30 面骰一次 vs 20 面骰两次取最大（平局 Bob

记 Alice 的点数为 $A$，Bob 两次为 $B_1,B_2$，Bob 的有效点数为 $\max\{B_1,B_2\}$。

按 $A>20$ 与 $A\le 20$ 分情况：

$$P(A>20)=\frac{1}{3},\quad P(A\le 20)=\frac{2}{3}.$$

• 若 $A>20$，则 Alice 必胜。
• 若 $A\le 20$，条件在 $A=a$ 时 Alice 获胜需要 $\max\{B_1,B_2\}<a$，即 $B_1<a$ 且 $B_2<a$，概率为 $\left(\frac{a-1}{20}\right)^2$。

并且在 $A\le 20$ 条件下，$A$ 在 1..20 上均匀，所以

$$P(\text{赢}\mid A\le 20)=\frac{1}{20}\sum_{a=1}^{20}\left(\frac{a-1}{20}\right)^2=\frac{247}{800}.$$

合并得

$$P(\text{赢})=\frac{1}{3}+\frac{2}{3}\cdot\frac{247}{800}=\frac{647}{1200}.$$

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Original Explanation

Let $W$ be the event Alice wins. Then we can use the Law of Total Probability to condition on whether or not Alice's roll is larger than $20$. Let $A$ be the value Alice rolls. Furthermore, let $B_1$ and $B_2$ be the values that Bob rolls on the first and second roll. We have that $$\mathbb{P}[W] = \mathbb{P}[W \mid A > 20] \mathbb{P}[A > 20] + \mathbb{P}[W \mid A \leq 20] \mathbb{P}[A \leq 20]$$ It is clear to see that $$\mathbb{P}[A > 20] = \frac{1}{3}$$ as this accounts for $10$ of $30$ values. Furthermore, $$\mathbb{P}[W \mid A > 20] = 1$$ as this is strictly larger than any value Bob can roll.

For the other case, we compute $$\mathbb{P}[W \mid A \leq 20] = \displaystyle \sum_{a=1}^{20} \mathbb{P}[\text{max}\{B_1,B_2\} < a] \mathbb{P}[A = a \mid A \leq 20] = \sum_{a=1}^{20} \mathbb{P}[B_1 < a]^2 \mathbb{P}[A = a \mid A \leq 20]$$ This is the event of interest as we want Alice to have a strictly higher value than Bob. We simplify the maximum statement by noting that $B_1$ and $B_2$ are IID and that $\text{max}\{B_1,B_2\} < a$ is equivalent to saying that $B_1 < a, B_2 < a$.

We know $$\mathbb{P}[A = a \mid A \leq 20] = \frac{1}{20}$$ as we know $A \leq 20$, so there are $20$ equally-likely values. Furthermore, we have that $$\mathbb{P}[B_1 < a] = \frac{a-1}{20}$$ as we want $B_1$ to take some value between $1$ and $a - 1$, inclusive of both. Therefore, $$\mathbb{P}[W \mid A \leq 20] = \frac{1}{20} \sum_{a=1}^{20} \left(\frac{a-1}{20}\right)^2 = \frac{247}{800}$$ Combining these, we have that $$\mathbb{P}[W] = \frac{1}{3} + \frac{2}{3} \cdot \frac{247}{800} = \frac{647}{1200}$$