### Lunchtime Sports Science: Introducing tanh5

As I mentioned in a previous article on ratings systems, the log5 estimate for participant 1 beating participant 2 given respective success probabilities $$p_1, p_2$$ is
\begin{align}
p &= \frac{p_1 q_2}{p_1 q_2+q_1 p_2}\\
&= \frac{p_1/q_1}{p_1/q_1+p_2/q_2}\\
\frac{p}{q} &= \frac{p_1}{q_1} \cdot \frac{q_2}{p_2}
\end{align} where $$q_1=1-p_1, q_2=1-p_2, q=1-p$$.

Where does this come from? Assume that both participants each played average opposition. In a Bradley-Terry setting, this means
\begin{align}
p_1 &= \frac{R_1}{R_1 + 1}\\
p_2 &= \frac{R_2}{R_2 + 1},
\end{align} where $$R_1$$ and $$R_2$$ are the (latent) Bradley-Terry ratings; the $$1$$ in the denominators is an estimate for the average rating of the participants they've played en route to achieving their respective success probabilities.

In a Bradley-Terry setting, it's true that the product of the ratings in the entire pool is taken to equal 1. But participants don't play themselves! Thus, if participant 1 played every participant but itself, the average opponent would have rating $$R$$, where $$R_1 \cdot R^{n-1} = 1$$. Here $$n$$ is the number of participants in the pool.

Our strength estimate for the average opponent faced is then
\begin{align}
R &= {R_1}^{-\frac{1}{n-1}}.
\end{align}
There are two extreme cases. If $$n=2$$, then $$R = \frac{1}{R_1}$$; as $$n \to +\infty$$, $$R \to 1$$.

The limit extreme case is log5; the $$n=2$$ extreme case I call tanh5. We compute
\begin{align}
p_1 &= \frac{R_1}{R_1 + 1/R_1}\\
p_2 &= \frac{R_2}{R_2 + 1/R_2}\\
\textrm{tanh5} = p &= \frac{ \sqrt{p_1 q_2} }{ \sqrt{p_1 q_2} + \sqrt{q_1 p_2} }.
\end{align}
Why tanh5? We can think of log5 as derived from the logistic function by setting $$\log(R)=0$$ for the opponent's rating; analogously, tanh5 is derived from the hyperbolic tangent function by setting $$\log(R)=0$$ for the opponent's rating.

Note that we have a spectrum of estimates corresponding to each value for $$n$$, but these are the two extremes. This also gives a new spectrum of activation functions for neural networks, but I'll explore this application later.

### 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…