### Poisson Games and Sudden-Death Overtime

Let's say we have a game that can be reasonably modeled as two independent Poisson processes with team $$i$$ having parameter $$\lambda_i$$. If one team wins in regulation with team $$i$$ scoring $$n_i$$, then it's well-known we have the MLE estimate $$\hat{\lambda_i}=n_i$$. But what if the game ends in a tie in regulation with each team scoring $$n$$ goals and we have sudden-death overtime? How does this affect the MLE estimate for the winning and losing teams?

Assuming without loss of generality that team $$1$$ is the winner in sudden-death overtime. As we have two independent Poisson processes, the probability of this occurring is $$\frac{\lambda_1}{\lambda_1 + \lambda_2}$$. Thus, the overall likelihood we'd like to maximize is $L = e^{-\lambda_1} \frac{{\lambda_1}^n}{n!} e^{-\lambda_2} \frac{{\lambda_2}^n}{n!} \frac{\lambda_1}{\lambda_1 + \lambda_2}.$ Letting $$l = \log{L}$$ we get $l = -{\lambda_1} + n \log{\lambda_1} - {\lambda_2} + n \log{\lambda_2} - 2 \log{n!} + \log{\lambda_1}-\log({\lambda_1 + \lambda_2}).$ This gives \eqalign{ \frac{\partial l}{\partial \lambda_1} &= -1+\frac{n}{\lambda_1}+\frac{1}{\lambda_1}+\frac{1}{\lambda_1 + \lambda_2}\\ \frac{\partial l}{\partial \lambda_2} &= -1+\frac{n}{\lambda_2}+\frac{1}{\lambda_1 + \lambda_2}. } Setting both partials equal to $$0$$ and solving, we get \eqalign{ (n-\hat{\lambda_1})(\hat{\lambda_1}+\hat{\lambda_2})+\hat{\lambda_2} &= 0\\ (n-\hat{\lambda_2})(\hat{\lambda_1}+\hat{\lambda_2})-\hat{\lambda_2} &= 0, } and so \eqalign{ \hat{\lambda_1} &= (n+1) \frac{2n}{2n+1}\\ \hat{\lambda_2} &= n \frac{2n}{2n+1}. } For example, if both teams score $$3$$ goals in regulation and team $$1$$ wins in sudden-death overtime, our MLE estimates are $$\hat{\lambda_1} = 3\frac{3}{7}, \hat{\lambda_2} = 2\frac{4}{7}$$.

Intuitively this makes sense, because $$2n$$ goals were scored in regulation time, hence we "expect" that the overtime goal occurred around a fraction $$\frac{1}{2n}$$ of regulation, so team $$1$$ scored $$n+1$$ goals in about $$\frac{2n+1}{2n}$$ regulation periods and team $$2$$ scored $$n$$ goals in about $$\frac{2n+1}{2n}$$ regulation periods. The standard Poisson process MLE estimates here coincide with the estimates we derived above.

Does this work in practice? Yes! I tested it on my NCAA men's lacrosse model, and it increased the out-of-sample testing accuracy by 0.5%. Surprisingly large for such a small change!

### A Bayes' Solution to Monty Hall

For any problem involving conditional probabilities one of your greatest allies is Bayes' Theorem. Bayes' Theorem says that for two events A and B, the probability of A given B is related to the probability of B given A in a specific way.

Standard notation:

probability of A given B is written $$\Pr(A \mid B)$$
probability of B is written $$\Pr(B)$$

Bayes' Theorem:

Using the notation above, Bayes' Theorem can be written: $\Pr(A \mid B) = \frac{\Pr(B \mid A)\times \Pr(A)}{\Pr(B)}$Let's apply Bayes' Theorem to the Monty Hall problem. If you recall, we're told that behind three doors there are two goats and one car, all randomly placed. We initially choose a door, and then Monty, who knows what's behind the doors, always shows us a goat behind one of the remaining doors. He can always do this as there are two goats; if we chose the car initially, Monty picks one of the two doors with a goat behind it at random.

Assume we pick Door 1 and then Monty sho…

### 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) = …

### 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 <- gs ~ year+field+d_div+o_div+game_length+(1|offense)+(1|defense)+(1|game_id) fit <- glmer(model, data=g, verbose=TRUE, family=poisson(link=log) ) 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); off…