抽帽子：付 1 元可再抽一次

最优策略为“阈值停”：若抽到 $\ge x$ 则停，否则继续。

计算可得最优整数阈值为 $x=87$，对应的期望净收益为

$$\frac{1209}{14}\approx 86.36.$$

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Original Explanation

We want to determine an optimal threshold $x$, where, if AJ's draw is at least $x$, then he will not play another round. Notice that, if AJ draws again, without taking into consideration the $1 penalty, then AJ's expected payoff is the same as the previous round. So, we can write the following expression, where$\mathbb{E}[A]$ denotes the total winnings.

$$\begin{aligned} \mathbb{E}[A] & = - 1 + \sum_{i = x}^{100} \frac{i}{100} + \sum_{i = 1}^{x - 1} \frac{\mathbb{E}[A]}{100} \\ & = -1 + \frac{(100 - x + 1)(100 + x)}{200} + \frac{x - 1}{100} \mathbb{E}[A] \end{aligned}$$

Let's isolate $\mathbb{E}[A]$.

$$\begin{aligned} \mathbb{E}[A] & = \frac{(101 - x)(100 + x) - 200}{2(101-x)} \end{aligned}$$

We now wish to determine

$$x_\text{optimal} = \underset{x}{\text{arg max}} \left\{\frac{(101 - x)(100 + x) - 200}{2(101-x)} \right\}.$$

Let's take the derivative with respect to $x$.

$$\frac{d}{dx} \mathbb{E}[A] = \frac{10001 - 202 x + x^2}{4(101-x)^2}$$

Setting $\frac{d}{dx} \mathbb{E}[A] = 0$ and solving for $x$, we find

$$\begin{aligned} 10001 - 202 x_\text{optimal} + x_\text{optimal}^2 & = 0 \\ x_\text{optimal} & = \frac{202 \pm \sqrt{800}}{2} \\ & = 101 \pm 10 \sqrt{2} \end{aligned}$$

Since $x \leq 100$,

$$x_\text{optimal} = 101 - 10 \sqrt{2} \approx 86.9.$$

Our optimal integer $x$ is then either 86 or 87. Plugging both into $\mathbb{E}[A]$, we find $\mathbb{E}[A]$ is maximized when $x = 87$ and has value $\frac{1209}{14}$.