HMMT 二月 2003 · COMB 赛 · 第 3 题

HMMT February 2003 — COMB Round — Problem 3

Daniel and Scott are playing a game where a player wins as soon as he has two points more than his opponent. Both players start at par, and points are earned one at a time. If Daniel has a 60% chance of winning each point, what is the probability that he will win the game?

解析

Daniel and Scott are playing a game where a player wins as soon as he has two points more than his opponent. Both players start at par, and points are earned one at a time. If Daniel has a 60% chance of winning each point, what is the probability that he will win the game? Solution: 9 / 13 Consider the situation after two points. Daniel has a 9 / 25 chance of winning, Scott, 4 / 25, and there is a 12 / 25 chance that the players will be tied. In the latter case, we revert to the original situation. In particular, after every two points, either the game 1 returns to the original situation, or one player wins. If it is given that the game lasts 2 k rounds, then the players must be at par after 2( k − 1) rounds, and then Daniel wins with probability (9 / 25) / (9 / 25 + 4 / 25) = 9 / 13. Since this holds for any k , we conclude that Daniel wins the game with probability 9 / 13.