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)

(2)

(3)

(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

2^{nd} 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.
[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.
No comments:
Post a Comment