### Power Rankings: Looking at a Very Simple Method

One of the simplest and most common power ranking models is known as the Bradley-Terry-Luce model, which is equivalent to other famous models such the logistic model and the Elo rating system. I'll be referring to "teams" here, but of course the same ideas apply to any two-participant game.

Let me clarify what I mean when I use the term "power ranking". A power ranking supplies not only a ranking of teams, but also provides numbers that may be used to estimate the probabilities of various outcomes were two particular teams to play a match.

In the BTL power ranking system we assume the teams have some latent (hidden/unknown) "strength" $$R_i$$, and that the probability of $$i$$ beating $$j$$ is $$\frac{R_i}{R_i+R_j}$$. Note that each $$R_i$$ is assumed to be strictly positive. Where does this model structure come from?

Here are three reasonable constraints for a power ranking model:
1.  If $$R_i$$ and $$R_j$$ have equal strength, the probability of one beating the other should be $$\frac{1}{2}$$.
2. As the strength of one team strictly approaches 0 (infinitely weak) with the other team fixed, the probability of the other team winning strictly increases to 1.
3. As the strength of one team strictly approaches 1 (infinitely strong) with the other team fixed, the probability of the other team winning strictly decreases to 0.
Note that our model structure satisfies all three constraints. Can you think of other simple model structures that satisfy all three constraints?

Given this model and a set of teams and match results, how can we estimate the $$R_i$$. The maximum-likelihood estimators are the set of $$R_i$$ that maximizes the probability of the observed outcomes actually happening. For any given match this probability of team $$i$$ beating team $$j$$ is $$\frac{R_i}{R_i+R_j}$$, so the overall probability of the observed outcomes of the matches $$M$$ occurring is $\mathcal{L} = \prod_{m\in M} \frac{R_{w(m)}}{R_{w(m)}+R_{l(m)}},$ where $$w(m)$$ is then winner and $$l(m)$$ the loser of match $$m$$. We can transform this into a sum by taking logarithms; $\log\left( \mathcal{L} \right) = \log\left(R_{w(m)}\right) - \log\left(R_{w(m)}+R_{l(m)}\right).$ Before going further, let's make a useful reparameterization by setting $$e^{r_i} = R_i$$; this makes sense as we're requiring the $$R_i$$ to be strictly positive. We then get $\log\left( \mathcal{L} \right) = r_{w(m)} - \log\left(e^{r_{w(m)}}+e^{r_{l(m)}}\right).$ Taking partial derivatives we get \begin{eqnarray*}
\frac{\partial \log\left( \mathcal{L} \right)}{\partial r_i} &=& \sum_{w(m)=i} 1 - \frac{e^{r_{w(m)}}}{e^{r_{w(m)}}+e^{r_{l(m)}}} + \sum_{l(m)=i} - \frac{e^{r_{l(m)}}}{e^{r_{w(m)}}+e^{r_{l(m)}}}\\
&=& \sum_{w(m)=i} 1 - \frac{e^{r_i}}{e^{r_i}+e^{r_{l(m)}}} + \sum_{l(m)=i} - \frac{e^{r_i}}{e^{r_{w(m)}}+e^{r_i}}\\
&=&0.
\end{eqnarray*} But this is just the number of actual wins minus the expected wins! Thus, the maximum likelihood estimators for the $$r_i$$ satisfy $$O_i = E_i$$ for all teams $$i$$, where $$O_i$$ is the actual (observed) number of wins for team $$i$$, and $$E_i$$ is the expected number of wins for team $$i$$ based on our model. That's a nice property!

If you'd like to experiment with some actual data, and to see that the resulting fit does indeed satisfy this property, here's an example BTL model using NCAA men's ice hockey scores. You can, of course, actually use this property to iteratively solve for the MLE estimators $$R_i$$. Note that you'll have to fix one of the $$R_i$$ to be a particular value (or add some other constraint), as the model probabilities are invariant with respect to multiplication of the $$R_i$$ by the same positive scalar.

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### 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 …