HMMT 二月 2021 · COMB 赛 · 第 2 题
HMMT February 2021 — COMB Round — Problem 2
题目详情
- Ava and Tiffany participate in a knockout tournament consisting of a total of 32 players. In each of 5 rounds, the remaining players are paired uniformly at random. In each pair, both players are equally likely to win, and the loser is knocked out of the tournament. The probability that Ava and a Tiffany play each other during the tournament is , where a and b are relatively prime positive integers. b Compute 100 a + b .
解析
- Ava and Tiffany participate in a knockout tournament consisting of a total of 32 players. In each of 5 rounds, the remaining players are paired uniformly at random. In each pair, both players are equally likely to win, and the loser is knocked out of the tournament. The probability that Ava and a Tiffany play each other during the tournament is , where a and b are relatively prime positive integers. b Compute 100 a + b . Proposed by: Sheldon Kieren Tan Answer: 116 Solution: Each match eliminates exactly one player, so exactly 32 − 1 = 31 matches are played, each ( ) 32 32 · 31 of which consists of a different pair of players. Among the = = 496 pairs of players, each pair 2 2 is equally likely to play each other at some point during the tournament. Therefore, the probability 31 1 that Ava and Tiffany form one of the 31 pairs of players that play each other is = , giving an 496 16 answer of 100 · 1 + 16 = 116.