The first player plays the rarest every second game.

## Solution

According to the condition, the second player played 21 games, so there were at least 21 games in total. Out of every two consecutive games, the first player must participate in at least one, which means there were no more than $2 \cdot 10+1=21$ games. Therefore, a total of 21 games were played, and the second player participated in each of them. In 10 games, he faced the first player, and in the remaining 11 games, he faced the third player.

## Answer

11 games.
3 [Pairings and Groupings; Bijections $]

Three people are playing table tennis, with the player who loses a game giving up their place to the player who did not participate in it. In the end, it turned out that the first player played 10 games, the second - 21. How many games did the third player play?

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