Problem 10 · AMC 8 Stretch
Stretch
Counting & Probability
visual-representationaccount-for-all-possibilities
Two evenly matched teams play a best-of-seven series (first to win \(4\) games wins). Each game is a coin flip. The chance the series lasts exactly \(7\) games is \(\frac{5}{16}\); the chance it lasts exactly \(6\) games is also \(\frac{5}{16}\). What is the probability the series lasts exactly \(7\) games? Give your answer as a fraction.
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Answer: 5/16
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Hint 1 of 4
Imagine the teams stubbornly play all \(7\) games even after the series is decided. There are \(2^7 = 128\) equally likely win/loss patterns.
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Hint 2 of 4
For the series to last exactly \(7\) games, the teams must be tied \(3\)-\(3\) after \(6\) games, then someone wins game \(7\). Count the ways to be tied \(3\)-\(3\).
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Hint 3 of 4
To be tied \(3\)-\(3\) after \(6\) games, pick which \(3\) of the \(6\) games team A won: that is \(\binom{6}{3} = 20\) ways. Game \(7\) can go either way, so \(20 \times 2 = 40\) full \(7\)-game outcomes out of \(128\).
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Approach: Count equally likely full-length patterns (visual representation of all cases)
- Imagine the teams play all \(7\) games no matter what; there are \(2^7 = 128\) equally likely win/loss patterns, and each probability is (count)\(/128\).
- The series reaches \(7\) games only if it is tied \(3\)-\(3\) after \(6\) games. The number of ways for team A to win exactly \(3\) of the first \(6\) is \(\binom{6}{3} = 20\), and game \(7\) can go either way, giving \(20 \times 2 = 40\) patterns.
- So \(P(7) = \frac{40}{128} = \frac{5}{16}\). (For the others: \(P(4) = \frac{2}{16}\), \(P(5) = \frac{4}{16}\), \(P(6) = \frac{5}{16}\), summing to \(1\).)
- A \(6\)-game and a \(7\)-game series are equally likely (\(\frac{5}{16}\) each): once the score is \(3\)-\(2\) after \(5\) games, game \(6\) either ends it or forces a game \(7\), each with probability \(\frac{1}{2}\).
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