JeffJor

Mr. Bertrand has (exactly - this needs to be included) two children (not twins, which is not quite the same as different ages). A gender, and a day of the week, that apply to at least one of his children have been written inside a sealed envelope. What is the probability that both children have that gender?

In this problem, we have no gender- or day-specific information. So the answer can only be the probability that he has two of the same gender. Which is 1/2.

Now open the envelope. If the answer changes to P based on what you see written, it has to change to the same P regardless of what you see written. Which means you didn't need to unseal the envelope; the answer was P before, not 1/2.

This is what Joseph Bertrand identified as his Box Paradox in 1889. That word was used to describe an actual contradiction, not a non-intuitive result. It disproves any answer except P=1/2. FOR ANY OF THESE PROBLEMS.

In fact, it is the same reason why the Monty Hall Problem's answer is what it is. Many "explanations" will claim that your original probability can't change, but never justify it. This is the justification - if it changes when one door is opened, it must change the same way when either door is opened.

And when you don't know, you have to assume it is 50:50. Otherwise you get different answers for different-but-equivalent information.

The reasons why the original problem is so confusing is the same reason why the Monty Hall is so confusing: people have different understandings of the question, and don't realize it in discussions.

Almost everybody understands the same problem, AND STILL GET DIFFERENT ANSWERS. If they don't understand it, they make pedantic arguments about Monty's motivations. All of which make the puzzle impossible to answer.

What they don't understand is probability. Probability is a measure of the information you lack about what causes a certain result to occur. That includes the physical details (where the prize is, what the genders are) but also the choices made for hidden reasons.

In the Monty Hall Problem, to reduce complexity, label the doors C (the contestant's original door), R (the door to its right, wrapping around if necessary), and L (the door to its left). What leads up to the game state at the time the decision to switch is made are (A) Where the prize was placed and (B) How Monty Hall chooses a door to open if the prize is behind C.

The naive answer is based on only (A). The two unopened doors (C and R, or C and L) started with the same probability. So they must now have the same probability, 1/2, right? No, wrong, because we need to take (B) into account. If the prize is behind R then the host had to open L. If the prize is behind L then the host had to open R. But if the prize is behind C then the host had to choose. Since we don't know how, we have to assume there was a 50% chance that he would choose R, and 50% for L. Once we see him open, say, R? This 50:50 reduces the probability that the prize is behind C, so switching becomes twice as likely to win.

The Two Child Problem works exactly the same way. What leads up to the point where we are asked for a probability is (A) the gender makeup of the family and (B) how the information came to us if there is a boy and a girl.

The naive answer is based on only (A). A mixed family is twice as likely as either two-of-a-kind family. So the probability of two-of-a-kind is 1/3, right? Wrong, unless we know WITH CERTAINTY that we could not have learned about the other gender. If we do not have that certainty, then just like with Monty Hall we have to assume that half of the time in a mixed we would have learned the other gender. This makes a mixed family half as likely as (A) alone would suggest; in other words, the same as two-of-a-kind.The answer is 1/2.

Joseph Bertrand pointed out, in 1889, why we need to take (B) into account. Martin Gardner, who originated the Two Child Problem, repeated it in 1959. In the same article where he introduced the predecessor to Monty Hall (called the Three Prisoners Problem), and explained why (B) is important. It should be embarrassing to anyone who thinks that the "Tuesday" variation's answer is 13/27. Because it was first mentioned at a puzzle convention named in honor of Martin Gardner and forgot his warning. Adding irrelevant information can't change the answer, and if you take (B) into account the answer doesn't change.

I am not making an assumption about the data-generating process in any of these questions. The only “assumptions” I make are that the information is true (so yes, the envelope in Q3 matches the family), that the information (whether or not it is sealed) is about one gender, and that information WILL NOT be sufficient to determine the complete makeup of the family. Suggesting otherwise is usually a sign that you have reached your answer first, and are trying to justify it by assertion.

Your error is that you seem to be deciding the answer is 1/3 first, forcing you to assume whatever makes that so. You literally said that when you said the probability must become 1/3.

I do look at this problem from the beginning with no assumptions. If you want to be pedantic, the probability space comprises the sample space (the set of possible outcomes), an event space (a set of subsets of the sample space with certain properties), and a probability function Pr(*) that maps each event in the event space to a number in [0,1]. The complete sample space is {BBb, BBg, BGb, BGg, GBb, GBg, GGb, GGg}, which I assume is what you want (B,G)^3 to mean. We then need the events {BBb, BBg}, {BGb, BGg}, {GBb, GBg}, and {GGb, GGg} to all have probability 1/4.

Since we make no assumptions about the how the lower-case letter is “generated” other than IT MUST MATCH ONE OF THE UPPER CASE ONES, we get that Pr({BBb} = Pr(GGg} = 1/4 and Pr({BBg}) = Pr({GGb}) = 0. What we need to determine is how we get Pr({BGb})+Pr({BGg}) = Pr({GBb})+Pr({GBg}) = 1/4.

