大满贯赛事

要解决这个问题，请尝试可视化锦标赛分组。第一轮共有128名选手参赛，其中64名选手晋级下一轮。每轮的玩家人数减半，直到进入最后一轮。

我们知道排名最高的玩家 $P_1$ 将始终赢得比赛。因此，第二高玩家 $P_2$ 进入决赛的机会取决于在任何早期回合中没有与 $P_1$ 交手。

我们在上文中指出，第一轮将有 128 名选手参赛。这 128 名球员将被分为两个小组，每组 64 名球员。 $P_2$ 在决赛前不与 $P_1$ 比赛的唯一原因是他们位于不同的小组。因此这个问题的答案就是找到$P(P_1 \textrm{ and } P_2 \ \textrm{are in different subgroups})$。

$P_2$在第一轮比赛中可能出战的选手有127名，其中64名与$P_1$不在同一小组。因此$P_1$和$P_2$进入决赛的概率是$\frac{64}{127}$

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Original Explanation

To solve this question try visualizing the tournament bracket. In the first round, 128 players will compete of which 64 will advance to the next round. Each round the number of players halves until we reach the final round.

We know that the highest-ranked player $P_1$ will always win the tournament. Therefore the second-highest player's, $P_2$, chance of making it to the finals is dependent upon not playing $P_1$ in any of the earlier rounds.

We noted above that in the first round 128 players will compete. These 128 players will be divided into two subgroups of 64 players each. The only way that $P_2$ doesn't play $P_1$ before the finals is if they are in different subgroups. Therefore the answer to this question is to find the $P(P_1 \textrm{ and } P_2 \ \textrm{are in different subgroups})$.

There are 127 possible players that $P_2$ can play in the first round, 64 of which are not in the same subgroup as $P_1$. Therefore the probability that $P_1$ and $P_2$ play in the finals is $\frac{64}{127}$