I used to be really good at basketball. *

Now in my old(er) age, I'm more of a "cagey veteran." Which means I do things like apply statistical analysis to my shooting strategy.

Something I've been thinking about recently: how does "winners outs" (make a basket and 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).

An example configuration:

- You make 50% off your 2 point shot attempts
- You shoot 33% from beyond the arc
- For the sake of simplicity, assume that there is no difference in the probability of getting an offensive rebound based on what kind of shot you take
- For the sake of simplicity, assume that the probability of a turnover is the same regardless of what kind of shot you are planning to take

In a full court game, you are entirely indifferent as to whether to take a 2 or a 3 - your expected points per shot is exactly 1: 50% × 2 points = 33% × 3 points = 1 expected point per shot attempt. (A decent counterpoint can be made 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", things change. If you "shoot the 2", you have a 50% chance not only of scoring 2 points, but also getting the ball back, and getting another chance to score again (and again, and again...) 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.

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. This means you'd expect to get to 20 points shooting just 2's at the same time you'd get to only 15 points shooting just 3's.

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!

What about other key numbers along the way? 15 points, which is 2 3's, or 3 2's? And does the other team's expected shooting ability factor in? And what if I'm feeling it today and shooting 40% from the 3? This all deserves some more analysis...and quite possibly even some coding(!)...a follow up post coming soon...!

OK, let's keep in 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 the 3. But is that right? 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 the half-court is kind of a genius!

* proof that I used to be good a basketball...and that my high school team unfortunately never got the memo about long shorts being the in thing...