I used to be good at basketball.*

I still love to play, but now in my old(er) age, I’m more of a “cagey veteran.” Which means I do things like applying statistical analysis to my shooting strategy.

A topic I’ve been thinking about recently: how does “winners outs” (when you make a basket you get the ball back) that is the common rule in a half court game affect the optimal shot taking strategy? Scoring *and earning another chance to score *is a different outcome than scoring and having the other team get the ball (like in a regular full-court game).

To explore the difference, let’s consider the example configuration:

- You make 50% off your 2 point shot attempts
- You shoot 33% from beyond the arc on 3 point shots

In addition, for the sake of simplicity, let’s assume:

- There is no difference in the probability of getting an offensive rebound based on what kind of shot you take
- The probability of a turnover is the same regardless of what kind of shot you are planning to take

In a full court game, with this configuration, you are entirely indifferent as to whether to take a 2 or a 3. You expect exactly 1 point per shot: 50% × 2 points = 33% × 3 points = 1.

Note: a decent counterpoint is that your “utility” of making a shot is not just the points you earn, but also the “glory”, and that making a 3 yields more, and thus even in this configuration, you’d shoot the 3 because of the potential “glory utility”…but we’ll leave that out of scope for now!

But in a half-court game, with “winner’s outs”, this calculation changes. If you “shoot the 2", you have a 50% chance of not only of scoring 2 points, but also getting the ball back, which means another chance to score (and another, and another…as long as you keep making shots) So your expected number of shot attempts per possession is a geometric series: 1 + .5 + (.5)^2 + (.5)^3 … — which means that you expect to have 2 possessions (1 / ( 1 — r) where r = .5), and thus score **2** points each time you take a 2 point shot in a winner’s outs game.

By contrast, if you “shoot the 3", you only expect 1.5 shots (1 / ( 1 — r) where r = .33). You still average 1 point per shot attempt, but that means only 1.5 expected points per possession following a strategy where you only shoot the 3.

This is huge! A 25% difference in the expected points per possession in the half-court game based on the same shot taking strategy that results in a wash for a full-court game.

But wait…we’re not done yet! Since we’re playing up until a point total (common half court games are played to 21) rather than a timed game, you need to consider how many more points you need to win. If you have 18 points, a single 3 can win the game (33% chance), whereas you’d need two 2-point baskets. (50% × 50% = 25% chance to win during this possession). So if you have 18 points, your best strategy is to shoot the 3!

OK, let’s keep this analysis going. In many games, 3s count as 2, and 2s count as 1. My gut has always been that this is unfair, and that when playing these kinds of games, I always shoot a “2” from behind the arc. But is that the right strategy? Let’s do the same Math with the same configuration: Shooting the 3: 2 pts * .33 pct * 1 / (1 — .33) shots = 1 expected point per possession. Shooting the 2: 1 pt * .5 pct * 1 / (1 — .5) shots = 1 expected point per possession. The same expected outcome! So in fact, whomever came up with the rule that 3s should be 2s and 2s should be 1s in a half-court game…is kind of a genius!

How does the other team’s expected shooting ability factor in? And what if I’m feeling hot from behind the arc today and shooting 40% from the 3? This all deserves some more analysis…and even some coding(!)…but we’ll stop this post here for now…and go outside to shoot some hoops!

* proof that I used to be good a basketball (swear that’s me!)…and that my high school team unfortunately never got the memo about long shorts being the in thing…