### Who Controls the Pace in Basketball, Offense or Defense?

During a recent chat with basketball analyst Seth Partnow, he mentioned a topic that came up during a discussion at the recent MIT Sloan Sports Analytics Conference. Who  controls the pace of a game more, the offense or defense? And what is the percentage of pace responsibility for each side? The analysts came up with a rough consensus opinion, but is there a way to answer this question analytically? I came up with one approach that examines the variations in possession times, but it suddenly occurred to me that this question could also be answered immediately by looking at the offense-defense asymmetry of the home court advantage.

As you can see in the R output of my NCAA team model code in one of my public basketball repositories, the offense at home scores points at a rate about $$e^{0.0302} = 1.031$$ times the rate on a neutral court, everything else the same. Likewise, the defense at home allows points at a rate about $$e^{-0.0165} = 0.984$$ times the rate on a neutral court; in both cases the neutral court rate is the reference level. Notice the geometric asymmetry; $$1.031\cdot 0.984 = 1.015 > 1$$. The implication is that the offense is responsible for about the fraction $\frac{(1.031-1)}{(1.031-1)+(1-0.984)} = 0.66$ of the scoring pace. That is, offensive controls 2/3 of the pace, defense 1/3 of the pace. The consensus opinion the analysts came up with at Sloan? It was 2/3 offense, 1/3 defense! It's nice when things work out, isn't it?

I've used NCAA basketball because there are plenty of neutral court games; to examine the NBA directly we'll have to use a more sophisticated (but perhaps less beautiful) approach involving the variation in possession times. I'll do that next, and I'll also show you how to apply this new information to make better game predictions. Finally, there's a nice connection to some recent work on inferring causality that I'll outline.

### 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 …