平均数猜数游戏

设两人的猜测为 $x$ 与 $y$。目标数为

$$T=0.8\cdot \frac{x+y}{2}=0.4(x+y).$$

在没有特殊规则时，如果对方固定为 $y>0$，你可以令自己的猜测满足 $x=T$，这样距离为 0，从而必胜。由 $x=0.4(x+y)$ 得到最佳反应

$$\boxed{x=\frac{2}{3}y}.$$

因此每个正数 $y$ 都有一个“赢面为 1 的”对应反应 $\tfrac{2}{3}y$，这导致标准版本里通过反复删除劣策略会把猜测往 0 推，极限是 $(0,0)$。

但本题额外规定 $(0,100)$ 时猜 100 获胜，使得“若相信对方会猜 0，则猜 100 反而保证获胜”。而一旦有人倾向于猜 100，对方的最佳反应又是猜 $\tfrac{2}{3}\cdot 100=66\tfrac{2}{3}$（使自己刚好等于目标）。

结论：特殊规则打破了简单的纯策略均衡直觉；更实用的表达是

• 对任意 $y\in(0,100]$，最优反应是 $x=\tfrac{2}{3}y$；
• 若你认为对方会猜 0，则你应猜 100（因为规则强行判你赢）。

真正的“最优策略”取决于你对对方行为的预期，而不存在一个对所有理性对手都稳健的单一固定点纯策略。

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英文解析

Let's say two people guess $x$ and $y$. The target number is

$$T=0.8\cdot \frac{x+y}{2}=0.4(x+y).$$

When there are no special rules, if the other party is fixed as $y>0$, you can make your guess meet $x=T$, so that the distance is 0, and you will win. Optimal response from $x=0.4(x+y)$

$$\boxed{x=\frac{2}{3}y}.$$

Therefore, each positive number $y$ has a "win-side" response of $\tfrac{2}{3}y$, which causes the standard version to push the guess to 0 by repeatedly deleting the inferior strategy, and the limit is $(0,0)$.

However, the additional requirement in this question is to guess 100 to win when $(0,100)$, so that "if you believe that the other party will guess 0, then guess 100 will guarantee to win". And once someone tends to guess 100, the best response is to guess $\tfrac{2}{3}\cdot 100=66\tfrac{2}{3}$ (make yourself exactly equal to the target).

Conclusion: Special rules break the simple pure strategy equilibrium intuition; the more practical expression is

• For any $y\in(0,100]$, the optimal response is $x=\tfrac{2}{3}y$;
• If you think the other person will guess 0, you should guess 100 (because the rules force you to win).

The real “optimal strategy” depends on your expectations of the other party's behavior, and there is no single fixed-point pure strategy that is robust to all rational opponents.