Optimal Putting in Disc Golf

May 20, 2021 in Games by hundalhh | Permalink

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 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.
  • If I do try to put it in the basket (choice 1), I will succeed with probability $p$, where $p$ depends only on distance to the basket.
  • If I do try to put it in the basket and fail to get it in the basket (choice 1), the probability that I will succeed on the second try is $q$ where $q$ is constant which does not depend on distance.
  • 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.

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: I get it the basket on the first throw!
  • outcome 1.2: I miss the basket, but get the disc in the basket on the second throw.
  • outcome 1.3: I miss twice, and get it on the third throw.

 

The probabilities for each of those outcomes are:  $p$, $(1-p) q$, and $(1-p)(1-q)$ respectively.

Let $a$ be the average number of throws for choice 1. Then $$\begin{aligned}a &= p\cdot 1 +(1-p)q\cdot 2 + (1-p)(1-q)\cdot 3 \\&= p + 2 q – 2 p q + 3 – 3 p – 3 q + 3 p q\\&=3 -2 p – q + p q.\end{aligned}$$

 

I should choose choice 1 if $ 2>a$.  This occurs when

$$\begin{aligned} 2 &> 3 -2 p – q + p q\\-1 &> -2 p – q + p q \\-1 &> (q-2) p – q  \\q -1 &> (q-2) p\\ \frac{q -1}{q-2} &< p   \\\frac{1-q}{2-q} &< p. \\ \end{aligned}$$

Now you can plug in various values for $q$ to find the lowest value for $p$ needed to go for it.

Probability of Success          Required
 After Missing             Probability of Success
                             on the first try
     100%                           0%
      99%                           1%
      95%                           5%
      90%                           9%
      85%                          13%
      80%                          17%
      75%                          20%
      50%                          33%
      0%                           50%

So, if you are 100% certain that you will put it in the basket on the second try, then you should use choice 1 (going for it) if $p>0$ (i.e. always).

If you are 90% certain that you will put it in the basket on the second try, then you should use choice 1 (going for it) if $p>0.09=9\%$.

$$ $$

a rule of thumb

A nice approximate rule of thumb is to go for it if the sum of $p$ and $q$ is more than 100%.

When I am 6 yards from the basket, I will get it in about 75% of the time (p=0.75), and if I miss, I will usually get it in 90% of the time.  The sum of 70% and 90% is 160%, so obviously, I should go for it.

When I am 9 yards from the basket, I will get it in about 20% of the time (p=0.20), and if I miss, I will usually get it in 85% of the time.  The sum of 20% and 85% is 105%, so it seems like I should go for it.

If the basket is on a hill or if it is windy, then the disc will sometimes roll a fair distance if I miss.  In that case, $q$ might be only 75%. The rule of thumb is that $p+q$ should be at least 100% to go for it, so according to the rule of thumb, I would need $p$ to be at least 25% to go for it.  On a windy day, that’s like 6 yards for me.