猜数博弈

把 B 的策略设为：猜 $n$ 的概率为 $q_n$（$q_n\ge 0$ 且 $\sum_{n=1}^{100} q_n=1$）。

若 A 选择数字 $n$，则 B 的期望收益为

$$\mathbb{E}[\text{payoff}\mid n]=n\,q_n.$$

A 会选择使该期望最小的 $n$，因此 B 要最大化

$$\max_{\{q_n\}}\ \min_{1\le n\le 100} (n q_n).$$

最优做法是把所有 $nq_n$ 拉平为同一常数 $c$：令

$$q_n=\frac{c}{n}.$$

由概率和为 1 得

$$\sum_{n=1}^{100} \frac{c}{n}=1\Rightarrow c=\frac{1}{H_{100}},\quad H_{100}=\sum_{n=1}^{100}\frac{1}{n}.$$

于是对任意 $n$ 都有 $nq_n=c$，A 无论选什么都只能给出期望 $c$。

因此 B 的最优策略是 $\boxed{q_n\propto 1/n}$，游戏的公平价格（B 的保证期望收益）为

$$\boxed{\frac{1}{H_{100}}}\approx 0.193\ \text{美元}.$$

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英文解析

Set B's strategy as: guessing $n$ with probability $q_n$ (where $q_n \ge 0$ and $\sum_{n=1}^{100} q_n = 1$).

If A chooses the number $n$, B's expected payoff is

$$\mathbb{E}[\text{payoff}\mid n]=n\,q_n.$$

A will choose the $n$ that minimizes this expected value, so B must maximize

$$\max_{\{q_n\}}\ \min_{1\le n\le 100} (n q_n).$$

The optimal approach is to flatten all $nq_n$ to the same constant $c$: let

$$q_n=\frac{c}{n}.$$

From the condition that probabilities sum to 1, we have

$$\sum_{n=1}^{100} \frac{c}{n}=1\Rightarrow c=\frac{1}{H_{100}},\quad H_{100}=\sum_{n=1}^{100}\frac{1}{n}.$$

Thus, for any $n$, $nq_n=c$, meaning A can only achieve an expected value of $c$regardless of their choice.

Therefore, B's optimal strategy is $\boxed{q_n\propto 1/n}$, and the fair price of the game (B's guaranteed expected payoff) is

$$\boxed{\frac{1}{H_{100}}}\approx 0.193\ \text{dollars}.$$