A tennis puzzle

A puzzle that occurred to me watching Wimbledon this week:

A tennis match goes to five sets. The number of games one of the players wins in each set forms an arithmetic series. Given that the two players won the same number of games in total, who won the match? How many possible solutions are there?
For clarity: a set is first to six with a two-game lead. If any of the first four sets is unresolved by game 12, a tie-break is played, which counts as a game in this puzzle. There is no tie-break in the fifth set, which is played to a finish (and so could end, for example, 70-68.)

What I like about this puzzle is that the first ‘solution’ I came across, didn’t work. In fact, it gave the wrong result. I tried:
6-(0, 1, 2, 3 or 4)
7-(5 or 6)

… meaning the two indeterminate sets would need to account for eleven games for the second player, which can’t happen.

I had made the same mistake that @Mathistopheles and @msjwright2 would go on to make when I set it on the twitters.

If the common difference is +1, the series player would need to score 4, 5, 6, 7 and 8 — starting from 3 would leave the match lost in the third set; starting from 5 is no good because you can’t score 8 in the fourth set.

The first two sets must end 4-6, 5-7; the third must end 6-X, where X = 0, 1, 2, 3, or 4; the fourth ends 7-Y, where Y is 5 or 6; and the final set ends either 8-6 or, the option I forgot, 8-10.

The 8-6 case, as we saw, doesn’t work; for 8-10, though, the series player has a total of 30 games, and his opponent has won 6+7+10 = 23 games in the first, second and fifth sets. To score a total of seven in the other two sets, they can score either 1 and 6 or 2 and 5. So two of the solutions are:
6-4 7-5 1-6 6-7 10-8
6-4 7-5 2-6 5-7 10-8

In both cases, the non-series player wins.

It’s also possible to have an arithmetic series with a common difference of -1. In this case, the only possible set scores for the series player are 7, 6, 5, 4 and 3. The final three sets have to end 5-7, 4-6 and 3-6, accounting for 19 games — leaving six for the opponent to win in the first two sets. These must end either 7-5 6-1 or 7-6 6-0, giving us two more solutions:
5-7 1-6 7-5 6-4 6-3
6-7 0-6 7-5 6-4 6-3

Again, the non-series player wins.

Lastly, as @Mathistopheles points out, the common difference could be 0. The only possibility there is 6, and the other player needs to win 30 games. For a match to go to a tie-break, each player must win two sets; the non-series player must win his 7-6, accounting for 14 games, and leaving 16 to distribute between the others. The only way to do that is for the series player to win two of the first four sets 6-4, and lost the last one 6-8. That gives a further six solutions, all to do with permuting the first four set results:

4-6 4-6 7-6 7-6 8-6
4-6 7-6 4-6 7-6 8-6
7-6 4-6 4-6 7-6 8-6
4-6 7-6 7-6 4-6 8-6
7-6 4-6 7-6 4-6 8-6
7-6 7-6 4-6 4-6 8-6

In all six of these cases, the non-series player wins.

So, there are ten solutions, all of which are won by the non-series player.

@Mathistopheles came up with a nice extension:

In addition, the series player wins the first point of each set with an ace. What was the score?

The rules are that service alternates between the players from one game to the next.

That implies that each set has an even number of games in it; the only such scoreline was the second one mentioned above: 6-4 7-5 2-6 5-7 10-8.

The very same @Mathistopheles has posted another extension here, complete with entirely plausible back-story!

Thanks to @msjwright2, @mathistopheles, @opettajaH and @MB_Whitworth for having a go at this puzzle and pointing out my oversights and errors!

Edited 2015-07-06. I had confounded @mathistopheles with @mathalicious. Apologies to both.

Edited 2015-07-09 to add link to @mathistopheles’s new extension.

Edited 2017-02-07 to fix a spelling error.


Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.


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