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Marginal vs. Average Efficiency

· 4 min read

Much of the sports analytics movement has centered around arguments like:

  1. A is more effective B on average by a certain metric.
  2. Thus, we should be doing more of A and less of B.

A, B can be passing and running, 3's and 2's, targeting your WR1 vs. WR2, fastball vs. other pitches etc. This has generally been my mental model too and directionally I agree with most of these takes. However, after reading a paper by Skinner and Goldman (2015) recently I was reminded that 1 alone is insufficient to prove 2 because it conflates marginal and average efficiency - what we care about most is whether an incremental rate of increase in A is worth more than an incremental rate of increase in B (marginal efficiency) rather than whether across all instances of A and B, A is worth more than B on average (average efficiency).

Consider a modified version of the toy example Skinner provides. Let p be the fraction of attempted shots that are 3's and all other 1 - p shots attempted are 2's. Let's further assume that each 3pt shot attempted is worth 2 - p on average and the points per 2pt shot attempted is a stable 1.0. That is, if you rarely if ever attempt 3's you will score close to 2 - 0 = 2 points per 3 (intuitively, you're only occasionally taking the most wide open 3's from the very best shooter so you're hitting around 66% of them.) and if you're attempting a 3 on every possession you will score close to 2 - 1 = 1 point per 3 (intuitively, the defense has entirely sold out to the 3-point line so you're taking a lot of contested 3's at a rate of around 33%). The expected value of this strategy is: P(3 pt) * EV[3 pt] + P(2 pt) * EV[2 pt] = p * (2 - p) + (1 - p) * 1.0

Notably, there is a diminishing marginal return to attempting 3's, but at no point is an average 3 less efficient than an average 2. Yet, the optimal solution1 is when p = 0.5 for an EV of 1.25, so in this toy example we should be taking exactly half our shots from 3 and half from 2. Counterintuitively, at this point in time an average 3 is worth 1.5 points, so still worth much more than a 2-point shot worth 1.0 points. The mental model set up at the beginning would suggest we should still be shooting way more 3's. However, if we increased our 3pt attempt rate to say 75%, the expected value of this strategy would be around 1.19, markedly less than a 50% 3pt attempt rate. The solution occurs precisely when the marginal efficiency of a 3pt attempt is equal to the marginal efficiency of a 2pt attempt.2

Phrased differently, to answer the question of whether we should increase our rate of A, we should focus more on the extra instances of A/B we'd be adding/removing and the effect on the rest of our strategy, rather than the average outcome of how we're currently using each. An argument like "when isolating from the top of the key, a player can replace some of his pull-up 2's with stepback 3's without sacrificing spacing or altering the defensive context" is more persuasive than "3's are worth more than 2's on average so let's take more 3's." Context matters.

In practice, this doesn't always make an enormous difference to the average efficiency argument for 2 reasons:

  1. The difference between A and B for the biggest inefficiencies is large enough that the distinction between marginal and average efficiency is more of a rounding error.
  2. Defenses, and more generally responses to changes in strategy, take longer to react to changes in strategy than offenses take to change strategy, so performing A at a higher rate does not decrease A's success enough to justify a large marginal decay.

However, as league contexts become more efficient and defensive strategies catch up with offensive ones, this may change substantially. At worst, this is a reminder to couch arguments about strategic changes in terms of incremental value add as opposed to generic averages we've observed.

  1. The expected value of this strategy is: P(3pt) (EV per 3pt) + P(2pt) * (EV per 2pt) = p * (2 - p) + (1 - p) * 1. 0.5 was found by taking the derivative of this function and setting it to 0↩
  2. When the derivative of p * EV[2 pt] equals the derivative of (1 - p) * EV[3 pt]. This is equivalent to marginal efficiency because it measures the probability normalized change in EV from a marginal change in p.↩