Talk:Monty Hall problem/Matt

From Wikipedia, the free encyclopedia

A little. I was bamboozled by your "singular" and "specific" and "average" players, but I think perhaps we are talking about the same thing after all. I think that while you are couching the description in terms of a "specific player" to which a particular scenario happens (such as the host opening door 3), I prefer to describe it as a conditional probability, and say for example, "given that the host opens door 3", which is IME a more conventional way of expressing it -- but let's not worry about that if we're really talking about the same thing.

Yes, I think we're talking about the same thing. -- Rick Block (talk) 04:18, 29 February 2008 (UTC)

It is trivially false that "a given player's odds at the point of deciding whether to switch are the same as the overall odds". Consider how would you determine this probability by experiment. You would put the same player in the exactly the same situation repeatedly. Car behind the same door every time, player chooses the same door every time, host opens the same door every time. If you change any of the parameters then you are no longer talking about a "given player at the point of deciding", you've changed into some "overall" probability calculation over many different players experiencing different conditions. Then, every time the player would (assuming we want the "winning by switching" probability) always switch (as discussed before, we are only interested in the single-strategy probabilities), and every time it would be the same outcome: he would always win, or he would always lose, depending on whether he chose the car. Therefore the probability is, as I said, either zero or one.

I disagree with how you've set up your experiment. There are only 6 possible combinations of initial picks and doors the host opens. The experiment is to run several thousand iterations, with the initial player pick equally distributed among the three doors. The "given player, at the point of deciding" means "given that the player initially selects door X and the host opens door Y" (where X is 1,2, or 3, and Y is 1,2 or 3 - if X is 1 then Y is 2 or 3, if X is 2 then Y is 1 or 3, and if X is 3 then Y is 1 or 2). These subsets of the data can have different observed probabilities of winning by switching. In the problem statement we're placed into one of these subsets. We can measure the probability of each of these subsets over enough independent iterations. IMO, a "well formed" problem statement would ensure this experiment would end up with equal probabilities for each of the 6 subsets. The potential difference is between the total probability of winning by switching over all iterations, vs. the probability of winning in one of these 6 subsets. -- Rick Block (talk) 04:18, 29 February 2008 (UTC)
OK, fair enough, we just have different interpretations of "a given player's odds at the point of deciding whether to switch". What you're describing is an experiment to discover, as I would put it, the probability of winning by switching given that... (whatever condition it might be). That's fine, I have no problem with this. Matt 14:32, 29 February 2008 (UTC).

There certainly are conditional (i.e. "given that") probability questions that do not have an answer of 2/3, but this is unavoidable in any formulation. For example, we could ask "what is the probability of winning by switching given that the car is behind door 1 and the player chooses door 2" Answer: one. I am not clear on what basis the "given that host chooses door X" condition is supposed to have a special significance that all the other numerous conditions that one could come up with (presumably) don't. Matt 03:24, 29 February 2008 (UTC).

