The Ping Pong Puzzle

Three friends (A, B and C) are playing ping pong. They play the usual way: the winner stays on, and the loser waits his/her turn again. At the end of the day, they summarize the number of games that each of them played:

A played 10
B played 15
C played 17.

Who lost the second game? 

This person played & lost the second game! 

Participant of the Second Game!

How games were played?

A played 10, B played 15 and C played 17 games. So total number of presences are 10 + 15 + 17 = 42. Every 2 presences form a game. Hence, the number of games played are 42/2 = 21.

Let's take into consideration the minimum number of games that a player can play. For that, he need to loose every game that he has played. That is, if he has played first game then he must have out in second but replaced looser of second in third game. In short, he must have played odd numbered of games like 1,3,5,7,9,11,13,15,17,19,21.That's 11 games in total.

And if he had made debut in second game then he must had played even numbered games like 2,4,6,8,10,12,14,16,18,20. That's 10 games in total.

Since, in the case only A has played 10 games, he must have made debut in second game where he lost that game to make comeback in 4th game thereby replacing looser of third game.