LAST EDITED ON 11-15-10 AT 05:27 PM (EST)

I have the probability of all male/all female teams as double what you posted, that is 1/231. At the risk of turning this into a math thresd...

You are correct in saying that there are 462 ways to choose a 5 member yellow team from 11 players. However, there are two situations in which the teams can break along gender lines:

a) the yellow team is all men AND a man sits out. Now there are six yellow teams that are all men (i.e. one where each man is not on the yellow team). Hence, the probability that the yellow team is all men is 6/462. However, no matter which five men are on the yellow team, the remaining man must be chosen to sit out when the blue team is chosen. The probability of this is 1/6. Hence the probability of (a) is (6/462)x(1/6)=1/462. I feel that this is the idea that you had in your argument Zoidberg.

b) The yellow team is all women. Since there are five women, there is exactly one way in which this condition is satisfied, making the probability of (b) 1/462.

Since (a) and (b) cannot be simultaneously satisfied by the choice of any one 5 player yellow team, the resulting probability of a men vs. women contest is 1/462 + 1/462 = 2/462 or 1/231.

Feel free to check this argument.

edited to fix formatting and phrasing