The host strategy for picking between the goat doors can affect the probability of winning by switching - in an identifiable subset of cases (i.e. given the door the player initially picks and the door the host opens). See my response to Father Goose's hypothetical question, at http://en.wikipedia.org/wiki/Talk:Monty_Hall_problem#Related_problem. -- Rick Block (talk) 04:18, 29 February 2008 (UTC)
In general, the answer to the question "what is the probability given that..." will be different from the answer to the question "what is the probability, full stop". The Monty Hall question asks "what is the probability, full stop". The fact that the answer might be different if the question was different is true but irrelevant. This issue is possibly muddled by the existing wording, which seems to imply that we know the host chooses door 3, and could conceivably (though in my view perversely) be construed as meaning "what is the probability, given that the host chooses door 3". The wording of the problem statement needs to be changed in this respect to make it clear that all doors are equivalent, thus eliminating any possibility that it could be misunderstood in this way.
In the actual Monty Hall question (the unconditional one), the host's preference does not matter. If the question is turned into a conditional one (specifically, one that conditions on the host opening a particular door) then the host's preference may matter. Again, this is true but irrelevant (though it could if desired be mentioned in a subsequent section, perhaps under "Sources of confusion").
The solution does not need to use the fact that the host chooses randomly between the goats because in the question that it is answering (the unconditional one) it doesn't matter. The answer, and the reason for the answer, is the same irrespective of whether the host chooses randomly. If the question instead asked for the probability given that the host opens door 3 (and also specified that the host may behave non-randomly), then the answer may be different. In that case the solution no longer applies because in the new problem space, the one that only considers cases where the host opens door 3, there may (if the host behaves non-randomly) no longer be a 1/3 chance that the player chooses the car. Matt 14:32, 29 February 2008 (UTC).
I think the problem is meant to be constructed so that the answer to "what is the probability, full stop" and "what is the probability given that... the player has picked a door and the host has opened some specific door" is the same (regardless of the player's pick and door the host opens). If it is the same, then the wording that apparently asks for the conditional answer is fine and, since the conditional answer is the same as the unconditional answer, there's only one answer (and everybody's happy). The point our other anonymous friend has made (and I think it's legitimate) is that the conditional analysis is inherently better. I don't have any idea why there's such continued resistance to phrasing our analysis as a conditional one. If we make sure the problem statement forces the conditional answer to be 2/3 for any initial player pick and door the host opens (which it currently does), this trivially shows the unconditional answer is also 2/3 (in fact, this is so trivial it isn't even worth mentioning and I think we want to avoid mentioning it this early in the article because of the whole can of worms the conditional vs. unconditional difference raises). Why not use a conditional analysis? -- Rick Block (talk) 17:03, 29 February 2008 (UTC)


At this stage in the article the explanation needs to be kept as simple as possible for the general reader, whose perseverance with explanations containing mathematical or logical content decreases exponentially with the length of the explanation. Why on earth would we want to use version 1, below, when we can use version 2?

Contents

[edit] Version 1

Problem

A thoroughly honest game-show host has placed a car behind one of three doors. There is a goat behind each of the other doors. You have no prior knowledge that allows you to distinguish among the doors. "First you point toward a door," he says. "Then I'll open one of the other doors to reveal a goat. After I've shown you the goat, you make your final choice whether to stick with your initial choice of doors, or to switch to the remaining door. You win whatever is behind the door." You begin by pointing to door number 1. The host shows you that door number 3 has a goat. Do the player's chances of getting the car increase by switching to Door 2?

The problem as generally intended also assumes that the particular door the host opens conveys no special information about whether the player's initial choice is correct. The simplest way to make this explicit is to add a constraint that the host will open one of the remaining two doors randomly if the player initially picked the car.

Solution

When the player is asked whether to switch (assuming the player initially picked door 1) there are three possible situations corresponding to the location of the car, each with probability 1/3:

  • The player originally picked the door hiding the car. In this case the game host opens door 2 half the time and door 3 half the time, so there are two subcases each with probability 1/6.
    • The player originally picked the car and the host opens door 2.
    • The player originally picked the car and the host opens door 3.
  • The car is behind door 2 and the host opens door 3.
  • The car is behind door 3 and the host opens door 2.

The host opens only one door, so only one of the first subcases with probability 1/6 and one other case with probability 1/3 apply. Staying wins in the 1/6 case where the player has initially picked the car while switching wins in the 1/3 case where the player has not. Switching wins twice as often as staying, so the probability of winning by switching is 2/3. In other words, a player who switches will win the car on average two times out of three.

[edit] Version 2

Problem

A game-show host has placed a car behind one of three doors, and a goat behind each of the other two doors. The player, who has no prior knowledge that allows him to distinguish between the doors, is asked to choose a door. The host then opens one of the other two doors to reveal a goat. The host, who knows the location of the car, then opens one of the other two doors, being sure to reveal a goat rather than the car. The player is asked if he would like to switch to the remaining unopened door. Do the player's chances of winning the car increase by switching?

Solution

There are three possible situations corresponding to the player's initial choice, each with probability 1/3:

  • The player originally picked the door hiding the car. The game host has shown one of the two goats.
  • The player originally picked the door hiding Goat A. The game host has shown the other goat.
  • The player originally picked the door hiding Goat B. The game host has shown the other goat.

If the player chooses to switch, he wins the car in the last two cases. A player choosing to stay with the initial choice wins in only the first case. Since in two out of three equally likely cases switching wins, the probability of winning by switching is 2/3. In other words, players who switch will win the car on average two times out of three.

