Pythagoras at the Bat

Abstract

The Pythagorean formula is one of the most popular ways to measure the true ability of a team. It is very easy to use, estimating a team’s winning percentage from the runs they score and allow. This data is readily available on standings pages; no computationally intensive simulations are needed. Normally accurate to within a few games per season, it allows teams to determine how much a run is worth in different situations. This determination helps solve some of the most important economic decisions a team faces: How much is a player worth, which players should be pursued, and how much should they be offered. We discuss the formula and these applications in detail, and provide a theoretical justification, both for the formula as well as simpler linear estimators of a team’s winning percentage. The calculations and modeling are discussed in detail, and when possible multiple proofs are given. We analyze the 2012 season in detail, and see that the data for that and other recent years support our modeling conjectures. We conclude with a discussion of work in progress to generalize the formula and increase its predictive power without needing expensive simulations, though at the cost of requiring play-by-play data.

Appendix

1.1 Calculating the Mean of a Weibull

Letting $\mu _{\alpha ,\beta ,\gamma }$ denote the mean of $f(x;\alpha ,\beta ,\gamma )$, we have

$$\begin{aligned} \mu _{\alpha ,\beta ,\gamma }&\ = \&\int \limits _\beta ^\infty x \cdot \frac{\gamma }{\alpha } \left( \frac{x-\beta }{\alpha }\right) ^{\gamma -1} e^{-((x-\beta )/\alpha )^\gamma }\mathrm{d} x\nonumber \\&= \int \limits _\beta ^\infty \alpha \frac{x-\beta }{\alpha } \cdot \frac{\gamma }{\alpha } \left( \frac{x-\beta }{\alpha }\right) ^{\gamma -1}e^{-((x-\beta )/\alpha )^\gamma }\mathrm{d} x\ +\ \beta . \end{aligned}$$

(22)

We change variables by setting $u = \left( \frac{x-\beta }{\alpha }\right) ^\gamma$. Then $\mathrm{d} u = \frac{\gamma }{\alpha } \left( \frac{x-\beta }{\alpha }\right) ^{\gamma -1}\mathrm{d} x$ and we have

$$\begin{aligned} \mu _{\alpha ,\beta ,\gamma }&\ = \&\int \limits _0^\infty \alpha u^{\gamma ^{-1}} \cdot e^{-u} \mathrm{d} u \ + \ \beta \nonumber \\&= \alpha \int \limits _0^\infty e^{-u} u^{1+\gamma ^{-1}} \frac{\mathrm{d} u}{u} \ + \ \beta \nonumber \\&= \alpha \varGamma (1+\gamma ^{-1}) \ + \ \beta .\end{aligned}$$

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1.2 Independence Test with Structural Zeros

We describe the iterative procedure needed to handle the structural zeros. A good reference is Bishop and Fienberg [2].

Let Bin($k$) be the $k$th bin used in the chi-squared test for independence. For each team’s incomplete contingency table, let $O_{r,c}$ be the observed number of games where the number of runs scored is in Bin($r$) and runs allowed is in Bin($c$). As games cannot end in a tie, we have $O_{r,r} = 0$ for all $r$.

We construct the expected contingency table with entries $E_{r,c}$ using an iterative process to find the maximum likelihood estimators for each entry. For $1 \le r,c \le 12$, let

$$\begin{aligned} E^{(0)}_{r,c} \ =\ \left\{ \begin{array}{lr} 1 &{} \ \mathrm{if}\ r \ne c\ \\ 0 &{} \ \mathrm{if}\ r = c, \end{array} \right. \end{aligned}$$

(24)

and let

$$\begin{aligned} X_{r,+}\ =\ \sum _{c} O_{r,c}, \ \ \ X_{c,+}\ =\ \sum _{r} O_{r,c}. \end{aligned}$$

(25)

We then have that

$$\begin{aligned} E^{(\ell )}_{r,c} \ =\ \left\{ \begin{array}{lr} E^{(\ell -1)}_{r,c}X_{r,+} / \sum _{c} E^{(\ell -1)}_{r,c}\ \ \ \mathrm{if }\ \ell \ \mathrm{is\ odd}\\ E^{(\ell -1)}_{r,c}X_{c,+} / \sum _{r} E^{(\ell -1)}_{r,c}\ \ \ \mathrm{if }\ \ell \ \mathrm{is\ even.} \end{array} \right. \end{aligned}$$

(26)

The values of $E_{r,c}$ can be found by taking the limit as $\ell \rightarrow \infty$ of $E^{(\ell )}_{r,c}$, and typically the convergence is rapid. The statistic

$$\begin{aligned} \sum _{r, c \atop r \ne c} \frac{(E_{r,c} - O_{r,c})^2}{E_{r,c}} \end{aligned}$$

(27)

follows a chi-square distribution with $(11-1)^2 - 11 = 89$ degrees of freedom.

1.3 Linearizing Pythagoras

Unlike the argument in Sect. 7, we do not assume knowledge of multivariable calculus and derive the linearization using just single variable methods. The calculations below are of interest in their own right, as they highlight good approximation techniques.

