rates of scoring – the odds ratio method and the Poisson distribution

this was motivated by how to estimate expected goals scored in the recent World Cup, based on runsgoals-scored and runsgoals-allowed rates versus average competition. In other words, when Germany (great offense, great defense) plays Brazil (great offense, great defense), how do you estimate the number of goals each team will score?

Suppose the scoring distribution is a Poisson distribution (that’s not entirely true, but it’s a useful approximation for this exercise), and the rate of scoring per unit time is $r$. Then, over an interval of time $\Delta t$, the probability to get n scores will be,

$P(n|r) = \frac{(r \Delta t)^n e^{-r \Delta t}}{n!}$

Now, consider a team with scoring rate per unit time $r_1$, playing a team allowing a scoring rate per unit time of $r_2$, when playing in an environment with a overall scoring rate of $r_a$. If the time interval $\Delta t$ is small enough then there are only two possibilities, n=0 or n=1. Then the probability for one score in the interval $\Delta t$ will be,

$\mathrm{prob} = x/(1+x) \\ x = (1-e^{-r_1 \Delta t})/e^{-r_1 \Delta t} \times (1-e^{-r_2 \Delta t})/e^{-r_2 \Delta t} \times e^{-r_a \Delta t}/(1-e^{-r_a \Delta t})$

In other words team 1 scores zero times with probability $e^{-r_1 \Delta t}$ and one times with probability $1-e^{-r_1 \Delta t}$. Similarly for team 2 and for the environmental average, and I have applied the odds-ratio method to get the probability for one score in an interval $\Delta t$ when team 1 faces team 2.

Since I am taking the limit when $\Delta t$ gets small, I can expand this to first order in $\Delta t$, which gives me

$\mathrm{prob} = \frac{r_1 r_2}{r_a} \Delta t = \frac{r_1}{r_a} \frac{r_2}{r_a} r_a \Delta t$

The Poisson distribution comes about from a binomial distribution where the probability in any trial is small and the number of trials is large in such a way that the product remains finite. This is the case here, so the resulting distribution would be a Poisson distribution with the new scoring rate per unit time, $\frac{r_1}{r_a} \frac{r_2}{r_a} r_a$, or in other words if team 1 scores at n times league average, and team 2 gives up scores at m times league average, then when they face each other team 1 will score at a rate $n \times m \times \mathrm{avg}$.