Matt 20:32, 29 February 2008 (UTC)

...(Oh, and don't crucify me if version 2 needs fine-tuning... I knocked it up very quickly!) Matt 21:05, 29 February 2008 (UTC) —Preceding unsigned comment added by 81.156.127.87 (talk)
It does need tweaking to avoid the ambiguity of whether the host knows where the car is and will never reveal it. That's the key condition that changes the odds from 50-50 to 2/3.--Father Goose (talk) 00:23, 1 March 2008 (UTC)
[I have removed my previous response as it was unsatisfactory. I will have another think] Matt 02:24, 1 March 2008 (UTC).
See revised wording above. Does that fix it? Matt 02:31, 1 March 2008 (UTC).
I see your point with the now-removed "unsatisfactory response", but the revised wording helps to make it clearer, so yes, it is better. However, I'm not sure this wording necessarily avoids the need to use a conditional analysis just because it doesn't mention door numbers. Rick?--Father Goose (talk) 07:13, 1 March 2008 (UTC)
Right. The point is not whether the doors are named. Consider host one who flips a coin to decide which of two goat doors to open vs. host two who always opens the rightmost door if he can. Any problem statement that does not include the equal goat constraint permits both of these and the solution for host two (but not host one) is different depending on whether the host has opened a rightmost or leftmost door. Whether the solution description is conditional or unconditional and whether we name the doors or not, we really, really want this constraint.
Is version 1 significantly harder to understand the version 2? Can we perhaps make it easier to follow somehow? -- Rick Block (talk) 14:35, 1 March 2008 (UTC)
[Written before reading Rick's latest response above, which I will respond to later if this doesn't cover it.] As far as I can tell, every statement in the solution is correct, the logic is correct, and the whole leads correctly to the correct conclusion that a player who switches has a 2/3 chance of winning the car. Of course, if anyone can identify a false statement or a logical flaw then I will be more than happy to reconsider my opinion! Incidentally, in case this point got lost in the earlier verbiage, another advantage of the Version 2 wording (apart from its simplicity and clarity) is that it works no matter how the host decides which goat to reveal when he has a choice. It seems intuitively obvious, to me at least, that this shouldn't doesn't matter, but in the "probability given that the player chooses door X and the host opens door Y" interpretation it does. (However -- and I hesitate to mention this -- there is a potential problem, in both versions, that the literal answer to the question "Do the player's chances of getting the car increase by switching?" is arguably "it depends: the probability will either change from zero one or from one to zero, depending on whether he originally chose the car". But we don't want to go there, do we?) Matt 14:43, 1 March 2008 (UTC). —Preceding unsigned comment added by 86.146.47.203 (talk)
Sorry about these piecemeal responses; I just overlapped with an edit conflict with Rick's post. Just a couple more points. Rick says "Whether the solution description is conditional or unconditional and whether we name the doors or not, we really, really want this constraint [i.e. the equal goat constraint]", but I'm not sure that's correct. As I understand it, the "equal goat constraint" is required only if the problem is considered a conditional one that includes a "host opens door X" condition (it makes no difference if the doors are numbered, or we refer to them as rightmost, leftmost, or whatever -- the principle's the same). I believe that most people's understanding of the "vanilla" problem is that all the doors are equivalent, and that we really don't care which one the player chooses or which one the host opens. I believe that this is the interpretation that we should use in the initial explanation. It is all that is required to get across the key point that the probability is 2/3 not 1/2. I believe that the consideration of "what is the probability given that (whatever)" is unnecessary at this stage, and should be convered in a later section. Matt 15:21, 1 March 2008 (UTC). —Preceding unsigned comment added by 86.146.47.203 (talk)
I really hope this doesn't come across as squabbling, but I think you're not quite getting it. I agree with your statement that most people's understanding of the problem is that all the doors are equivalent, and that we really don't (shouldn't) care which one the player chooses or which one the host opens. The issue is that this is not true unless we include the equal goat constraint. Without it, if we have the (perverse) host who always opens the rightmost door if he can, over 9000 trials of this game (with the car randomly placed and players initially picking randomly between doors) we'll have approximately:
Number of trials Initial pick Host opens Switching wins
1000 Door 1 Door 2 100%
2000 Door 1 Door 3 50%
1000 Door 2 Door 1 100%
2000 Door 2 Door 3 50%
1000 Door 3 Door 1 100%
2000 Door 3 Door 2 50%
I think what you're saying is that we should ignore this and consider only the overall number which works out to 6000 trials resulting in winning by switching and say "players win 2/3 of the time by switching". However, with this host, whether we expect them to be or not, whether we number them or not, whether we consider it a conditional problem or not, I think nearly anyone looking at this table would conclude the doors are NOT equivalent. By assuming the doors ARE equivalent, what we're actually doing is assuming the host is picking randomly among two goat doors (!). I.e. the only way to make this table look like this (which I think is what is meant by saying the doors are equivalent):
Number of trials Initial pick Host opens Switching wins
1500 Door 1 Door 2 2/3
1500 Door 1 Door 3 2/3
1500 Door 2 Door 1 2/3
1500 Door 2 Door 3 2/3
1500 Door 3 Door 1 2/3
1500 Door 3 Door 2 2/3
is to include the constraint. Do you disagree with this? -- Rick Block (talk) 16:14, 1 March 2008 (UTC)


