85. Tournaments. The so-called cup system for determining the winner from 8 athletes consists of dividing the players into pairs by drawing lots. 4 matches determine four winners, who participate in the second round; the third round of the competition is the final. The winner of the final match receives the first prize, and their opponent receives the second prize. Let's assume that each player has a certain strength (similar to how each object has a certain weight) and that a stronger player always wins against a weaker one (similar to how a heavier object always outweighs a lighter one if placed on different scales). Under these assumptions, the process described above is suitable for determining the champion, as the winner will indeed be stronger than all their opponents; however, the second place is not always occupied by the second strongest player.

What is the probability that the second participant in the final match is indeed worthy of the second prize?
85. The question should be understood as follows: in a cup system, it can happen that a player who deserves the second prize, because he is stronger than all the other players except the champion, meets the champion before the final, i.e., in the case of eight players, in the first or second round; in such a case, the second prize will be awarded unfairly. This can only happen when, during the draw to determine the partners, the player who deserves the silver medal, i.e., the second place, ends up in the same subgroup of four players as the champion.

Let's denote the champion by $\Psi$, and the "true" silver medalist by $C$; since in the group that $Y$ is in, there are only 3 spots, each of which $C$ can occupy, out of the seven possible spots for $C$, 4 are favorable to him, and 3 are not. The probability that $C$ will receive the second prize is, therefore, $4 / 7 = 0.5714 \ldots$, i.e., slightly more than $57\%$.