30 面骰 vs 20 面骰：高者胜且输家付赢家点数（平

令 Alice 掷出 $A$、Bob 掷出 $B$，Alice 的收益 $P$ 为：若 Alice 赢则 $+A$，若输（含平局）则 $-B$。

按 $A\le 20$ 与 $A>20$ 分情况：

• $P(A\le 20)=\frac{2}{3}$。此时若忽略平局偏置，赢输对称；但平局（概率 $1/20$）判 Bob 赢，且平局值在 1..20 上等可能，因此该情形下 Alice 的期望为

$$\mathbb{E}[P\mid A\le 20]= -\frac{1+20}{2}\cdot\frac{1}{20}= -\frac{21}{40}.$$

• $P(A>20)=\frac{1}{3}$。此时 Alice 一定赢，且 $A$ 在 21..30 上均匀，所以

$$\mathbb{E}[P\mid A>20]=\frac{21+30}{2}=\frac{51}{2}.$$

合并：

$$\mathbb{E}[P]= -\frac{21}{40}\cdot\frac{2}{3}+\frac{51}{2}\cdot\frac{1}{3}=\frac{163}{20}=8.15.$$

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Original Explanation

Let $P$ be the payout, while $A$ and $B$ are the values Alice and Bob roll. The key here is to condition on whether or not $A \leq 20$. Namely, by Law of Total Expectation, we have that

$$\mathbb{E}[P] = \mathbb{E}[P \mid A \leq 20] \mathbb{P}[A \leq 20] + \mathbb{E}[P \mid A > 20]\mathbb{P}[A > 20]$$

We quickly see that $\mathbb{P}[A \leq 20] = \frac{2}{3}$, as this accounts for $20$ of the $30$ values that can appear. The expected payout for Alice in this case would be $0$ if ties were not settled in Bob's favor. Ties happen with probability $\frac{1}{20}$ in this case, as the first roll is completely arbitrary and the second roll just needs to match the first value. Given a tie occurs, it is equally likely to be any of the $20$ values. Therefore,

$$\mathbb{E}[P \mid A \leq 20] = -\frac{1 + 20}{2} \cdot \frac{1}{20} = -\frac{21}{40}$$

If $A > 20$, occurring with probability $\frac{1}{3}$, then Alice is guaranteed to win. Her expected payout in this case then is

$$\frac{21 + 30}{2} = \frac{51}{2}$$

Combining this, we see that

$$\mathbb{E}[P] = -\frac{21}{40} \cdot \frac{2}{3} + \frac{51}{2} \cdot \frac{1}{3} = 8.15$$