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国际象棋锦标赛

Chess Tournament

专题
Brainteaser / 脑筋急转弯
难度
L3

题目详情

一场国际象棋锦标赛有 128 名选手，且每名选手的等级分都互不相同。比赛采用单败淘汰赛赛制。假设等级分更高的选手总能战胜等级分更低的对手，且胜者晋级下一轮。问等级分最高和次高的两名选手在决赛相遇的概率是多少？

A chess tournament has 128 players, each with a distinct rating. The tournament is set up as a single-elimination bracket. Assume that the player with the higher rating always wins against a lower rated opponent and that the winner proceeds to the subsequent round. What is the probability that the highest rated and second-highest rated players will meet in the final?

解析

由于等级分最高的选手一定能一路获胜，等级分第二高的选手也一定会一路获胜，直到碰到第一名为止。

因此，这两人只有在被分到对阵表的不同半区时，才会在决赛相遇。

  • 整个 128 人签表分成两个半区，每个半区 64 个位置；
  • 先固定第一名的位置后，第二名在剩下的 127 个位置中等概率落位；
  • 其中有 64 个位置位于另一半区。

所以所求概率为 64127.\frac{64}{127}.

答案是 64127.\boxed{\frac{64}{127}}.

Original Explanation

In a single-elimination bracket with 128 players (272^7), the highest-rated player will win every match they play. The second-highest rated player will also win every match until they face the highest-rated player.

The only way they meet in the final is if they are placed in opposite halves of the bracket.

  • Each half has 64 slots.
  • Once the top player is placed, there are 127 remaining slots for the second-highest player.
  • To meet in the final, the second-highest player must be in the opposite half (64 slots).

Hence, the probability that the top two meet in the final is 64127\frac{64}{127}