> 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.

> What they don't understand is probability.

Exactly. As the first reply to the first comment explains “The problem is that we don't know p(you're told at least one is a girl | they aren't both girls).”

It’s funny that the same commenter who writes that “to get the right answer you must be careful about conditional probabilities” finds that doing so is “splitting hairs about something boring and irritating.”