Problem 39 · AMC 8 Stretch
Stretch
Counting & Probability
binomial-probabilityor-process-addconsidering-extreme-cases
Two equally good teams play a best-of-three series (first to win 2 games wins the series). Each game is a coin-flip (each team wins with probability \(\tfrac{1}{2}\)). Find the probability that Team A wins the series.
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Answer: 1/2
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Hint 1 of 4
Each game is independent with probability \(1/2\) for Team A. Team A wins the series the moment it reaches 2 wins.
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Hint 2 of 4
The series lasts 2 or 3 games. A clean way to list outcomes: write the game-by-game winners until someone reaches 2 wins.
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Hint 3 of 4
List all the ways the series can go and add the chances — or notice the two teams are equally good, so by symmetry each is equally likely to win the whole thing.
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Approach: Case-listing, with a symmetry shortcut
- Each game is independent, Team A winning with probability \(\frac{1}{2}\).
- Method 1 (list the ways A wins, A must win the last game): A wins in 2 games (A A) is \(\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}\). A wins in 3 games: 2 patterns (A B A and B A A), each \(\left(\frac{1}{2}\right)^3 = \frac{1}{8}\), totaling \(\frac{1}{4}\).
- Add: \(\frac{1}{4} + \frac{1}{4} = \frac{1}{2}\).
- Method 2 (symmetry): the two teams are equally good, so A and B are equally likely to win the series, giving A's chance as exactly \(\frac{1}{2}\) — matching Method 1.
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