# Why doesn't the solution to the Monty Hall problem make sense?

#### sysprog

This seems to me like it might make the MH game problem reasonings clearer for some people:

After a person mis-evaluates, and says either that the chances after the goat is revealed go to 50-50, or that both of the doors retain their original 1/3 chance, or that, for whatever reason, he doesn't improve his chances by switching, present a modified version of the problem as follows:

The contestant picks a door. 1/3 chance per door of the car being behind it.
Monty says: I'm in an especially generous mood today, so I'm going to make you an offer: you can have both of the other two doors in exchange for your chosen door.

Should the contestant switch?

I think most people will recognize that the 2 doors between them have a 2/3 chance compared to the original 1/3 chance, so most people will say the contestant should switch.

So let's say in this version of the game that the contestant trades his 1 door for the 2 doors.

Monty then says: I'm going to open one of your 2 doors first. If neither door has the car, I don't care which one of them I open, but if one of them has the car, I'm going to open first the only one that doesn't. Either way, I'm going to open one of your doors that doesn't have the prize, whether it's the only one available for that or not, and I'm not going to open your originally chosen door right now.

The contestant isn't too worried about this, because he knows that only one of the doors could have the car anyway. He's anxious to see whether his 2/3 chance will pay off, but that's all that worries him in the game at that moment.

So Monty opens one of the contestant's 2 doors, and reveals a goat. No-one, including the contestant, is surprised, but a murmer comes from the audience.

Monty says: You traded in your 1/3 chance for what was then, at least then, a 2/3 chance. Now there are only 2 doors remaining. If you want to, I'll let you trade back for your original door."

The contestant becomes flustered, because for a moment, the audience murmers more loudly.

Should he switch back?

Monty then offers him 10% of the car's value to switch. If he's thinking his chances are "1 door 1/3" as at the start, or that they went up to 2/3 when he was allowed to switch for 2 doors, but now that there are only 2 unopened doors, his chances are now 50-50, the 10% cash should tip the scales for him, so he might think he should switch back.

Should he switch back?

I think most people will realize that, given what Monty said, the opening of a non-winning door of the 2 doors switched for won't diminish the 2/3 chance. The contestant knew when he switched for the 2 doors that at least one had to be a non-winner, because there's only 1 car in the game.

If a question respondent who thought in the original problem that there was no advantage to switching, thinks that in this version there's an advantage to not switching back, ask him to reconsider, after closer examination, whether in the original problem switching doors after the reveal is equivalent to not switching back after the reveal in the second version.

If he doesn't think so, ask him to imagine the game played both ways with the car behind the same door, and the same door originally chosen in each game. Why should the game-1 contestant not switch, if the game-2 contestant was right to switch before a door was opened and would be mistaken to switch back afterward?

#### PeroK

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The genders of a person's children can't affect the form of the information he states,
For example, if you were in a society (and there are plenty of these still alive and kicking on Earth even in the 21st Century) where men generally are ashamed to have daughters and want only sons, then this would affect the information they give you. In such a society they would talk only about their sons and not their daughters.

Therefore, when the information is unrequested, the problem becomes one of statistics - or data gathering.

I agree that it's not so clear cut with the children question, but for other types of questions it might be much stronger (*). So, it's a bad policy to assume that information given unsolicited is the same as information given to direct questions.

In any case, almost all of the errors caused in these problems are as a result of the information being given unsolicited. If you stick to the information being obtained by direct question, then there are no such issues.

(*) To give an example. Suppose you ask Ms Jones:

Are you a mountaineer? Yes
Have you climbed the Matterhorn? Yes.

Then, let's assume that 1% of climbers who have climbed the Matterhorn have also climbed Everest. Then, there is a 1% chance that Ms Jones has climbed Everest.

But, if you are talking to Ms Jones and she says:

I'm a mountaineer and I've climbed the Matterhorn.

Then, there is strong probability that if she had climbed Everest she would have told you this instead of telling you about the Matterhorn.

In this case, the probability that Ms Jones has also climbed Everest is potentially much lower than 1%. And, it cannot be determined by probability theory directly, as it involves assumptions about psychology.

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#### JeffJo

Therefore, when the information is unrequested, the problem becomes one of statistics - or data gathering.
Why? Any biases you find using statistics will apply only to the society you used to gather those statistics. I can postulate another society with the opposite biases. And since I didn't say what society this parent belongs to, you can't claim that the one you use is a better choice.

I asked a hypothetical question. Not one about a society you will choose. And before you say there is no such thing as a purely hypothetical question, you have to assume one to have the gender probabilities be equal and independent. Because neither are true if you resort to statistics.

I agree that it's not so clear cut with the children question, but for other types of questions it might be much stronger (*). So, it's a bad policy to assume that information given unsolicited is the same as information given to direct questions.
Which is why I didn't make any such assumption. Yes, I got the same answer as someone who did, but that does not mean I made that assumption. In fact, the only assumption I made was that there was no pre-determined form to the information.

And this has a direct application to the Monty Hall Problem. We don't know that Monty doesn't favor the door he opened. We also don't know that he doesn't favor the other door, and only opened the one he did because he had to. So we can only model his choice as equiprobable. Which is not the same thing as saying that we know he chose with equal probability.

And one reason this is important, is because this is not an equivalent problem:
So let's say in this version of the game that the contestant trades his 1 door for the 2 doors.

Monty then says: I'm going to open one of your 2 doors first. If neither door has the car, I don't care which one of them I open, but if one of them has the car, I'm going to open first the only one that doesn't. Either way, I'm going to open one of your doors that doesn't have the prize, whether it's the only one available for that or not, and I'm not going to open your originally chosen door right now.
The probability that this contestant gets the car is 2/3, regardless of how Monty Hall chooses a door in the case he says he doesn't care about. Because it is determined before the reveal. But if Monty Hall opens a door before the switch is offered, and we model his choice of the door he actually opened as a probability of Q, then the probability the contestant wins after switching is 1/(1+Q).