We assume there is some exponent $\gamma$ such that the winning percentage, $\mathrm WP$, is

$$\begin{aligned} \mathrm WP\ = \ \frac{\mathrm{RS}^\gamma }{\mathrm{RS}^\gamma +\mathrm{RA}^\gamma }, \end{aligned}$$

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with $\mathrm{RS}$ and $\mathrm{RA}$ the total runs scored and allowed. We multiply the right hand side by $(1/\mathrm{RS}^\gamma )/(1/\mathrm{RS}^\gamma )$ and write $\mathrm{RA}^\gamma$ as $\mathrm{RS}^\gamma - (\mathrm{RS}^\gamma - \mathrm{RA}^\gamma )$, and find

$$\begin{aligned} \mathrm WP&\ = \&\frac{1}{1 + \frac{\mathrm{RA}^\gamma }{\mathrm{RS}^\gamma }} \ = \ \left( 1 + \frac{\mathrm{RA}^\gamma }{\mathrm{RS}^\gamma }\right) ^{-1} \ = \ \left( 1 + \frac{\mathrm{RS}^\gamma - (\mathrm{RS}^\gamma -\mathrm{RA}^\gamma )}{\mathrm{RS}^\gamma }\right) ^{-1} \nonumber \\&= \left( 1 + 1 - \frac{\mathrm{RS}^\gamma -\mathrm{RA}^\gamma }{\mathrm{RS}^\gamma }\right) ^{-1} \nonumber \\&= \left( 2 \cdot \left( 1 - \frac{\mathrm{RS}^\gamma -\mathrm{RA}^\gamma }{2\mathrm{RS}^\gamma }\right) \right) ^{-1} \nonumber \\&= \frac{1}{2} \left( 1 - \frac{\mathrm{RS}^\gamma -\mathrm{RA}^\gamma }{2\mathrm{RS}^\gamma }\right) ^{-1}; \end{aligned}$$

(29)

notice we manipulated the algebra to pull out a 1/2, which indicates an average team; thus the remaining factor is the fluctuations about average.

We now use the geometric series formula, which says that if $|r| < 1$ then

$$\begin{aligned} \frac{1}{1-r} \ = \ 1 + r + r^2 + r^3 + \cdots .\end{aligned}$$

(30)

We let $r = (\mathrm{RS}^\gamma -\mathrm{RA}^\gamma )/2\mathrm{RS}^\gamma$; since runs scored and runs allowed should be close to each other, the difference of their $\gamma$ powers divided by twice the number of runs scored should be small. Thus $r$ in our geometric expansion should be close to zero, and we find

$$\begin{aligned} \mathrm WP&\ = \&\frac{1}{2}\left( 1 + \frac{\mathrm{RS}^\gamma -\mathrm{RA}^\gamma }{2\mathrm{RS}^\gamma } + \left( \frac{\mathrm{RS}^\gamma -\mathrm{RA}^\gamma }{2\mathrm{RS}^\gamma }\right) ^2 + \left( \frac{\mathrm{RS}^\gamma -\mathrm{RA}^\gamma }{2\mathrm{RS}^\gamma }\right) ^3 + \cdots \right) \nonumber \\&\approx 0.500 + \frac{\mathrm{RS}^\gamma -\mathrm{RA}^\gamma }{4\mathrm{RS}^\gamma }. \end{aligned}$$

(31)

We now make some approximations. We expect $\mathrm{RS}^\gamma -\mathrm{RA}^\gamma$ to be small, and thus $\frac{\mathrm{RS}^\gamma -\mathrm{RA}^\gamma }{2\mathrm{RS}}$ should be small. This means we only need to keep the constant and linear terms in the expansion. Note that if we only kept the constant term, there would be no dependence on points scored or allowed!

We need to do a little more analysis to obtain a formula that is linear in $\mathrm{RS}- \mathrm{RA}$. Let $\mathrm{R}_\mathrm{total}$ denote the average number of runs scored per team in the league. We can write $\mathrm{RS}= \mathrm{R}_\mathrm{ave}+ x_s$ and $\mathrm{RA}= \mathrm{R}_\mathrm{ave}+ x_a$, where it is reasonable to assume $x_s$ and $x_a$ are small relative to $\mathrm{R}_\mathrm{total}$. The Mean Value Theorem from Calculus says that if $f(x) = (\mathrm{R}_\mathrm{total}+x)^\gamma$, then

$$\begin{aligned} f(x_s) - f(x_a) \ = \ f'(x_c) (x_s - x_a), \end{aligned}$$

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where $x_c$ is some intermediate point between $x_s$ and $x_a$. As $f'(x) = \gamma (\mathrm{R}_\mathrm{total}+x)^{\gamma -1}$, we find

$$\begin{aligned} \mathrm{RS}^\gamma - \mathrm{RA}^\gamma&\ = \&f(x_s) - f(x_a) \ = \ f'(x_c) (x_s - x_a) \ = \ \gamma (\mathrm{R}_\mathrm{total}+x_c)^{\gamma -1}(\mathrm{RS}- \mathrm{RA}), \nonumber \\ \end{aligned}$$

(33)

as $x_s - x_a = \mathrm{RS}- \mathrm{RA}$. Substituting this into (31) gives

$$\begin{aligned} \mathrm WP&\ \approx \&0.500 + \frac{\gamma (\mathrm{R}_\mathrm{total}+x_c)^{\gamma -1} (\mathrm{RS}- \mathrm{RA})}{4\mathrm{RS}^\gamma } \ = \ 0.500 + \frac{\gamma (\mathrm{R}_\mathrm{total}+x_c)^{\gamma -1}}{4\mathrm{RS}^\gamma } (\mathrm{RS}-\mathrm{RA}).\nonumber \\ \end{aligned}$$

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We make one final approximation. We replace the factors of $\mathrm{R}_\mathrm{total}+x_c$ in the numerator and $\mathrm{RS}^\gamma$ in the denominator with $\mathrm{R}_\mathrm{total}^\gamma$, the league average, and reach

$$\begin{aligned} \mathrm WP\ \approx \ 0.500 + \frac{\gamma }{4\mathrm{R}_\mathrm{total}} (\mathrm{RS}- \mathrm{RA}). \end{aligned}$$

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Thus the simple linear approximation model reproduces the result from multivariable Taylor series, namely that the interesting coefficient $\mathrm{B}$ should be approximately $\gamma /(4\mathrm{R}_\mathrm{total})$.