THE BOOK--Playing The Percentages In Baseball

Wednesday, March 02, 2011

How much skill is there at free throws?

The answer is 25 free throw attempts.

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All data courtesy of Justin at http://www.basketball-reference.com

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I selected all player seasons with at least 100 free throw attempts since 1987.  There were 5067 such seasons.  The average number of free throws per player season was 249.

I calculated the success rate (free throws made divided by attempted) for each player and found the standard deviation.  The observed standard deviation of success rate was .090 (simple average) or .087 (weighted by attempts).

What we also want to determine is what is the expected spread of the random variation.  Simply put, it would be sqrt(.25*.75/249) = .027, if each and every player took 249 attempts.  If I calculate it on a player-by-player basis and: (a) do a simple average, I get .030, (b) do a weighted average based on attempts, I get .027.

That is, if every single player had equal talent, we’d expect to observe a distribution of success rates centered at 75%, with one standard deviation of 3%.  What we instead observe is one standard deviation of 9%.  This is a huge indication that there is talent at throwing free throws.  ALOT of talent.

Now, what can we do with this?  Plenty!  Remember this equation:
standard deviation of observed ^ 2 = standard deviation of true talent ^ 2 + standard deviation of random variation ^ 2

Or:
sd(obs)^2 = sd(true)^2 + sd(random)^2

We have this information:
.090^2 = sd(true)^2 + .030^2
or
.087^2 = sd(true)^2 + .027^2

That gives us an sd(true) of .085 or .083.

r^2 = sd(true)^2 / sd(obs)^2

That gives us an r-squared of .90 in either case.(*) That is, given 249 attempts, the observed success rate is 90% explained by the true talent of the player, and 10% by random variation.

(*) Alternatively, you can do this using z-scores.  .090/.030 is 3.18 (after rounding) and .087/.027 is also 3.18.  r-squared = 1 - 1/zScore^2 = 1 - 1/3.18^2 = .901.  If you are in school doing z-scores, just remember that “z-score is my friend”.

In order to have 50% of the variation explained, you’d have to have 25 attempts (.10/.90*249).

Our regression equation is therefore:
regression amount = 25 / (25 + Attempts)

If you have someone with 250 attempts, you regress his performance 10%.  If you have someone with 25 attempts, you regress 50%.

In practicality, you rarely need to regress.  For example, if we look at all players with at least 2500 free throw attempts, the regression rate would be at most 1%.

The #1 observed is Steve Nash at 90.4%.  With 3063 attempts, we’d regress him to 90.3%.

The two worst free throw shooters by far are Ben Wallace, observed at 41% and Shaq, observed at 53%. Wallace goes up to 42% (mostly because of rounding).  Shaq stays at 53%.

The key point is that once someone takes 25 shots, what you’ve observed is already half real.