The reason the MHP continues to baffle people, even after explanations like sysprog's, is because that explanation is incorrect. It gets the right answer, because we must model Q=1/2, but it is not based on a mathematically-correct solution method. And it teaches an incorrect method, that you can ignore how information is acquired.

#### PeroK

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Why? Any biases you find using statistics will apply only to the society you used to gather those statistics. I can postulate another society with the opposite biases. And since I didn't say what society this parent belongs to, you can't claim that the one you use is a better choice.
Your asssertion that the problem is well-posed demands that there cannot be any significant factors that are unknown. You say that the answer is 1/2 and I say the problem is ill-posed. And, in general, unsolicited information is different (whether you like it or not) from information obtained from a direct question. That's a basic fact. There can be no argument about that.

I looked at the Wikipedia link and it provided the problem in a different format. It says you are simply told:

Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?

Your view if I understand it, is that the answer is unequivocally 1/2.

Whereas, my answer is that it depends how this question was arrived at.

a) If Mr Smith would have been rejected if he had two girls, hence the question always relates to boys, then the answer is 2/3.

b) If the question could equally well have been "... at least one of them is a girl ...", then the answer is 1/2.

This is critical and, ironically, is analogous to the difference between Monty Hall and the alternative. The Monty Hall problem depends not just on the evidence presented to your eyes but on knowing what options Monty has.

Likewise, the two-child problem depends on how the knowledge that Mr Smith has at least one boy was arrived at.

Your problem therefore, is that IF the "Mr Smith" question was generated by process a), then your answer of 1/2 is wrong. It's only correct IF the question was generated by process b). And, since the Wikipedia page doesn't specify how the question was arrived at, there is no definite answer and the problem is ill-posed.

#### PeroK

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PS In fact, the Wikipedia article says exactly what I am saying:

"The Boy or Girl paradox surrounds a set of questions in probability theory which are also known as The Two Child Problem,[1] Mr. Smith's Children[2] and the Mrs. Smith Problem. The initial formulation of the question dates back to at least 1959, when Martin Gardner published one of the earliest variants of the paradox in Scientific American. Titled The Two Children Problem, he phrased the paradox as follows:

• Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls?
• Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?
Gardner initially gave the answers 1/2 and 1/3, respectively, but later acknowledged that the second question was ambiguous.[3] Its answer could be 1/2, depending on what more information is available beyond that you found out just that one child was a boy. The ambiguity, depending on the exact wording and possible assumptions, was confirmed by Bar-Hillel and Falk,[4] and Nickerson.[5]"

...

This question is identical to question one, except that instead of specifying that the older child is a boy, it is specified that at least one of them is a boy. In response to reader criticism of the question posed in 1959, Gardner agreed that a precise formulation of the question is critical to getting different answers for question 1 and 2. Specifically, Gardner argued that a "failure to specify the randomizing procedure" could lead readers to interpret the question in two distinct ways:

• From all families with two children, at least one of whom is a boy, a family is chosen at random. This would yield the answer of 1/3.
• From all families with two children, one child is selected at random, and the sex of that child is specified to be a boy. This would yield an answer of 1/2.[4][5]
Grinstead and Snell argue that the question is ambiguous in much the same way Gardner did.[12]

#### JeffJo

Your asssertion that the problem is well-posed demands that there cannot be any significant factors that are unknown.
Where did I say "well posed?" I don't even think you can define what that means in general. But my point is quite the opposite of what you just said. The entire point of Probability is to model the unknown aspects of a problem. So the presence of unknown aspects is not an obstacle. And if there are no such unknown aspects, you have a deterministic problem.

The issue here is that you have decided that some categories of unknown information are unacceptable for treatment with probability. What I and trying to say is that a question is solvable if you can reasonably model the ambiguous aspects of it with probability.

You say that the answer is 1/2 and I say the problem is ill-posed.
I say that any answer except 1/2 is unacceptable, because it produces a variation of Bertrand's Box Paradox (not the Problem, the Paradox that Bertrand identified and that Jason Rosenhouse described in his book). And I go on to demonstrate a reasonable probability model that shows the answer can be 1/2. I do this by modeling the part that you consider to be ill-posed with probability.

And, in general, unsolicited information is different (whether you like it or not) from information obtained from a direct question. That's a basic fact.
It is also the point that I am trying to make. You can't assume that unsolicited information is an answer to any specific question, but you can treat the information content in the unsolicited statement as an element of an event space.

Your view if I understand it, is that the answer is unequivocally 1/2.
My point is that when the information is unsolicited, or stated without an indication of how or if it was solicited, then any answer other than 1/2 leads to a logical paradox and so is unacceptable.

My point is that I can justify 1/2 as an answer, by modeling what information was given with probability. Not that I know why the information was given.

My point is that the technique I use to do this is exactly what is needed to correctly solve the MHP. (As opposed to asserting "the other two doors start with 2/3 probability, so the chance that switching wins after the reveal must also be 2/3" which is incorrect math.)

Whereas, my answer is that it depends how this question was arrived at.
And the answer to "Will this coin land on Heads or Tails?" depends on how much torque you applied when you flipped it, among other factors. If you know these factors, the coin flip becomes deterministic and you can calculate the result. If you don't know these factors, you use probability.

If you don't know why the information was given in the TCP, you can use probability the same way.

