### A Simple Estimate for Pythagorean Exponents

Given the number of runs scored and runs allowed by a baseball team, what's a good estimate for that team's win fraction? Bill James famously came up with what he called the "Pythagorean expectation" $w = \frac{R^2}{R^2 + A^2},$ which can also be written as $w = \frac{{(R/A)}^2}{{(R/A)}^2 + 1}.$ More generally, if team $$i$$ scores $$R_i$$ and allows $$A_i$$ runs, the Pythagorean estimate for the probability of team $$1$$ beating team $$2$$ is $w = \frac{{(R_1/A_1)}^2}{{(R_1/A_1)}^2 + (R_2/A_2)^2}.$ We can see that the estimate of the team's win fraction is a consequence of this, as an average team would by definition have $$R_2 = A_2$$. Now, there's nothing magical about the exponent being 2; it's a coincidence, and in fact is not even the "best" exponent. But what's a good way to estimate the exponent? Note the structural similarity of this win probability estimator and the Bradley-Terry estimator $w = \frac{P_1}{P_1+P_2}.$ Here the $$P_i$$ are what we could call the "Bradley-Terry power" of the team. This immediately suggests one way to estimate the expectation model's exponent - fit a Bradley-Terry model, then fit the log-linear regression $$\log(P_i)$$ vs $$\log(R_i/A_i)$$. The slope of this regression will be one estimate for the expectation exponent.

How well does this work? I get 1.727 for MLB in 2014. The R code and data files for MLB and other sports may be found in my GitHub repo.

### 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…

### 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…

### Gambling to Optimize Expected Median Bankroll

Gambling to optimize your expected bankroll mean is extremely risky, as you wager your entire bankroll for any favorable gamble, making ruin almost inevitable. But what if, instead, we gambled not to maximize the expected bankroll mean, but the expected bankroll median?

Let the probability of winning a favorable bet be $$p$$, and the net odds be $$b$$. That is, if we wager $$1$$ unit and win, we get back $$b$$ units (in addition to our wager). Assume our betting strategy is to wager some fraction $$f$$ of our bankroll, hence $$0 \leq f \leq 1$$. By our assumption, our betting strategy is invariant with respect to the actual size of our bankroll, and so if we were to repeat this gamble $$n$$ times with the same $$p$$ and $$b$$, the strategy wouldn't change. It follows we may assume an initial bankroll of size $$1$$.

Let $$q = 1-p$$. Now, after $$n$$  such gambles our bankroll would have a binomial distribution with probability mass function \[ \Pr(k,n,p) = \binom{n}{k} p^k q^{n-k…