倒霉的 7 II

分情况讨论第一次点数 $a$：

• $a=1$：再掷期望为

$$\frac{1}{6}(-1)+\frac{2+3+4+5+6}{6}=\frac{19}{6}>1,$$

应再掷。

• $a=2$：再掷有 $1/3$ 概率（掷到 5 或 6）亏 2，否则得到 3..6。期望

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

应再掷。

• $a=3$：再掷有 $1/2$ 概率（掷到 4..6）亏 3，否则得到 4..6。期望

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

不再掷。对 $a\ge 3$ 同理更不应再掷。

因此策略：第一次为 1 或 2 则再掷，否则停。总体期望为

$$\frac{\frac{19}{6}+\frac{14}{6}+3+4+5+6}{6}=\frac{47}{12}.$$

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Original Explanation

Condition on the value of the first roll. 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 a total of $2, 3, \dots, 6$, each with probability $\dfrac{1}{6}$. Therefore our expected payout if we roll again is $\dfrac{1}{6} \cdot (-1) + \dfrac{2+3+4+5+6}{6} = \dfrac{19}{6} > 1$. 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$3, 4, 5,$or$6$with equal probability$\dfrac{1}{6}$. Therefore, the expected value upon rolling again is$\dfrac{1}{3} \cdot (-2) + \dfrac{3+4+5+6}{6} = \dfrac{7}{3} > 2$. This implies we should roll.

If we roll a $3$, then with probability $\dfrac{1}{2}$ we lose $3 (rolling 4 or more). Otherwise, we earn$4, 5,$or$6$each with probability$\dfrac{1}{6}$. Therefore, our expected profit upon rolling again is$\dfrac{1}{2} \cdot (-3) + \dfrac{4 + 5 + 6}{6} = 1 < 3$. Therefore, for any value$3$or more, we should not roll again. This implies that our expected value on this game with this strategy is$\dfrac{\frac{19}{6} + \frac{14}{6} + 3 + 4 + 5 + 6}{6} = \dfrac{47}{12}$.