If a team finishes a 4-game season with a 3-1 record (3 wins, 1 loss), they have a winning percentage of 75%. But what is (was) the probability that they would win any one game? There are 4 ways that a team can win 3 and lose 1: WWWL, WWLW, WLWW, & LWWW. If P=probability of winning one game, then the probability of a 3-1 season is: 4 x (P^3) x (1-P) [For some reason, I can't get LaTeX to work] If P=0.75, that comes out to 0.421875. Of course, with a 75% probability of winning one game, there is also a 31.64 chance of finishing 4-0, a 21.1% chance of finishing 2-2, a 4.69% chance of finishing 1-3, and a 0.39% chance of finishing 0-4. Can I reverse engineer that formula and set it = 0.75: 4 x (P^3) x (1-P) = 0.75 Then P^3 - P^4 = 0.75/4 I don't know how to solve that for P or even if that would solve the problem.