递增随机数游戏：先手胜率

设游戏在第 $k$ 次抽样时首次失败（即 $X_1<\cdots<X_{k-1}$ 但 $X_k<X_{k-1}$）。

对连续分布，$k$ 个样本的相对大小顺序等可能。事件“前 $k-1$ 个严格递增”意味着在 $k$ 个数的全排列中，前 $k-1$ 个必须是递增的那一种顺序，占比 $1/(k-1)!$。在此条件下，要在第 $k$ 次失败等价于 $X_k$ 不是这 $k$ 个数中的最大值，共有 $k-1$ 种可能位置，因此

$$\mathbb{P}(L=k)=\frac{k-1}{k!}.$$

你在 $k$ 为偶数时获胜（对手抽到了失败那步），所以

$$\mathbb{P}(\text{Win})=\sum_{j=1}^{\infty}\mathbb{P}(L=2j)=\sum_{j=1}^{\infty}\frac{2j-1}{(2j)!}.$$

并且

$$\frac{2j-1}{(2j)!}=\frac{1}{(2j-1)!}-\frac{1}{(2j)!},$$

因此

$$\mathbb{P}(\text{Win})=\left(\frac{1}{1!}-\frac{1}{2!}\right)+\left(\frac{1}{3!}-\frac{1}{4!}\right)+\cdots =1-e^{-1}.$$

答案：$\boxed{1-e^{-1}}\approx 0.6321$。

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英文解析

Suppose the game fails for the first time at the $k$-th draw (i.e., $X_1 < \cdots < X_{k-1}$but $X_k < X_{k-1}$).

For a continuous distribution, the relative ordering of $k$ samples is equally likely. The event "the first $k-1$ are strictly increasing" implies that among all $k!$ permutations of the $k$ numbers, the first $k-1$ must be in increasing order, which occurs with probability $1/(k-1)!$. Under this condition, failing at the $k$-th draw is equivalent to $X_k$not being the maximum of these $k$numbers; there are $k-1$possible positions for the maximum, so

$$\mathbb{P}(L=k)=\frac{k-1}{k!}.$$

You win when $k$ is even (the opponent draws the failure step), so

$$\mathbb{P}(\text{Win})=\sum_{j=1}^{\infty}\mathbb{P}(L=2j)=\sum_{j=1}^{\infty}\frac{2j-1}{(2j)!}.$$

And

$$\frac{2j-1}{(2j)!}=\frac{1}{(2j-1)!}-\frac{1}{(2j)!},$$

therefore

$$\mathbb{P}(\text{Win})=\left(\frac{1}{1!}-\frac{1}{2!}\right)+\left(\frac{1}{3!}-\frac{1}{4!}\right)+\cdots =1-e^{-1}.$$

Answer: $\boxed{1-e^{-1}}\approx 0.6321$.