Monday, May 21, 2018

Quota System Vs. Stableford System



The Quota System and the Stableford System are two common methods for tournament scoring.  In the Quota System a player is given a quota equal to 36 minus his handicap (e.g., a 15 handicap would have a quota of 21).  As shown in the table below, a player earns points based on his gross score on each hole.  A player’s tournament score is the total of points earned minus his quota.  For example, if a player earned 24 points with a quota of 21, his tournament score would be +3.  In the Stableford System, a player earns points on his net score as shown in the table. The total number of points earned is his tournament score.

Table
Points for Quota System and Stableford System

Quota System
Stableford System
Gross Double Eagle
5
Net Double Eagle
5
Gross Eagle
4
Net Eagle
4
Gross Birdie
3
Net Birdie
3
Gross Par
2
Net Par
2
Gross Bogey
1
Net Bogey
1
Gross Double Bogey
0
Net Double Bogey
0

It would seem a system based on gross scores would not necessarily produce the same winners as one based on net scores.  A closer analysis reveals that is not the case except in one instance.  (For simplicity, eagles and double eagles have been excluded from the proof.)
The Quota System is described by:

                 Eq. 1)           Q= 3∙(Xs + Xn) + 2∙ (Ps + Pn) +1∙(Bs +Bn) – (36-H)

Where,
Q= Quota Score
Xs = Number of Birdies on stroke holes
Xn = Number of Birdies on non-stroke holes
Ps = Number of Pars on stroke holes
Pn = Number of Pars on non-stroke holes
Bs = Number of Bogeys on stroke holes
Bn = Number of Bogeys on non-stroke holes

The Stableford System is described by:

Eq. 2)      S = 4∙Xs + 3∙Xn + 3∙Ps +2∙Pn + 2∙Bs + 1∙Bn + 1∙Ds + 0∙Dn + 0∙Ts

Where,
              S = Stableford Score
              Ds =   Number of Double Bogeys on stroke hole
              Dn = Number of Double Bogeys on non-stroke hole
              Ts = Number of Triple Bogeys on stroke holes

Now a player’s handicap must equal:

Eq. 3)     H = Xs +Ps + B+ Ds +Ts

Substituting equation 3) into equation 2 a player's Stableford Score is:

Eq. 4)    S = H + 3∙Xs + 3X+ 2∙Ps + 2∙Pn + 1∙Bs +1∙Bn + 1∙Ds + 0∙Dn + 0∙Ts

The difference between a player's Stableford score and his Quota score is shown in equation 5):

              Eq. 5)    S – Q = H + 3∙(Xs + Xn) 2∙(Ps + Pn) + 1∙(Bs +Bn) – (3∙(Xs +Xn) + 2∙(Ps + Pn) + 1∙(Bs +Bn) + 1∙Ds +       
                                              Ts - (36 – H)

                      = 36 - Ts

In summary, a player’s Stableford score will be 36 points higher than his Quota score minus the number of holes where a player scored a triple bogey on a stroke hole.  The appendix gives an example that includes the case where a player’s score is his adjusted gross score minus his handicap.  Where no player scores a triple bogey with a stroke (i.e., Ts=0) all three scoring system yield the same order of finish and margin of victory. 

When one player has a triple bogey instead of a double bogey on a stroke hole stroke (player D on the par 4 ninth hole), his Quota score  remains unchanged.  His Stableford score is reduced by one point and his adjusted score minus his handicap is increased by one.  The Quota System gives a slight edge to players more likely to have a triple bogey.

 Which scoring system should be preferred?  Since a player must post his adjusted score, a player’s adjusted score minus his handicap would seem to be the simplest.  Under the World Handicap system, the Adjusted score minus handicap will always produce the same results as the Stableford System.  A player’s adjusted score on a par 72 course is ((72 + (36-Stableford Points) + Handicap). Therefore, the Stableford System is acceptable since it is easy to use and generates a players adjusted score.  It is not possible to translate a Quota Score to a player’s adjusted score.  This means a player should also keep his adjusted score as well as his Quota Score. Having to keep two sets of scores should eliminate the Quota System from consideration.