My choice of the word "equivalent" was unfortunate. If the host is biased then the doors are in that sense no longer equivalent; yes I agree with that, and I apologise that I caused you to waste your time going down that avenue. Way back at the start of this discussion I wrote out a table of all possible scenarios of car placement, initial player choice, host choice, and outcome if the player switches, along with their probabilities, both for the biased and unbiased hosts. Though I haven't checked all your figures(!), this should translate exactly to your tables. So, I hope I am "getting it".

As I understand it, the main motivation behind your proposed change is that you want the solution to apply to every specific player at the point he switches -- for example, to the player who has chosen door 1 and encounters the host opening door 3. To me, a question about a specific player at the point of switching, in a specific realisation of the game, inevitably leads to the answer "the probability that he will win by switching is zero or one, depending on whether he originally chose the car". Given the problem statement -- in both versions -- it seems incontrovertible that the outcome is already decided at this point: there's no uncertainty left, and hence no non-trivial probability calculation that can be performed.

Yes, I want the solution to apply to every specific player at the point he switches - but only given the information available at that point from the problem statement. In each instance the player wins or loses by switching, but we can't know which unless we are also given the placement of the car (which we're not). We're not analyzing the problem across all possible "givens" or any specific case given all possible knowledge, only the known givens explicit in the statement of the problem. In any conventional version, at the critical point ("do you want to switch") the player has picked a specific door and the host has opened some other specific door. Even if we don't know which specific doors these are, if it matters I think our analysis must take these into consideration.

A "general" question about the probability given that the host chooses door 3 (i.e. considering all the possible games where this scenario occurs) is a perfectly legitimate one, with a non-trivial answer that depends on the host's preference. However, I don't think that this is the question that anyone has in mind when considering the "vanilla" Monty Hall problem.

I'd rephrase this to "no one posing the Monty Hall problem means for this (general) probability to differ given which door the host opens or which door the player initially picks". This "general" probability given which door the player initially picks and which door the host opens is precisely what I mean by the "player specific" probability given this set of circumstances (isn't this how probability is defined?).

Since we don't really want a conditional question ("given that the host chooses door 3"), and since the question about a specific player in a specific realisation of the game has a trivial and uninteresting answer, what's left? The only think I can think of is to look at the game as a "whole" (for want of a better way of expressing it), and treat the fact that the host chooses (say) door 3 just as a particular circumstance that the player, with some probability, may or may not encounter.

Again, I don't think we want to avoid a conditional answer but rather want the conditional answer (or the "general" probability) to be the same regardless of which door the player picks and which door the host opens. If this is true, then we can legitimately treat any setup (player picks door 1, host opens door 3) as equivalent to any other. To do this, all we need to do is keep the equal goat door constraint (!).

Unfortunately, this analysis also highlights another problem with the wording, which I mentioned above; namely, the question "Do the player's chances of getting the car increase by switching?"; Answer: they might change from zero to one or from one to zero. This leads me to...

... Suggestion ...

How about if (in the version 2 wording) we changed "Do the player's chances of getting the car increase by switching?" to "Is the player's best strategy to switch or to stick with his initial choice?" Will that help at all?

I don't think so, since this is only a problem if we assume we know where the car is (which I think we cannot).

