A Six Game Series

**The probability that the NL wins in 6 games = P(NL wins exactly 3 out of the first 5 games) · P(NL wins the sixth game) = (0.3125)·(0.5000) = 0.15625**
Games 1-5 (order does not matter)	Game 6

P(NL wins exactly 3 of 5 games) = C[5,3]·[(1/2)^3]·[(1/2)^2] = [(5·4·3)/(3·2·1)]·(1/8)·(1/4) = (10)·(1/8)·(1/4) = 10/32 = 0.3125	P(NL wins) = 1/2 = 0.5000

**The probability that the AL wins in 6 games = P(AL wins exactly 3 out of the first 5 games) · P(AL wins the sixth game) = (0.3125)·(0.5000) = 0.15625**
Games 1-5 (order does not matter)	Game 6

P(AL wins exactly 3 of 5 games) = C[5,3]·[(1/2)^3]·[(1/2)^2] = [(5·4·3)/(3·2·1)]·(1/8)·(1/4) = (10)·(1/8)·(1/4) = 10/32 = 0.3125	P(AL wins) = 1/2 = 0.5000

The probability that the series ends in 6 games =
P(NL wins in 6 games) + P(AL wins in 6 games) = 0.15625 + 0.15625 = 0.3125

Return to the analytical solution.

Or go straight to the explanation page for a 7 game series.