PS In fact, the Wikipedia article says exactly what I am saying:

"The Boy or Girl paradox surrounds a set of questions in probability theory which are also known as The Two Child Problem,[1] Mr. Smith's Children[2] and the Mrs. Smith Problem. The initial formulation of the question dates back to at least 1959, when Martin Gardner published one of the earliest variants of the paradox in Scientific American. Titled The Two Children Problem, he phrased the paradox as follows:

• Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls?
• Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?
Gardner initially gave the answers 1/2 and 1/3, respectively, but later acknowledged that the second question was ambiguous.[3] Its answer could be 1/2, depending on what more information is available beyond that you found out just that one child was a boy. The ambiguity, depending on the exact wording and possible assumptions, was confirmed by Bar-Hillel and Falk,[4] and Nickerson.[5]"
Since I wrote much of that, I suggest to you that it is not contradicting what I am trying to tell you now. The reason I didn't continue, in Wikipedia, with the more detailed explanation I provide here, is because many people do not want to accept it. But there are references I could have used.

Yes, the problem is ambiguous. So is "If I flip a coin, will it land on Heads, or on Tails?" The entire purpose of probability is to handle such ambiguous situations.

With the coin, the answer isn't "Heads" or "Tails," it is the ambiguous statement "50% chance for Heads, and 50% chance for Tails." That is also the answer if we are told the coin is unfair, but not told how unfair or in which direction. This is because probability is not a property of the system, as you are trying to make it out to be. The system itself contains enough information to be deterministic. Probability is a property of what we know - or more specifically, what we don't know - about the deterministic parts of that system.

With the ambiguous question "Why do we know Mr. Smith has a boy?", the answer can only be "we can't state the absolute probability, but if it is P for a man with BB, then we must use P/2 for a man with BG or GB, P/2 for knowing he has a girl if he has BG or GB, and P for knowing he has a girl if he has GG.

I DON'T KNOW THAT THOSE PROBABILITIES ARE CORRECT IN THE ACTUAL SYSTEM. But I can't assume anything else. And regardless of what I assume, I can show that any answer other than the one I get when I use those probabilities is incorrect.

#### PeroK

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"Why do we know Mr. Smith has a boy?",
You could have asked him. In fact, you could have asked him if he had two girls and he could have said "no".

There's a similar issue if three coins are tossed and you are told that the first and the third are the same. The probability that the middle coin is the same then depends on the process by which that information was obtained. And, crucially, whether the second coin has already been looked at.

In this situation, you cannot make any meaningful progress until you resolve how the available information was generated.

Anyway, the Wikipedia article explains clearly that everyone involved in the research accepts the fundamental ambiguity in the question and the two possible interpretations.

#### stevendaryl

Staff Emeritus
The issue here is that you have decided that some categories of unknown information are unacceptable for treatment with probability. What I and trying to say is that a question is solvable if you can reasonably model the ambiguous aspects of it with probability.
I suppose you can model anything unknown using probabilities. But in general, you can't get a unique answer for some uncertainties.

Suppose a man comes up to you and tells you: "I have two children. At least one of them is a boy."

You can model it using probabilities this way:
• A = He tells you that he has two children, at least one of which is a boy.
• B = He has two boys
• C = He has one boy and one girl.
• D = He has two children, at least one of which is a boy = B or C.
So you're trying to figure out P(B|A), the probability of B being true given that A is true. Note that A and D are different events: It's possible for D to be true while A is not true.

Using Bayesian probabilities, we can calculate:

$P(B|A) = \dfrac{P(A|B) P(B | B or C)}{P(A|B) P(B | B or C) + P(A|C) P(C | B or C)}$

To solve the problem, you need to know the conditional probabilities:
1. P(B | B or C)
2. P(C | B or C)
3. P(A | B)
4. P(A | C)
I would say that you don't know any of those conditional probabilities. You might assume that each time someone has a baby, the baby is equally likely to be a boy or a girl. But that doesn't imply that all four of these possibilities are equally likely: (1) Two boys, (2) two girls, (3) oldest is a boy, youngest is a girl, (4) oldest is a girl, youngest is a boy. They may not be equally likely, because maybe the decision to have a second child depends on the sex of the first child. Maybe a couple wants a boy, and so if they have a boy first try, they stop with one child. If their first baby is a girl, they try a second time. If that's their plan, then two boys would have probability 0.

You also don't know the conditional probabilities 3 or 4.

Even if we know that the man is truthful, we haven't specified the rules for how he reports on his children. Maybe he follows this rule:

If he has two boys, he would say "I have two children, both boys".
If he has two girls, he would say "I have two children, both girls".
If he has one of each, he would say "I have two children, at least one of them is a boy"

If he follows that rule, then you know exactly that he has one boy and one girl. So P(A|B) = 0 and P(A|C) = 1.

So that gets to the question of "ill-posed" problems. A problem is ill-posed (I think this is what it means) if it is impossible to solve without making additional assumptions that were not implied by the original statement of the problem.

#### JeffJo

JeffJo said:
"Why do we know Mr. Smith has a boy?"
You could have asked him. In fact, you could have asked him if he had two girls and he could have said "no".
Or you could just know he participates in Boy Scout events. There are lots of ways, many of which don't involve asking him vague questions. The point you are (purposely?) ignoring is that we don't know how. So speculation is irrelevant.

What you are also ignoring, is that we don't need to know , in order to assess a probability for the general case where we have this information. We do need to know it for a specific situation, but we don't have that information.

I suppose you can model anything unknown using probabilities. But in general, you can't get a unique answer for some uncertainties.
And my point is that, with the information we have, any answer other than 1/2 creates a paradox that makes that answer unacceptable. While a simple, and reasonable, application of the Principle of Indifference makes the answer 1/2.

