Comment by rubergly

    Let's try a simpler problem. Suppose we know that a 
    certain man has two children and we also know that the
    older one is a boy. In this case we would say that the
    probability that the other child is a boy is 1/2. After
    all, the sex of one child is independent of the sex of
    the other child. That the older child is a boy has no
    bearing on the sex of the younger child.

    Now suppose we know simply that a man has two children
    and that one of them is a son. This time we would reason
    that there is no possibility that the person has two
    girls. It follows that the sexes of his two children,
    ordered from oldest to youngest, are either BB, BG or GB.
    Since these cases are equally likely, and since only one
    of them involves having two boys, we would say the
    probability that the man has two boys is 1/3.

Maybe I'm missing something, but this seems fundamentally wrong. Pr(two boys | older child = boy) is not equal to Pr(two boys | one+ child = boy)? Why is time so special? Could we not order them based on their height, and assert Pr(two boys | taller child = boy) = 1/2. Or, if there were a scale of masculinity, order them based on that and assert Pr(two boys | manlier child = boy) = 1/2, where manlier child = boy <==> one+ child = boy, so Pr(two boys | manlier child = boy) = Pr(two boys | one+ child = boy) (contradiction).