1971 AHSME Problems/Problem 23

Problem

Teams $A$ and $B$ are playing a series of games. If the odds for either to win any game are even and Team $A$ must win two or Team $B$ three games to win the series, then the odds favoring Team $A$ to win the series are

$\textbf{(A) }11\text{ to }5\qquad \textbf{(B) }5\text{ to }2\qquad \textbf{(C) }8\text{ to }3\qquad \textbf{(D) }3\text{ to }2\qquad \textbf{(E) }13\text{ to }6$

Solution

We have two cases: one where $A$ wins the first game and the other where $A$ loses.

$\underline{\text{Case 1:}}$ In the $\tfrac12$ chance that $A$ wins the first game, $A$ simply needs to win at least $1$ of the next $3$ games. We see that the probability of $A$ losing the next $3$ games is $(\tfrac12)^3=\tfrac18$, so, by complementary counting, the probability that $A$ wins at least $1$ of the next $3$ games is $1-\tfrac18=\tfrac78$.

$\underline{\text{Case 2:}}$ In the $\tfrac12$ chance that $A$ loses the first game, both teams need to win $2$ games, so $A$'s advantage completely disappears. Thus, the proability that $A$ wins the series from here is $\tfrac12$.

Combining the information from the two cases, we see that $A$'s probability of winning the series is $\tfrac12 \cdot \tfrac78 + \tfrac12 \cdot \tfrac12 = \tfrac7{16} + \tfrac4{16} = \tfrac{11}{16}$. Thus, our answer is $\boxed{\textbf{(A) } 11 \text{ to } 5}$.

See Also

1971 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
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