Handicap Allowances for Eclectic Golf Tournaments

In an eclectic golf tournament, players play the same course more than once and select the best score on each hole as the score for the competition.  If played a full handicap, the competition would favor the high handicap player.  To demonstrate this, assume two players have the probabilities for scoring shown in
Table 1.

Table 1

Probability of Scoring Relative to Par on Each Hole

Handicap	Score Relative to Par
	-1	0	+1	+2
Low Handicap	.1	.8	.1	.0
High Handicap	.0	.2	.6	.2

Over eighteen holes, the low handicap player would average even par.  The high handicap player would average 18 over par for a difference between players of 18 strokes.    There handicaps would be approximately 18 strokes different.[1]  Now assume each player is allowed to play two rounds and take his best score.  The new probabilities for each hole are shown in Table 2.

Table 2

Probability of Scoring Relative to Par on Each Hole after Two Rounds

Handicap	Score Relative to Par
	-1	0	+1	+2
Low Handicap	.19	.80	.01	.00
High Handicap	.00	.36	.60	.04

The low handicap player would average 3.24 strokes under par while the high handicap player would average 12.24 strokes over par.  The difference between players is now 15.48 strokes.  To make the competition fair, the handicaps should be multiplied by a factor of .86 (i.e., 15.48/18).

For a four-ball eclectic the percentage allowance should be even lower.  Assume that both players get an identical partner for a four-ball eclectic.  The probabilities of each score after two rounds for both teams are shown in Table 3.

Table 3

Probability of Scoring Relative to Par on Each Hole after Two Rounds of Four-Ball

Handicap	Score Relative to Par
	-1	0	+1	+2
Low Handicap Team	.3439	.6560	.0001	.0000
High Handicap Team	.0000	.5904	.4080	.0016

The low handicap team would average 6.19 strokes under par and the high handicap team would average 7.37 strokes over par.  The difference in team scores would be 13.56 strokes.  To make the competition fair for four-ball, the allowance should be 75 percent (i.e., 13.56/18).

The allowances of 86 percent for a stroke play eclectic and 75 percent for a four-ball eclectic were derived using a very simple model.  To see if these allowances are reasonable, a small four-ball eclectic tournament is analyzed.  The allowance for each player in the competition was 90 percent of his handicap.  The tournament handicaps and results from the first and second day for two flights are presented in the Appendix.

In a perfectly equitable competition, scores should not be correlated with handicaps.  When Day 1 Scores were regressed against Total Handicap, the estimated equations were:

Flight 1                  Day 1 Score = 64.7 - .004·Total Handicap

Flight 2                  Day 1 Score = 72.1 - .250·Total Handicap

In Flight 1, the coefficient of Total Handicap variable was both small and not statistically significant
(t-statistic = -.01).  This finding implies taking 90 percent of each player’s handicap led to a fair competition.  A one-day four-ball competition is similar to a two-day stroke play eclectic.  Basically, one best score out of two is chosen.  So the theoretical allowance of 86 percent is clearly in the ball park.  In Flight 2, both the coefficient of the Total Handicap variable and its t-statistic (t=-.9) were larger.  While the coefficient is not statistically significant at the 95 percent level of confidence, it does indicate equity could be improved if the allowance was less than 90 percent.

   

The second day results were:

Flight 1                  Day 2 Score = 60.8 - .086·Total Handicap (t-statistic = -.59)

Flight 2                  Day 2 Score = 67.1 - .257·Total Handicap (t=statistic = -1.30)

The equations indicate the higher handicap teams are favored, but neither coefficient is significant at the 95 percent level of confidence.  The equation for Flight 1 predicts the highest handicap team should have a score 1.3 strokes lower that the lowest handicap team (i.e., the difference in team handicaps multiplied by .086).   If the 75 percent allowance was used, the high handicap team would lose three more strokes than the low handicap team.  On average, this should lead to a reduction in the highest handicap team’s score of 1.5 strokes.  In essence, the advantage of the high handicap team is wiped out when the 75 percent allowance is used.  In Flight 2, the regression equation predicts a 2.5 stroke advantage for the highest handicap team.   Using the 75 percent allowance would reduce the score of the highest handicap team by 1.7 strokes on average.  While the highest handicap team would still have an advantage, it would be less than one stroke.

Of the top three finishers in each flight, only one came from the top half (i.e., the lower handicap teams) the flight.   Clearly, 90 percent did not produce an equitable competition.  The lesson here is Tournament Committees should study the effect of allowances on tournament outcomes.  Experiments with different allowances should proceed until equitable competition is achieved.  

   

Appendix

Flight Scores and Handicaps

Tournament Results-Flight 1

Team	Total Handicap	Day 1	Day 2
1	11	67	62
2	12	60	57
3	16	67	60
4	21	64	59
5	21	69	63
6	23	63	56
7	24	68	59
8	24	62	58
9	25	62	57
10	26	65	60

Tournament Results-Flight 2

Team	Total Handicap	Day 1	Day 2
1	26	62	59
2	26	65	61
3	28	67	60
4	29	66	58
5	30	69	63
6	30	63	60
7	34	61	56
8	33	65	57
9	35	66	61
10	36	60	57

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[1] The USGA’s “bonus for excellence” and possible differences in scoring variances would have small effects on the handicaps.  The effects, however, are small and are neglected in this analysis.

