HMMT 二月 2010 · COMB 赛 · 第 9 题
HMMT February 2010 — COMB Round — Problem 9
题目详情
- [ 7 ] Rosencrantz and Guildenstern are playing a game where they repeatedly flip coins. Rosencrantz wins if 1 heads followed by 2009 tails appears. Guildenstern wins if 2010 heads come in a row. They will flip coins until someone wins. What is the probability that Rosencrantz wins?
解析
- [ 7 ] Rosencrantz and Guildenstern are playing a game where they repeatedly flip coins. Rosencrantz wins if 1 heads followed by 2009 tails appears. Guildenstern wins if 2010 heads come in a row. They will flip coins until someone wins. What is the probability that Rosencrantz wins? 2009 2 − 1 Answer: We can assume the first throw is heads (because neither player can win starting 2008 3 · 2 − 1 from a string of only tails). Let x be the probability that Rosencrantz wins. Let y be the probability that Rosencrantz wins after HT. Whenever there is a string of less than 2009 tails followed by a heads, the heads basically means the two are starting from the beginning, where Rosencrantz has probability x of winning. 1 1 We also know that x = y (1 − ). This is because from the initial heads there is a (1 − ) chance 2009 2009 2 2 Rosencrantz doesn’t lose, and in this case the last two flips are HT, in which case Rosencrantz has probability y of winning. 1 If the first two throws are HT, there is a chance Rosencrantz wins; otherwise, there is eventually 2008 2 a heads, and so we are back in the case of starting from a heads, which corresponds to x . Therefore, 1 1 y = + x (1 − ). Putting this together with the previous equation, we get: 2008 2008 2 2 ( ) ( ) 1 1 1 x = + x (1 − ) 1 − 2008 2008 2009 2 2 2 ( ) ( ) 2008 2009 1+2 x − x 2 − 1 = ⇒ x = 2008 2009 2 2 ( ) 4017 4017 2009 2008 2009 = ⇒ 2 x = x 2 − 2 − 2 + 1 + 2 − 1 2009 2 − 1 = ⇒ x = , 2009 2008 2 +2 − 1 2009 2009 2 − 1 2 − 1 so the answer is = . 2009 2008 2008 2 +2 − 1 3 · 2 − 1