Optimal Putting in Disc Golf – Revised

April 23, 2022 in Games by hundalhh | Permalink

About a year ago, I posted a short article about optimal putting in disc golf.  I gave an approximate rule of thumb to use at the end of the article, but it turns out that there is a better thumb rule, so I’m revising the post accordingly below.

I often have to make a decision when putting at the edge of the green during disc golf.  Should I try to put the disc in the basket (choice 1) or should I just try to put the disc near the basket (choice 2)? If I try to put it in the basket and I miss, sometimes the disc will fly or roll so far way that I miss the next shot.

In order to answer this question, I created a simplified model.  Assume:

  • If I do try to put it in the basket (choice 1), I will succeed with probability $p_1$.
  • If I do try to put it in the basket (choice 1), fail to get it in the basket, and then fail again on my second try, then I will always succeed on the third try.
  • If I don’t try to put the disc in the basket (choice 2), I will land near the basket and get it in the basket on the next throw.

Using these assumptions, I can compute the average number of throws for each choice.

For choice 2, I will always use two throws to get the disc in the basket.

For choice 1, there are three possible outcomes:

  • outcome 1.1: With probability $p_1$, I get it the basket on the first throw!
  • outcome 1.2: With probability $p_2$, I miss the basket, but get the disc in the basket on the second throw.
  • outcome 1.3: With probability $p_3$, I miss twice, and get it on the third throw.

I am assuming that if I miss twice, I will always get it on the third try, so

$$p_1 + p_2 + p_3=1.$$

Let $a$ be the average number of throws for choice 1. Then $$a = p_1\cdot 1 +p_2\cdot 2 +p_3\cdot 3.$$

I should choose choice 1 (go for it) if I average fewer throws than choice 2, i.e. if $a<2$.  This occurs when

$$\begin{aligned}2 >& p_1\cdot 1 +p_2\cdot 2 +p_3\cdot 3 \\2 (p_1+p_2+p_3) >&  p_1\cdot 1 +p_2\cdot 2 +p_3\cdot 3\\p_1\cdot 2+p_2\cdot 2+p_3\cdot 2 >&  p_1\cdot 1 +p_2\cdot 2 +p_3\cdot 3\\ p_1 >& p_3.\end{aligned}$$

So, I should choose choice 1 if $p_1> p_3$.  In words,

 

“Go for it if the probability of getting it on the first try is greater than the probability of missing twice”.