I really enjoy disc golf. This year I have done the 9 hole Circleville course in State College Pennsylvania (USA) about 50 times (usually two or three times per outing) and I’ve gotten 3 under par at least three times, but I have never gotten four under par. I would love to have a decent estimate of the likelihood of getting four under par. There is a way to estimate this probability with polynomials.
Two Holes
Suppose that on hole one you have a 30% chance of birdie (one under par and a score of -1) and a 70% chance of getting par (a score of 0). Suppose that on hole 2, you have a 20% chance of birdie and an 80% chance of par. What is the probability of each possible outcome after completing the first two holes?
- You might get two birdies which is a total of -2. The probability of that is 0.3 times 0.2 = 0.06 = 6%.
- You might get a par followed by a birdie for a total of -1. The probability of that is 0.7 times 0.2 = 0.14 = 14%.
- You might get a birdie followed by a par for a total of -1. The probability of that is 0.3 times 0.8 = 0.24 = 24%.
- The last possibility is that you get two pars. The probability of that result is 0.7 times 0.8 = 0.56 = 56%.
(Technical note: we are assuming that the performance on each hole is statistically independent of the performance on the other holes.)
Notice that the probabilities of getting a score -2, -1, or 0, are 6%, 38%, and 56% respectively. (The 38% comes from adding 14% to 24%).
Perhaps surprisingly, these three probabilities can be calculated with polynomials. If we expand
$$( 0.3\, x + 0.7) (0.2\, x + 0.8), \quad\mathrm{then\ we\ get\ } $$
$$\begin{aligned} ( 0.3\, x + 0.7) (0.2\, x + 0.8)&= 0.3 x\, (0.2 \,x + 0.8) + 0.7(0.2\, x + 0.8) \\&= 0.06 \,x^2 + 0.24 \,x + 0. 14\, x + 0.56 \\&= 0.06\, x^2 + 0.38\, x + 0.56. \end{aligned}$$
Nine Holes
I can use the same method to estimate the probability of getting four strokes below par on the 9 hole Circlesville course. Let’s suppose that the probability of getting a birdie on any given hole is given by the table below. (We will also optimistically assume that you always get par or a birdie on every hole.)
$$\begin{array}{cc}
\text{Hole} & \text{Birdie Probability} \\
\hline
1 & 0.04 \\
\hline
2 & 0.1 \\
3 & 0.03 \\
4 & 0.4 \\
5 & 0.25 \\
6 & 0.12 \\
7 & 0 \\
8 & 0 \\
9 & 0.3 \\
\end{array}$$
Now the corresponding polynomial is
$$\begin{aligned}p(x) = &(0.96 + 0.04 x) (0.9 + 0.1 x) (0.97 + 0.03 x) (0.6 + 0.4 x)\\ &\quad\times (0.75
+ 0.25 x) (0.88 + 0.12 x)(0.7 + 0.3 x) .\end{aligned}$$
We can use Wolfram Alpha to expand $p(x)$ thusly
$$\begin{aligned} p(x) = 0.2323& + 0.4062 x + 0.2654 x^2 + 0.0823 x^3 + 0.0128 x^4 \\ +&0.00098 x^5 +0.000034344 x^6 + 4.32\cdot\, 10^{-7} x^7 \end{aligned}$$
We can conclude that my most likely result is 1 under par (40.6%) and the probability that I will get exactly 4 under par over 9 holes is about 1.28%.
(What is p(x)? Suppose that a tournament sponsor will give you $1 for getting par, x dollars for getting 1 below par, x^2 dollars for getting 2 below par, …, and x^9 dollars for getting 9 below par, then p(x) is the expected value for playing 9 holes in the tournament.
(You don’t actually need polynomials. In reality you are just doing convolution of the coefficients of the polynomials when you are multiplying the polynomials. It is not very difficult to modify this algorithm to account for bogies.)