HMMT 十一月 2010 · 团队赛 · 第 4 题

HMMT November 2010 — Team Round — Problem 4

[ 6 ] In terms of k , for k > 0, how likely is it that after k minutes Sherry is at the vertex opposite the vertex where she started? Circles in Circles

解析

[ 6 ] In terms of k , for k > 0, how likely is it that after k minutes Sherry is at the vertex opposite the vertex where she started? 1 1 Answer: + Take p ( n ) from the last problem. By examining the last move of Sherry, the k 6 3( − 2) probability that she ends up on the original vertex is equal to the probability that she ends up on the 1 − p ( n ) top vertex, and both are equal to for n ≥ 1. 2 From the last problem, p ( n ) p ( n + 1) = 1 − 2 ( ) 2 1 2 p ( n + 1) − = − p ( n ) − 3 2 3 ( ) n 2 1 2 2 1 and so p ( n ) − is a geometric series with ratio − . Since p (0) = 0, we get p ( n ) − = − − , or 3 2 3 3 2 ( ) n 2 2 1 that p ( n ) = − − . 3 3 2 Now, for k ≥ 1, we have that the probability of ending up on the vertex opposite Sherry’s initial vertex ( ) ( ) ( ) k k 1 − p ( k ) 1 1 2 2 1 1 1 1 1 1 after k minutes is = − − − = + − = + . k 2 2 2 3 3 2 6 3 2 6 3( − 2) Circles in Circles