> [...] imagine that instead of starting with 100 balls, you start with 101 balls in a row. Pick a ball at random. Then color the balls to the left of it green

> [...] imagine that instead of starting with 100 balls, you start with 101 balls in a row. Pick a ball at random. Then color the balls to the left of it green and the ones to the right of it red. Throw that ball away, leaving 100 balls. > Then pick a second ball at random. That ball corresponds to the first ball in the original problem. The problem tells you that you picked a red ball, so it was to the right of the ball you threw away. Now pick a third ball. This ball is either to the left of the first ball, between the first ball and the second, or to the right of the second. In two of the three possibilities, the third ball is red. So the probability that the ball is red is 2/3. I think that explanation will confuse a lot of people. Let's say the first ball picked is ball 94 (numbering from 1 to 101 left to right). So 1-93 end up green and 95-101 are red. Let's say the second ball is 98. The third ball then must be from one of these three intervals: green 1-93, red 95-97, red 99-101. Those intervals have lengths 93, 3, and 3. OK, so now we've got three intervals which sounds like the situation the explanation is talking about, and two of those three give a red ball, so the probability the second ball is red is 2/3. But that is clearly wrong, since the probability of the ball being picked from an interval is proportional to the length of the interval. There is a 93/99 chance that ball 3 is green and only a 6/99 chance it is red for these particular intervals. It needs to be made clear that what you have to look at is all possible ways to pick the three balls. Maybe something like this. I'm going to call the position of the ball that divides red and green D, which is an integer in [1,101], P1 the position of the first colored ball picked, also an integer in [1,101] not equal to D, and P2 the position of the second colored ball, also an integer in [1,101] not equal to D or P1. Imagine a list of every possible legal (D, P1, P2) 3-tuple. Note that each consists of 3 distinct integers in [1,101]. Partition that list into several sublists, where two of the 3-tuples, (D, P1, P2) and (D', P1', P2') are in the same sublist if and only if they contain the same numbers. E.g., (5, 2, 12) and (2, 5, 12) would be in the same sublist. Each sublist will have 6 of the 3-tuples, one for each permutation of the three distinct numbers shared by all the 3-tuples in that sublist. Each of those 6 permutations corresponds to a different ordering of D, P1, and P2. For example (5, 2, 12) corresponds to the ordering P1 < D < P2 and (2, 5, 12) corresponds to D < P1 < P2. Here are the 6 possible orderings, and what the outcome is when the balls are drawn: D < P1 < P2 P1 red, P2 red D < P2 < P1 P1 red, P2 red P1 < D < P2 P1 green, P2 red P1 < P2 < D P1 green, P2 green P2 < D < P1 P1 red, P2 green P2 < P1 < D P1 green, P2 green In the 3 cases where P1 is red, P2 is red in 2 of them and green in 1. That gives a 2/3 chance of P2 being red if P1 was red. This is true in all of the partitions of our list of all legal(D, P1, P2), so is true in all cases.