Skip to main content

Probability and Cumulative Dice Sums

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 \[ \frac{\partial W}{\partial P}(P=1) = \frac{ge}{4}.\] Our estimate for the run value of a win now follows by setting \[\frac{\Delta W}{\Delta P} = \frac{ge}{4} \] giving \[ \Delta W = 1 = \frac{ge}{4} \Delta P.\] What is \(\Delta P\)? Well \(P = R/A\), where \(R\) is runs scored over the season and \(A\) is runs allowed over the season. We're assuming this is a league average team and asking how many more runs they'd need to score to win an additional game, so \(A\) is actually fixed at \(L\), the league average number of runs scored (or allowed). This gives us \[1 = \frac{ge}{4} \Delta P = \frac{ge\Delta R}{4L}.\] Now \(L/g = l\), the league average runs per game, so we arrive at the estimate \[\Delta R = \frac{4l}{e}.\] In the specific case of MLB, we have \(e = 1.8\) and \(l = 4.3\), giving that a win is approximately \(\Delta R = 9.56\) runs.

Bill James originally used the exponent \(e=2\); in this case the formula simplifies to \(\Delta R = 2l\), i.e. we get the particularly simple result that a win is equal to approximately twice the average number of runs scored per game.

Applying this estimate to the NBA, a win is approximately \( \Delta R = \frac{4\cdot 101}{16.4} = 24.6\) points. Similarly, we get the estimates for a win of 4.5 goals for the NHL and 5.1 goals for the Premier League.

Comments

  1. I think you've assigned the incorrect goals/win to the wrong league. NHL I think is 5.1 and Premier League is 4.5. Thanks for sharing!!

    ReplyDelete

Post a Comment

Popular posts from this blog

Simplified Multinomial Kelly

Here's a simplified version for optimal Kelly bets when you have multiple outcomes (e.g. horse races). The Smoczynski & Tomkins algorithm, which is explained here (or in the original paper): https://en.wikipedia.org/wiki/Kelly_criterion#Multiple_horses Let's say there's a wager that, for every $1 you bet, will return a profit of $b if you win. Let the probability of winning be \(p\), and losing be \(q=1-p\). The original Kelly criterion says to wager only if \(b\cdot p-q > 0\) (the expected value is positive), and in this case to wager a fraction \( \frac{b\cdot p-q}{b} \) of your bankroll. But in a horse race, how do you decide which set of outcomes are favorable to bet on? It's tricky, because these wagers are mutually exclusive i.e. you can win at most one. It turns out there's a simple and intuitive method to find which bets are favorable: 1) Look at \( b\cdot p-q\) for every horse. 2) Pick any horse for which \( b\cdot p-q > 0\) and mar...

Probability and Cumulative Dice Sums

Let a die be labeled with increasing positive integers \(a_1,\ldots , a_n\), and let the probability of getting \(a_i\) be \(p_i>0\). We start at 0 and roll the die, adding whatever number we get to the current total. If \({\rm Pr}(N)\) is the probability that at some point we achieve the sum \(N\), then \(\lim_{N \to \infty} {\rm Pr}(N)\) exists and equals \(1/\rm{E}(X)\) iff \((a_1, \ldots, a_n) = 1\). The direction \(\implies\) is obvious. Now, if the limit exists it must equal \(1/{\rm E}(X)\) by Chebyshev's inequality, so we only need to show that the limit exists assuming that \((a_1, \ldots, a_n) = 1\). We have the recursive relationship \[{\rm Pr}(N) = p_1 {\rm Pr}(N-a_1) + \ldots + p_n {\rm Pr}(N-a_n);\] the characteristic polynomial is therefore \[f(x) = x^{a_n} - \left(p_1 x^{(a_n-a_1)} + \ldots + p_n\right).\] This clearly has the root \(x=1\). Next note \[ f'(1) = a_n - \sum_{i=1}^{n} p_i a_n + \sum_{i=1}^{n} p_i a_i = \rm{E}(X) > 0 ,\] hence this root is als...

Mixed Models in R - Bigger, Faster, Stronger

When you start doing more advanced sports analytics you'll eventually starting working with what are known as hierarchical, nested or mixed effects models . These are models that contain both fixed and random effects . There are multiple ways of defining fixed vs random random effects , but one way I find particularly useful is that random effects are being "predicted" rather than "estimated", and this in turn involves some "shrinkage" towards the mean. Here's some R code for NCAA ice hockey power rankings using a nested Poisson model (which can be found in my hockey GitHub repository ): model The fixed effects are year , field (home/away/neutral), d_div (NCAA division of the defense), o_div (NCAA division of the offense) and game_length (number of overtime periods); offense (strength of offense), defense (strength of defense) and game_id are all random effects. The reason for modeling team offenses and defenses as random vs fixed effec...