In order to make the answer become 1/3 in Q1 and Q2, as you assert, we must assume that we do know how the lower-case letter is generated. In Q1 and Q2 we must assume it is an answer to “is there a girl/boy.” This makes one half of each pair 1/4, and the other 0. If we make no assumption, then the Principle of Indifference (literally, that we make no assumptions to distinguish functionally equivalent outcomes) says Pr({BGb}) = Pr({BGg} = Pr({GBb}) = Pr({GBg}} = 1/8. This makes one answer:

A1 = Pr({GGg}) / [Pr({BGg}) + Pr({GBg}) + Pr({GGg})] = (1/4) / [1/8 + 1/8 + 1/4] = 1/2

Yes, this is a variation of the Monty Hall Problem. Most "solutions" to it are really just explanations for how it can make sense. The mathematical solution follows the outline I used above. It is based on the probability, if the door you choose has the prize (compare to a mixed-gender family), that Monty will open door X or door Y (i.e., the other two) as determined PRIOR TO it happening. If you assume it is 100% for the door he did open, which is only determined AFTER HAVING SEEN IT, like you want to assume in Q1 and Q2 that only a girl/boy can be mentioned, then the answer is that switching does not matter. It is only if you use the Principle of Indifference – meaning each has a 50% chance – that the answer is that switching wins 2/3 of the time.

Bertrand's Box Paradox, which I wrote about in my own comment, applies to it. The upshot is that probability is not based on which prize placements _could_ lead the current game state, it is the set of all possible game states. Lets assume that the contestant starts off with door #3.

Case 1: The prize is behind door #1, and the host must open door #2. Probability 1/3.

Case 2: The prize is behind door #2, and the host must open door #1. Probability 1/3.

Case 3: The prize is behind door #3, and the host has a choice. Case 3A: The host opens door #1. Probability Q/3. Case 3B: The host opens door #2. Probability (1-Q)/3.

If the host actually opens door #1, the probability that door #2 has the prize is (Case 2)/(Case 2 + Case 3A) = (1/3)/(1/3+Q/3) = 1/(1+Q).

If the host actually opens door #2, the probability that door #1 has the prize is (Case 1)/(Case 1 + Case 3B) = (1/3)/(1/3+(1-Q)/3) = 1/(2-Q).

My point is that, since you get to see which door is opened, 2/3 is correct only if you assume Q=1/2. We aren't told what Q is, but we must assume it is 1/2 because otherwise the answer is different depending on which door is chosen.

Q1: "A family has two children. You're told that at least one of them is a girl. What's the probability both are girls?"

Q2: "A family has two children. You're told that at least one of them is a boy. What's the probability both are boys?"

Note that these are symmetric problems, and must have the same answer.

Q3: "A family has two children. You're told that a gender, that applies to at least one, is written inside a sealed envelope. What's the probability both have that gender?"

In Q3, we have no information. So the answer is the proportion of two-child families that are single gendered. That is, 1/2.

But if we open the envelope, and read what is written inside, the problem becomes either Q1 or Q2. Which have the same answer. So we don't have to open it; whatever the answer to Q1 and Q2 is, opening the envelope in Q3 make its answer the same. If that answer is 1/3, we have a paradox. The answer has to be 1/2 of we don't look.

This is what is known as "Bertrand's Box Paradox." Well, if we add a fourth box to his problem, with one gold and one silver coin. I realize that in modern times the problem itself is called the paradox, but what Bertrand actually wrote (edited to this problem) was "How can it be that opening the envelope suffices to change the probability from 1/2 to 1/3?"

The resolution is that probability must be based on the full set of possibilities, not the possibilities that _could_ result from the full set of _states._ These are the possibilities for this problem:

1) BB and you are told that there is at least one boy. 2A) BG and you are told that there is at least one boy. 2B) BG and you are told that there is at least one girl. 3A) GB and you are told that there is at least one boy. 3B) GB and you are told that there is at least one girl. 4) GG and you are told that there is at least one girl.

Each numbered case has a prior probability of 1/4. Let's say the "A" subcases have a probability of Q/4, so the "B" subcases have a probability of (1-Q)/4.

The answer to the first problem is the probability of case 1, which is 1/4, divided by the total probability of cases 1, 2A, which is (1+2Q)/4. That's 1/(1+2Q).

The answer to the second problem is the probability of case 4, divided by the total probability of cases 4, 2B, and 3B. Which is (3-2Q)/4.

Bertrand's paradox, stated another way, is that these must be equal, but can only be equal if Q=1/2 and both answers are 1/2.

The problem is ambiguous, due to under specification. That means that neither #1 nor #2 is "actually equivalent" to "at least one child is a boy," and more information is needed to construct a probability space.

#1 is "When both genders are known, and boys are preferred in the description, at least one is a boy." The preference is what makes the answer 1/3, and assuming it adds information to the problem.

#2 can be "When only one gender is known, and how we know it is uncorrelated with either possibility, at least one is a boy." But it can also be "When both genders are known, and the description reflects the probability of that gender being chosen at random from the two, at least one is a boy." In both cases, the answer is 1/2.

But being under-specified does not mean the question can't be answered, it just requires applying a reasonable assumption instead of an unreasonable one. #1 is very unreasonable since it adds information, #2 is close, but #3 is best.

And the proof is Bertrand's Box Paradox. That name does not properly refer to a probability problem, it applies to how to make this reasonable assumption.

"Mr. Jones has exactly two children. I have written the gender, of at least one, inside this sealed envelope. What is the probability that both children have that gender?"

If you were to open the envelope, and see the word "boy," the problem becomes the same as the one under discussion. If it can be answered, that answer is correct here as well. But it is an equivalent problem if you see the word "girl," and again the answer must be the same. If 1/3 is an acceptable answer, it means that 1/3 of all two-child families have two of the same gender, and 2/3 have mixed genders.

But that is a contradiction. We know that the split is 1/2:1/2. So the assumption, that 1/3 is a reasonable answer, is disproven. Now, that does not mean that the information came to us via #2 or #3, it just means we can't assume that it was #1.

Most often, the same logic is used for the Monty Hall Problem, it is just applied backwards.