Suppose a man comes up to you and tells you: "I have two children. At least one of them is a boy."

You can model it using probabilities this way:
• A = He tells you that he has two children, at least one of which is a boy.
• B = He has two boys
• C = He has one boy and one girl.
• D = He has two children, at least one of which is a boy = B or C.
...

Note that A and D are different events: It's possible for D to be true while A is not true.
Indeed. That's part of my point. But I prefer:
• B2 = He has two boys.
• B1G1 = He has one boy and one girl
• G2 = He has two girls.
• ALOB = He tells you that he has at least one boy.
• ALOG = He tells you that he has at least one girl.
• SE = He tells you something else (including nothing) about genders.
Can we agree that {B2,B1G1,G2}x{ALOB,ALOG,SE} is a nine-element partition of the sample space, given that he has two children?

Then, since we don't know why he would say this, the only things we can assume about probabilities are:
• Pr(B2) = P2(G2) = 1/4.
• Pr(B1G1) = 1/2.
• Pr(ALOB|B2) = Pr(ALOB or ALOG|B1G1) = Pr(ALOG|GG). Call this value Q.
• Pr(ALOB|B1G1) = Pr(ALOG|B1G1) = Q/2.
• Pr(ALOB|G2) = Pr(ALOG|B2) = 0.
From this, we can deduce Pr(B2|ALOB)=1/2.
I would say that you don't know any of those conditional probabilities.
And there is a difference between:
1. Knowing a situation-specific conditional probability, that applies to one instance of a random procedure, and...
2. Assigning a conditional probability to the general case of that random procedure.
You are saying we don't know #1. I agree. I'm solving a question about #2. I never said I know the conditional probabilities that apply specifically to your Mr. Smith. I said the Principle of Indifference applies to the population of all Mr. Smiths who end up in a similar situation.

Similar example: Some studies claim to show a correlation between how you hold a fair coin before flipping it, and the result of the flip. If true, that still doesn't change the answer to "What is the probability that flipping this fair coin will result in Heads?" The answer, IN GENERAL, is 1/2, You don't know how I intend to hold it, and you can only assume there is a 50% chance either way. The fact that the information exists cannot affect the probability you use IF YOU DON'T KNOW IT.

#### sysprog

And this has a direct application to the Monty Hall Problem. We don't know that Monty doesn't favor the door he opened. We also don't know that he doesn't favor the other door, and only opened the one he did because he had to. So we can only model his choice as equiprobable. Which is not the same thing as saying that we know he chose with equal probability.
In the original problem, the reasonable understanding of the rules for Monty's options is not at issue -- we know that Monty will not open the contestant's door, and that he will not reveal the car -- if he does either of those, the game is over -- and we reasonably assume that, as far as the contestant knows, if Monty has a choice, he will make it in a manner that does not allow the contestant to distinguish cases in which Monty has a choice from those in which he doesn't.

If, among the non-chosen doors, he always (one of the 2 extrema of non-random choices, the other of which is the equally definite never) opens the 1st unchosen door when he has a choice, that matters if and only if the contestant knows that, and it means that if Monty opens the 2nd unopened door, the contestant knows for certain that he did it because he had no choice but to do so, wherefore the car must be behind the 1st unopened door, so he should switch, but if Monty opens the 2nd unopened door, the contestant knows that Monty had a choice. which means that both the unchosen doors have goats behind them, wherefore the car must be behind his originally chosen door.
This is critical and, ironically, is analogous to the difference between Monty Hall and the alternative. The Monty Hall problem depends not just on the evidence presented to your eyes but on knowing what options Monty has.
It's not a reasonable problem if the reasonable assumptions aren't made. We know that Monty doesn't open the contestant's door, or the door with car behind it, or there wouldn't be a problem. What we don't know, is which of the 2 doors Monty will open. The only reasonable assumption is that the contestant has no basis for determining, based on which of the 2 doors Monty opens, whether that was the only one he could have opened, or was one of 2.

If the assumption is to be made that the contestant can determine, by which door was opened, either that it was the only possibility, in which case the contestant knows that he wins by switching, or that it was 1 of 2 possibilities, in which case the contestant knows that he wins by not switching, the problem wouldn't have been posed. That seems obvious enough to me to take the contrary assumption for granted.
And one reason this is important, is because this is not an equivalent problem:
sysprog said:
So let's say in this version of the game that the contestant trades his 1 door for the 2 doors.

