HMMT 二月 2004 · COMB 赛 · 第 5 题
HMMT February 2004 — COMB Round — Problem 5
题目详情
- A best-of-9 series is to be played between two teams; that is, the first team to win 5 games is the winner. The Mathletes have a chance of 2 / 3 of winning any given game. What is the probability that exactly 7 games will need to be played to determine a winner?
解析
- A best-of-9 series is to be played between two teams; that is, the first team to win 5 games is the winner. The Mathletes have a chance of 2 / 3 of winning any given game. What is the probability that exactly 7 games will need to be played to determine a winner? Solution: 20 / 81 If the Mathletes are to win, they must win exactly 5 out of the 7 games. One of the 5 games they win must be the 7th game, because otherwise they would win the tournament before 7 games are completed. Thus, in the first 6 games, the Mathletes must win 4 games and lose 2. The probability of this happening and the Mathletes winning the last game is [ ] ( ) ( ) ( ) ( ) 4 2 6 2 1 2 · · · . 2 3 3 3 Likewise, the probability of the other team winning on the 7th game is [ ] ( ) ( ) ( ) ( ) 4 2 6 1 2 1 · · · . 2 3 3 3 Summing these values, we obtain 160 / 729 + 20 / 729 = 20 / 81 .