You won 56 points. Your opponent won 48. Who won the match?
You are Blue. Winning 56 points to 48 looks convincing. Open the scorebook: it preserves something that the total has thrown away.
The total is hiding the winner
Reveal one set at a time. Then compare a second match with exactly the same point and game totals.
Points won in the entire match
Who would you give the trophy to from this scoreboard?
An invented best-of-three match. A set needs at least six games and a two-game margin. Every game is won to love; there are no tie-breaks or unfinished games.
One token is one game won to love: four points for the winner, none for the other player. B or A names the winner. Within each set, wins alternate first, followed by the remaining games.
Both complete scorebooks
An invented best-of-three match. A set needs at least six games and a two-game margin. Every game is won to love; there are no tie-breaks or unfinished games.
In every pair of numbers, Blue is on the left and Amber on the right.
First match
| Set | Games | Points won in the entire match |
|---|---|---|
| Set 1 | 6 : 0 | 24 : 0 |
| Set 2 | 4 : 6 | 16 : 24 |
| Set 3 | 4 : 6 | 16 : 24 |
| Total | 14 : 12 | 56 : 48 |
Match over. Amber won two sets, despite winning fewer points and fewer games.
Second match
| Set | Games | Points won in the entire match |
|---|---|---|
| Set 1 | 6 : 4 | 24 : 16 |
| Set 2 | 2 : 6 | 8 : 24 |
| Set 3 | 6 : 2 | 24 : 8 |
| Total | 14 : 12 | 56 : 48 |
Now Blue wins. Each player has exactly as many points and games as before. Their distribution across sets has changed.
These are points won, not a score of 40–15
Here, each rally won counts as one point. The umpire’s calls of 15, 30 and 40 name the score within a game; they are not quantities to add together.
Our invented match is deliberately easy to audit: every game is won to love, with four consecutive points. Each coloured token in the scorebook is therefore worth exactly four points. A real game can last longer; this is an assumption of our puzzle.
A first-set rout buys no head start in the next
In the first match, Blue takes the opening set 6–0, then loses the next two 4–6. Blue leads in total games: 6 + 4 + 4 = 14, against 0 + 6 + 6 = 12. Multiply by four and you get 56–48.
But this is a best-of-three match. Blue has won one set; Amber has won two. The six games from the first set cannot be carried into the third. Winning a set 6–0 earns exactly as many sets as winning it 6–4: one.
Keep all 104 points. Change the scorebook.
In the second match, Blue’s set scores are 6–4, 2–6, 6–2. Blue still has 14 games and 56 points. Amber still has 12 games and 48 points. Now Blue has the two winning sets.
We have not corrected an umpire’s mistake or swapped the column headings. We have constructed a different legal order for the same fourteen blue and twelve amber game wins. In each set, the last token really does finish the set; neither match should have stopped earlier.
Try it on paper
Cut out 14 blue and 12 yellow tokens, or label them B and A. Lay out both scorebooks. Within each set, alternate wins first, then add the remaining tokens for that set’s winner. Each token stands for four consecutive points won.
You now have two stories with identical totals. Counting the points again will not tell you who gets the trophy. You need to keep the boundaries between sets.