Eighteen x 18 feet = 18%

The Tour finally has gone to statistical record-keeping that allows us to see how good … “These Guys Are Good” … really are.

Let’s take one of my intermediate benchmarks in testing… 18 footers. The Tour “make %” average from 18 feet is actually 17%, but as we modify upward all Tour averages to account for the repeated same distance (but different line) putts effect, we boost that up to 18% in our models. If we were to boost for “reading break” advantage, the number goes to approximately 21%. But for this example, we’re sticking at 18% for our benchmark probability.

Now, suppose a Tour player (who was a composite, or average of them all) hit all 18 greens, and had eighteen straight 18 footers for birdie. This would of course be a (near) impossibility, but it’s illuminative for our purposes, so imagine it happens. How many birdie putts should the Pro make?

Simply put, 18% of 18 = 3.24, so on average, the number would be… Three (and maybe Four on a good day). But, will he repeatedly make 3? Nope.

The way to estimate the “spread” of results, is simple mathematics, courtesy of a Calvinist minister turned Mathematician named Jacob Bernoulli long ago… (you can learn about it here… http://en.wikipedia.org/wiki/Bernoulli_distribution).

If we plug the probability (0.18), and trials (18) into the formula, it yields these probabilities (rounded)…

0 makes… 3%

1 make… 11%

2 makes… 21%

3 makes… 24%

4 makes… 20%

5 makes… 12%

6 makes… 6%

7 makes… 2%

8 or more… 1%

So the probability of making 7 or more is about the same as making 0. About 3% each. But both are fairly likely, at any given round.

Oh yeah, the odds on making 17 of 18? Three in a trillion… try to make that many!

What does the above tell us? It tells us that it is not possible to find causality in a single round, for an “input” factor, like for example, changing the putter used. You need statistical data that is of significance… enough to give a level of confidence to your assertion. Just as you wouldn’t flip a coin three times, observe two heads occurances, and declare that two-thirds of all future flips would be heads with that coin.

The make values here are quite a spread… graphically, it looks like this… and on another post, I’ll show how many rounds are necessary, and why, to draw valid inference from the data.