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One of the simplest and most common power ranking models is known as the Bradley-Terry-Luce model, which is equivalent to other famous models such the logistic model and the Elo rating system. I'll be referring to "teams" here, but of course the same ideas apply to any two-participant game.
Let me clarify what I mean when I use the term "power ranking". A power ranking supplies not only a ranking of teams, but also provides numbers that may be used to estimate the probabilities of various outcomes were two particular teams to play a match.
In the BTL power ranking system we assume the teams have some latent (hidden/unknown) "strength" \(R_i\), and that the probability of \(i\) beating \(j\) is \( \frac{R_i}{R_i+R_j} \). Note that each \(R_i\) is assumed to be strictly positive. Where does this model structure come from?
Here are three reasonable constraints for a power ranking model:
Let me clarify what I mean when I use the term "power ranking". A power ranking supplies not only a ranking of teams, but also provides numbers that may be used to estimate the probabilities of various outcomes were two particular teams to play a match.
In the BTL power ranking system we assume the teams have some latent (hidden/unknown) "strength" \(R_i\), and that the probability of \(i\) beating \(j\) is \( \frac{R_i}{R_i+R_j} \). Note that each \(R_i\) is assumed to be strictly positive. Where does this model structure come from?
Here are three reasonable constraints for a power ranking model:
- If \(R_i\) and \(R_j\) have equal strength, the probability of one beating the other should be \( \frac{1}{2}\).
- As the strength of one team strictly approaches 0 (infinitely weak) with the other team fixed, the probability of the other team winning strictly increases to 1.
- As the strength of one team strictly approaches 1 (infinitely strong) with the other team fixed, the probability of the other team winning strictly decreases to 0.
Note that our model structure satisfies all three constraints. Can you think of other simple model structures that satisfy all three constraints?
Given this model and a set of teams and match results, how can we estimate the \(R_i\). The maximum-likelihood estimators are the set of \( R_i \) that maximizes the probability of the observed outcomes actually happening. For any given match this probability of team \( i \) beating team \( j \) is \( \frac{R_i}{R_i+R_j} \), so the overall probability of the observed outcomes of the matches \( M \) occurring is \[ \mathcal{L} = \prod_{m\in M} \frac{R_{w(m)}}{R_{w(m)}+R_{l(m)}},\] where \( w(m) \) is then winner and \( l(m) \) the loser of match \( m \). We can transform this into a sum by taking logarithms; \[ \log\left( \mathcal{L} \right) = \log\left(R_{w(m)}\right) - \log\left(R_{w(m)}+R_{l(m)}\right).\] Before going further, let's make a useful reparameterization by setting \( e^{r_i} = R_i \); this makes sense as we're requiring the \( R_i \) to be strictly positive. We then get \[ \log\left( \mathcal{L} \right) = r_{w(m)} - \log\left(e^{r_{w(m)}}+e^{r_{l(m)}}\right).\] Taking partial derivatives we get \begin{eqnarray*}
\frac{\partial \log\left( \mathcal{L} \right)}{\partial r_i} &=& \sum_{w(m)=i} 1 - \frac{e^{r_{w(m)}}}{e^{r_{w(m)}}+e^{r_{l(m)}}} + \sum_{l(m)=i} - \frac{e^{r_{l(m)}}}{e^{r_{w(m)}}+e^{r_{l(m)}}}\\
&=& \sum_{w(m)=i} 1 - \frac{e^{r_i}}{e^{r_i}+e^{r_{l(m)}}} + \sum_{l(m)=i} - \frac{e^{r_i}}{e^{r_{w(m)}}+e^{r_i}}\\
&=&0.
\end{eqnarray*} But this is just the number of actual wins minus the expected wins! Thus, the maximum likelihood estimators for the \( r_i \) satisfy \( O_i = E_i \) for all teams \( i \), where \( O_i \) is the actual (observed) number of wins for team \( i \), and \( E_i \) is the expected number of wins for team \( i \) based on our model. That's a nice property!
If you'd like to experiment with some actual data, and to see that the resulting fit does indeed satisfy this property, here's an example BTL model using NCAA men's ice hockey scores. You can, of course, actually use this property to iteratively solve for the MLE estimators \( R_i \). Note that you'll have to fix one of the \( R_i \) to be a particular value (or add some other constraint), as the model probabilities are invariant with respect to multiplication of the \( R_i \) by the same positive scalar.
\frac{\partial \log\left( \mathcal{L} \right)}{\partial r_i} &=& \sum_{w(m)=i} 1 - \frac{e^{r_{w(m)}}}{e^{r_{w(m)}}+e^{r_{l(m)}}} + \sum_{l(m)=i} - \frac{e^{r_{l(m)}}}{e^{r_{w(m)}}+e^{r_{l(m)}}}\\
&=& \sum_{w(m)=i} 1 - \frac{e^{r_i}}{e^{r_i}+e^{r_{l(m)}}} + \sum_{l(m)=i} - \frac{e^{r_i}}{e^{r_{w(m)}}+e^{r_i}}\\
&=&0.
\end{eqnarray*} But this is just the number of actual wins minus the expected wins! Thus, the maximum likelihood estimators for the \( r_i \) satisfy \( O_i = E_i \) for all teams \( i \), where \( O_i \) is the actual (observed) number of wins for team \( i \), and \( E_i \) is the expected number of wins for team \( i \) based on our model. That's a nice property!
If you'd like to experiment with some actual data, and to see that the resulting fit does indeed satisfy this property, here's an example BTL model using NCAA men's ice hockey scores. You can, of course, actually use this property to iteratively solve for the MLE estimators \( R_i \). Note that you'll have to fix one of the \( R_i \) to be a particular value (or add some other constraint), as the model probabilities are invariant with respect to multiplication of the \( R_i \) by the same positive scalar.
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