HMMT 二月 2006 · 冲刺赛 · 第 39 题

HMMT February 2006 — Guts Round — Problem 39

[15] A fat coin is one which, when tossed, has a 2 / 5 probability of being heads, 2 / 5 of being tails, and 1 / 5 of landing on its edge. Mr. Fat starts at 0 on the real line. Every minute, he tosses a fat coin. If it’s heads, he moves left, decreasing his coordinate by 1; if it’s tails, he moves right, increasing his coordinate by 1. If the coin lands on its edge, he moves back to
If Mr. Fat does this ad infinitum , what fraction of his time will he spend at 0? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . th IX HARVARD-MIT MATHEMATICS TOURNAMENT, 25 FEBRUARY 2006 — GUTS ROUND ∞ ∑ 3 k + 1 k +1

解析

A fat coin is one which, when tossed, has a 2 / 5 probability of being heads, 2 / 5 of being tails, and 1 / 5 of landing on its edge. Mr. Fat starts at 0 on the real line. Every minute, he tosses a fat coin. If it’s heads, he moves left, decreasing his coordinate by 1; if it’s tails, he moves right, increasing his coordinate by 1. If the coin lands on its edge, he moves back to 0. If Mr. Fat does this ad infinitum , what fraction of his time will he spend at 0? 1 Answer: 3 Solution: For n ∈ Z , let a be the fraction of the time Mr. Fat spends at n . By n symmetry, a = a for all n . n − n 2 2 5 For n > 0, we have a = a + a , or a = a − a . This Fibonacci-like n n − 1 n +1 n +1 n n − 1 5 5 2 recurrence can be solved explicitly to obtain | n | −| n | a = α · 2 + β · 2 n for all n ∈ Z . Now we also have ∑ a = 1 , n n ∈ Z β so we better have α = 0, so that a = β and a = . Now we also have a = 0 ± 1 0 2 ∑ 2 2 1 1 a + a + , so β = . This matches perfectly with a = 1 . − 1 1 n n ∈ Z 5 5 5 3