Number of ways that a football league can complete with all teams having the same number of points

TL;DR

This work tackles the problem of counting how many complete football leagues with $n$ teams can finish with all teams tied on points under a double round-robin with $M = n(n-1)$ matches and a 3-point win system. It first demonstrates a brute-force approach over all $3^M$ outcomes and then develops symmetry- and constraint-based pruning to drastically reduce the search space, enabling exact counts up to $n=8$ and revealing a connection to Eulerian digraphs in the all-no-draw case when $P_1 = 3n-3$. The key innovations include combining home/away matches to reduce inter-team outcomes to six tuples, constructing a unique, permutable representation for team-1 results, and using rigorous bounds on score totals to prune branches, complemented by a recursive simulation for the remaining matches. The resulting counts are cataloged in OEIS as $A380592$, and the study demonstrates how symmetry, combinatorial structures, and graph-theoretic concepts can jointly illuminate the combinatorics of fair leagues, with practical computation up to $n=8$ and implications for larger $n$ requiring further optimization.

Abstract

When $n$ teams play in a football league with home and away matches against every opponent there are $M = n \cdot (n-1)$ matches. There are 3 possible match results: a victory is awarded 3 points, a draw 1 point and 0 points for a defeat. Hence we have $3^M$ possible outcomes. In this paper the number of ways is determined that a football league can complete with all teams having the same number of points. An algorithm that works until $n=8$ is presented.

Figures (1)

• Figure 1: Example of a league with 4 teams: The arrow from $A$ to $B$ means team $A$ wins at home 3 points against team $B$. 7 of the 12 matches are won by the home team (shown by the red arrows) and 5 are won by the away team. In-degree and out-degree of all vertices are equal.