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掷骰：先出 12 还是先出连续两个 7

Dice Question

专题
Probability / 概率
难度
L4

题目详情

两名玩家反复掷两枚标准六面骰。

  • 玩家 A 赌“先出现点数和为 12”。
  • 玩家 B 赌“先出现连续两个 7”。

一直掷到某事件首次发生为止。问：A 获胜的概率是多少？

Two players bet on rolls of two standard six-sided dice. Player A bets that a sum of 12 will occur first. Player B bets that two consecutive 7s will occur first. They keep rolling until one of these events occurs. What is the probability that A wins?

解析

设状态：

  • 12（吸收态，A 赢），
  • 7–7（吸收态，B 赢），
  • SS（一般态：尚未出现连续 7，且未出现 12），
  • 77（上一轮刚掷出 7，但还没连续）。

记 aSa_S 为从 SS 出发 A 获胜的概率。

每轮：

  • 掷出 12 概率 1/361/36，
  • 掷出 7 概率 6/366/36，
  • 其余概率 29/3629/36。

方程：

{a12=1,a7−7=0,aS=136⋅1+636a7+2936aS,a7=136⋅1+636⋅0+2936aS.\begin{cases} a_{12}=1,\quad a_{7-7}=0,\\ a_S=\tfrac{1}{36}\cdot 1+\tfrac{6}{36}a_7+\tfrac{29}{36}a_S,\\ a_7=\tfrac{1}{36}\cdot 1+\tfrac{6}{36}\cdot 0+\tfrac{29}{36}a_S. \end{cases}

解得 aS=713a_S=\boxed{\frac{7}{13}}。

Original Explanation

Define states:

  • 12 (absorbing, A wins),
  • 7–7 (absorbing, B wins),
  • S (start/general state, no consecutive 7 yet, no 12),
  • 7 (one 7 was just rolled, but not consecutive 7s yet).

We want aSa_S, the probability that starting in state S leads to 12 before 7–7.

Using transition probabilities:

  • Rolling 12: 1/361/36,
  • Rolling 7: 6/366/36,
  • Rolling neither 7 nor 12: 29/3629/36.

Set up: {a12=1,a7−7=0,aS=136⋅1+636 a7+2936 aS,a7=136⋅1+636⋅0+2936 aS.\begin{cases} a_{12} = 1,\quad a_{7-7} = 0,\\ a_S = \tfrac{1}{36}\cdot 1 + \tfrac{6}{36}\,a_7 + \tfrac{29}{36}\,a_S,\\ a_7 = \tfrac{1}{36}\cdot 1 + \tfrac{6}{36}\cdot 0 + \tfrac{29}{36}\,a_S. \end{cases} Solving gives aS=713.a_S = \frac{7}{13}.

Hence, the probability that A (sum of 12 first) wins is 7/137/13.

(Alternative via conditional probability: the same result 7/137/13.)