1971 AHSME Problems/Problem 23
Problem
Teams and are playing a series of games. If the odds for either to win any game are even and Team must win two or Team three games to win the series, then the odds favoring Team to win the series are
Solution
We have two cases: one where wins the first game and the other where loses.
In the chance that wins the first game, simply needs to win at least of the next games. We see that the probability of losing the next games is , so, by complementary counting, the probability that wins at least of the next games is .
In the chance that loses the first game, both teams need to win games, so 's advantage completely disappears. Thus, the proability that wins the series from here is .
Combining the information from the two cases, we see that 's probability of winning the series is . Thus, our answer is .
See Also
1971 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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