Skip to main content

### A Stupid and Strange Way of Looking at Sports Power Ratings that could be Smart and Useful

As I've mentioned previously, a common method used in sports for estimating game outcomes known as log5 can be written $p = \frac{p_1 q_2}{p_1 q_2+q_1 p_2}$ where $$p_i$$ is the fraction of games won by team $$i$$ and $$q_i$$ is the fraction of games lost by team $$i$$. We're assuming that there are no ties. What's the easiest way to derive this estimate? Here's one argument. Assume team $$i$$ has a probability $$p_i$$ of beating an average team (a team that wins half its games). Now imagine that this means for any given game the team has some "strength" sampled from [0,1] with median $$p_i$$ and that the stronger team always wins. Thus, the probability that team 1 beats team 2 is $p = \int_0^1 \int_0^1 \! \mathrm{Pr}(p_1 > p_2) \, \mathrm{d} p_1 \mathrm{d} p_2 .$ This looks complicated, but but with probability $$p_1$$ team 1 is stronger than an average team and with probability $$p_2$$ team 2 is stronger than an average team. From this perspective the log5 estimate is just the Bayesian probability that team 1 will be stronger than an average team while team 2 will be weaker than an average team, conditional on either team 1 being stronger than an average team and team 2 weaker than an average team, or team 1 weaker than an average team and team 2 stronger than an average team. In these cases it's unambiguous which team is stronger. The cases where the strength of both teams is stronger or weaker than an average team (the ambiguous cases) are thus discarded.

How could this be useful? Instead of ignoring the ambiguous outcomes when estimating the outcome probabilities under this "latent strength" model, we could instead determine which probability distributions best fit the outcome distributions for a given league! Furthermore, this allows us to cohesively put a power rating system into a Bayesian framework by assigning to each team a Bayesian prior strength distribution. These priors could either be uninformative or informative using e.g. preseason rankings.

### Comments

1. This is interesting and seems to be closely related to the Bradley-Terry ranking model. I think the missing link in either of these approaches is connecting p_i with underlying cumulative strength of the team's individual players.

1. Yes, a Bayesian strength Bradley-Terry-Luce model would have $p= \frac{S_1}{S_1+S_2},$ where $$S_i$$ is sampled from some probability distribution. As I've mentioned in a previous article, these approaches may be related by setting $$S_i = \frac{p_i}{q_i}$$.

### A Bayes' Solution to Monty Hall

For any problem involving conditional probabilities one of your greatest allies is Bayes' Theorem. Bayes' Theorem says that for two events A and B, the probability of A given B is related to the probability of B given A in a specific way.

Standard notation:

probability of A given B is written $$\Pr(A \mid B)$$
probability of B is written $$\Pr(B)$$

Bayes' Theorem:

Using the notation above, Bayes' Theorem can be written: $\Pr(A \mid B) = \frac{\Pr(B \mid A)\times \Pr(A)}{\Pr(B)}$Let's apply Bayes' Theorem to the Monty Hall problem. If you recall, we're told that behind three doors there are two goats and one car, all randomly placed. We initially choose a door, and then Monty, who knows what's behind the doors, always shows us a goat behind one of the remaining doors. He can always do this as there are two goats; if we chose the car initially, Monty picks one of the two doors with a goat behind it at random.

Assume we pick Door 1 and then Monty sho…

### What's the Value of a Win?

In a previous entry I demonstrated one simple way to estimate an exponent for the Pythagorean win expectation. Another nice consequence of a Pythagorean win expectation formula is that it also makes it simple to estimate the run value of a win in baseball, the point value of a win in basketball, the goal value of a win in hockey etc.

Let our Pythagorean win expectation formula be $w=\frac{P^e}{P^e+1},$ where $$w$$ is the win fraction expectation, $$P$$ is runs/allowed (or similar) and $$e$$ is the Pythagorean exponent. How do we get an estimate for the run value of a win? The expected number of games won in a season with $$g$$ games is $W = g\cdot w = g\cdot \frac{P^e}{P^e+1},$ so for one estimate we only need to compute the value of the partial derivative $$\frac{\partial W}{\partial P}$$ at $$P=1$$. Note that $W = g\left( 1-\frac{1}{P^e+1}\right),$ and so $\frac{\partial W}{\partial P} = g\frac{eP^{e-1}}{(P^e+1)^2}$ and it follows \[ \frac{\partial W}{\partial P}(P=1) = …

### Behind the Speadsheet

In the book "The Only Rule Is It Has to Work: Our Wild Experiment Building a New Kind of Baseball Team", Ben Lindbergh and Sam Miller recount a grand adventure to take command of an independent league baseball team, with the vision of trying every idea, sane or crazy, in an attempt to achieve a winning edge. Five infielders, four outfielders, defensive shifts, optimizing lineups - everything.

It was really an impossible task. Professional sports at every level are filled with highly accomplished and competitive athletes, with real lives and real egos. Now imagine walking in one day and suddenly trying to convince them that they should be doing things differently. Who do you think you are?

I was one of the analysts who helped Ben and Sam in this quest, and I wanted to write some thoughts down from my own perspective, not as one of the main characters, but as someone more behind the scenes. These are some very short initial thoughts only, but I'd like to followup with some…