📂Sabermetrics

Pythagorean Winning Percentage Derivation

Formulas

Let’s assume we have a team from a certain sports league. The Team Scores $S$ and Team Allows $A$ are random variables that each follow a Weibull distribution, $$\begin{align*} S & \sim \text{Weibull} \left( \alpha_{S} , \beta , \gamma \right) \\ A & \sim \text{Weibull} \left( \alpha_{A} , \beta , \gamma \right) \end{align*}$$ and are also independent of each other independently. The team’s expected winning percentage $p$ is given with respect to $\gamma > 0$ as follows. $$p_{\gamma} = {{ \mu_{S}^{\gamma} } \over { \mu_{S}^{\gamma} + \mu_{A}^{\gamma} }}$$ Here, $\mu_{S} := E (S)$ and $\mu_{A} := E (A)$ represent the expected score and expected allows, respectively.

Derivation ¹

Strategy: This is a statistical derivation of the Pythagorean winning percentage. It’s straightforwardly deduced through the joint probability density function. The function $\Gamma : \mathbb{R} \to \mathbb{R}$ represents the gamma function.

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Mean and variance of the Weibull distribution: A probability distribution named Three-parameter Weibull Distribution has the probability density function as follows, with the scale parameter $\alpha > 0$, location parameter $\beta > 0$, and shape parameter $\gamma > 0$. $$f(x) = {{ \gamma } \over { \alpha }} \left( {{ x-\beta } \over { \alpha }} \right)^{\gamma-1} e^{- \left( (x - \beta) / \alpha \right)^{\gamma}} \qquad , x \ge \beta$$ When $X \sim \text{Weibull} (\alpha, \beta, \gamma)$, its mean and variance are as follows. $$\begin{align*} E(X) =& \alpha \Gamma \left( 1 + {{ 1 } \over { \gamma }} \right) + \beta \\ \operatorname{Var} (X) =& \alpha^{2} \left[ \Gamma \left(1 + {{ 2 } \over { \gamma }} \right) - \left( \Gamma \left( 1 + {{ 1 } \over { \gamma }} \right)^{2} \right) \right] \end{align*}$$

$$\begin{align*} \mu_{S} =& E \left( S \right) = \alpha_{S} \Gamma \left( 1 + \gamma^{-1} \right) + \beta \\ \mu_{A} =& E \left( A \right) = \alpha_{A} \Gamma \left( 1 + \gamma^{-1} \right) + \beta \end{align*}$$ If we denote the population means of $S$ and $A$ as $\mu_{S}$ and $\mu_{A}$ respectively, then the first parameters of the Weibull distribution $\alpha_{S}$, $\alpha_{A}$ $$\begin{align*} \alpha_{S} =& {{ \mu_{S} - \beta } \over { \Gamma \left( 1 + \gamma^{-1} \right) }} \\ \alpha_{A} =& {{ \mu_{A} - \beta } \over { \Gamma \left( 1 + \gamma^{-1} \right) }} \end{align*}$$ are represented as above, and for the sake of simplification in the derivation, let’s define $\alpha$ as follows. $${{ 1 } \over { \alpha^{\gamma} }} = {{ 1 } \over { \alpha_{S}^{\gamma} }} + {{ 1 } \over { \alpha_{A}^{\gamma} }} = {{ \alpha_{S}^{\gamma} + \alpha_{A}^{\gamma} } \over { \alpha_{S}^{\gamma} \alpha_{A}^{\gamma} }}$$

