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I was thinking of the following problem related to discrete math. Assume that we have n teams scheduled for a round robin tournament. For any given round in the tournament, how many possible win-loss outcomes are there?

For example, if we have the following teams: A B C D, then we have 12 possible outcomes for a given round:

A defeats B, C defeats D

A defeats B, D defeats C

A defeats C, B defeats D

A defeats C, D defeats B

B defeats A, C defeats D

B defeats A, D defeats C

B defeats C, A defeats D

B defeats C, D defeats A

C defeats A, B defeats D

C defeats A, D defeats B

C defeats B, D defeats A

C defeats B, A defeats D

I would like a to find this number as a function of n.

M.H.
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  • What you are counting as "outcomes" is really a mix of the way the round is scheduled and the binary outcomes of those matches. In the case of four teams a round consists of two matches, so we can choose the opponents in three ways (choose one match for $A$ and the rest is determined) and then $2^2$ for the outcomes of the two matches, Now $3\cdot 2^2 = 12$ as you observed by exhaustive listing. – hardmath Jun 29 '16 at 20:47

2 Answers2

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I will first assume that $n=2m$. There are

$$\frac{(2m)!}{2^mm!}=(2m-1)!!$$

ways to divide $2m$ teams into $m$ pairs. In each pair either team can win, so for each possible pairing of teams to form a round there are $2^m$ possible outcomes. The total number of possible outcomes is therefore

$$\frac{(2m)!}{2^mm!}\cdot 2^m=\frac{(2m)!}{m!}=\binom{2m}mm!\;.$$

If $n=2m+1$, there are $n$ ways to choose the team that gets a bye in a round, and everything else remains as before, so there are

$$\frac{(2m+1)(2m)!}{m!}=\frac{(2m+1)!}{m!}=\binom{2m+1}m(m+1)!$$

possible outcomes.

Community
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Brian M. Scott
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If you imagine lining the teams up and taking the first to beat the second, the third to beat the fourth, etc., there are $n!$ ways to line them up. You can interchange pairs in $m!$ ways and get the same pairing result, so there are $\frac {n!}{\frac n2!}$ possible results. For your case $\frac {4!}{2!}=12$

Ross Millikan
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