Predicting the Overall Winner of a Golf Tournament When Players Compete in Flights

Here is a probability problem that arose during a recent golf tournament.  The field of 36 two-man teams was divided into six flights.  The lower-handicap players were assigned to the first three flights and played from the blue tees.  The higher-handicap players were assigned to the last three flights and played from the white tees.  The blue tees had a Course Rating one stroke higher than the white tees.

 

The competition was conducted over two rounds.  The first round was four-ball and the second round was Pinehurst.  The competition was conducted at full handicap for all players using Stableford scoring.   That is, there was no 90 percent reduction in handicap for the four-ball competition nor were handicaps adjusted for the difference in Course Ratings in accordance with Section 3-5 of the USGA Handicap System.  Given these conditions of the competition, what is the probability the overall winner (i.e., the highest Stableford point total) would come from a blue tee flight?

To make this problem analytically tractable, it is assumed all of the blue tee players are ten handicaps and all of the white tee players are twenty handicaps.  If the USGA recommendation for four-ball was used, the blue tee players would have their handicaps reduced to nine (i.e., 90 percent of their Course Handicap).  Similarly, the white tee players would have their handicaps reduced to eighteen.  The blue tee players would have their handicaps increased by one-stroke (the difference in Course Ratings between tees).   That would bring the blue tee players back to ten handicaps.  The white tee players would be eighteen handicaps under USGA guidelines, but under the conditions of the competition they are allowed two extra strokes and play as twenty handicaps.

 

The two extra handicap strokes should lead to a 2-point increase in the Stableford score of the white tee teams in the four-ball competition.   On the first and second stroke allocation holes, the white tee players would get an additional handicap stroke.  Unless both players had a quadruple bogey on these holes, the additional strokes would lead to additional Stableford points.[1]

In the Pinehurst Competition, the teams from the blue tees should have received an additional handicap stroke under Section 3-5 of the USGA Handicap System.   Since these teams did not receive the Section 3-5 adjustment, their Stableford score will be one Stableford point lower—unless the team scored a triple bogey on the eleventh stroke allocation hole.

The next assumption is that if the USGA recommendations were followed, the average team score in each flight would be equal.   Without the adjustments, the white tee flight should have an average Stableford point total three points higher than the blue tee flight (2 points in four-ball and 1 point in the Pinehurst competition). 

The probability of a blue tee team winning depends upon the distribution of scores within each group.  The wider the distribution, the better chance a blue tee team has of overcoming the three point difference in average scores.  For this analysis, it is assumed the standard deviation of team scores is three points for both flights.  That is, approximately two-thirds of the team scores are within plus or minus three points of the average. The probabilities of scores can be estimated using the standard normal distribution table as shown below in Table 1.[2]

Table 1

Probability of Team Scores

Average	+1	+2	+3	+4	+5	+6	+7
.135	.1240	.1052	.0823	.0542	.0332	.0186	.0150

For a blue tee team to be the over-all winner, one of the combinations of outcomes shown in Table 2 must occur.  (Note:  Ties between a blue tee team and a white tee team are not considered a win for the blue tee team.)

Table 2

Winning Combinations for a Blue Tee Team to be the Overall Winner

Blue Tee  Outcome	Probability	White Tee Outcome	Probability	Probability of Both Outcomes
+7	.238	No Team > +3	.098	.023
≥+6	.459	No Team > +2	.017	.006
≥+5	.712	No Team > +1	.001	.001

The probabilities of each outcome will depend upon the number of players in each flight.  The tournament in question had 18 teams in each flight so that number will be used.  The probability of at least one Blue Tee scores +7 is one minus the probability that there are no +7 team scores ((1- (1 - 0.015)¹⁸ = .238).  The probabilities of at least one blue tee team scores +6 or greater or +5 or greater are calculated similarly.

For a blue team scoring +7 to win outright, no white tee team can score higher than +3.  The probability of a no white tee team scoring  greater than +3  is the probability of a team scoring +3 or less raised to the eighteenth power (i.e., (1-(.0542 +.0186 + .0150))¹⁸ which equals .098.  Similarly, the probability of all white tee teams scoring +2 or less is .017 and the probability all teams scoring +1 or less is .001.

The probability of a blue team winning with a +7 is the probability of it scoring +7 times the probability no white tee team scoring +3 or greater.  That probability is .023 as shown in Table 2.  The probability of a blue tee team being the overall winner is the sum of probabilities for the three ways it can win shown in Table 2.  That probability is .03.  In other words, a blue tee team should be expected to win once in every 33 years.

Whether the format of this competition is strongly discriminatory against blue tee players rests upon the validity of the assumptions.  Is there a three point difference in average scores between blue tee and white tee teams?  Is the standard deviation for both tees equal to 3.0 points?  Is there a large difference in Course Handicaps between blue and white tee players?  How often are blue tee teams the over-all winner?  A check of tournament statistics can determine if these assumptions are realistic and if the blue tee teams are at the disadvantage claimed here.

 

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[1] If the partnership had different handicaps (i.e., were not both twenty handicaps), the two stroke advantage could lead to an increase in their Stableford score of between zero and four points.  See Dougharty, Laurence, The Four-Ball Adjustment Under Section 3-5, www.ongolfhandicaps.com, September 15, 2014.

[2] For example, the probability of a team scoring +2 is the area under the standard normal curve from 1.5/3 to 2.5/3.  This would be .3085 - .2033 = .1052.