HMMT 二月 2005 · COMB 赛 · 第 5 题

HMMT February 2005 — COMB Round — Problem 5

Doug and Ryan are competing in the 2005 Wiffle Ball Home Run Derby. In each round, each player takes a series of swings. Each swing results in either a home run or an out, and an out ends the series. When Doug swings, the probability that he will hit a home run is 1/3. When Ryan swings, the probability that he will hit a home run is 1/2. In one round, what is the probability that Doug will hit more home runs than Ryan hits?

解析

Doug and Ryan are competing in the 2005 Wiffle Ball Home Run Derby. In each round, each player takes a series of swings. Each swing results in either a home run or an out, and an out ends the series. When Doug swings, the probability that he will hit a home run is 1/3. When Ryan swings, the probability that he will hit a home run is 1/2. In one round, what is the probability that Doug will hit more home runs than Ryan hits? Solution: 1 / 5 Denote this probability by p . Doug hits more home runs if he hits a home run on his first try when Ryan does not, or if they both hit home runs on their first try and Doug 1 1 1 hits more home runs thereafter. The probability of the first case occurring is · = , 3 2 6 p p 1 1 1 and the probability of the second case occurring is · · p = . Therefore p = + , 3 2 6 6 6 1 which we solve to find p = . 5