Saturday, July 4, 2020

Eliminating the Blind Draw with Mr. Par

This blog has previously examined ways to eliminate the blind draw (Eliminating the Blind Draw,” October 22, 2019) when threesomes competed against foursomes. The post concluded giving the threesome a phantom player who always scores par (Mr. Par) led to equitable competition if the Course Rating and Par were not too far apart.  Since the publication of the post, the World Handicap System has made par and not the Course Rating the target score of the handicap system.  This eliminates the caveat about par and the Course Rating mentioned in the previous post.

The previous post was based on theory and urged playing groups to provide actual tournament results to verify the equity of using Mr. Par.  One group, playing two-bests balls of four, has used Mr. Par over the season.  The question addressed here is whether actual tournament results match up with what theory suggests?

What does theory suggest?  Let’s assume you are the captain of a threesome and have the choice of selecting a handicap player at random or taking Mr. Par.  You realize the handicap player will only have a net score below par twenty percent of the time.[1]  If the game was aggregate four- ball, Mr. Par would be the clear choice.  In two-best balls of four, however, the handicap player may contribute some net birdies while Mr. Par will not. If you take the handicap player, the standard deviation of the team score will be higher than with Mr. Par.   That means you will have a better chance of coming in first, but also a better chance of coming in last.  Since you are not risk averse, you opt for the handicap player.  Let’s look at the data to see if you made the right choice.

Empirical Test - A group plays two-best balls of four format and assign Mr. Par to the threesome.  At the end of the season, the mean score and standard deviations of threesomes and foursomes are computed (Team scores are presented in the Appendix):

              M3 = Mean team score of threesome relative to par = -12.09

              S3 = Standard deviation of threesomes scores = 3.49

              N3 = Sample size of threesomes = 23

              M4 = Means team score of foursomes relative to par = -12.67

              S4 = Standard deviation of foursome scores = 4.77

              N4 = Sample size of foursomes = 101

The null hypothesis is that the difference in the true means of each distribution (threesomes and foursomes) is zero.  The alternative hypothesis is the difference is not zero.  These hypotheses constitute a two tailed test.  The null hypothesis will be rejected if the absolute difference between the sample means is too large. 

The standard error of the sampling distribution is given by:

              SE = sqrt((S32/N3) + (S442/N4)) = 0.87

The t statistic is then: 

              t = (M4 – M3)/SE = (-12.09 – (-12.67))/ .87 = 0.67

The t-Distribution Calculator for 51 degrees of freedom shows the probability of the t-value less than -.067 and greater than 0.67 is 0.50.[2]  Therefore, the null hypothesis that the difference between the two means (threesomes and foursomes) is zero cannot be rejected.

If the threesome could deduct one stroke from its score, the t statistic would be 0.48 ((12.67-13.09)/0.87)).  The probability of a t-value less than -.48 and greater than 0.48 is .68.  Due to the small sample size it would be difficult to choose between giving a threesome zero or minus one stroke on statistical grounds.  It is clear, however, that either of these two options would be equitable.

Other Considerations – The smaller standard deviation for threesomes indicate it is more difficult for a threesome to go low. A look at the winners of competitions bears this out.   The data shows 14 competitions where threesomes competed against foursome.  The table below presents the probability of each type winning assuming threesomes and foursomes had an equal chance.


Probability of Winning


The table shows the threesomes are expected to win four times and foursomes 10 times.  Threesome had two outright firsts and one tie for first.  Had threesomes been given an additional stroke, they would have had three wins which is close to what was predicted.  As expected, threesomes do not go as low as foursomes, but they also do not go as high.  Threesomes only had one tie for last. 

Whether the threesome captain made a good choice in selecting the handicap player depends in part on the payoff structure.[3]  If the payoff structure pays a winning team the same regardless of its composition, a threesome’s smaller chance of winning is offset by a 33 percent larger payoff per player.  Even if the payoffs are per player, a threesome with Mr. Par may have a better chance of being in the money even though they do not win.  The examination of expected payoffs is beyond the scope of this study, but a cursory review indicates if payoffs are per man, neither threesomes nor foursomes have a significant advantage.  

In formatting a competition, there should be no team that realizes it has little or no chance before teeing off (e.g., a team of three low handicappers and Mr. Par in a two-best ball of four game).  If in the game under consideration, the handicaps among teams are evenly spread, threesomes with Mr. Par (possibly with a stroke deduction) and foursomes should have an equitable competition.  Importantly, this conclusion should hold for all courses regardless of the Course Rating



Team Scores

Three -10 -10-10  -4-15-14
Three   -9    -14-14
Three   -7      
Four-16 -13 -8-11-11-5-8 
Four-10 -13 -7-9-10-3  
Four     -8-7-3 



Three-13 -16 -14-19-14 -12-17
Three-10 -14 -14 -11   
Three    -9     
Four -10-11-9-12-14 -14-11-14
Four -8-9-7 -14 -14 -12
Four   -7 -11 -11  
Four   -4   -10 

[1] Under the USGA Handicap System (UHS), a player had a net score below the Course Rating about 20-25 percent of the time.  Under the World Handicap System (WHS), a player’s eight best score are used in calculating his Index instead of his ten best in the under the UHS.  This lowering of a player’s Index under the WHS is offset by the elimination of the Bonus for Excellence (.96).  The WHS now makes “par” the standard of performance and not the Course Rating.  Therefore, a player will have a net score of par or better approximately twenty percent of the time.

[2] The degrees of freedom is estimated by:

              DF = (S32/N3 + S42/N4)2/ ((S32/N3)2/(N3 -1)) +( (S42/N4)2/(N4 -1)) = 57

[3] Weather is also a factor.   If the weather is bad, most players will have net scores over par, which would make the consistent Mr. Par a better choice.     

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