Monty then says: I'm going to open one of your 2 doors first. If neither door has the car, I don't care which one of them I open, but if one of them has the car, I'm going to open first the only one that doesn't. Either way, I'm going to open one of your doors that doesn't have the prize, whether it's the only one available for that or not, and I'm not going to open your originally chosen door right now.
The probability that this contestant gets the car is 2/3, regardless of how Monty Hall chooses a door in the case he says he doesn't care about. Because it is determined before the reveal.
It's not determined before the reveal any more than it is in the original problem -- the contestant who switched to the 2 doors before the reveal is allowed to switch back to his 1 original door after the reveal, which is equivalent to sticking with the same door after the reveal in the original problem, and in this 2nd version of the problem, the contestant that switched to the 2 doors is allowed to stick with the 1 of those doors that remains after the reveal, which is equivalent to the contestant in the original problem being allowed to switch after the reveal.
But if Monty Hall opens a door before the switch is offered,
It doesn't really matter when the reveal is done. It's just flash meant to obscure the fact that the contestant's original door had only 1/3 chance at the outset, and the other 2/3 chance is still behind the set of doors other than the one originally chosen. The reveal does nothing to change that. The contestant already knew that at least one of the non-chosen doors had a goat. All the reveal really tells the contestant is that of the 2 originally unopened not-chosen doors, that together contained 2/3 chance at the outset, one of the 2 doors no longer holds any of that 2/3 chance, so that means the other one holds the 2/3 chance by itself.
and we model his choice of the door he actually opened as a probability of Q,
That requires an unreasonable interpretation of the problem. Monty's not going to give away where the car is by letting the contestant know whether the door opened was the only option or was instead 1 of 2. People who get it wrong aren't tripping up over that.
then the probability the contestant wins after switching is 1/(1+Q).
The chance, from the contestant's perspective, is 2/3, if Monty always chooses randomly, or always chooses in any way that doesn't clue in the contestant about a bias, whenever (2 out of 3 times) he has a choice -- "Monty has 1 option" is twice as likely as "Monty has 2 options" -- in either case he exercises an option to open 1 non-car door out of the 2 doors, and it is not the case that by doing that, he changes what the contestant knows about Monty's options, or about the likelihood of the contestant's original choice having been the best one.
The reason the MHP continues to baffle people, even after explanations like sysprog's, is because that explanation is incorrect.
I disagree with that diagnosis. I've seen a lot of good illustrations and explanations that are not incorrect, that nevertheless fail to persuade people who are truly recalcitrant about this. I've never seen anywhere else an illustration quite like that one, in which Monty's rules for what he's doing are explicitly stated to the contestant by Monty, instead of merely being presented, explicitly or implicitly, as part of the problem, but that doesn't convince me that it will win anyone over. I do think that the illustration might make it easier for some people to see why the 2/3 probability for the unopened door not originally chosen is correct.
It gets the right answer, because we must model Q=1/2, but it is not based on a mathematically-correct solution method.
I use the reasonable assumption that the contestant can't tell by which door is opened, anything about whether it's the only option or not. To make that more explicit in the illustration, Monty says that if he has a choice of doors he doesn't care (i.e. doesn't have a bias regarding) which one he opens.

My illustration didn't say anything more than that about what you're calling Q. It did present the contestant being confronted by a 1-car 2-door scenario that Monty nudges the contestant to assume, but does not expressly state, might indicate a 1/2 probability. but in both the original problem, and in my second version of it, the 2-door not-originally-chosen subset always contains 2/3 of the probability, and the reveal does not affect anything about that, except which member of that subset the 2/3 aggregates behind. My illustration, by placing the contestant's possession of the aggregation ahead of his final choice, tends, I think, to obscure that fact less than it is, at least for some recipients of the problem, obscured in the original problem statement.

In my illustration, Monty vocally stating his choice rules to the contestant, I think, makes them more conspicuous than they are when placed in the prologue of the problem statement.
And it teaches an incorrect method, that you can ignore how information is acquired.
I didn't teach an incorrect method. I just provided yet another illustration of the unchanging fact that the 1-door subset consisting of the originally-chosen door, persists in containing 1/3 chance, while the 2-door subset, consisting of the 2 not originally-chosen doors, persists in containing 2/3 of the chance, even after that subset's number of unopened-door members is reduced to 1. In my illustration, Monty says he won't reveal the prize, and he'll definitely open a door, and he won't open the originally-chosen door. Those are the same conditions as in the original problem. That should be enough for the contestant, and the problem recipient, to infer that the 1 unopened-door member of the original 2-door 2/3-chance-containing subset at time of inquiry contains the same 2/3 chance that the original 2 members of that original 2-door 2/3-chance-containing subset originally between them contained.

My point is that the technique I use to do this is exactly what is needed to correctly solve the MHP. (As opposed to asserting "the other two doors start with 2/3 probability, so the chance that switching wins after the reveal must also be 2/3" which is incorrect math.)
It's not incorrect math. Your abbreviated version skips some details that the reasoning I presented didn't skip. I'll elaborate on my reasoning here, with the hope that you'll see that it's not incorrect, despite not being the same methodologically as yours.

The placement of the car behind 1 door and the goats behind the 2 others creates a hidden partitioning (p1) of the set of 3 doors into 2 subsets: 1 subset (p1s1) consisting of 1 member door behind which is the car and another subset (p1s2) consisting of 2 member doors behind each of which is a goat.

The original choosing of a door by the contestant creates a non-hidden second partitioning (p2), also into 2 subsets, also 1 having 1 member and 1 having 2 members, the 1-member subset (p2s1) consisting of the 1 door chosen, and the 2 member subset (p2s2) consisting of the 2 doors not chosen.

The subset in p2 that has 1 member, p2s1, has 1/3 chance of being identical to p1s1, the subset in p1 that has 1 member, i.e. the originally chosen door has a 1/3 chance of the car being behind it.

The subset in p2 that has 2 members, p2s2, has 2/3 chance of including p1s1, the subset in p1 that has 1 member, i.e. the 2 doors not originally chosen have between them a 2/3 chance of the car being behind one or the other of them.

Given the fact that it is known from the moment of the creation of the 2nd partitioning that there is at least 1 member of p2s2 that is also a member of p1s2, i.e. that at least 1 of the 2 doors not chosen has a goat behind it, provided that the content of neither partitioning's 1st subset is directly disclosed, the revealing of what is behind 1 of the members of the 2nd subset in the 2nd partitioning, thereby revealing it to be also a member of the 1st partition's 2nd subset, does not further than that unhide the 1st partitioning.

It does not render it inferable, unless we make the unreasonable assumption that which of the members of p2s2 is revealed to also be a member of p1s2 somehow discloses whether it is the only member of p2s2 that is a member of p1s2 or is 1 of 2 members of p2s2 that are also members of p1s2.