[Noticed your new suggestion below; I'll comment when I've had a chance to study it.] Matt 20:01, 1 March 2008 (UTC).

It's not really a suggestion, just more background. -- Rick Block (talk) 23:37, 1 March 2008 (UTC)
BTW - the logical flaw in this version (version 2) is the unqualified statement with equal probability. We haven't named the doors and haven't said which one the host opened, but it's clear from the problem description that there are three doors and we can tell them apart. In trots our perverse host who always opens the rightmost door if he can (we haven't kicked him out of the building and banished him to Monty Hell yet!). Since it's unqualified, I think the statement with equal probability should mean with equal probability regardless of which door the host opens (this may be debatable, but I think an unqualified statement of probability should apply to any scenario clearly within the bounds of the problem statement). And I suspect you know where this is going - with this host, if the rightmost door is open the probabilities of the 3 cases are 1/2, 1/4, and 1/4 respectively, while if the leftmost door is open the probabilities are 0, 1/2, and 1/2.
I've thought of another potential way out of our dilemma which I'll describe at the main talk page. -- Rick Block (talk) 02:13, 2 March 2008 (UTC)


[Not yet digested your most recent comments, immediately above and on the main talk page. I wrote this offline in response to your comments timestamped 23:37, 1 March 2008. I'll post this now anyway and take a proper look at the other stuff later.]

Great stuff. So at the point of switching, the car is, for our purposes, still a fuzzy probabilistic thing (even though in the "real world" it is behind exactly one door with certainty). Having in this way disposed of my objection, the question "Do the player's chances of getting the car increase by switching?" can non-trivially refer to a specific player at the point after the host has opened the door.

OK, so starting with the wording in version 2 above, can we add the "host chooses evenly" assumption and add a note to the solution explaining why (under the above interpretation) it doesn't work if the host behaves non-randomly? My brain is a bit too fuzzy right now to figure out how this might be worded, but on reflection something might come to me, or maybe you can think of some good wording?

This way we might be able to retain the elegance, simplicity and symmetry of the version 2 solution. Casual readers who are happy with just getting a flavour of the solution can zip over it and say "OK, I get the general idea", but an explanation of the consideration we've been discussing is there for the more technical readers. For those readers, it also makes more of a feature of why (in our interpretation) it matters that the host chooses evenly (not terribly clear in Version 1, where it's not obvious why the wording is more complicated than many of the popular explanations that appear elsewhere). Matt 03:11, 2 March 2008 (UTC).

[edit] Variant 3

I think this problem statement has the properties you're looking for, but it is distinctly WP:OR (I've never seen it posed this way anywhere).

Problem

A thoroughly honest game-show host has placed a car behind one of three doors. There is a goat behind each of the other doors. You have no prior knowledge that allows you to distinguish among the doors. "First you point toward a door," he says. "Then I'll put a curtain in front of the other two doors, open one and bring out a goat from behind the curtain. After I've shown you the goat, you make your final choice whether to stick with your initial choice of doors, or to switch to the remaining door I haven't opened. You win whatever is behind the door." You begin by pointing to door number 1. The host puts up the curtain, opens a door behind the curtain, and brings out a goat. Do the player's chances of getting the car increase by switching to the unopened door behind the curtain?

Solution

There are three possible situations corresponding to the player's initial choice, each with probability 1/3:

  • The player originally picked the door hiding the car. Both goats were behind the curtain and the game host has shown one of them.
  • The player originally picked the door hiding Goat A. The game host has shown the other goat.
  • The player originally picked the door hiding Goat B. The game host has shown the other goat.

If the player chooses to switch, he wins the car in the last two cases. A player choosing to stay with the initial choice wins in only the first case. Since in two out of three equally likely cases switching wins, the probability of winning by switching is 2/3. In other words, players who switch will win the car on average two times out of three.

The critical difference between this setup and the standard one is that the player cannot distinguish which door the host opens. In the standard setup, whether we name them or not, the player can tell the difference. Here, the player can't tell the difference and has no way to know. This makes an analysis based on the host's choice of door invalid, like an analysis of the standard problem based on knowledge of where the car is (i.e uses knowledge unavailable to the player).

