Skip to main content

Poisson Games and Sudden-Death Overtime

Lunchtime Sports Science: Fitting a Bradley-Terry Model

Power rankings are game rankings that also allow you to estimate the likely outcome if two opponents were to face each other. One of the simplest of these models is known as the Bradley-Terry-Luce model (or commonly, Bradley-Terry). The idea is that each player \( i \) is assumed to have an unknown rating \( R_i \). If players \( i \) and \( j \) compete, the probability that \( i \) wins under this model is expected to be about \[ \frac{R_i}{R_i + R_j}. \] This model is very popular for hockey and other games; one commonly seen version is called KRACH.

Let's fit a Bradley-Terry model to the current season of NCAA D1 men's hockey. The Frozen Four starts on Thursday, April 11, so you'll get to see how well your predictions do.

You'll need to have R installed. Once R is installed, install the "BradleyTerry2" package that's freely available for R (thanks to Heather Turner and David Firth). To do this, start R and run the following command; you'll have to pick a source.
Next, download two files from my hockey GitHub - R code that fits a basic Bradley-Terry model and a data file containing the NCAA D1 men's hockey game results going back to 1998.

Make sure both files are in the same directory and run the R code. That's it, you've built a power ranking using a Bradley-Terry model. You should get output that looks like this:

                            ability      s.e.
Quinnipiac              1.687042594 0.5678939
Massachusetts-Lowell    1.480098569 0.5872701
Minnesota               1.428503522 0.5638946
Yale                    1.115338226 0.5641414
Miami                   1.114307264 0.5479346
Notre Dame              1.109912670 0.5523657
Boston College          1.091836391 0.5815233
St. Cloud State         1.079314965 0.5573018

The order should be the same as USCHO's KRACH rankings.

How do we use these ability estimates to predict game outcomes? These values are the logarithms of the ratings I've mentioned above, so first apply the exponential to get the rating, then the estimated winning probability is the team's rating divided by the sum of the team and opponent ratings. For the teams in the Frozen Four we get a power rating of \( e^{1.687} = 5.40 \) for Quinnipiac and \( e^{1.079} = 2.94 \) for St. Cloud State, so we estimate the probability of Quinnipiac beating St. Cloud State to be about \[ \frac{5.40}{5.40+2.94} = 0.65. \] What's your estimate for Massachusetts-Lowell beating Yale?


Popular posts from this blog

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) = …

Solving a Math Puzzle using Physics

The following math problem, which appeared on a Scottish maths paper, has been making the internet rounds.

The first two parts require students to interpret the meaning of the components of the formula \(T(x) = 5 \sqrt{36+x^2} + 4(20-x) \), and the final "challenge" component involves finding the minimum of \( T(x) \) over \( 0 \leq x \leq 20 \). Usually this would require a differentiation, but if you know Snell's law you can write down the solution almost immediately. People normally think of Snell's law in the context of light and optics, but it's really a statement about least time across media permitting different velocities.

One way to phrase Snell's law is that least travel time is achieved when \[ \frac{\sin{\theta_1}}{\sin{\theta_2}} = \frac{v_1}{v_2},\] where \( \theta_1, \theta_2\) are the angles to the normal and \(v_1, v_2\) are the travel velocities in the two media.

In our puzzle the crocodile has an implied travel velocity of 1/5 in the water …