Question on an old Manhattan challenge problem

[ilakshmir]

Given below is a question from the Manhattan archive of challenge problems.

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%

I have attached the given solution at the end of my note.

I agree only partly with the given solution. I concur that there are 40 ways in which 7 matches would occur before the winner is picked. But I do not think the sample space (all possible outcomes) is 128 (derived in Manhattan's solution) for this problem. Here is why:

All possible outcomes should include only cases corresponding to a maximum of 4, 5, 6, 7 matches being played. This works out to 2*(1+4+10+20) = 70 outcomes only. So in my opinion, the probability is (70-40)/70 = 3/7.

Consider 2 outcomes for 7 matches A& B:

Outcome: 1
A: W W W W L L L
B: L L L L W W W

Outcome: 2
A: W W W W W W W
B: L L L L L L L

Both these outcomes are identical in reality because the series will end after the 4 match. Manhattan's solution (2^7=128) ends up counting such outcomes multiple times. Please refer to their answer below and let me know your thoughts!

Manhattan's Answer

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.

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

40/128 = 31.25%

Thus the probability that the World Series will last fewer than 7 games is 100% - 31.25% = 68.75%.

The correct answer is D.

[StaceyKoprince]

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Thanks!