Originally Posted by

**SlipEternal** For best out of 5:

A wins 3-0: $\dbinom{2}{0}(0.6)^3(0.4)^0 = 0.216$

A wins 3-1: $\dbinom{3}{1} (0.6)^3(0.4)^1 = 0.2592$

A wins 3-2: $\dbinom{4}{2} (0.6)^3(0.4)^2 = 0.20736$

B wins 2-3: $\dbinom{4}{2} (0.6)^2(0.4)^3 = 0.13824$

B wins 1-3: $\dbinom{3}{1} (0.6)^1(0.4)^3 = 0.1152$

B wins 0-3: $\dbinom{2}{0}(0.6)^0(0.4)^3 = 0.064$

In general, if A wins a best of $2n-1$ with a game record of n to k, then you know A won the last game (the n-th game). Prior to that game, you know A won (n-1) games and B won k games in any order. If team A wins with probability a and team B wins with probability b, the probability of the outcome that A wins n-k is:

$\dbinom{n-1+k}{k}a^nb^k$