What, if any, is the relationship between Monty Hall and the Boy or Girl puzzles? The two puzzles seem like they share a very similar structure and a very similar piece of intuition (or counter-intuition). I wonder if anyone has put more thought into clarifying what exactly the relationship is between the two puzzles?

Here are the hopefully-unambiguous statements of each puzzle:

Monty Hall puzzle: At a game-show, there are three doors. Behind one lies the grand prize. The contestant first selects one door but it isn't opened for her. The host then always opens one of the other two doors, always revealing that it is empty. The host then gives the contestant of sticking with the door she first chose or switching to the remaining unopened door. Should she switch?
Boy or Girl puzzle: Consider all two-child families in the world that contain at least one boy. Pick at random one such family. What is the probability that this family contains two boys?

 A: The problems can be restated to make the answers less counterintuitive.
Monty Hall: Of the two doors that Monty does not open there is exactly one that is hiding a goat, so what is the probability that this is the door you picked, when given that there is probability of two-thirds that it was the door you picked.

*

*You had an unbiased choice of three doors.   One hid the car, and two hid the goats.   Picking a goat hiding door means you win a car if you switch.

Boy-Girl: From the families of two children, the one you pick has at least one boy, so what is the probability that the family has one boy, when given that of the four equally-possible types of families, one has two boys, two have one boy pair, and the other is the type you did not pick (ie two-girls).

*

*There are four equally likely family structures: $\{BB, BG, GB, GG\}$. You know that you did not pick the later, so...


*By excluding one type of family structure, you made an unbiased choice from three family structures.   One with two boys, two with one boy (who is the eldest in one type, or youngest in the other).
There is no deep coincidence here.   They are both essentially the same situation.   This is just obscured by the original counterintuitive phrasing of the problems.
A: Has anyone compared these two (I call them the MHP and TCP, respectively)? I certainly have.
The short answer is that, as you have worded them, they are different. But a more common wording of the TCP - which is different than yours but too often interpreted as being the same - they are both variations of Joseph Bertrand's famous Box Problem. Other similar problems include the Three Prisoners, Three Pancakes, and/or Three Cards. You can look them up on the internet, but they are all essentially the same thing. There is even a similar situation in card games like Bridge, where experts use what they call the Principle of Restricted Choice to help estimate who has a missing card.
What is common in all these situations, is that there can be more information in what we learn than is immediately obvious. Because its importance isn't obvious, it can be either omitted in the problem statement, like you did in your MHP, or strongly worded to prevent it, like you did in your TCP. I call it "hidden information." What makes it important, is if it could have taken a different form in the situation we are asked to provide a probability for, than in others.
This may be easiest to explain with the MHP. Your statement of the problem left out the fact that we actually see which doors are chosen at each point. Say, as in Marilyn vos Savant's version, that the contestant chose Door #1 and the host opened Door #3. I'm not saying these specific door numbers matter - they don't - but the fact that the host theoretically could have opened door #2 does matter. So "Opened #3" is the hidden information, and we have to consider the cases where he could have opened #2 instead.
If the prize is behind Door #2, he would have been forced to open Door #3; that is, a 100% chance. If the prize is behind Door #1, he must choose between two equivalent doors, and there only is a 50% chance he would have opened Door #3. It is the ratio of these two numbers - 100% to 50% - that makes it twice as likely that we are in the first case instead of the second!
Note that most solutions you will see say that switching has a 2/3 chance because the 1/3 chance that your original choice had the prize can't change. THIS IS WRONG; or at least, incomplete. This probability can change based on what door we see opened, but only if the host is biased when he makes the choice. What if, when he can choose between #2 and #3, he chooses #2 100% of the time? Then the only way he will ever open #3 is if the car is behind #2 and the 1/3 probability has changed to 0. It changes to 1/2 if he opens #2. But note that the weighted average of these two results - that is, the probability if you ignore the source of this hidden information - has to remain 1/3.
No matter how it is worded, the TCP is usually interpreted as you stated it. But it is not often worded so strongly. Martin Gardner (Scientific American's "Mathematical Games" columnist) popularized it in 1959, with the problem statement "Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?" He answered "1/3," but later acknowledged that the problem was ambiguous. He received many letters pointing out that the gender you know is hidden information. If you don't know how you learned the fact "at least one is a boy," you have to consider whether it would have been possible to learn "at least one is a girl."
If you decide it would have been possible, the answer to the TCP is 1/2: there was a 25% chance that Mr. Smith has two boys and you could only learn about a boy, and while there was a 50% chance he has a boy and a girl, there is only a 25% chance that he has a boy and a girl AND the gender you would know is "boy."
