倒霉的 7 I

若第一次掷到 3、4、5、6，再掷必然使和 $\ge 7$，只会导致付出第一次点数，因此应直接停。

只需比较第一次为 1 或 2 时是否再掷：

• 若第一次为 1：再掷仅在第二次为 6 时亏 1（概率 $1/6$），其余得到 1..5。再掷期望

$$\frac{1}{6}(-1)+\frac{1+2+3+4+5}{6}=\frac{7}{3}>1,$$

所以第一次为 1 时应再掷。

• 若第一次为 2：若第二次为 5 或 6（概率 $1/3$）则亏 2，否则得到 1..4。再掷期望

$$\frac{1}{3}(-2)+\frac{1+2+3+4}{6}=1<2,$$

所以第一次为 2 时不再掷。

于是策略为：第一次为 1 则再掷，否则停。总体期望为

$$\frac{\frac{7}{3}+2+3+4+5+6}{6}=\frac{67}{18}.$$

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Original Explanation

The first observation that can be made is that if we roll a $3, 4, 5,$ or $6$, we should not consider rolling again. This is because we can only improve our stance by rolling a number larger, but the sum is guaranteed to go over $7$ in any of these cases if our first roll is any of these numbers. Therefore, we should look at just rolling a $1$ or $2$. If we roll a $1$, we either lose $1$ with probability $\dfrac{1}{6}$ (if we roll a $6$ on the second roll). Otherwise, we make $1, 2, \dots, 5$, each with probability $\dfrac{1}{6}$. Therefore our expected payout if we roll again is $\dfrac{1}{6} \cdot (-1) + \dfrac{1+2+3+4+5}{6} = \dfrac{7}{3} > 1$. This implies we should roll again if we receive a $1$. If we roll a $2$, then with probability $\dfrac{1}{3}$ we lose $2$ (if we roll a 5 or 6). Otherwise, we receive $1,2,3,$ or $4$ with equal probability $\dfrac{1}{6}$. Therefore, the expected value upon rolling again is $\dfrac{1}{3} \cdot (-2) + \dfrac{1+2+3+4}{6} = 1 < 2$. This implies we should not roll again and keep the two. As a result, we keep all values that aren't $1$, and otherwise, we roll a $1$ again. This yields an expected value of $\dfrac{\frac{7}{3} + 2 + 3 + 4 + 5 + 6}{6} = \dfrac{67}{18}$.