### Elo's Rating System as a Forgetful Logistic Model

Elo's rating system became famous from its use in chess, but it and variations are now used in sports like the NFL to eSports like League of Legends. It also was infamously used on various "Hot or Not" type websites, as shown in this scene from the movie "Social Network":

Of course, there's a mistake in the formula in the movie!

What is the Elo rating system? As originally proposed, it presumes that if two players A and B have ratings $$R_A$$ and $$R_B$$, then the expected score of player A is $\frac{1}{1+10^{\frac{R_B-R_A}{400}}}.$ Furthermore, if A has a current rating of $$R_A$$ and plays some more games, then the updated rating $${R_A}'$$ is given by $${R_A}' = R_A + K(S_A-E_A)$$, where $$K$$ is an adjustment factor, $$S_A$$ is the number of points scored by A and $$E_A$$ was the expected number of points scored by A based on the rating $$R_A$$.

Now, the expected score formula given above has the same form as a logistic regression model. What's the connection between the two? One answer is that Elo's rating system is a type of online version of a logistic model. An online algorithm is an algorithm that only sees each piece of data once. As applied to a statistical model, it's a model with parameter estimates that are updated as new data comes in, but not refitting on the entire data set. It can also be considered a memoryless model; it has "forgotten" the old data and only knows the current parameter estimates. The appeal of such models is that they're extremely efficient, can operate on enormous data sets and parameter estimates can be updated in real-time.

Okay, let's say we have a "forgetful" logistic model. Can we derive an updating rule, and does it look like Elo's? I'm going to give one possible derivation under the simplifying assumption that games are won or lost, with no ties.

We don't know how many games A and B had previously played, so let's assume they both had previously played $$n$$ games and have just played $$m$$ additional games between them, with A scoring $$S_A$$ points. That means they've both played $$n+m$$ games, but we're just going to forget this again, so let's adjust everything so that they end up with $$n$$ games. One way to do this is to normalize $$n$$ and $$m$$ so that they sum to $$n$$, thus $$n$$ becomes $$n\frac{n}{n+m}$$ and $$m$$ becomes $$m\frac{n}{n+m}$$.

We're now assuming they had each played $$n\frac{n}{n+m}$$ games in the past, have just played $$m\frac{n}{n+m}$$ additional games and A scored $$S_A \frac{n}{n+m}$$ points (it has to be adjusted, too!) in those games.

Again, we're memoryless, so we don't know how strong the opponents were that each had played in the past, so we're going to assume that they had both played opponents that were as strong as themselves and had won half and lost half of those games. After all, people generally prefer to play competitive games.

Define $$d$$ by $${R_A}' = R_A + d$$ and let $$c = R_A - R_B$$; we also require that $${R_B}' = R_B - d$$. The log-likelihood $$L$$ of A having scored $$S_A \frac{n}{n+m}$$ points is $\frac{-2 n^2}{n+m}\log(1+e^{-d}) -\frac{n^2}{n+m}d-\frac{m n}{n+m}\log(1+e^{-c-2d}) - \frac{(m-S_A)n(c+2d)}{n+m}.$ Factoring out the constant term $$n/(n+m)$$ simplifies this to $L = -2 n\log(1+e^{-d}) - n d - m \log(1+e^{-c-2d}) - (m-S_A)(c+2d).$ Taking the partial derivative of $$L$$ with respect to $$d$$ we get
\begin{align}
\frac{\partial L}{\partial d} &= 2n \frac{e^{-d}}{1+e^{-d}} -n + 2m \frac{e^{-c-2d}}{1+e^{-c-2d}}-2(m-S_A) \\
&= -n\frac{1-e^{-d}}{1+e^{-d}} + 2 S_A - 2m\frac{1}{1+e^{-c-2d}} \\
&= -n\tanh(d/2) + 2 S_A - 2m\frac{1}{1+e^{-c-2d}}.
\end{align} What is $$m\frac{1}{1+e^{-c-2d}}$$? This is actually just $${E_A}'$$, the expected score for A when playing B for $$m$$ games, but assuming the updated ratings for both players. Finally, setting $$\frac{\partial L}{\partial d} = 0$$, we get $n\tanh(d/2) = 2(S_A - {E_A}')$ and hence $\tanh(d/2) = \frac{2}{n} (S_A - {E_A}').$ Assuming $$n$$ is large relative to $$S_A - {E_A}'$$, we have $$\tanh(d/2) \approx d/2$$ and $${E_A}' \approx E_A$$. This is Elo's updating rule in the form $d = \frac{4}{n} (S_A - E_A ).$ If we rescale with the constant $$\sigma$$, the updating rule becomes $d = \frac{4\sigma }{n} (S_A - E_A ).$ We also now see that the adjustment factor $$K = \frac{4\sigma }{n}$$.

1. Well done and neat post.
Also, served to remind me just how much I have forgotten.
As soon as "partial derivative of L wrt to d" came up, I realized I would have to relearn derivative / power rules to follow. Ha.

2. With "A having scored S_A*n/(n+m) points", do you mean A scoring that amount of points out of m*n/(n+m) games after playing n*n/(m+n) games and having their ratings adjusted to R_a+d?

### 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 …