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骰子均衡：完美点数

Dicey Equilibrium: The Perfect Roll

专题
Strategy / 策略
难度
L6

题目详情

有 100 枚骰子，第 ii 枚骰子是公平的，点数为 11 到 ii。

Nature 等概率随机选一枚骰子并掷一次，最终结果为 k∈{1,…,100}k\in\{1,\ldots,100\}。

三位玩家在知道先前玩家选择的情况下，依次从 {1,…,100}\{1,\ldots,100\} 选择一个目标数。最终离 kk 最近的玩家获胜。

问：子博弈完美均衡（subgame–perfect equilibrium）下三位玩家分别选择什么数？他们各自获胜概率是多少？

英文原题

In a game with 100 dice, where die ii is fair with faces numbered 11 through ii, Nature rolls a die chosen uniformly at random, resulting in outcome kk from {1,…,100}\{1, \ldots, 100\}. Three players, knowing previous picks, sequentially choose a target number from {1,…,100}\{1, \ldots, 100\}. The player with the number closest to kk wins. What choices form the subgame–perfect equilibrium and what are their respective win probabilities?

解析

令随机结果为 KK。由于先等概率选骰子再掷点，有

P(K=k)=1100∑i=k1001i,k=1,…,100.\mathbb{P}(K=k)=\frac{1}{100}\sum_{i=k}^{100}\frac{1}{i},\quad k=1,\ldots,100.

该分布明显偏向小数值，其中位数为 k=19k=19（即 P(K≤19)≈0.5015\mathbb{P}(K\le 19)\approx 0.5015）。

用逆推（每一步都选择能最大化自己赢面的位置）可得到子博弈完美均衡选择为

(y1,y2,y3)=(19,7,39).(y_1,y_2,y_3)=(19,7,39).

在该三点下，按“最近者获胜”的划分区间（若出现到两个点距离相同，则两者平分该点的胜率），边界在中点 13 与 29：

  • 玩家 2（选 7）在 k≤12k\le 12 时胜，k=13k=13 时与玩家 1 平分；
  • 玩家 1（选 19）在 14≤k≤2814\le k\le 28 时胜，并与玩家 2 在 k=13k=13 平分、与玩家 3 在 k=29k=29 平分；
  • 玩家 3（选 39）在 k≥30k\ge 30 时胜，k=29k=29 时与玩家 1 平分。

把上述区间上的 P(K=k)\mathbb{P}(K=k) 相加，可得三者胜率约为

(p1,p2,p3)≈(0.25864, 0.38052, 0.36084).\boxed{(p_1,p_2,p_3)\approx(0.25864,\ 0.38052,\ 0.36084)}.

英文解析

Make the random result KK. Due to the probability of choosing the dice first and then throwing the point, there are

P(K=k)=1100∑i=k1001i,k=1,…,100.\mathbb{P}(K=k)=\frac{1}{100}\sum_{i=k}^{100}\frac{1}{i},\quad k=1,\ldots,100.

This distribution is clearly biased towards decimal values, with a median of k=19k=19 (i.e. P(K≤19)≈0.5015\mathbb{P}(K\le 19)\approx 0.5015).

With reverse inference (choosing the position that maximizes your winning side at each step), you can get the perfect equilibrium choice for the sub-game

(y1,y2,y3)=(19,7,39).(y_1,y_2,y_3)=(19,7,39).

Under the three points, the boundary is divided between the "nearest winner" (if the distance to the two points is the same, the winning rate of the point is divided equally between the two points), and the boundary is at the midpoint 13 and 29:

  • Player 2 (Option 7) wins at k≤12k\le 12 and is split equally with Player 1 at k=13k=13;
  • Player 1 (Option 19) wins at 14≤k≤2814\le k\le 28 and is split equally with Player 2 at k=13k=13 and with Player 3 at k=29k=29;
  • Player 3 (Option 39) wins at k≥30k\ge 30 and is split equally with Player 1 at k=29k=29.

Adding up the P(K=k)\mathbb{P}(K=k) in the above interval, the win rate of the three is about

(p1,p2,p3)≈(0.25864,\0.38052,\0.36084).\boxed{(p_1,p_2,p_3)\approx(0.25864,\0.38052,\0.36084)}.