This is a homework question, but I've clearly made a mistake that I can't point out...

Nadal and Federer play in a grand slam tournament. They have to win three sets to win. (Best of five). How many possible games are there?

There are 2 options with only three sets: NNN and FFF.

I figured that for the options with 4 sets, e.g. NNFN or FNFF.
I did 4C3 - 2, because NNNF and FFFN are not possible. 4C3 - 2 = 2

Then there are also the games with 5 sets, like NNFFN or FNFNF.
I figured that this should be 5C3 - 4, because NNNFF, FNNNF, FFFNN, NFFFN are not possible. This gives me 6.

All of this added up is 11. The answer book said 20. I can't figure out where I messed up.

  • $\begingroup$ Why is the number of games finite? A set has to be won by $2$ games. If the players trade victories they can play forever. Isner/Mahut went to $183$ games. $\endgroup$ – lulu Nov 1 '15 at 15:10
  • $\begingroup$ In the case 4, you forgot $FNNN$ and $NFFF$ $\endgroup$ – mrprottolo Nov 1 '15 at 15:10
  • $\begingroup$ @lulu I think the problem was poorly worded. It looks like we are counting the number of possible arrangements of sets. $\endgroup$ – N. F. Taussig Nov 1 '15 at 15:11
  • $\begingroup$ @N.F.Taussig Oh, of course. thanks. $\endgroup$ – lulu Nov 1 '15 at 15:14
  • $\begingroup$ FNNN and NFFF are possible, since the game stops after 3 sets won by the same person. NNNF continues after three games won by Nadal, and is not a possible arrangement. $\endgroup$ – Eowyn12 Nov 1 '15 at 15:19

To count the games of length $4$, let us count the games of length $4$ where F wins, and multiply by $2$.

To win in $4$, $F$ has to win the last set, and two of the first three sets. Which two? They can be chosen in $\binom{3}{2}$ ways. Double. We get $6$.

Foe length $5$, again count the ways F can win and double. F has to win the fifth set, and two of the first four. Which two? They can be chosen in $\binom{4}{2}$ ways. When we double we get $12$.


I believe the intent of the question is to cover all possible scenarios, and the ans of $20$ is correct.

To start with, suppose Nadal wins: Instead of counting ways of winning in $3/4/5$ sets,

realize that he has to win $3$ sets placed somewhere in $5$, hence $\dbinom53$ = 10 ways.

Double it to cover the possibilities of Federer winning


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.