A Five Game Series

**The probability that the NL wins in 5 games = P(NL wins exactly 3 out of the first 4 games) · P(NL wins the fifth game) = (0.2500)·(0.5000) = 0.1250**
Games 1-4 (order does not matter)	Game 5

P(NL wins exactly 3 of 4 games) = C[4,3]·[(1/2)^3]·(1/2) = [(4·3·2)/(3·2·1)]·(1/8)·(1/2) = (4)·(1/8)·(1/2) = 1/4 = 0.2500	P(NL wins) = 1/2 = 0.5000

**The probability that the AL wins in 5 games = P(AL wins exactly 3 out of the first 4 games) · P(AL wins the fifth game) = (0.2500)·(0.5000) = 0.1250**
Games 1-4 (order does not matter)	Game 5

P(AL wins exactly 3 of 4 games) = C[4,3]·[(1/2)^3]·(1/2) = [(4·3·2)/(3·2·1)]·(1/8)·(1/2) = (4)·(1/8)·(1/2) = 1/4 = 0.2500	P(AL wins) = 1/2 = 0.5000

The probability that the series ends in 5 games =
P(NL wins in 5 games) + P(AL wins in 5 games) = 0.1250 + 0.1250 = 0.2500

Return to the analytical solution.

Or go straight to the explanation page for a 6 game series.