Manhattan Challenge Problem of the week! - 05/28/07

Baseball's World Series matches 2 teams against each other in a best-of-seven series. The first team to win four games wins the series and no subsequent games are played. If you have no special information about either of the teams, what is the probability that the World Series will consist of fewer than 7 games?

(A) 12.5%
(B) 25%
(C) 31.25%
(D) 68.75%
(E) 75%

Answer: D

OE - In order to determine the probability that the World Series will last fewer than 7 games, we can first determine the probability that the World Series WILL last exactly 7 games and then subtract this value from 1.

In order for the World Series to last exactly 7 games, the first 6 games of the series must results in 3 wins and 3 losses for each team.

Let's analyze one way this could happen:

Game 1 -----Game 2 -----Game 3 -----Game 4 -----Game 5 -----Game 6
T1 Wins-----T1 Wins-----T1 Wins-----T1 Loses-----T1 Loses-----T1 Loses

There are many other ways this could happen. Using the permutation formula, there are 6!/(3!)(3!) = 20 ways for the two teams to split the first 6 games (3 wins for each).

There are then 2 possible outcomes to break the tie in Game 7. Thus, there are a total of 20 × 2 = 40 ways for the World Series to last the full 7 games.

The probability that any one of these 40 ways occurs can be calculated from the fact that the probability of a team winning a game equals the probability of a team losing a game = 1/2.

Given that 7 distinct events must happen in any 7 game series, and that each of these events has a probability of 1/2, the probability that any one particular 7 game series occurs is

(1/2) ^ 7 = 1/ 128

Since there are 40 possible different 7 game series, the probability that the World Series will last exactly 7 games is:

40 * 1/128 = 40/ 128 = .3125 = 31.255%

Thus the probability that the World Series will last fewer than 7 games is:

100% - 31.25% = 68.75%.

The correct answer is D.