### When is a Lead Safe in the NBA?

Assuming two NBA teams of equal strength with $$t$$ seconds remaining, what is a safe lead at a prescribed confidence level? Bill James has a safe lead formula for NCAA basketball, and the topic has been addressed by other researchers at various levels of complexity, e.g. Clauset, Kogan and Redner.

I'll present a simple derivation. Start by observing there are about 50 scoring groups per team per game (scoring groups include all baskets and free throws that occur at the same time), with each scoring group worth about two points. Assume scoring events by team are Poisson distributed with parameter $$\lambda = \frac{50\cdot t}{48\cdot 60} = \frac{t}{57.6}$$. Using a normal approximation, the difference of these two distributions is normal with mean 0 and variance $$\sqrt{2}\lambda$$, giving a standard deviation of $$0.1863\sqrt{t}$$.

Using this approximation, what is a 90% safe lead? A 90% tail is 1.28 standard deviations, $$1.28\cdot 0.1863\sqrt{t} = 0.2385\sqrt{t}$$ scoring groups. As a scoring group is about two points, this means a 90% safe lead, assuming a jump ball, is about $$0.477\sqrt{t}$$ points (Clauset et. al. obtained $$0.4602\sqrt{t}$$). For example, a safe lead at halftime is approximately $$0.477 \sqrt{24\cdot 60} = 18.1$$ points.

Next - adjustments for possession arrow and shot clock time; validity of approximation; adjusting for team strengths.

1. Can you explain a bit better what you mean by scoring groups ?

### 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…

### Mixed Models in R - Bigger, Faster, Stronger

When you start doing more advanced sports analytics you'll eventually starting working with what are known as hierarchical, nested or mixed effects models. These are models that contain both fixed and random effects. There are multiple ways of defining fixed vs random random effects, but one way I find particularly useful is that random effects are being "predicted" rather than "estimated", and this in turn involves some "shrinkage" towards the mean.

Here's some R code for NCAA ice hockey power rankings using a nested Poisson model (which can be found in my hockey GitHub repository):
model <- gs ~ year+field+d_div+o_div+game_length+(1|offense)+(1|defense)+(1|game_id) fit <- glmer(model, data=g, verbose=TRUE, family=poisson(link=log) ) The fixed effects are year, field (home/away/neutral), d_div (NCAA division of the defense), o_div (NCAA division of the offense) and game_length (number of overtime periods); off…

### 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) = …