Robin Round Tournament 1: All Tied Up
What is a necessary condition so that there is a possibility of an \(n\)-way tie in the orderings of a round-robin tournament with no draws?
Answer
In a round-robin tournament of \(n\) teams (no draws), each of the \(\binom{n}{2}\) matches produces exactly one win and one loss. Therefore, the total number of wins across all teams is
\[
\binom{n}{2} \;=\; \frac{n(n-1)}{2}.
\]
If an \(n\)-way tie is possible, each team must finish with the same number of wins. Let this common number of wins be \(k\). Then
\[
n \times k \;=\; \frac{n(n-1)}{2}
\quad\Longrightarrow\quad
k \;=\; \frac{n-1}{2}.
\]
For \(k\) to be an integer, \(n-1\) must be even, which means \(\boxed{n \text{ is odd}}\). Thus, a necessary condition for an \(n\)-way tie is that \(n\) be an odd number.