最多掷三次骰子的最优停策略

第三次若到达，期望为

$$\frac{1+2+3+4+5+6}{6}=3.5.$$

第二次看到 $x$：若继续的期望是 3.5，则当 $x>3.5$（即 $4,5,6$）应停，否则继续。于是第二次后的期望为

$$\frac{3}{6}\cdot 3.5 + \frac{1}{6}(4+5+6)=4.25.$$

第一次看到 $x$：与 4.25 比较，若 $x>4.25$（即 $5,6$）应停，否则继续。于是从开始的期望为

$$\frac{2}{6}\cdot 4.25 + \frac{1}{6}(5+6)=\frac{14}{3}\approx 4.67.$$

最优策略：

• 第一次掷到 5 或 6 就停；
• 第二次掷到 4、5、6 就停；
• 否则掷到第三次并接受结果。

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Original Explanation

• Final stage (3rd roll): If you get to the 3rd roll, the expected value of that single roll is $$\frac{1+2+3+4+5+6}{6} = 3.5.$$

• Second roll: If you see a value $x$ on the second roll, compare $x$ to the expected value if you continue (which is 3.5 for the third roll).

  • If $x > 3.5$ (i.e., $x = 4,5,6$), stop and take $x$.
  • If $x \le 3.5$ (i.e., $x = 1,2,3$), continue.
  • The expected value after the second roll is thus $$\frac{3}{6}\times 3.5 + \frac{1}{6}\times (4+5+6) = 4.25.$$

• First roll: Compare the first-roll $x$ to 4.25.

  • If $x > 4.25$ (i.e., $x=5$ or $6$), stop and take $x$.
  • If $x \le 4.25$ (i.e., $x=1,2,3,4$), continue.
  • The expected value from the first roll is then $$\frac{2}{6}\times 4.25 + \frac{1}{6}\times (5+6) = \frac{14}{3} \approx 4.67.$$

Thus, the game is worth $\tfrac{14}{3}\approx 4.67$.
Optimal strategy:

• Stop if the roll is 5 or 6 on the first roll,
• Stop if the roll is 4, 5, or 6 on the second roll,
• Take whatever the third roll is if you reach that point.