OK, I'm seeing it now. Generalizing, imagine N+1 jars each containing N balls. The jars are numbered 0 through N. The jar numbered k has k red balls and N-k gre

OK, I'm seeing it now. Generalizing, imagine N+1 jars each containing N balls. The jars are numbered 0 through N. The jar numbered k has k red balls and N-k green balls. We have N+1 people, also numbered 0 through N. Person k is holding jar k. We ask each person to draw a ball at random from their jar. If they draw a green ball they take their jar and leave. If they draw a red ball they stay. We expect on average (N+1)/2 people will remain. We now ask them to draw another ball. The expected number of red balls on this second drawing is (N+1)/3. The expected number of green balls on this second drawing is (N+1)/6, half the expected number of red balls. That's because on the first drawing, the more green balls in your jar the more likely you will draw green on the first round and thus take your jar and go home. The second round is biased toward people who have jars with more red balls. Here's the math behind those results. The probability that person k makes it to round 2 is k/N. The expect number that will make it is sum {k=0, N) = (N+1)/2. Once in round 2 the probability they draw red is (k-1)/(N-1). The probability they draw green is (N-k)/(N-1). The probability then that they make it to the second round and then draw red is k/N (k-1)/(N-1). The probability that they make it to the second round and then draw green is k/N (N-k)/(N-1). The expected number of reds we see in the second round is sum {k=1, N} k (k-1) / N / (N-1). The expected number of greens is sum {k=1, N} k (N-k) / N / (N-1). Use the well known formulas 1 + 2 + 3 + ... + M = M (M+1) / 2 and 1^2 + 2^2 + ... + M^2 = M (M+1) (2M+1) / 6 and simplify and you'll get the (N+1)/3 expected red and (N+1)/6 expected green I stated above.