A simple answer, then:Â Â Â
Just use the (obvious)
points per match rate?
(as was used in France in April 2020, as an ad hoc solution)
\[
\textrm{rate} = {\textrm{points won} \over \textrm{matches played}}
\]
This at least would help to address the ‘games in hand’ problem.
But we can do better!
The key is to recognise that some matches are harder than others. Not all matches are equal, when ‘weighed’ as opportunities to gain league points.
- a match against a strong team is only a small opportunity
- a match against a weak team is a bigger opportunity
- (in normal times, for most teams) a match played at home is a bigger opportunity than a match played away
An example: Premier League after 5 matches
September 2017
Everton had won one match, and drawn one (of the five played).
The official table had Everton in 18th place.
Facts neglected in the official table:
- Everton had already played 4 of the top 5 teams
- 3 of Everton’s 5 matches were away from home
The Solution
The alt-3 method
Adjusted league table, fully respecting:
The principles of a good solution are best considered in a simpler setting.
Let’s simplify, for now, by considering a single round-robin tournament with only binary match outcomes (one point for a win).
Consider a generalized version of the points-per-match rate:
\[
\textrm{rate} = {\textrm{points won} \over \textit{effective } \textrm{matches played}}
\]
with
\[
\textrm{effective matches played by team } i = \sum_{j \textrm{ played}} e_{ij} / \bar{e}_{i.},
\] with the notion that \(e_{ij}\) is a probability that \(i\) beats \(j\) (or equivalently expected points for \(i\) when playing \(j\)).
The simple ‘points per match’ measure is the special case with all of the \(e_{ij}\) equal to \(1/2\).
How to determine expectations (or probabilities) \(e_{ij}\)?
Simple points per match has a very appealing aspect: probabilities \(e_{ij} = 1/2\) imply no distinction between teams.
But the assumption of \(e_{ij} = 1/2\) for all matches is (almost always) in conflict with match results already known.
This is resolved by seeking probabilities \(e_{ij}\) which are as uninformative as possible, while preserving agreement with match results already seen.
Consider first the situation at the end of the season.
Given only the points totals \(p_i\) for all teams, what is the most uninformative (or ‘least prejudiced’) set of probabilities \(e_{ij}\) that matches those points totals exactly, i.e., such that
\[
\sum_j e_{ij} = p_i, \textrm{ for all } i
\] ?
A natural mathematical formulation of ‘most uninformative’ is to maximize the total entropy of probabilities \(e_{ij}\), \[
\textrm{total entropy} = - \sum_{i, j} [e_{ij}\log e_{ij} + (1 - e_{ij})\log(1 - e_{ij})]
\]
Under the (essential!) constraint that end-of-season points totals are matched exactly, entropy is maximized by
\[
e_{ij} = {s_i \over s_i + s_j}
\] with the \(s_1,s_2,\ldots\) determined uniquely to match the team-by-team points totals.
This solution is well known to statisticians as the Bradley-Terry model (with its parameters determined by maximum likelihood).
And the connection with maximum entropy is also well known (e.g., Joe, 1988, Annals of Statistics).
Hence, during the season: use probabilities \(e_{ij} = t_i / (t_i + t_j)\), with the ‘team strengths’ \(t_i\) determined uniquely by points totals to date.
(sometimes called a ‘retrodictive’ Bradley-Terry model)
Two attractive features of the ‘retrodictive’ Bradley-Terry model
Not so well known?
- Exact equivalence of
- points per effective match played
- projected points per match over the whole season
- Formal indifference to future schedule of matches
- expected points-rate contribution from team \(i\)’s next match is the same as the current points rate \(\bar e_{i.}\), regardless of which opponent is played.
The second of these is an essential fairness property in round-robin leagues.
