HMMT 十一月 2020 · 冲刺赛 · 第 9 题

HMMT November 2020 — Guts Round — Problem 9

[7] A fair coin is flipped eight times in a row. Let p be the probability that there is exactly one pair of consecutive flips that are both heads and exactly one pair of consecutive flips that are both tails. If a p = , where a, b are relatively prime positive integers, compute 100 a + b . b . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMO 2020, November 14–21, 2020 — GUTS ROUND Organization Team Team ID#

解析

[7] A fair coin is flipped eight times in a row. Let p be the probability that there is exactly one pair of consecutive flips that are both heads and exactly one pair of consecutive flips that are both tails. If a p = , where a, b are relatively prime positive integers, compute 100 a + b . b Proposed by: Yannick Yao Answer: 1028 Solution: Separate the sequence of coin flips into alternating blocks of heads and tails. Of the blocks of heads, exactly one block has length 2, and all other blocks have length 1. The same statement applies to blocks of tails. Thus, if there are k blocks in total, there are k − 2 blocks of length 1 and 2 blocks of length 2, leading to k + 2 coins in total. We conclude that k = 6, meaning that there are 3 blocks of heads and 3 blocks of tails. 2 The blocks of heads must have lengths 1 , 1 , 2 in some order, and likewise for tails. There are 3 = 9 ways to choose these two orders, and 2 ways to assemble these blocks into a sequence, depending on 8 whether the first coin flipped is heads or tails. Thus the final probability is 18 / 2 = 9 / 128.