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# Round robin tournament

Problem : Suppose there are $${k}$$ teams playing a round robin tournament; that is, each team plays against all the other teams and no game ends in a draw.Suppose the $${i^{th}}$$ team loses $${l_{i}}$$ games and wins $${w_{i}}$$ games.Show that

$${{\displaystyle}{\sum_{i=1}^{k}{l_i^{2}}}}$$ = $${{\displaystyle}{\sum_{i=1}^{k}{w_i^{2}}}}$$

Solution : Each team plays exactly one match against each other team.

Consider the expression $$\displaystyle{\sum_{i=1}^{k} l_i^{2} – {w_i^2} = \sum_{i=1}^{k}(l_i + w_i)(l_i – w_i) }$$

Since each team plays exactly k-1 matches and no match ends in a draw, hence number of wins plus numbers of loses of a particular team is k-1 (that is the number of matches it has played). In other words $$l_i + w_i = k-1$$ for all i (from 1 to k).

Hence

$$\displaystyle{\sum_{i=1}^{k} l_i^{2} – {w_i^2} }$$
$$\displaystyle{= \sum_{i=1}^{k}(l_i + w_i)(l_i – w_i) }$$
$$\displaystyle{= \sum_{i=1}^{k}(k-1)(l_i – w_i) }$$
$$\displaystyle{= (k-1)\left( \sum_{i=1}^{k} l_i – \sum_{i=1}^{k} w_i\right) }$$

But $$\displaystyle{ \sum_{i=1}^{k} l_i = \sum_{i=1}^{k} w_i }$$ (as total number of loses = total number of matches = total number of wins; as each match results in a win or lose of some one)

Hence $$\displaystyle{= (k-1)\left( \sum_{i=1}^{k} l_i – \sum_{i=1}^{k} w_i\right) = (k-1) \times 0 = 0 }$$

Therefore $$\displaystyle{\sum_{i=1}^{k} l_i^{2} – {w_i^2} = 0 }$$ implying $${{\displaystyle}{\sum_{i=1}^{k}{l_i^{2}}}}$$ = $${{\displaystyle}{\sum_{i=1}^{k}{w_i^{2}}}}$$

Proved. ## By Ashani Dasgupta

Pursuing Ph.D. in Geometric Group Theory at University of Wisconsin, Milwaukee