I am distinctly not suggesting we use this formulation, but offer it only as a means of furthering the understanding. -- Rick Block (talk) 19:02, 1 March 2008 (UTC)

[edit] Version 4

(Draft for comments -- Matt 14:47, 2 March 2008 (UTC).)

Problem

[Blurb introducing the version of the problem that we're using for the purposes of exposition.]

A game-show host has placed a car behind one of three doors, and a goat behind each of the other two doors. The player, who has no prior knowledge that allows him to distinguish between the doors, is asked to choose a door. The host, who knows the location of the car, then opens one of the other two doors, being sure to reveal a goat rather than the car. The player is asked if he would like to switch to the remaining unopened door. Do the player's chances of winning the car increase by switching?

The problem as generally intended also assumes that, for the purposes of calculating the player's chances, the opening of the door conveys no information about the location of the car. The simplest way to make this explicit is to add a constraint that the host will open one of the remaining two doors randomly if the player initially picked the car. [ref. later section]

Solution

There are three possible situations corresponding to the player's initial choice, each with probability 1/3:

  • The player originally picked the door hiding the car. The game host has shown one of the two goats.
  • The player originally picked the door hiding Goat A. The game host has shown the other goat.
  • The player originally picked the door hiding Goat B. The game host has shown the other goat.

If the player chooses to switch, he wins the car in the last two cases. A player choosing to stay with the initial choice wins in only the first case. Since in two out of three equally likely cases switching wins, the probability of winning by switching is 2/3. In other words, players who switch will win the car on average two times out of three, and players who stay with their original choice will win on average one time out of three.

Host behaviour when the player initially chooses the car

A subtlety arises in respect of the host's behaviour when the player initially chooses the car and the host is required to choose between two goats. In an overall sense, the host's behaviour here does not matter: whatever he does, two out of three players who switch will still win the car. However, as far as a specific player at the point of switching is concerned – that is, once we know which which door the player has chosen and which the door the host has opened – it does matter.

For example, suppose it is known that the host always chooses door 3 if he can, otherwise he chooses door 2 if he can, otherwise he chooses door 1. Suppose that, at the point of switching, we know that the player chose door 1 and the host opened door 3. The position of the car, which we assume was originally randomly placed behind one of the doors, is still unknown. In this setup the player's chance of winning if he switches is actually 1/2, not 2/3. If, in an otherwise identical setup, the host opens door 2 rather than door 3, then the player's chance of winning by switching is 1 (i.e. certainty). The first scenario (host opens door 3) occurs 2/3 of the time, and the second (host opens door 2) occurs 1/3 of the time. The overall probability of winning by switching (averaged across many players) is therefore 2/3*1/2 + 1/3*1 = 2/3, yet the individual player probabilities vary. Essentially this because the host's behaviour is conveying information about the position of the car; specifically, if he chooses door 2 then he is revealing that the car is not behind door 3.

If, however, it is known that the host chooses uniformly randomly between the goats, then the probability of winning by switching is 2/3 for all players regardless of which door the host opens; the host's choice conveys no information about the position of the car. If the host's behaviour (whether he chooses randomly or has a preference for a particular door) is not known then it is not possible to calculate an individual player's probability of winning at the point of switching in this way: there is insufficient information. However, the average over all players is still 2/3.

Under the assumption that we know which door the host opened, the host's random behaviour is implicit in the solution statement: "There are three possible situations corresponding to the player's initial choice, each with probability 1/3". If the host chooses non-randomly then these scenarios are no longer equally likely when considering a player who encounters the host opening a particular door.

[edit] Version 5

The Monty Hall problem is a puzzle involving probability loosely based on the American game show Let's Make a Deal. The name of the puzzle comes from the show's host, Monty Hall. The problem is also called the Monty Hall paradox; it is a veridical paradox in the sense that the solution is counterintuitive.

In the problem, a game-show host places a valuable prize (such as a car) behind one of three closed doors, and booby prizes (such as goats) behind the other two. The player is asked to choose a door, after which the host opens one of the other two doors to reveal a goat. The player is then asked if he wants to switch from his original choice to the other unopened door. The puzzle asks whether the player should do so.

Because there is no way for the player to know which of the two unopened doors is the winning door, many people assume that each door has an equal probability and conclude that switching does not matter. In fact, in the usual interpretation of the problem, the player should switch: a player who switches has a 2/3 chance of winning the car, and a player who doesn't has a 1/3 chance.

