I guess I'm late to the party, but here's my take. There are 100 balls. First the host chooses N uniformly from [0, 100] (unknown to us) and makes N of them re

I guess I'm late to the party, but here's my take. There are 100 balls. First the host chooses N uniformly from [0, 100] (unknown to us) and makes N of them red and 100-N of them green. But it's annoying to paint 100 balls, right? So he does it this way instead: the host takes 100 white balls and puts them all in a row. Then he takes a black ball and puts it between balls N and N+1, and he declares to himself that all the white balls to the left of the black one will be "red," and all to the right will be "green." Then he has us randomly select two white balls without replacement, and depending on which side of the black ball they were, he'll paint just those two balls red or green accordingly before showing us! Since we never see the other 98 unpainted white balls, we'll never know the difference between this setup and the original. (I think it's intuitively obvious that this setup is mathematically equivalent to the original, i.e. generates the same distribution, so I won't try to justify that rigorously.) Actually though, the host does it slightly differently, again without changing the underlying distribution: you see, he actually starts with 101 white balls, all in a row, and he begins by randomly selecting one of them and paints it black, and then proceeds as above. But this gives the whole problem away: there are 101 white balls in a row, and three are drawn uniformly at random without replacement. The first one is declared "black," and the other two are red or green depending on their positions relative to the black one. All three-ball sequences are equally likely, so by symmetry, the probability that the black ball is between the second and third is exactly 1/3. The other two balls, the ones we draw, are different colors if and only if the black ball is between them, so the probability that we get two balls of the same color is exactly 2/3. That's the answer: whatever is the color of the first ball we draw, the second one has that color with probability 2/3. Thus, the second ball is red with probability 2/3.