### 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!

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### 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) = … ### Solving a Math Puzzle using Physics The following math problem, which appeared on a Scottish maths paper, has been making the internet rounds. The first two parts require students to interpret the meaning of the components of the formula $$T(x) = 5 \sqrt{36+x^2} + 4(20-x)$$, and the final "challenge" component involves finding the minimum of $$T(x)$$ over $$0 \leq x \leq 20$$. Usually this would require a differentiation, but if you know Snell's law you can write down the solution almost immediately. People normally think of Snell's law in the context of light and optics, but it's really a statement about least time across media permitting different velocities. One way to phrase Snell's law is that least travel time is achieved when \[ \frac{\sin{\theta_1}}{\sin{\theta_2}} = \frac{v_1}{v_2},$ where $$\theta_1, \theta_2$$ are the angles to the normal and $$v_1, v_2$$ are the travel velocities in the two media.

In our puzzle the crocodile has an implied travel velocity of 1/5 in the water …