Eliminating the Blind Draw-Continued

In commenting on a previous post (“Eliminating the Blind Draw' October 31, 2012 now published with the date October 22, 2019), an astute reader suggested an alternative method:

My club had been messing around with options to handle this problem until I suggested that we should try using the second best score as the score of the third player. For example, in Stableford Scoring if Player A gets 2 points, Player B gets 2 points, and Player C gets 1 point, then the team gets 6 points.
This has now been accepted by all the informal groups in the club as results seem to suggest that teams of 3 win a prize in proportion to the number of teams of 3 or 4 entered. The number of prizes is in proportion to the size of the field.  I would be interested to know if you thought there was any mathematical support for this.

Let’s look at a simplified example to get a preliminary judgment of equity.  Assume only four scores are possible (3 = net birdie, 2 = net par, 1 = net bogey, and 0 = net double bogey or worse).  Further assume the probability of each score is the same (i.e., 1/4).  The question is whether the expected second best score of the threesome is greater, equal, or less than the expected score of a fourth player.  Column 4 of Table 1 shows the probability of each outcome for the threesome.[1]

Table 1

Probability of Outcomes

(1)  Highest	(2)  Second Best	(3)  Third Best	(4) Probability
3	3	3	1/64
3	3	2	3/64
3	3	1	3/64
3	3	0	3/64
3	2	2	3/64
3	2	1	6/64
3	2	0	6/64
3	1	1	3/64
3	1	0	6/64
3	0	0	3/64
2	2	2	1/64
2	2	1	3/64
2	2	0	3/64
2	1	1	3/64
2	1	0	6/64
2	0	0	3/64
1	1	1	1/64
1	1	0	3/64
1	0	0	3/64
0	0	0	1/64

The expected second best score is the sum of the value of each score multiplied by the probability of each score occurring:

                Expected Second Best = 3·(10/64) +2·(22/64) + 1·(22/64) + 0·(10/64) =  96/64 = 1.5

The expected score of the fourth player is:

                Expected Score of the Fourth Player = 3·(1/4) + 2·(1/4) + 1·(1/4) + 0·(1/4) = 1.5

Under all of the assumptions made, using the second best score would be equitable.  This result is not surprising.  The expected second best score is also the expected average score of any player.   (It would not be equitable under a modified Stableford System where the difference between hole scores is not constant.  For example, assume a net birdie is now worth 5 points while all the points for the other scores remain the same.   The expected second best score would be 1.81.  The expected score of the fourth player would be 2.00.) 

To get a more realistic estimate of any advantage, the number of possible hole scores is expanded as shown in Table 2.  Since not all hole scores are equally likely, however, Table 2 presents reasonable probability estimates for an easy course (Course Rating less than par) and a more difficult course.

Table 2

Probability of Hole Scores (Stableford Points) for Easy and Difficult Courses

Course	Net Eagle(4)	Net Birdie(3)	Net Par(2)	Net Bogey(1)	Net Double Bogey or Worse(0)	Expected Score Par = 72
Easy	.10	.30	.40	.15	.05	67.5
Difficult	.05	.20	.45	.20	.10	73.8

Using the same methodology presented in the simplified example, the expected score of the second best hole score of the threesome and the expected score of the fourth player can be estimated.  Those holes scores are shown in Table 3.

Table 3

Expected Holes Scores

Course	2nd  Best	Fourth Player	Threesome Advantage per Hole
Easy	2.27	2.25	0.02
Difficult	1.91	1.90	0.01

Table 3 shows a small edge for the threesome on both courses.   Given all of the uncertainties embedded in the model (e.g., probability assumptions, assumption that all players have identical scoring patterns and variances), such a small edge is not significant.  And given the random nature of team scoring, such a small edge would be hard to notice without a large number of trials.  If threesomes are winning in proportion to their numbers as the reader suggests, there is no reason to believe his method is not equitable. [2]

The one remaining problem is how prize money is distributed.  If first place wins the same amount regardless of team size, the threesomes have a big edge.  A threesome is just as likely to win, but only has to split the prize money three ways instead of four.  Since the reader was clever enough to devise an equitable competition, I am sure he has solved this problem too.

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[1] The probability of an outcome is the probability of any one outcome multiplied by the number of ways that outcome can be achieved.  For example, the outcome 3,3,2 can be achieved in three ways—Player A gets a 2 and the other players get  3s, Player B gets a 2 and the other players get 3s, Player C gets a 2 and the other players get 3s).   The probability of any one outcome is 1/4·1/4·1/4 = 1/64.  Therefore, the probability a team has two 3s and a 2 is 3x1/64 =3/64. 

[2]  Probabilities for courses can be constructed where there is a significant difference in hole scores between the second best score and the score of the fourth player.  A course where net bogey and net double bogey are the likeliest scores will favor the foursome.  A course where net eagle and net birdie are the likeliest scores would favor the threesome.  Neither course is likely to exist in the real world.