Last month we asked you to find optimal strategies in basketball. Before we dive in, congratulations to our two winners, Elizabeth Zarek, who solved part 1 by considering extreme strategies, and Jelmer H, who solved part 2 by finding the Nash equilibrium between the offense and the defense. See our solution to part 2 below which also takes this approach.
You’re a basketball coach, preparing your defense for the big game against the Spiders. On each play, you can either set your defense to guard the layup (2 pts) or the three-pointer (3 pts). Historically, the Spiders try for layups 60% of the time, and three-pointers 40% of the time.
- If they try the layup, it goes in 30% of the time if you defend it, and 90% of the time if you don’t.
- Likewise, if they try the three-pointer, it goes in 25% of the time if you defend it, and 50% of the time if you don’t.
If you’re trying to minimize the amount of points the Spiders score per play, how often should you defend the three-pointer?
As a coach setting up our defense against the Spiders, we should choose whatever strategy that leads to the lowest number of expected points for the Spiders.
Let’s call the fraction of the time we set the defense up for a 2 or 3 pointer, $d_2$ and $d_3$ respectively, and the fraction of the time the Spiders shoot 2 or 3 pointers, $s_2$ and $s_3$ respectively.
Since the defense and offense each have two choices for how to set up a play, there are four possible defense vs. offense match-ups.
Each one contributes an expected number of points to the offense. For example, one match-up is where the Spiders shoot a 2 pointer while the defense sets up to guard a 2 pointer. This situation has probability $s_2d_2$ to happen and, when it does, the shot (worth 2 points) has a $30\%$ chance to go in. This contributes $2\times\frac{3}{10}s_2d_2$ to the expected number of points per play.
Going on like this, and adding up all four possibilities, the expected number of points per play is
$$ \langle\text{points}\rangle = 2\times \frac{3}{10} s_2d_2 + 2\times \frac{9}{10}s_2d_3 + 3\times \frac{1}{4} s_3d_3 + 3\times\frac{1}{2} s_3d_2. $$
The problem tells us that $s_2 = \frac{6}{10}$ and $s_3 = \frac{4}{10}$. Also, the defense always sets up for a 2 or a 3 pointer, so $d_2 + d_3 = 1$.
Making use of these facts, we can examine the expected number of points in terms of $d_2$ alone:
$$ \langle\text{points}\rangle=1.38-0.42d_2 . $$
The lower we set $d_2$, the more points the Spiders will score. This makes some sense — the Spiders shoot 2 pointers $60\%$ of the time, so if we always set up for the wrong shot, we should expect a poor outcome.
What’s surprising is that we minimize the number of points that the Spiders score per play by always setting up to guard the 2 pointer. When we set $d_2$ to $1$, we minimize the expected number of Spider points per play at $0.96$.
Likewise, if we set $d_2$ to $0$ (i.e. set up to defend the 3 pointer every time), the Spiders’ expected points per play will be maximized at $1.38$ .
So, our optimal defensive strategy is to defend the 3 pointer $0\%$ of the time.
Seeing your strategy, the Spiders adjust theirs to get an edge. Now, you’re the one who needs to switch things up! Given that the shot percentages are what they are, and knowing that the Spiders will counter-adjust to whatever new strategy you choose—what should your new strategy be?
How often should you defend the three-pointer versus guarding the layup?