I went back and changed the simulation a little to factor in different number of strokes taken.

Here’s what the 2009 Sample Looked like

Average number of Putts: 925.05
Standard Deviation of Putts: 207.60
Average %: 87.1%
Standard Deviation %: 1.47%

Then I ran 5 trials using 200 putters, with a normal distributed number of putts around the average of 925, and standard deviation of 207.6 making on average 87.1% of their putts. Here are the results:

Average %: 87.0%
Standard Deviation 1.11%

As you can see randomly, there is slightly smaller variation. Obviously this still doesn’t account for factors like different green difficulty and other things that might make the standard deviation higher.

I think overall there is a slight difference between the best and worst putters on tour, but most of what we see is just random.



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6 responses to “IS PUTTING RANDOM? PART 2

  1. N.

    In your previous post you had

    Simulated stdev: 9.5/925 = 1.03%
    Actual stdev: 12/925 = 1.30% [1]

    Now that you varied the number of attempts

    Newly simulated stdev: 1.11%
    Actual stdev: 1.47%

    How did the sample stdev increase from 1.3% to 1.47%? What am I misreading here?

    [1] “The average player in 2009 made about 86% of his putts from inside 10 feet and took about 925 attempts on the season. Adjusting everyone out to 925 putts, the minimum was 765 and max was 836 with a standard deviation of almost 12.”

  2. N.

    Also, if randomness gives you a stdev of 1.11% and the actual stdev is 1.47%, then we can say that randomness accounts for 1.11^2/1.47^2 = 57% of the variance of the sample. In other words, if randomness were removed, then the variance, which is the square of the standard deviation, would decrease by 57%. (That is the Bienaymé formula)

    So, rewording your last sentence, most (57%) of the variance we see is just random.

    Additionally, according to these stats if there’s a “good” putter with say a putting stat in the 84th percentile, then there’s an 18% chance that he’s actually a bad putter (with a true putting stat < 50th percentile). This of course ignores different courses, tee-times, etc.: if we took those into account the 18% would obviously by higher.

    (To get the 18% I had to sum up probabilities using an integral.)

  3. Thanks for that explanation in the second comment. Definitely didn’t learn that in stats 101, but I figured there was some way to figure that out.

    I think that I actually might have used 2009 in this one and 2010 in the first post. Probably should have double checked that before I posted, but I just wanted to do a quick post.

  4. N.

    I’m sorry, I had a typo in my first comment, it should’ve begun:

    Simulated stdev: around 9.25/925 = 1.00%

    I wrote 9.5/925. Oops!


    By the way, that 9.25/925 figure can be computed exactly (see ): Assuming a player has 925 attempts and a 87.1% chance of making each attempt, his standard deviation is precisely sqrt(.871*(1-.871)/925) = 1.102%.

    I quickly redid the simulation on this page* by varying the number of attempts using a normal distribution with mean=925.05 and stdev=207.6, and got 1.126%. So this post should say:

    Average %: 87.1%
    Standard Deviation 1.126%

    *I just typed in
    PDF[NormalDistribution[925.05, 207.60], x]*
    Sqrt[.871*(1 – .871)/x], {x, 0, 2000}]]”
    into Mathematica and pressed enter. Mathematica is really good at approximating integrals, so the answer (1.126%) is exact up to the decimal places given.

  5. N.

    Ooops, I should’ve checked my math…

    I should’ve used the formula N[Sqrt[Integrate[
    PDF[NormalDistribution[925.05, 207.60], x]*.871*(1 – .871)/x, {x,
    1, 2000}]]]
    and gotten 1.135%, not 1.126%. I can gladly provide an explanation of the formula if you want: it’s based on being able to add variances.

    So, 1.135% should be the value your simulation would spit out if you ran it forever.

  6. N.

    Wrong again!

    If you ran your simulation forever with perfect precision, the standard deviation would actually diverge (since eventually there would be someone with zero attempts, and this would contribute an infinite amount to the standard deviation). The 1.135% figure assumes everyone has a positive number of putting attempts.

    Sorry for spamming your post just based on hundredths of a percentage point 😛

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