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.

[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.

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