Three friends ($A$, $B$ and $C$) are playing ping pong. They play the usual way: two play at a time, the winner stays on, and the loser waits his/her turn again. At the end of the day, they summarise the number of games that each of them played:
$A$ played $10$ games
$B$ played $15$ games
$C$ played $17$ games
Who lost the second game?
There were 21 matches played. The rules ensure that ...
... each player gets to play at least in every other match: Whoever isn't playing will replace the loser of the ongoing match in the next match.
This means that ...
... A didn't play in the first match. If he had, he would have played at least 11 matches. He came in for the loser of the first match. He also lost every game. Had he won a match, he would have played two consecutive matches and therefore played at least eleven. A's participation pattern is:
-x-x-x-x-x-x-x-x-x-x-
So ...
... A was in the second match and he lost it.
The player who lost the second game is player
A
The total number of games is
$\frac{10+15+17}{2} = 21$
Since you can only wait for one game, the minimum of games every player plays is
$10$
This means that:
A had to lose every game, starting with the second game. If he would have played (and lost) the first game, he would have played 11 games.
The answer is
A lost the second game
explanation incoming.
To figure this out
First we need to find the total number of games played. Since two play pingpong at a time, we find this easily enough by adding up the total of all games played, 42, and dividing by 2 to get 21 total games.
Now, logically the more a person wins the more games they get to play because they stay on after they've won to take on a new challenger. Therefore, the highest number of games a player could play is 21, if they won the first game and then won every game thereafter (up to the last game, I guess, where it doesn't matter if they won or not).
On the flipside, if a player lost every game they played, they would only be able to play in every other game. They play, lose, sit one out. Play again, lose, sit one out. If they played in the first game, they would have played in all odd numbered games 1 to 21, which is 11 games total. If they didn't play in the first game, they would have played all even numbered games, 10 games total. Since A only played 10 games, we can conclude that A played the second game and lost every game they played!!! Therefore, A lost the second game.
Now, a slightly more difficult question, how many games each did A, B, and C, win?
We know that A lost 10 games.
We also know that B and C played 11 games against each other. This means that B played 4 games against A, which B won, and C played 6 games against A, which C won.
In order to have played 4 games against A, B must have won the four games against C prior to accepting A as a challenger. So B won at least 8 games.
Likewise, C beat B at least 6 times prior to playing against A, so B lost 8 games and won 6, and the fate of the last game cannot be known from the given information.
So the final score is A-0, B-8 (or 9), and C-12 (or 13).
