掷或拿 II

设阈值为 $n$。

• 每次 roll 命中 $\ge n$ 的概率为 $p=\frac{21-n}{20}$，命中所需 roll 次数期望为 $1/p=\frac{20}{21-n}$。
• 一次“成功兑现循环”平均消耗动作数为 $\frac{20}{21-n}+1$（roll 若干次 + 1 次 take）。
• 命中后点数的条件期望为 $\frac{20+n}{2}$。

因此期望收益为

$$f(n)=\frac{100}{\frac{20}{21-n}+1}\cdot\frac{20+n}{2}.$$

比较相邻整数可得最优阈值为 $n=6$，此时期望收益为

$$\frac{300}{7}\cdot 13=\frac{3900}{7} \approx 557.143.$$

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Original Explanation

First, let's get a baseline to compare to, the simplest strategy is to roll and always take the value that shows up. We will be able to perform this 50 times (as each is one action) and the expected value per roll is $10.5 = \frac{(1+2+\ldots +20)}{20}$, so our expected payout would be $50 \cdot 10.5 = 525$ with this strategy. Now, let's write the expected payoff as a function of $n$. Given we accept any value at least $n$, the expected value we roll given we accept is $\frac{20+n}{2}$. Out of our 20 values there are $(21-n)$ values that are at least value $n$ (note: $n > 0$), so the probability on each roll that we obtain a value at least $n$ is $\frac{21-n}{20}$. This probability is constant between rolls as we have a fair die, meaning the expected number of rolls to obtain a value at least $n$ is $\frac{20}{21-n}$. However, whenever we roll a value at least $n$ we also must claim it which takes up 1 action, so the expected number of turns needed to roll and claim the money is $\frac{20}{21-n} + 1$. Thus on average, we are able to roll and claim the money $\frac{100}{\frac{20}{21-n} + 1}$ times in the game, as it takes us that many turns on average to roll and claim and we have 100 total turns. Lastly, this implies our expected payout is

$$f(n) = \frac{100}{\frac{20}{21-n} + 1} \cdot \frac{20+n}{2}$$

To find the $n$ maximizing this, one can treat $f$ as continuous and use the derivative of it to find the optimal $n$. The details of taking the derivative messy and not enlightening, so the steps are excluded. However, after using the basic rules and simplifying,

$$f'(n) = \frac{50}{(n-41)^2} \cdot \left(n^2 - 82n + 461\right) = 0$$

The roots of the polynomial are

$$n_{1,2} = 41 \pm 2\sqrt{305}$$

The root adding $2\sqrt{305}$ is larger than 20, so

$$n^* = 41 - 2\sqrt{305}$$

must be the maximizer. As $17 < \sqrt{305} < 18$ and our optimal $n$ must be an integer, we can test $n = 5, 6, 7$ to see which gives us the largest expected payout.

Plugging all three of these in reveals

$$n = 6$$

maximizes $f(n)$ with payout $\frac{3900}{7}$.