Now, it’s time to calculate the expected winning percentage. In most sports, a win is defined as the event where the score $S$ is greater than the allows $A$, hence the expected winning percentage is essentially $P \left( S > A \right)$. If the probability density functions of $S$ and $A$ are $f_{S}$ and $f_{A}$ respectively, following the assumption that $S$ and $A$ are independent, their joint probability density function is $f_{S} f_{A}$. $$\begin{align*} & P \left( S > A \right) \\ =& \int_{\beta}^{\infty} \int_{\beta}^{x} f_{S} (x) f_{A} (y) dy dx \\ =& \int_{\beta}^{\infty} \int_{\beta}^{x} {{ \gamma } \over { \alpha_{S} }} \left( {{ x-\beta } \over { \alpha_{S} }} \right)^{\gamma-1} e^{- \left( (x - \beta) / \alpha_{S} \right)^{\gamma}} {{ \gamma } \over { \alpha_{A} }} \left( {{ y-\beta } \over { \alpha_{A} }} \right)^{\gamma-1} e^{- \left( (y - \beta) / \alpha_{A} \right)^{\gamma}} dy dx \\ =& \int_{0}^{\infty} \int_{0}^{x} {{ \gamma } \over { \alpha_{S} }} \left( {{ x } \over { \alpha_{S} }} \right)^{\gamma-1} e^{- \left( x / \alpha_{S} \right)^{\gamma}} {{ \gamma } \over { \alpha_{A} }} \left( {{ y } \over { \alpha_{A} }} \right)^{\gamma-1} e^{- \left( y / \alpha_{A} \right)^{\gamma}} dy dx \\ =& \int_{0}^{\infty} {{ \gamma } \over { \alpha_{S} }} \left( {{ x } \over { \alpha_{S} }} \right)^{\gamma-1} e^{- \left( x / \alpha_{S} \right)^{\gamma}} \left[ \int_{0}^{x} {{ \gamma } \over { \alpha_{A} }} \left( {{ y } \over { \alpha_{A} }} \right)^{\gamma-1} e^{- \left( y / \alpha_{A} \right)^{\gamma}} dy \right] dx \\ =& \int_{0}^{\infty} {{ \gamma } \over { \alpha_{S} }} \left( {{ x } \over { \alpha_{S} }} \right)^{\gamma-1} e^{- \left( x / \alpha_{S} \right)^{\gamma}} \left[ 1 - e^{- \left( x / \alpha_{A} \right)^{\gamma}} \right] dx \\ =& 1 + \int_{0}^{\infty} {{ \gamma } \over { \alpha_{S} }} \left( {{ x } \over { \alpha_{S} }} \right)^{\gamma-1} e^{- \left( x / \alpha_{S} \right)^{\gamma}} \left[ - e^{- \left( x / \alpha_{A} \right)^{\gamma}} \right] dx \\ =& 1 - \int_{0}^{\infty} {{ \gamma } \over { \alpha_{S} }} \left( {{ x } \over { \alpha_{S} }} \right)^{\gamma-1} \exp \left( - x^{\gamma} \left( {{ 1 } \over { \alpha_{S}^{\gamma} }} + {{ 1 } \over { \alpha_{A}^{\gamma} }} \right) \right) dx \\ =& 1 - \int_{0}^{\infty} {{ \gamma } \over { \alpha_{S} }} \left( {{ x } \over { \alpha_{S} }} \right)^{\gamma-1} \exp \left( - \left( {{ x } \over { \alpha }} \right)^{\gamma} \right) dx \\ =& 1 - {{ \alpha^{\gamma} } \over { \alpha_{S}^{\gamma} }} \int_{0}^{\infty} {{ \gamma } \over { \alpha }} \left( {{ x } \over { \alpha }} \right)^{\gamma-1} e^{- \left( x / \alpha \right)^{\gamma} } dx \\ =& 1 - {{ \alpha^{\gamma} } \over { \alpha_{S}^{\gamma} }} \cdot 1 \\ =& 1 - {{ 1 } \over { \alpha_{S}^{\gamma} }} {{ \alpha_{S}^{\gamma} \alpha_{A}^{\gamma} } \over { \alpha_{S}^{\gamma} + \alpha_{A}^{\gamma} }} \\ =& 1 - {{ \alpha_{A}^{\gamma} } \over { \alpha_{S}^{\gamma} + \alpha_{A}^{\gamma} }} \\ =& {{ \alpha_{S}^{\gamma} } \over { \alpha_{S}^{\gamma} + \alpha_{A}^{\gamma} }} \\ =& {{ \left( \mu_{S} - \beta \right)^{\gamma} } \over { \left( \mu_{S} - \beta \right)^{\gamma} + \left( \mu_{A} - \beta \right)^{\gamma} }} \end{align*}$$ Here, $\beta$ represents the minimum value between allows and scores, so it’s reasonable to set $\beta = 0$, leading to the following result. $$P \left( S > A \right) = {{ \mu_{S}^{\gamma} } \over { \mu_{S}^{\gamma} + \mu_{A}^{\gamma} }}$$

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