Only three totals survive from an afternoon of table tennis. Are they enough to reconstruct one particular defeat?
The Second Defeat
Riddle statement
Three friends—A, B and C—spend the afternoon playing table tennis.
Two play each game while the third waits. The winner stays at the table; the loser gives up their place to the waiting player. There are no draws, and they follow this rule throughout.
At the end, they count how many games each has played:
- A played 10.
- B played 15.
- C played 17.
Who lost the second game of the afternoon?
Show solution
Solution
A lost the second game.
The players’ totals add to 42. Two people play each game, so there were
(10 + 15 + 17) ÷ 2 = 21 games.
A played 10 and sat out 11. Nobody can sit out two consecutive games: the waiting player must enter the next one.
Separating A’s eleven breaks therefore requires at least ten games played by A. That uses all ten appearances. There is none left to put before the first break, after the last break, or between two already separated breaks.
A’s sequence is forced:
sit out, play, sit out, play, …, sit out.
A played game 2 and sat out game 3. Since the winner stays at the table, A must have lost game 2.
The totals do not tell us every result. They tell us just enough: A has the largest number of breaks possible, so their placement leaves no room for choice.