Wherefore, the opening of a goat door in the situation does not alter the probability distribution of the 1st subset in the 1st partitioning over the 2nd partitioning, i.e. the subset in p2 that has 1 member, p2s1, continues to have 1/3 chance of being identical to p1s1, the subset of p1 that has 1 member, i.e. the originally chosen door continues to have a 1/3 chance of the car being behind it, and the subset in p2 that has 2 members, p2s2, the content behind both of which was initially obscured and the content behind 1 of which has been shown to make it a member of p1s2, continues to have 2/3 chance of including p1s1, the subset in p1 that has 1 member, i.e. the 2 doors not originally chosen continue to have between them a 2/3 chance of the car being behind one or the other of them, even though 1 of them has been eliminated as a possibility.

#### stevendaryl

Staff Emeritus
And my point is that, with the information we have, any answer other than 1/2 creates a paradox that makes that answer unacceptable. While a simple, and reasonable, application of the Principle of Indifference makes the answer 1/2.
I don't think that the principle of indifference can give you an answer. Like I said, we haven't stated the key facts:
1. What is the probability of a random person having 2 boys versus 2 girls versus 1 each?
2. For each combination of children, what is the probability that someone would utter the statement "I have two children, at least one of which is a boy"?
Unlike flipping a coin, where there is a certain "symmetry" between the heads and tails outcomes, I just don't see any kind of symmetry among the possibilities that would warrant any particular answer to questions 1 & 2. We can certainly make a guess about the answers to 1 & 2, and then calculate a probability based on that guess, but there is no principled reason for making one guess over another.

Indeed. That's part of my point. But I prefer:
• B2 = He has two boys.
• B1G1 = He has one boy and one girl
• G2 = He has two girls.
• ALOB = He tells you that he has at least one boy.
• ALOG = He tells you that he has at least one girl.
• SE = He tells you something else (including nothing) about genders.
Can we agree that {B2,B1G1,G2}x{ALOB,ALOG,SE} is a nine-element partition of the sample space, given that he has two children?

Then, since we don't know why he would say this, the only things we can assume about probabilities are:
• Pr(B2) = P2(G2) = 1/4.
Maybe that's a reasonable assumption, but it is an assumption. What if, as I said, the man wanted a boy and would have stopped at one child if that one child had been a boy? That's a possibility. What's the weight that you would give to that possibility? There is no principled way of giving one weight over another. If that possibility were true, then the probability of two boys would be zero.

Maybe you can say that it's far-fetched to consider such bizarre twists. I guess I would agree. But to me, there is a distinction between (1) solving a problem based on the information given, and (2) solving a problem based on the information given plus auxiliary reasonable assumptions.

#### JeffJo

Since the others don't seem to want to consider what I say, as opposed to parroting what they think must be correct, I'm bot going to keep repeatgingmyself after this reply.
In the original problem, the reasonable understanding of the rules for Monty's options is not at issue -- we know that Monty will not open the contestant's door, and that he will not reveal the car -- if he does either of those, the game is over -- and we reasonably assume that, as far as the contestant knows, if Monty has a choice, he will make it in a manner that does not allow the contestant to distinguish cases in which Monty has a choice from those in which he doesn't.
The part I put in red is the critical part. It is what I have been trying to say, about the TCP, so itis unclear why you think you need to explain it to me.

In the MHP, contestant does not know how Monty Hall chooses a door to open when there are two goat doors he could open. The contestant does not assume that, however Monty Hall chooses, it results in a 50%:50% split between the two doors. The contestant only assumes that there is no information that could allow him, THE CONTESTANT, to consider either door more, or less, likely to be opened than the other. So when he models the system, he assumes that from his perspective, not the host's, that probability split must be 50%:50%. The result is that the probability that switching wins is 1/(1+Q) where Q=1/2, the split factor he assumed. The incorrect answer, that switching can't matter, comes from implicitly assuming Q=1.

In the TCP, the problem solver does not know why he was told that there was at least one boy. So the solver does not know why the given information should be "boy" instead of "girl" when there is one of each in the family. The solver does not assume that, however the information came to be given, it results in an even split between the two genders. doors. The solver only assumes that there is no information that could allow him, THE SOLVER, to consider either gender, or less, likely to be given. So when he models the system, he assumes that from his perspective, not anybody else's, that probability split must be even. (A similar argument applies to the form "at least one X" over all gender combinations.) The result is that the probability of two boys is 1/(1+2*Q) where Q=1/2, the split factor he assumed. The incorrect answer, that this probability is 1/3, comes from implicitly assuming Q=1.

It doesn't really matter when the reveal is done
Yes, it does. If you choose to a switch before a door is opened, you are trading the 1/3 prior probability of winning with the original choice for the 2/3 prior probability that it loses. If you trade after, you are trading the Q/(1-Q) conditional probability that the original door wins, given what was opened, for the 1/(1-Q) conditional probability that it loses. If Q=1/2, the values are the same, but the reason for those values are different. The point is that only one these different reasons is correct for the original problem. Getting the right result does not make the other valid, and people do recognize your reasons are wrong.

I don't think that the principle of indifference can give you an answer.
Then we have different ideas of what the PoI is. To me, it means that if there are n mutually exclusive cases whose union is an event with probability P, and they are indistinguishable except by name to an observer, then to that observer each has probability P/n.

Like I said, we haven't stated the key facts:
1. What is the probability of a random person having 2 boys versus 2 girls versus 1 each?
2. For each combination of children, what is the probability that someone would utter the statement "I have two children, at least one of which is a boy"?
This is a puzzle, so we assume genders are equiprobable and independent. So Pr(2 boys)=Pr(2 girls)=2*Pr(1 each). And I thought I made it clear that I don't know, and don't care, what that probability is. But that I can only consider it to be the same as the probability that someone would utter the statement "I have two children, at least one of which is a girl." If you feel differently, please explain how you justify that.

