7. Anchurian Checkers ${ }^{1}$ (from 7th grade, 1 point). A tournament of Anchurian Checkers is held in several rounds. If an even number of players participate in a round, they are divided into random pairs. If the number of players is odd, random pairs are selected by lot, and one player remains without a pair and does not participate in the round. A draw cannot occur; the loser in each pair is eliminated from the tournament, while the winners and the player without a pair, if there is one, advance to the next round, which is conducted according to the same rules. This continues until only two players remain, who play the final round, that is, the last game, which determines the winner of the tournament.[^5]

XIV Correspondence Internet Olympiad in Probability Theory and Statistics for Schoolchildren

(c) Moscow Center for Continuous Mathematical Education, 2021

26 participants arrived at the Anchurian Checkers tournament, and all of them play equally well, meaning that in any game played by any two of them, the chances of the opponents are equal. Among the players are Denis and Oleg. Find the probability that they will play against each other.
Solution. Let there be $n$ participants in the tournament. As a result of each match, one participant is eliminated, so there will be a total of $n-1$ matches. At the same time, no pair can play two matches, since the loser is eliminated. Therefore, there will also be $n-1$ game pairs throughout the tournament. In total, from $n$ participants, $C_{n}^{2}$ pairs of players can be formed, but only $n-1$ of these possible pairs will be game pairs, that is, they will play against each other. Due to the randomness of the selection and the equal strength of the opponents in each match, the probability that two predetermined participants will end up in one of the game pairs is $\frac{n-1}{C_{n}^{2}}=\frac{2(n-1)}{n(n-1)}=\frac{2}{n}$. Substituting $n=26$, we get that Denis and Oleg will end up in one of the game pairs with a probability of $\frac{1}{13}$.

Answer: $\frac{1}{13}$.