A Study of Fairness in Fourball Golf Competition

Abstract

The golf handicap system is designed to be fair in single player games. The same system is also applied to team games with two or more players per team. In this study the fairness of the particular team golf game known as fourball match play was analyzed as well as measures to improve it. By using historical player data, software was produced to simulate games. The level of fairness was estimated given the handicaps of the players. It was determined that:

• Players with high handicap have high variance and therefore also an advantage.

• Teams should seek to optimize the spread in handicaps by pairing with someone as close to 11 or 29 handicap strokes apart and as far from 0 or 18 as possible.

• The handicap allowance introduced by the United States Golf Association to penalize high handicappers in team golf is not an effective tool to handle the unfairness.

• The tiebreaker rule is a simple way to lower the general unfairness by roughly 70%.

Appendix

1.1 Derivation of Fig. 6

The games [hcp hcp] vs [0 0] were analyzed for $0 \leq \mathrm{hcp} \leq 18$ with variance set to

$$\mathrm{Var}(\mathrm{hcp}) = \frac{6.7671 \cdot e^{0.038313 \cdot \mathrm{hcp}}+2.3733}{18}$$

((11))

according to Eq. 2. The mean was set to

$$E(\mathrm{hcp}) = \frac{1.0887\cdot \mathrm{hcp}}{18}$$

((12))

according to Eq. 1 with the last term removed for simplification. The a and b parameters in the uniform probability distributions $\mathrm{U}(a(\mathrm{hcp}),b(\mathrm{hcp}))$ were thereby given by solving Eqs. 6 and 7.

$$a(\mathrm{hcp}) = E(\mathrm{hcp})-\sqrt{3\cdot \mathrm{Var}(\mathrm{hcp})} \\[10pt] \label{(13)}$$

((13))

$$b(\mathrm{hcp}) = E(\mathrm{hcp})+\sqrt{3\cdot \mathrm{Var}(\mathrm{hcp})}$$

((14))

The team distributions were the first order statistics of the individual distributions. They can be seen graphically in the bottom row of Fig. 5 with triangle shape. The left plot represents the holes where Team A got strokes from handicap, thus increasing their chances. The right plot represents the remaining holes where Team A got no strokes but still suffered from being weaker players, thus lowering their chances. The distribution for holes without strokes was

$$f_{X}(x,\mathrm{hcp}) = \left\{ \begin{array}{rl} \frac{2b}{(b-a)^2}-\frac{2}{(b-a)^2}x & \mathrm{if } \;a \leq x \leq b \\ 0 & \quad\mathrm{else} \end{array} \right.$$

((15))

Team A’s probability of winning the holes where no strokes were given was

$$\begin{array}{l}p(\mathrm{hcp})\\ = P(X(\mathrm{hcp})<X(0)) \\ = \int\nolimits^{b(0)}_{a(\mathrm{hcp})}f_{X}(x,\mathrm{hcp}) \left( \int\nolimits^{b(0)}_{x}f_{X}(y,0) dy \right)dx \\ = \frac{1}{b(0)(b(\mathrm{hcp})-a(\mathrm{hcp}))^2} \\ \cdot \left[ \frac{b(0)b(\mathrm{hcp})}{2}x - \frac{b(0)+2b(\mathrm{hcp})}{4}x^2 + \frac{b(\mathrm{hcp})+2b(0)}{6b(0)}x^3 - \frac{1}{8b(0)}x^4 \right]^{b(0)}_{b(\mathrm{hcp})}. \end{array}$$

((16))

On the holes where strokes were given the distribution was shifted to the left by one stroke.

$$\hat{a}(\mathrm{hcp}) = a(\mathrm{hcp})-1$$

((17))

$$\hat{b}(\mathrm{hcp}) = b(\mathrm{hcp})-1$$

((18))

The distribution for holes with strokes was

$$f_{\hat{X}}(x,\mathrm{hcp}) = \left\{ \begin{array}{rl} \frac{2\hat{b}}{(\hat{b}-\hat{a})^2}-\frac{2}{(\hat{b}-\hat{a})^2}x & \mathrm{if }\; \hat{a} \leq x \leq \hat{b} \\ 0 & \mathrm{else } \end{array} \right.$$

((19))

