HMMT 二月 2015 · COMB 赛 · 第 3 题
HMMT February 2015 — COMB Round — Problem 3
题目详情
英文原题
- Starting with the number 0, Casey performs an infinite sequence of moves as follows: he chooses anumber from { 1 , 2 } at random (each with probability ) and adds it to the current number. Let p 1
2 mbe the probability that Casey ever reaches the number m . Find p − p .
20 15
解析
英文解析
- Starting with the number 0, Casey performs an infinite sequence of moves as follows: he chooses anumber from { 1 , 2 } at random (each with probability ) and adds it to the current number. Let p 1
2 mbe the probability that Casey ever reaches the number m . Find p − p .
20 15
Answer: We note that the only way n does not appear in the sequence is if n − 1 and then n +111
( )220
1 2 1 2
appears. Hence, we have p = 1, and p = 1 − p for n > 0. This gives p − = − p − ,
0 n n − 1 n n − 1
2 3 2 3
so that
( )
2 1 1 np = + · − ,
3 3 2 nso p − p is just
20 15
1 − ( − 2) 115 = .
20 20
3 · 2 2