选对罐子

先算后验：

$$P(A\mid\text{抽到 1})=\frac{\frac12\cdot\frac{4}{10}}{\frac12\cdot\frac{4}{10}+\frac12\cdot\frac{3}{10}}=\frac{4}{7}, \quad P(B\mid\text{抽到 1})=\frac{3}{7}.$$

若继续抽同一只：

• 若在 A：剩余为 3 个 1 与 6 个 10（共 9 个），期望 $\frac{3\cdot 1+6\cdot 10}{9}=7$。
• 若在 B：剩余为 2 个 1 与 7 个 10（共 9 个），期望 $\frac{2+70}{9}=8$。

综合期望：

$$7\cdot\frac{4}{7}+8\cdot\frac{3}{7}=\frac{52}{7}\approx 7.4286.$$

若改抽另一只则期望更低，因此最优是“不换”，期望收益为 $52/7$。

---

Original Explanation

Evaluate the expected payout if you switch urns versus if you stay with your current urn. This can be split into $2$ cases depending on which urn you select the first chip from. We can assign Urn $A$ to be the urn that started with $4$ $(\$1$ chips) and Urn $B$ be the one that started with $3$ $(\$1$ chips). Define the random variable $X$ as the urn that you selected your first chip from. We can say that $\mathbb{P}[X = A] = \dfrac{4}{3}\mathbb{P}[X = B]$ as the Urn $A$ started with $4$ $(\$1$ chips) while Urn $B$ has $3$. Therefore, as those 2 values must sum to $1$, $\mathbb{P}[X = A] = \dfrac{4}{7}$ and $\mathbb{P}[X = B] = \dfrac{3}{7}$.

Now we can consider $2$ more cases being if we switch urns or stay with the same urn. Denote $P_0$ be the payout if we stay with the same urn. If we don't switch, then:

$$\mathbb{E}[P_0] = \mathbb{E}[P_0 \mid X = A]\mathbb{P}[X = A] + \mathbb{E}[P_0 \mid X = B]\mathbb{P}[X = B]$$

If $X = A$, then after drawing one $(\$1$ chip), there are $6$ $(\$10$ chips) and $3$ $(\$1$ chips) left, so $\mathbb{E}[P_0 \mid X = A] = \dfrac{63}{9} = 7$. Similarly, there would be $7$ $(\$10$ chips) and $2$ $(\$1$ chips) after the first draw if $X = B$, so $\mathbb{E}[P_0 \mid X = B] = \dfrac{72}{9} = 8$. Plugging these values in yields: $$\mathbb{E}[P_0] = 7 \cdot \dfrac{4}{7} + 8 \cdot \dfrac{3}{7} = \dfrac{52}{7}$$

Let $P_1$ be your profit if you switch. Similarly, we have: $$\mathbb{E}[P_1] = \mathbb{E}[P_1 \mid X = A]\mathbb{P}[X = A] + \mathbb{E}[P_1 \mid X = B]\mathbb{P}[X = B]$$

If $X = A$, then by switching, we pick from an untouched urn $B$, so $\mathbb{E}[P_1 \mid X = A] = \dfrac{73}{10} = 7.3$. Similarly, if $X = B$, then we pick from an untouched urn $A$, so $\mathbb{E}[P_1 \mid X = B] = \dfrac{64}{10} = 6.4$.

Therefore, $$\mathbb{E}[P_1] = 6.4 \cdot \dfrac{3}{7} + 7.3 \cdot \dfrac{4}{7} = \dfrac{242}{35} < \dfrac{245}{35} = 7$$

Therefore, you should not switch, and your expected payout is $\dfrac{52}{7}$.