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 $\boxed{\textbf{(A) } 11 \text{ to } 5}$ .

1971 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1971 AHSME Problems/Problem 23

Problem

Solution

See Also