#### sysprog

In the MHP, contestant does not know how Monty Hall chooses a door to open when there are two goat doors he could open. The contestant does not assume that, however Monty Hall chooses, it results in a 50%:50% split between the two doors. The contestant only assumes that there is no information that could allow him, THE CONTESTANT, to consider either door more, or less, likely to be opened than the other. So when he models the system, he assumes that from his perspective, not the host's, that probability split must be 50%:50%. The result is that the probability that switching wins is 1/(1+Q) where Q=1/2, the split factor he assumed. The incorrect answer, that switching can't matter, comes from implicitly assuming Q=1.
I think that's a misdiagnosis. I think it's more likely that the most common reason for the incorrect 50-50 answer comes from people who see two doors, one of which has the car, and the other of which has the goat, and are thereby persuaded to regard the situation as the same as if those were the initial conditions and they had no basis for preferring 1 door over the other.

The correction of that, although it can be done by analysis of conditional probabilities, doesn't have to be: in this problem, the rules, inclusive of the reasonably assumed one, allow the contestant to simply recognize that the reveal doesn't change the 1/3 probability of his original choice, wherefore the original 2/3 probability for the 2 other doors must be the same 2/3 between them, including after the opening of 1 of them.
If you choose to a switch before a door is opened, you are trading the 1/3 prior probability of winning with the original choice for the 2/3 prior probability that it loses. If you trade after, you are trading the Q/(1-Q) conditional probability that the original door wins, given what was opened, for the 1/(1-Q) conditional probability that it loses. If Q=1/2, the values are the same, but the reason for those values are different. The point is that only one these different reasons is correct for the original problem. Getting the right result does not make the other valid, and people do recognize your reasons are wrong.
In this problem, the necessity of differentiating between the prior and the conditional probabilities is illusory. There's nothing I'm claiming to be wrong with the conditional probability analysis that you presented; it's sufficient, but not necessary, for arriving at the correct 2/3 answer. You have not shown your reasoning to be necessary; you've shown it only to be sufficient. You persistently claim that my reasoning is wrong despite it producing the correct answer. If my reasoning is wrong, then there must be some set of conditions, consistent with the problem definition, under which it could produce an incorrect answer.

It appears to me that you are asserting that conditional probability analysis is both sufficient and necessary, while you are demonstrating only that it's sufficient.

I think my reasoning is sufficient, and I think I've adequately demonstrated that. I also think you would have to justify your claim that it's insufficient or invalid or incorrect, by more than bare assertion, to show that it's not a legitimate counterexample that falsifies your apparent claim that the conditional probability analysis you present is necessary for legitimately and correctly arriving at the correct answer.

#### DaveE

I'm wondering if this thread might actually go on forever.
Not complaining or criticizing, just curious.
What started as a fascinating (albeit, well worn) probability puzzle is slowly morphing into a study in human psychology.

#### PeroK

Homework Helper
Gold Member
2018 Award
It appears to me that you are asserting that conditional probability analysis is both sufficient and necessary, while you are demonstrating only that it's sufficient.

I think my reasoning is sufficient, and I think I've adequately demonstrated that. I also think you would have to justify your claim that it's insufficient or invalid or incorrect, by more than bare assertion, to show that it's not a legitimate counterexample that falsifies your apparent claim that the conditional probability analysis you present is necessary for legitimately and correctly arriving at the correct answer.
Absolutely. This is fundamentally the issue. Most of us can explain it with different emphases: Bayesian, frequentist, whatever. And see what's good or interesting about another person's analysis. Even provide more than one alternative analysis ourselves!

There's more than one way to make a jam sandwich, as they say!

Gold Member

#### stevendaryl

Staff Emeritus
Then we have different ideas of what the PoI is. To me, it means that if there are n mutually exclusive cases whose union is an event with probability P, and they are indistinguishable except by name to an observer, then to that observer each has probability P/n.
Yes, and that principle does not give you a unique answer to the question "What is the probability that a man has two boys, given that he says that he has two children, at least one boy?"

This is a puzzle, so we assume genders are equiprobable and independent. So Pr(2 boys)=Pr(2 girls)=2*Pr(1 each).
That does not follow from the principle of indistinguishability. Don't you agree that it is possible that someone prefers boys? Or that they would like to have one of each? Assuming that every time a baby is born, it is equally likely to be a boy or a girl does not imply that Pr(2 boys)=Pr(2 girls)=2*Pr(1 each).

What you're doing is making up additional (reasonable, but additional) assumptions.

#### stevendaryl

Staff Emeritus
Yes, and that principle does not give you a unique answer to the question "What is the probability that a man has two boys, given that he says that he has two children, at least one boy?"
I'm actually not sure if there is a clear distinction between (A) applying the principle of indifference and (B) making auxiliary assumptions in order to solve a problem.

#### sysprog

Yes, and that principle does not give you a unique answer to the question "What is the probability that a man has two boys, given that he says that he has two children, at least one boy?"
That does not follow from the principle of indistinguishability. Don't you agree that it is possible that someone prefers boys? Or that they would like to have one of each? Assuming that every time a baby is born, it is equally likely to be a boy or a girl does not imply that Pr(2 boys)=Pr(2 girls)=2*Pr(1 each).

What you're doing is making up additional (reasonable, but additional) assumptions.
"What is the probability that a man has two boys, given that he says that he has two children, at least one boy?"

