### Closed Under Means

Here's a nice little problem from the 13th All Soviet Union Mathematics Olympiad.
Given a set of real numbers $$S$$ containing 0 and 1 that's closed under finite means, show that it contains all rational numbers in the interval $$\left[0,1\right]$$.
This isn't a difficult problem, but there's a particularly nice solution.

First observe that if $$x\in S$$ then both $$\frac{x}{4}$$ and $$\frac{3x}{4}$$ are in $$S$$; average $$\{0,x\}$$ to get $$\frac{x}{2}$$, average $$\{0, \frac{x}{2}\}$$ to get $$\frac{x}{4}$$, average $$\{\frac{x}{2}, x\}$$ to get $$\frac{3x}{4}$$.

We could show any rational number $$\frac{m}{n}$$ with $$1\leq m < n$$ is in $$S$$ if we had $$n$$ distinct elements from $$S$$ that summed to $$m$$. Let's exhibit one.

Start with an array with $$m$$ 1s on the left, $$n-m$$ 0s on the right. Repeatedly replace adjacent $$x,y$$ values with $$\frac{3(x+y)}{4}, \frac{(x+y)}{4}$$, where either $$x=1,y\neq1$$ or $$x\neq 0, y=0$$, until there is one 0 and one 1 left. We can do this in exactly $$n-2$$ steps, each resulting number is guaranteed to be in $$S$$ by the above note, and each number is guaranteed to be distinct by construction!

Examples:

$$\frac{1}{3}: \left[1,0,0\right] \to \left[\frac{3}{4},\frac{1}{4},0\right]$$

$$\frac{2}{5}: \left[1,1,0,0,0\right] \to \left[1,\frac{3}{4},\frac{1}{4},0,0\right] \to \left[1,\frac{3}{4},\frac{3}{16},\frac{1}{16},0\right]$$

### 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…