The Angry Statistician

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) ...

Given the number of runs scored and runs allowed by a baseball team, what's a good estimate for that team's win fraction? Bill James famously came up with what he called the " Pythagorean expectation " \[w = \frac{R^2}{R^2 + A^2},\] which can also be written as \[w = \frac{{(R/A)}^2}{{(R/A)}^2 + 1}.\] More generally, if team \(i\) scores \(R_i\) and allows \(A_i\) runs, the Pythagorean estimate for the probability of team \(1\) beating team \(2\) is \[w = \frac{{(R_1/A_1)}^2}{{(R_1/A_1)}^2 + (R_2/A_2)^2}.\] We can see that the estimate of the team's win fraction is a consequence of this, as an average team would by definition have \(R_2 = A_2\). Now, there's nothing magical about the exponent being 2; it's a coincidence, and in fact is not even the "best" exponent. But what's a good way to estimate the exponent? Note the structural similarity of this win probability estimator and the Bradley-Terry estimator \[ w = \frac{P_1}{P_1+P_2}.\] Here ...

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 ...

Assuming two NBA teams of equal strength with \(t\) seconds remaining, what is a safe lead at a prescribed confidence level? Bill James has a safe lead formula for NCAA basketball , and the topic has been addressed by other researchers at various levels of complexity, e.g. Clauset, Kogan and Redner . I'll present a simple derivation. Start by observing there are about 50 scoring groups per team per game (scoring groups include all baskets and free throws that occur at the same time), with each scoring group worth about two points. Assume scoring events by team are Poisson distributed with parameter \(\lambda = \frac{50\cdot t}{48\cdot 60} = \frac{t}{57.6}\). Using a normal approximation, the difference of these two distributions is normal with mean 0 and variance \(\sqrt{2}\lambda\), giving a standard deviation of \(0.1863\sqrt{t}\). Using this approximation, what is a 90% safe lead? A 90% tail is 1.28 standard deviations, \(1.28\cdot 0.1863\sqrt{t} = 0.2385\sqrt{t}\) scoring g...