possible states of the world, given your information at the moment the decision is needed.

There was no handwaving in my description of the solution. I told you explicitly that we assume that the choice of the player is independent of the location of the car and therefore that we can, and shall, condition on the choice of the player, since anyway we are only interested in the probability of possible locations of the car at the moment the player has to make his decision and given the information which he then possesses. Absolutely sensible, absolutely normal. Richard Gill (talk) 14:26, 2 December 2012 (UTC)

PS This is the principled way to solve the problem from the perspective of probability theory. From that perspective, giving a simple solution is simply plain wrong. It only coincidentally happens to give the right numerical answer, and anyway, it does not target what the probabilist wants to know, according to his own "rule book". Note: this is nothing to do with Morgan and nothing to do with biased hosts. Selvin 1975b gave the conditional solution in response to complaints from correspondents that his initial 1975a solution was wrong (that is: the derivation was wrong). Amusingly though, he does not admit it was wrong, and in the same article reproduces the "combined doors" simple solution provided by Monty Hall himself. So he just includes the conditional solution to satisfy the nit-pickers. On the other hand he emphasizes that all his solutions use the unbiased host assumption. Richard Gill (talk) 08:14, 2 December 2012 (UTC)

Satisfying nit pickers is fine but we must make sure that we pick all the nits. Martin Hogbin (talk) 10:14, 2 December 2012 (UTC)

No. No own research. We have to survey the literature. It Martin Hogbin has problems with how modern scholars teach probability theory to their students, he had better get his objections published. If anyone takes any notice of them, then in about 10 years Misplaced Pages can start to take account of them too. Richard Gill (talk) 14:32, 2 December 2012 (UTC)

Popular sources for conditional solutions do seem to feel a strong need to motivate the approach. Especially since they have already, typically, explained some simple solutions.

Oh well. Misplaced Pages is not a text book. We just survey what's out there: motivation is not relevant. Richard Gill (talk) 08:08, 1 December 2012 (UTC)

Rick, I have challenged you on several occasions to give sources that say the conditional solutions are necessary in the case where the host is stated to choose evenly between legal doors. I think you may have found one, maybe two. Most other sources which give conditional solutions explain and justify them by considering the case where the host is biased or they do not mentioning the host's strategy. Martin Hogbin (talk) 23:30, 1 December 2012 (UTC)

Just about every beginner's text book in probability theory gives the conditional solution without motivation and only for the symmetric case because it is the natural principled way to solve the problem from the point of view of probability theory. See Selvin (1975b). Richard Gill (talk) 08:18, 2 December 2012 (UTC)

All taken from the same original source. Martin Hogbin (talk)

That's what you say. Everyone who has studied probability theory can solve MHP quite on their own without any need for any source, and they'll almost all solve it the same way that Selvin did in his second, 1975b, note. It's the result of routinely writing down the natural assumptions and then routinely calculating the thing you have to calculate, from first principles. Richard Gill (talk) 14:32, 2 December 2012 (UTC)

But it was a good idea to put the "300 repetitions" into the conditional probability diagram, and take out my splitting into 6 equally likely cases! Richard Gill (talk) 08:13, 1 December 2012 (UTC)

I do not think so, Why 300 and not 6 as I suggested? Martin Hogbin (talk) 23:31, 1 December 2012 (UTC)

There are two common ways to make probability calculations intuitive, both of them are based on converting fractions to whole numbers, one goes for large numbers, the other goes for small.

Large numbers: people think of probabilities in terms of what would happen in many (imaginary) repetitions. Emphasis on many. For this example, thinking of 300 or 600 repetitions is convenient. How many times would each outcome, roughly, occur? About 100 times this, about 100 times that. The story has to have large numbers to become realistic. Talking about what we would expect in 6 repetitions doesn't make sense. Anything could happen. We certainly wouldn't expect the four distinct probability 1/6 outcomes each to occur exactly once and the single probability 1/3 outcome to occur exactly twice.

Yes I know that, but 600 is not a large number it is a quite small and arbitrary number. We give the impression to our readers that if we repeated the game 600 times we would get the exact results shown, which is , of course, wrong. After a large number of plays the results would tend towards the fractions shown but there is nothing special about 600. To show a total of 600 plays with exact figures is misleading. Martin Hogbin (talk) 10:41, 2 December 2012 (UTC)

The text talks about 300 now, and says "about 100 ..." and so on. Not misleading at all, and easy to undertand. Richard Gill (talk) 14:16, 2 December 2012 (UTC)

Small numbers: when all distinct outcomes (cases) have the same probability, then probabilities are found by counting all cases favourable to the event in question. This is the "table of six" solution which TotalClearance discovered some weeks ago, and which I subsequently found in a number of notable literature references. I recall that Martin and Nijdam both thought it was completely crazy. But anyway, it is now in the article. Richard Gill (talk) 07:55, 2 December 2012 (UTC)

The table of six would be a good idea if it were indeed a table of six equally likely possibilities, but it is not. It shows exactly the same event twice in order to make the table work.

It shows one event of probability 1/3 split in two equal parts, e.g. by imagining a fair coin toss being added to the story. This makes the table work and it is both intuitively and mathematically correct. It helps to have a bit of imagination. Richard Gill (talk) 14:16, 2 December 2012 (UTC)

So both approaches are now included. Richard Gill (talk) 07:55, 2 December 2012 (UTC)

What we have is neither fish nor fowl. It does not show the results of a large number of games or an number of equally likely events. Martin Hogbin (talk) 10:41, 2 December 2012 (UTC)

Martin, you clearly haven't read the relevant section recently. It's all there, it's all carefully explained. Hopefully other people than you will find it illuminating, even if you don't. Richard Gill (talk) 14:16, 2 December 2012 (UTC)

Vos Savant, Morgan, and the media furore

I did not get much response last time I mentioned this subject but it is quite a large part of the subject and should have a section in the article. I will start one. I am not fussed about the section title. Martin Hogbin (talk) 12:38, 2 December 2012 (UTC)

I have added a section. What do you think? Do we have a useable picture of vos Savant anywhere?Martin Hogbin (talk) 14:21, 2 December 2012 (UTC)

There is Vos Savant and the media furore. A completely different affair was Morgan's fuss about vos Savant, which is already covered extensively in the article, and which (IMHO) needs to be downplayed. Richard Gill (talk) 14:35, 2 December 2012 (UTC)

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