Question

Dave 掷一枚公平六面骰，直到第一次出现六点为止。Linda 独立地掷一枚公平六面骰，也直到第一次出现六点为止。设互质正整数 $m$ 和 $n$ 满足 $\frac{m}{n}$ 等于 Dave 与 Linda 的掷骰次数相差不超过一的概率。求 $m + n$ 。

Dave rolls a fair six-sided die until a six appears for the first time. Independently, Linda rolls a fair six-sided die until a six appears for the first time. Let $m$ and $n$ be relatively prime positive integers such that $\frac{m}{n}$ is the probability that the number of times Dave rolls his die is equal to or within one of the number of times Linda rolls her die. Find $m + n.$

答案

答案：41

知识点：几何分布等比数列对称性

难度评级：2560

解答：

某人的第一次六点出现在第 $k$ 次的概率为 $p_k = \left(\frac{5}{6}\right)^{k-1} \cdot \frac{1}{6}$ 。两人次数相同的概率为 $\sum_{k=1}^{\infty} p_k^2 = \frac{1}{36} \cdot \frac{1}{1 - \frac{25}{36}} = \frac{1}{11}.$

Linda 恰好比 Dave 多掷一次的概率为 $\sum_{k=1}^{\infty} p_k p_{k+1}$ $= \frac{5}{6} \sum_{k=1}^{\infty} p_k^2$ $= \frac{5}{66}$ ，由对称性，Dave 恰好比 Linda 多掷一次也有相同概率。

总概率为 $\frac{1}{11} + 2 \cdot \frac{5}{66} = \frac{6 + 10}{66} = \frac{8}{33}$ ，所以 $m + n = 8 + 33 = 41$ 。

The probability that a player's first six appears on roll $k$ is $p_k = \left(\frac{5}{6}\right)^{k-1} \cdot \frac{1}{6}.$ The probability of a tie is $\sum_{k=1}^{\infty} p_k^2 = \frac{1}{36} \cdot \frac{1}{1 - \frac{25}{36}} = \frac{1}{11}.$

The probability that Linda needs exactly one more roll than Dave is $\sum_{k=1}^{\infty} p_k p_{k+1}$ $= \frac{5}{6} \sum_{k=1}^{\infty} p_k^2$ $= \frac{5}{66},$ and by symmetry the same holds with the players swapped.

The total probability is $\frac{1}{11} + 2 \cdot \frac{5}{66} = \frac{6 + 10}{66} = \frac{8}{33},$ so $m + n = 8 + 33 = 41.$

2009 AIME II 第 8 题

答案：41

难度评级：2560

解答：

其他年份的第 8 题