A widely known statement of the problem appeared in a letter to Marilyn vos Savant's Ask Marilyn column in Parade (vos Savant 1990). When the problem and the solution appeared, approximately 10,000 readers, including several hundred mathematics professors, wrote to the magazine claiming the published solution was wrong. Some of the controversy was because the Parade statement of the problem fails to fully specify the host's behavior and is thus technically ambiguous. However, even when given completely unambiguous problem statements, explanations, simulations, and formal mathematical proofs, many people still meet the correct answer with disbelief.

[edit] Problem and solution

[edit] Problem

The original[?] statement of the problem in the Ask Marilyn column in Parade was as follows:

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?

This formulation leaves critical aspects of the procedure unstated, making the problem mathematically ambiguous. In the usual interpretation some additional assumptions are included:

  • The player has no prior knowledge of where the car is located.
  • The host will always open another door and always make the offer to switch (rather than, for example, just if the player originally chose the car).
  • The host is constrained to always open a door revealing a goat (rather than it being merely fortuitous that he does so).
  • The particular door the host opens conveys no information about whether the player's initial choice is correct. The simplest way to make this explicit is to add a constraint that the host will open one of the remaining two doors randomly if the player initially picked the car.

As explained [ref], if the last condition is not present then the answer depends on whether we are asking the question about players in general, or about a specific player at the point of switching given knowledge of the door that the player chose and the door that the host opened. With this condition present, the answer and the method of solution are identical in both cases.

[edit] Solution

When the player is asked whether to switch there are three possible situations corresponding to the player's initial choice, each with probability 1/3:

  • The player originally picked the door hiding the car. The game host has shown one of the two goats.
  • The player originally picked the door hiding Goat A. The game host has shown the other goat.
  • The player originally picked the door hiding Goat B. The game host has shown the other goat.

If the player chooses to switch, the player wins the car in the last two cases. A player choosing to stay with the initial choice wins in only the first case. Since in two out of three equally likely cases switching wins, the probability of winning by switching is 2/3. In other words, players who switch will win the car on average two times out of three, and players who stay with their initial choice win on average one time out of three.

[edit] Version 6

[edit] Solution

The three possible locations of the car lead to three equally likely outcomes, each with probability 1/3:

  • The player originally picked the door hiding the car. The game host has shown one of the two goats.
  • The player originally picked the door hiding Goat A. The game host has shown the other goat.
  • The player originally picked the door hiding Goat B. The game host has shown the other goat.

Players who choose to switch win the car in the last two cases. Players choosing to stay with the initial choice win in only the first case. Since in two out of three equally likely cases switching wins, the probability of winning by switching is 2/3. In other words, players who switch will win the car on average two times out of three, and players who stay with their initial choice win on average one time out of three.

[edit] Conditional and unconditional analysis

The solution presented above is "unconditional" in the sense that it makes no reference to the particular door that the player chooses, or the particular door that the host opens. In effect, it looks at the situation facing an "average player", and finds that, on average, 2/3 of players who switch will win.

An alternative "conditional" analysis examines the situation facing a specific player at the point of switching, given knowledge of the door that the player chose and the door that the host opened (though not, obviously, the position of the car).

Provided it is known that the host chooses randomly between the two goats when the player chooses the car, the unconditional and conditional methods give exactly the same answer of 2/3. The problem is completely "symmetrical": the doors are unimportant, and an "average" player's chance of winning by switching is exactly the same as a specific player's once the doors are known.

If the host does not choose randomly between the two goats, then the unconditional analysis is still valid: on average 2/3 of players who switch will still win. However, a specific player's chance of winning by switching, given knowledge of the doors, may no longer be 2/3, and a conditional analysis is required.

For example,

[give an example of the conditonal analysis that gives a different answer].

[This next paragraph needs to be worded better...] —Preceding unsigned comment added by 86.142.109.205 (talk) 01:06, 8 March 2008 (UTC)

(Although the initial solution has been presented as the "unconditional" one, the letter of wording does strictly apply equally to the conditional case, provided that the host chooses equally between the goats, when, as we have said, the answer is always the same. However, the unconditional solution fails if applied to the conditional case when the host does not choose randomly, because the three cases may no longer be equally likely given knowledge of the doors.)