The probability of Team A winning the holes where they had the advantage of handicap strokes was

$$\begin{array}{ll} \displaystyle &\hat{p}(\mathrm{hcp}) = P(\hat{X}(\mathrm{hcp})<X(0)) \\ & = \displaystyle \int^{a(0)}_{\hat{a}(\mathrm{hcp})}f_{\hat{X}}(x,\mathrm{hcp})dx + \! \int^{\min(b(0),\hat{b}(\mathrm{hcp}))}_{a(0)}\!\!f_{\hat{X}}(x,\mathrm{hcp}) \left( \int^{b(0)}_{x}\!\!f_{X}(y,0) dy \right)dx \\ & = \displaystyle \frac{2}{(\hat{b}(\mathrm{hcp})-\hat{a}(\mathrm{hcp}))^2} \left( -\hat{b}(\mathrm{hcp})b(0) - \frac{b(0)^2}{2} - \hat{b}(\mathrm{hcp})\hat{a}(\mathrm{hcp}) + \frac{\hat{a}(\mathrm{hcp})^2}{2} \right) \\ &\quad+ \displaystyle \frac{1}{b(0)(\hat{b}(\mathrm{hcp})-\hat{a}(\mathrm{hcp}))^2} \displaystyle \cdot \left[ \frac{b(0)\hat{b}(\mathrm{hcp})}{2}x - \frac{b(0)+2\hat{b}(\mathrm{hcp})}{4}x^2 \right.\\ &\quad\left. \displaystyle + \frac{\hat{b}(\mathrm{hcp})+2b(0)}{6b(0)}x^3 - \frac{1}{8b(0)}x^4 \right]^{\min(b(0),\hat{b}(\mathrm{hcp}))}_{-b(0)} \end{array}$$

where $a(0)=-b(0)$ was used for simplification.

1.2 Derivation of Fig. 10

The games [hcp 0] vs [0 0] were analyzed for $0 \leq \mathrm{hcp} \leq 18$ where the probability distribution of Team B on all 18 holes was the first order statistic of the two $Y \in \mathrm{U}(-1,1)$ distributions. The density function was plotted in Fig. 9 with a shape of a triangle. The distribution was thus

$$f_{Y_{(1)}}(y) = \left\{ \begin{array}{rl} 0.5-0.5y & \mathrm{if} -1 \leq y \leq 1 \\ 0 & \mathrm{else} \end{array} \right.$$

((20))

Next to this distribution are the red plots that give us the distributions of Team A. They are both comprised of one rectangular and one triangular part. The left plot represents the holes where Player 1 gets strokes from handicap, thus increasing Team A’s chances. The right plot represents the remaining holes where Player 1 gets no strokes but still suffers from being a weaker player, thus lowering Team A’s chances. These two distributions are given below as $f_{\hat{X}}(x)$ and $f_{X}(x)$ respectively. The variable a is the rightward shift in Player 1’s distribution. It is defined as $a=\frac{\mathrm{hcp}}{18}$ with $0 \leq \mathrm{hcp} \leq 18$ and hence $0 \leq a \leq 1$.

$$f_{\hat{X}}(x) = \left\{ \begin{array}{rl} 0.5 & \mathrm{if}\; a-2 \leq x \leq -1 \\ \frac{a}{a+1} - \frac{1}{a+1} x & \mathrm{if }\; -1 \leq x \leq a \\ 0 & \mathrm{else} \end{array} \right.$$

((21))

$$f_{X}(x) = \left\{ \begin{array}{rl} 0.5 & \mathrm{if } \;-1 \leq x \leq a-1 \\ \frac{1}{2-a} - \frac{1}{2-a} x & \mathrm{if}\; a-1 \leq x \leq 1 \\ 0 & \mathrm{else} \end{array} \right.$$

((22))

The probability of Team A winning the holes where they had the advantage of handicap strokes was

$$\begin{aligned}\hat{p}(a) & = P(\hat{X}<Y_{(1)}) = \int\nolimits^{a}_{a-2}f_{\hat{X}}(x) \left( \int\nolimits^{1}_{x}f_{Y_{(1)}}(y) dy \right)dx \\ & \qquad = \frac{1}{48(a+1)} \left( 17+28a+6a^2-4a^3+a^4 \right).\end{aligned}$$

((23))

Similarly, the probability of winning the holes where no strokes were given was

$$\begin{aligned}p(a) = & P(X<Y_{(1)}) = \int\nolimits^{1}_{-1}f_{X}(x) \left( \int\nolimits^{1}_{x}f_{Y_{(1)}}(y) dy \right)dx \\ & = \frac{1}{16(2-a)} \left( 16-32a+24a^2-8a^3+a^4 \right).\end{aligned}$$

((24))

The simple test that $\hat{p}(1)=p(0)=0.5$ verified that [0 0] vs [18 0] was an even game.