三名枪手轮流射击：X 获胜概率

X 在第 1、4、7、… 次射击命中时获胜。命中发生在第 $k$ 次射击的概率为 $2^{-k}$。

因此

$$P(X\text{赢})=\sum_{j=0}^\infty 2^{-(3j+1)}=\frac{1/2}{1-1/8}=\frac{4}{7}.$$

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Original Explanation

Solution #1: We're looking for the distribution of the first success of repeated trials with probability $\frac{1}{2}$ of success. The distribution of the number of trials needed is $N \sim \text{Geom}\left(\frac{1}{2}\right)$. Therefore, $\mathbb{P}[N = k] = \frac{1}{2^k}$ for $k \geq 1$. If $X$ wins, then the shot is on one of trials $1, 4, 7, \ldots$, as it must cycle back to them. Therefore, the probability of interest is $$\sum_{k=0}^{\infty} \left(\frac{1}{2}\right)^{3k+1} = \frac{1}{2}\sum_{k=0}^{\infty} \frac{1}{8^k} = \frac{1}{2} \cdot \frac{1}{1-\frac{1}{8}} = \frac{4}{7}$$

Solution #2: Let's look at another method to arrive at this. We can call the probability of person $X$ winning $p$. Every outcome where person $Y$ has the opportunity to win is the same as person $X$ except scaled by a factor of half since person $Y$ must ensure person $X$ misses right before them, which occurs with a probability of $\frac{1}{2}$, so the probability for person $Y$ is $\frac{1}{2}p$.

By the same logic, the probability person $Z$ wins is $\frac{1}{4}p$. Adding all of these up yields $$p\left(1 + \frac{1}{2} + \frac{1}{4}\right) = 1 \iff p = \frac{4}{7}$$