2006 iTest Problems/Problem 16

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Problem

The Minnesota Twins face the New York Mets in the 2006 World Series. Assuming the two teams are evenly matched (each has a $.5$ probability of winning any game) what is the probability that the World Series (a best of 7 series of games which lasts until one team wins four games) will require the full seven games to determine a winner?

$\text{(A) }\frac{1}{16}\qquad \text{(B) }\frac{1}{8}\qquad \text{(C) }\frac{3}{16}\qquad \text{(D) }\frac{1}{4}\qquad \text{(E) }\frac{5}{16}\qquad$

$\text{(F) }\frac{3}{8}\qquad \text{(G) }\frac{5}{32}\qquad \text{(H) }\frac{7}{32}\qquad \text{(I) }\frac{9}{32}\qquad \text{(J) }\frac{3}{64}\qquad \text{(K) }\frac{5}{64}\qquad$

$\text{(L) }\frac{7}{64}\qquad \text{(M) }\frac{1}{2}\qquad \text{(N) }\frac{13}{32}\qquad \text{(O) }\frac{11}{32}\qquad \text{(P) }\text{none of the above}$


Solution

In order for the World Series to have seven games, each team must win exactly three times. The last game does not matter. There are $\tbinom{6}{3} = 20$ possibilities, and since all of the possibilities involve each team winning three times, the probability of having 7 games in the World Series is $20 \cdot \left( \frac12 \right)^6 = \boxed{\textbf{(E) }\frac{5}{16}}$.

See Also

2006 iTest (Problems)
Preceded by:
Problem 15
Followed by:
Problem 17
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