HMMT 十一月 2016 · 团队赛 · 第 5 题
HMMT November 2016 — Team Round — Problem 5
题目详情
- [ 5 ] Allen and Brian are playing a game in which they roll a 6-sided die until one of them wins. Allen wins if two consecutive rolls are equal and at most 3. Brian wins if two consecutive rolls add up to 7 and the latter is at most 3. What is the probability that Allen wins?
解析
- [ 5 ] Allen and Brian are playing a game in which they roll a 6-sided die until one of them wins. Allen wins if two consecutive rolls are equal and at most 3. Brian wins if two consecutive rolls add up to 7 and the latter is at most 3. What is the probability that Allen wins? Proposed by: Eshaan Nichani Answer: 5/12 Note that at any point in the game after the first roll, the probability that Allen wins depends only on the most recent roll, and not on any rolls before that one. So we may define p as the probability that Allen wins at any point in the game, given that the last roll was a 1, 2, or 3, and q as the probability that he wins given that the last roll was a 4, 5, or 6. Suppose at some point, the last roll was r ∈ { 1 , 2 , 3 } , and the next roll is r ∈ { 1 , 2 , 3 , 4 , 5 , 6 } . By the 1 2 definition of p , Allen wins with probability p . Furthermore, if r = r , which happens with probability 2 1 1 2 , Allen wins. If r ∈ { 1 , 2 , 3 } but r 6 = r , which happens with probability , neither Allen nor Brian 2 2 1 6 6 wins, so they continue playing the game, now where the last roll was r . In this case, Allen wins with 2 3 probability p . If r ∈ { 4 , 5 , 6 } , which happens with probability , neither Allen nor Brian wins, so they 2 6 continue playing, now where the last roll was r . In this case, Allen wins with probability q . Hence, 2 1 2 3 the probability that Allen wins in this case can be expressed as + p + q , and thus 6 6 6 1 2 3 p = + p + q 6 6 6 By a similar analysis for q , we find that 1 2 3 q = · 0 + p + q 6 6 6 1 1 1 Solving, we get p = and q = . Allen wins with probability p = if the first roll is 1, 2, or 3, and he 2 3 2 1 wins with probability q = if the first roll is 4, 5, or 6. We conclude that the overall probability that 3 5 1 1 he wins the game is p + q = . 2 2 12