Appendix
Scores Without Triple Bogeys with a Stroke

Adjusted Score – Course Handicap

P
1
2
3
4
5
6
7
8
9
Tot
Score
A(8)
5*
4
5*
5*
6*
5*
4*
6*
5*
45
37
B(6)
5*
4
5*
3
5*
4
5*
5*
5*
41
35
C(6)
5*
5
6*
3
5*
6
4*
6*
4*
44
38
D(7)
5*
4
6*
2
5*
3*
4*
6*
6*
41
34
S
Stableford

P(H)
1
2
3
4
5
6
7
8
9
Score
A(8)
2
1
3
1
1
2
3
2
2
17
B(6)
2
1
3
2
2
2
2
3
2
19
C(6)
2
0
2
2
2
0
3
2
3
16
D(7)
2
1
2
3
2
4
3
2
0 20

Quota

P(H)
1
2
3
4
5
6
7
8
9
Score
A(8)
1
1
2
0
0
1
2
1
1
9-10=-1
B(6)
1
1
2
2
1
2
1
2
1
13-12-=1
C(6)
1
0
1
2
1
0
2
1
2
10-12=-2
D(7)
1
1
1
3
1
3
2
1
0
13-11=2


Scores with Triple Bogeys with a Stroke
(Player D on hole No. 9)

Adjusted Score – Course Handicap

P(H)
1
2
3
4
5
6
7
8
9
Tot
Score
A(8)
5*
4
5*
5*
6*
5*
4*
6*
5*
45
37
B(6)
5*
4
5*
3
5*
4
5*
5*
5*
41
35
C(6)
5*
5
6*
3
5*
6
4*
6*
4*
44
38
D(7)
5*
4
6*
2
5*
3*
4*
6*
7*
42
35

Stableford
Stableford
P(H)
1
2
3
4
5
6
7
8
9
Score
A(8)
2
1
3
1
1
2
3
2
2
17
B(6)
2
1
3
2
2
2
2
3
2
19
C(6)
2
0
2
2
2
0
3
2
3
16
D(7)
2
1
2
3
2
4
3
2
0
19

Quota

P(H)
1
2
3
4
5
6
7
8
9
Score
A(8)
1
1
2
0
0
1
2
1
1
9-10=-1
B(6)
1
1
2
2
1
2
1
2
1
13-12=1
C(6)
1
0
1
2
1
0
2
1
2
10-12=-2
D(7)
1
1
1
3
1
3
2
1
0
13-11=2



4 comments:

  1. It should be worth noting that the rank order might not be the same if using the Modified Stableford Scoring System, since the scoring is not linear. The scoring system for MSS is:

    • Double Eagle: 8 points
    • Eagle: 5 points
    • Birdie: 2 points
    • Par: 0 points
    • Bogey: -1 point
    • Double Bogey or more: -3 points

    ReplyDelete
  2. We frequently play in groups of 3 somes and 4 somes and play a “team quota” . At the end of the matches, should we divide by 3 or 4 to determine winners or use the actual + or - from total quota?

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  3. Glad to chat your blog, I seem to be forward to more reliable articles and I think we all wish to thank so many good articles, blog to share with us. Grumpy Gopher

    ReplyDelete
  4. I would use the “team average” in deciding a winner. Under the World Handicap System, a player has about a 20 percent chance of having a positive quota score—i.e., having a net score below par. Therefore, on average, a foursome would have a lower team score than a threesome. For example, assume all players have a symmetrical quota score distribution with a mean of -2. The average team quota score from a large sample of threesomes would be -6. The average team quota score for a large sample of foursoursomes would be -8. That is, adding a fourth player decreases the team quota score. If the team score is adjusted by the number of players, the average team score would be -2 for both threesomes and foursomes which would be fair.
    Since net scores vary widely, the bias against foursomes may not be apparent in small samples. You might want to keep track of individual quota scores to see if the assumption about the average quota score is valid for your group.

    ReplyDelete