I don’t think it is any different. The knowledge that one child is a girl means that in reality you are only figuring the odds of the other child’s sex, since the two events are independent and there’s no indication in the question that order matters. In order to get to the 1 in 3 or 2 in 3 odds, a different question must be asked. For example if the question was "There are two children. At least one is a girl. What are the odds that the oldest is a girl and the youngest is a boy ? " ,the odds would be 1 in 3, since in this case, order does matter.
Odds have nothing to do with what you believe or know about a situation. If you are playing three card shuffle, no matter how firmly you believe the Ace is still on the table it is NOT, and your odds of finding it are zero.
Sorry, you stated the problem incompletely, and wound up with an incorrect statement. Odds are always a prediction of a future event (although we can consider calculating the odds of events that have or have not happened by pretending we know only what happened before the event).
If you are playing three card shuffle and you do not have any knowledge about whether or not the dealer cheats and you assume that the dealer does not cheat, your chances of finding the ace by picking a random card are one in three.
If you are playing three card shuffle and you presume or know that the dealer is always cheating and the ace is gone, you chances of finding it by picking a random card are indeed zero.
If you are playing three card shuffle and you presume that the probability of the dealer cheating and removing the ace is “P”, then your chances of finding the ace by picking a random card ar P/3.
jrf
Jon: I don’t think thats right - I don’t think you ever have 1 in 3 odds if you believe the dealer is not cheating (but in fact he is). I think that is just semantics though.
Ok, I’m wrong and Cecil is right. Yoop’s pestering me about making the program drop all the Bbs is what finally drove me over the edge. Obviously when I do that, I am left with 2/3s mixed and 1/3rd GG. If the computer says there is a B it will be right 2/3s of the time. Initially my reaction to this was to continue claiming that one of the mixed had to be eliminated, but I couldn’t think of a way to rationalize which one. So I threw out that idea, thought of a new one. This is just a thought experiment:
If I trolled through IRC chatrooms till I got 100 people in a single room who all had 2 children, I would ideally have 25 with 2 boys, 25 with 2 girls, and 50 with one of each. I ask everyone if they have a girl, those who say no leave the room. 25 people leave, 75 are left. Of those 75, 25 of course have two girls, 50 have a boy and a girl. I ask if the other child is a boy. 50 of course say yes, which is 2/3ds of the remaining population. For some reason thinking about it this way helped me.
Where I was really getting hung up was this: suppose in the original question they had said that their oldest child was a girl, instead of just that one of their children was a girl. Now certainly when you say that, there is only a 1 in 2 chance the other child is a boy. In the internet chat room, only 25 people will have a mixed pair in which the girl is older. It seemed to me that it didn’t matter whether you were told it was the oldest child because this information is not a determinant for the sex of the other child.
This is true of course, but by adding that the oldest child is the girl, we are actually changing the question. We are now asking what are the odds that the youngest in the family is a boy. The odds of this are 1 in 2.
Thanks for everyone’s help in shattering my hubris - to think I was smarter than Cecil!