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抛硬币：首次出现 HT 时掷出 T 的人获胜

Coin Toss Game

专题
Probability / 概率
难度
L4

题目详情

两名玩家 A 与 B 轮流抛一枚公平硬币（A 先手）。记录正反序列。

第一次出现相邻两次结果为 “HT” 时游戏结束，并且掷出 T 的那个人获胜。

问：A 获胜的概率是多少？

Two players, A and B, alternately toss a fair coin (A first, then B, etc.). We record the sequence of Heads (H) and Tails (T). The first time we see “HT” in consecutive tosses, the game ends and the person who tossed T wins. What is the probability that A wins?

解析

设 A 最终获胜概率为 PP。

A 首次抛掷：

  • 若为 T（概率 1/2），则角色对调，A 的胜率变为 1−P1-P。
  • 若为 H（概率 1/2），接下来 B 抛：若 B 抛到 T（概率 1/2），B 立刻获胜；若 B 抛到 H（概率 1/2），又回到“轮到 A、且上一枚为 H”的对称状态，可推得此分支下 A 的胜率为 1/31/3。

因此

P=12⋅13+12⋅(1−P)⇒P=49.P=\tfrac12\cdot\tfrac13+\tfrac12\cdot(1-P) \Rightarrow P=\frac{4}{9}.

Original Explanation

Let P(A)P(A) be the probability that A ultimately wins.

  • If A’s first toss is H, then with probability 1/21/2, B tosses T and B wins immediately. With probability 1/21/2, B also tosses H, and it’s like the roles are reversed. Solving shows P(A∣A tossed H)=13.P(A \mid \text{A tossed H}) = \frac{1}{3}.

  • If A’s first toss is T, effectively B is now in A’s original position, so P(A∣A tossed T)=1−P(A).P(A \mid \text{A tossed T}) = 1 - P(A).

From P(A)=12 (13)+12 (1−P(A)),P(A) = \tfrac{1}{2}\,\bigl(\tfrac{1}{3}\bigr) + \tfrac{1}{2}\,\bigl(1 - P(A)\bigr), we get P(A)=49.P(A) = \frac{4}{9}.