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?

Source: Puzzle from Alex Bellos in Guardian.

  • $\begingroup$ Eurgh, Alex Bellos. $\endgroup$ Oct 13, 2016 at 1:10

3 Answers 3


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:


So ...

... A was in the second match and he lost it.

  • $\begingroup$ answered at almost the same time! I liked the pattern you came up with to help visualize the answer, well done. $\endgroup$
    – MMAdams
    Oct 12, 2016 at 14:38
  • 1
    $\begingroup$ @MMAdams: But you beat me to it. I saw the "new answer" notification while I was proofreading and then posted anyway. $\endgroup$
    – M Oehm
    Oct 12, 2016 at 14:40

The player who lost the second game is player


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


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).

  • $\begingroup$ how do you do line breaks inside a spoiler? I'm sorry if it's kinda hard to read, it turned into one big block and I didn't mean it to. $\endgroup$
    – MMAdams
    Oct 12, 2016 at 15:23
  • 1
    $\begingroup$ Double space at the end of a line will make a soft line break without breaking the spoiler block. $\endgroup$
    – Ax.
    Oct 12, 2016 at 15:34
  • 1
    $\begingroup$ Or you can just add another >! at the start of the new paragraph. The block will still be one. $\endgroup$
    – Alenanno
    Oct 12, 2016 at 21:50
  • $\begingroup$ and the one we don't know the result for has to be the last one. every other game is determined. $\endgroup$
    – njzk2
    Oct 13, 2016 at 20:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.