Analyzing Lebron James’ Offensive Play

Where is Lebron James most effective on the court?

Based on 2015-2016 data, we obtained from the following data which tracks Lebron’s FG% based on defender distance:


From, we then obtained data of Lebron’s FG% based on his shot distance from the basket:


Based on this data, we generated tens of thousands of sample data points to perform a Monte Carlo simulation to obtain relevant probability density functions. We found that the joint PDF was a very lengthy expression(!):


Graphically, this is:


A contour plot of the joint PDF was computed to be:


From this information, we can compute where/when LeBron has the highest probability of making a shot. Numerically, we found that the maximum probability occurs when Lebron’s defender is 0.829988 feet away, while Lebron is 1.59378 feet away from the basket. What is interesting is that this analysis shows that defending Lebron tightly doesn’t seem to be an effective strategy if his shot distance is within 5 feet of the basket. It is only an effective strategy further than 5 feet away from the basket. Therefore, opposing teams have the best chance at stopping Lebron from scoring by playing him tightly and forcing him as far away from the basket as possible.


Optimal Positions for NBA Players

I was thinking about how one can use the NBA’s new SportVU system to figure out optimal positions for players on the court. One of the interesting things about the SportVU system is that it tracks player (x,y) coordinates on the court. Presumably, it also keeps track of whether or not a player located at (x,y) makes a shot or misses it. Let us denote a player making a shot by 1, and a player missing a shot by 0. Then, one essentially will have data in the form (x,y, \text{1/0}).

One can then use a logistic regression to determine the probability that a player at position (x,y) will make a shot:

p(x,y) = \frac{\exp\left(\beta_0 + \beta_1 x + \beta_2 y\right)}{1 +\exp\left(\beta_0 + \beta_1 x + \beta_2 y\right)}

The main idea is that the parameters \beta_0, \beta_1, \beta_2 uniquely characterize a given player’s probability of making a shot.

As a coaching staff from an offensive perspective, let us say we wish to position players as to say they have a very high probability of making a shot, let us say, for demonstration purposes 99%. This means we must solve the optimization problem:

\frac{\exp\left(\beta_0 + \beta_1 x + \beta_2 y\right)}{1 +\exp\left(\beta_0 + \beta_1 x + \beta_2 y\right)} = 0.99

\text{s.t. } 0 \leq x \leq 28, \quad 0 \leq y \leq 47

(The constraints are determined here by the x-y dimensions of a standard NBA court).

This has the following solutions:

x = \frac{-1. \beta _0-1. \beta _2 y+4.59512}{\beta _1}, \quad \frac{-1. \beta _0-28. \beta _1+4.59512}{\beta _2} \leq y

with the following conditions:


One can also have:

x = \frac{-1. \beta _0-1. \beta _2 y+4.59512}{\beta _1}, \quad y \leq 47

with the following conditions:


Another solution is:

x = \frac{-1. \beta _0-1. \beta _2 y+4.59512}{\beta _1}

with the following conditions:


The fourth possible solution is:

x = \frac{-1. \beta _0-1. \beta _2 y+4.59512}{\beta _1}

with the following conditions:


In practice, it should be noted, that it is typically unlikely to have a player that has a 99% probability of making a shot.

To put this example in more practical terms, I generated some random data (1000 points) for a player in terms of (x,y) coordinates and whether he made a shot from that distance or not. The following scatter plot shows the result of this simulation:


In this plot, the red dots indicate a player has made a shot (a response of 1.0) from the (x,y) coordinates given, while a purple dot indicates a player has missed a shot from the (x,y) coordinates given (a response of 0.0).

Performing a logistic regression on this data, we obtain that \beta_0 = 0, \beta_1 = 0.00066876, \beta_2 = -0.00210949.

Using the equations above, we see that this player has a maximum probability of 58.7149 \% of making a shot from a location of (x,y) = (0,23), and a minimum probability of 38.45 \% of making a shot from a location of (x,y) = (28,0).