### 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…

### Notes on Setting up a Titan V under Ubuntu 17.04

I recently purchased a Titan V GPU to use for machine and deep learning, and in the process of installing the latest Nvidia driver's hosed my Ubuntu 16.04 install. I was overdue for a fresh install of Linux, anyway, so I decided to upgrade some of my drives at the same time. Here are some of my notes for the process I went through to get the Titan V working perfectly with TensorFlow 1.5 under Ubuntu 17.04.

Old install:
Ubuntu 16.04
EVGA GeForce GTX Titan SuperClocked 6GB
2TB Seagate NAS HDD

New install:
Ubuntu 17.04
Titan V 12GB
/ partition on a 250GB Samsung 840 Pro SSD (had an extra around)
/home partition on a new 1TB Crucial MX500 SSD
New WD Blue 4TB HDD