You know that a couple has two children. You go to the couple's house and one of their children, a young boy, opens the door. What is the probability that the couple's other child is a girl?
If you list all possibilities for the sexes of two children, $BB, BG, GB, GG$, you see that $2$ of the $3$ pairs that have B (for boy) in them also have a girl, so the answer one could argue is $2/3$.
On the other hand, one could argue that the answer is $1/2$, since the probability that any one child is a girl is $1/2$, and intuitively (?) should be independent of the gender of its siblings.
Some background to possibly justify posting it here: the question was asked at an interview for an actuarial/insurance type position, and the interviewer said the answer was $2/3$, whereas my friend who was being interviewed (and has a masters in math) thought the answer was $1/2$, even after the interviewer explained his logic. My friend felt that the interviewer wasn't taking into account the fact that it is not equally likely that a boy will open the door in the $BB$ versus the $BG$ combination, and one has to take into account that fact. I have no idea which is the correct answer, both sound somewhat convincing to me (I have a Ph.D. in math, but I won't mention from where in an effort to avoid embarrassing my degree granting institution!). Anyways, any help would be appreciated and I apologize if this is too simple a question for this forum.
Best Answer
This was also posted on MathOverflow, where the answer of 1/2 was confirmed several ways. My favorite is this: There are really three independent "choices" are being made: the sex of the first (say, older) child, the sex of the younger child, and whether the older or younger child opens the door. So there are eight equally likely outcomes, and by considering them it's easy to get 1/2.
I want to mention a related game, which I believe I first saw in Martin Gardner's Aha! Gotcha.
I have three cards; one is black on both sides, one is red on both sides, and one is black on one side and red on the other. I will let you select a card at random and look at one side, and I'll bet you even money that the other side is the same color.
This sounds fair. Suppose you see black; now the card is either the black-red (which would win the game for you) or black-black (which wins for me), and at the outset, you were equally likely to have picked either one. But in fact I win the game 2/3 of the time. As with the puzzle, considering all possible outcomes, there are 6 total sides of cards. You are equally likely to be looking at any of the 3 black sides, but only one of them (the black side of the black-red card) wins the game for you. Another argument is to note that if you pick the black-black card, you are twice as likely to see a black side as if you pick the black-red card, so I am twice as likely to win.
I recommend you find this interviewer and offer to play this game with him. You could even sweeten the odds a little to make it sound more attractive. With high probability, you can clean him out. Don't forget to kick back a "finder's fee" to your friend, or at least a nice dinner (as consolation for the job).