A non-exhaustive list of what I think are reasonable assumptions not explicit in the problem statement is as follows:
1. the man is not lying or mistaken about the number or gender of his children.
2. When the man says he has 2 children he means exactly 2 (not more than 2).
3. There are exactly 2 genders of children: boys and girls.
4. The independent gender probability is 1/2 for each gender for any child the gender of which is not disclosed.
5. The man did not do anything to bias the likelihood of either gender.
6. The gender distribution between his 2 children did not determine or influence whether he said or did not say something about it.

If we ask the man, "why did you say 'at least one of them', instead of saying 'one of them'?", and he replies "because the 2nd of them isn't born yet, and I won't know whether it's a boy or not until the birth happens."

At that point, the problem is the same as it would be if the man had said "at least one of them, the older one, is a boy", and the correct answer to the question "what is the probability that both children are boys?" would be 1/2.

If, however, in answer to our question, he said "I only wanted to disclose that they weren't both girls", that would mean that there were 3 remaining possibilities for the gender distribution over his children, one of which is both boys, and 2 of which are 1 of each, and the correct answer to the question "what is the probability that both children are boys?" would be 1/3.
I'm actually not sure if there is a clear distinction between (A) applying the principle of indifference and (B) making auxiliary assumptions in order to solve a problem.
I think it's reasonable to say that (A) is an instance of (B).

#### Eric Bretschneider

In the scenario you described you can start the game with just two doors left, since the host opens one empty door before you even chose one. He could have done that before you enter the room and it wouldn't change anything, so in your described scenario the chance of getting the car should be 50-50. The original problem is based on the assumption that you have to choose one door before the host opens an empty one:

1. Choose a door
2. The host opens an empty door
3. You have to decide if you switch to third door or stick with your initial choice

In that case you can choose one of three doors: the car door $C$ or one of the empty doors $E1$, $E2$ - each of the choices is made with the same probability:

1 x 1/3) You choose $E1 \rightarrow$ the host opens $E2\rightarrow$ change = win, stick = lose
1 x 1/3) You choose $E2 \rightarrow$ the host opens $E1\rightarrow$ change = win, stick = lose
1 x 1/3) You choose $C \rightarrow$ the host opens an empty door $\rightarrow$ change = lose, stick = win

In two thirds of your initial choices changing leeds two winning the car, in one third you lose. Therefore you should change.
Your answer includes a dangerous and misleading compression of data. If you happen to choose the door with the car behind it, then there are two scenarios for the door that is opened, not one. Monty can choose either door without a car - there isn't just one door for him to choose. If you chose one of the two wrong doors then Monty opens the other wrong door, but there are two configurations for this as well.

Choice Car Open Change
1______1_____2______Lose
1______1_____3______Lose
1______2_____3______Win
1______3_____2______Win

2______1_____3______Win
2______2_____3______Lose
2______2_____1______Lose
2______3_____2______Win

3______1_____2______Win
3______2_____1______Win
3______3_____1______Lose
3______3_____2______Lose

If you work through the full probability chart, then there are a total of 12 scenarios

Your odds of winning are 50/50. Thus sanity is restored to the world.

#### Eric Bretschneider

Let's work this another way.

You choose a door, Monty opens another door (big surprise no car) and offers you the chance to change doors.

If the door you are offered has a 50% chance of having the car behind it, then the door you selected must also have a 50% chance of having the car behind it. The sum of probabilities is after all 100%.

If, as supposed by so many, the door you selected has a 1/3 (33.33%) chance of having the door behind it then the sum of probabilities is 83.33%. If you believe the door that was opened also had a 1/3 (33.33%) chance of having the car behind it, then the sum of probabilities is 116.66% which is also problematic.

#### stevendaryl

Staff Emeritus
Your answer includes a dangerous and misleading compression of data. If you happen to choose the door with the car behind it, then there are two scenarios for the door that is opened, not one. Monty can choose either door without a car - there isn't just one door for him to choose. If you chose one of the two wrong doors then Monty opens the other wrong door, but there are two configurations for this as well.

Choice Car Open Change
1______1_____2______Lose
1______1_____3______Lose
1______2_____3______Win
1______3_____2______Win

2______1_____3______Win
2______2_____3______Lose
2______2_____1______Lose
2______3_____2______Win

3______1_____2______Win
3______2_____1______Win
3______3_____1______Lose
3______3_____2______Lose

If you work through the full probability chart, then there are a total of 12 scenarios

Your odds of winning are 50/50. Thus sanity is restored to the world.
Think about what you're saying. You think that switching or keeping your first choice gives you equal probability of winning: 50/50. So there is no reason, according to you, to switch. You might as well just always keep your first choice.

So now think about what you're saying. If you're always going to keep your first choice, then there is no reason to pay any attention to what Monty Hall says. Ignore him. If you ignore him, then you win if your first pick was the one that had the car. What are the odds that your first choice had a car behind it? Remember, there are three doors to choose from. And you think that choosing one at random gives you a 50/50 chance of picking the door with the car?

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#### Eric Bretschneider

Think about what you're saying. You think that switching or keeping your first choice gives you equal probability of winning: 50/50. So there is no reason, according to you, to switch. You might as well just always keep your first choice.

So now think about what you're saying. If you're always going to keep your first choice, then there is no reason to pay any attention to what Monty Hall says. Ignore him. If you ignore him, then you win if your first pick was the one that had the car. What are the odds that your first choice had a car behind it? Remember, there are three doors to choose from. And you think that choosing one at random gives you a 50/50 chance of picking the door with the car?
That is the reason for my second post. If the door I am offered after one has already been opened has a 50/50 chance of having the car, then what is the sum of probabilities? Are you saying that the sum of probabilities doesn't have to be 100%?

When you say that Monty opens one door you and shows it to be empty, you are ignoring the fact that if you chose the correct door then he has two options and not just one.

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