### 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…

### 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 …