You are here:

- Home
- Science
- Mathematics
- Probability & Statistics
- golf probability

Advertisement

Not sure if you're a golf fan, but let's say prior to the final round of a 72 hole tournament, Player A is leading at -2, Player B is second at even par, and Player C is in third at +2. No one else is remotely close so assume only the 3 leaders have a chance. Assuming all 3 leaders have exactly equal skill, what is the probability that each will win the tournament and how would you calculate this?

According to Golf by the Numbers, a "cautious" golf will have approximately the following probabilities:

P1(1)=negligible

P1(2)=0.01

P1(3)=0.21

P1(4)=0.52

P1(5)=0.20

P1(6)=0.05

P1(7)=0.01

P1(8+)=negligible

A risky golfer has:

P2(1)=negligible

P2(2)=0.04

P2(3)=0.28

P2(4)=0.34

P2(5)=0.20

P2(6)=0.09

P2(7)=0.03

P2(8)=0.02

P2(9+)=negligible

The leading golfer (A) would play a conservative strategy (presumably) and the trailing golfer would probably need to play the risky strategy. The golfer in the middle might play the average of the two strategies.

Assuming par is 4 (this is irrelevant to the score, but gives us something to compare), we can compute the expected outcome for each player. For example, player A (playing the first set of probabilities, the cautious strategy) will have expected score:

-2 + (-2)×P1(2) + (-1)×P1(3) + (0)×P1(4) + (1)×P1(5) + (2)×P1(6) + 3×P1(7)

= -2 + (-2)×0.01 + (-1)×0.21 + (0)×0.52 + (1)×0.20 + (2)×0.05 + 3×0.01

= -1.9

Likewise, the expected outcome for C (playing risky) is:

+2 + (-2)×P2(2) + (-1)×P2(3) + (0)×P2(4) + (1)×P2(5) + (2)×P2(6) + 3×P2(7) = 2.19

This is called "risky" because the average outcome is actually worse (losing 0.19 on average) than the "cautious" strategy (which loses 0.1 on average). However, the average is not what golfer C is concerned with -- he's already four behind. He wants the best possible chance of passing B or even A.

You can compute the same for golfer B, playing the average of the two strategies, to get 0.145.

But, as I mentioned, the average does not tell the whole story. Golfer C has the worst score and his average is even worse compared to the others -- but a risky strategy has benefits because he is also more likely to get a very high score compared to the others (even if it is still unlikely) on this hole. To compute each probability, you have to break down all the cases. For example, there is the case that they all shoot par, which has probability:

P1(4)×P2(4)×(P1(4)+P2(4))/2 = 0.076024

You might just call this:

P(4,4,4) = 0.076024

where P(A,B,C) is the probability of getting that score for A, B, C. So in order for golfer A to lose, we would have to count all the cases where A+2>B or A+4>C, since A is 2 points ahead of B and four ahead of C.

There are many other cases, and we can compute them all:

P(2,2,2) = 0.00001

P(2,2,3) = 0.000098

P(2,2,4) = 0.000172

....

P(4,4,4) = 0.076024

....

P(6,6,6) = 0.000315

....

Now, that's not really going to help unless we sort out these cases. We have to decide who wins and add up those probabilities. In order for C to win, we must have C<A+4 and C<B+2, which leaves us with options like:

P(7,5,2) = 0.000085

There is also the possibility of a tie:

P(6,5,2) = 0.000425

That would have A and C tied at even, and B losing at +1.

Now, I won't list all the possible combinations, but there are really a few possible outcomes:

Using only the non-negligible outcomes (2-8), there are only three ways to get this:

P(6,4,2) = 0.000425

P(7,5,3) = 0.00049

P(8,6,4) = 0

So P(A=B=C) = 0.000425 + 0.00049 + 0 = 0.000915

This is the most likely outcome, as we should suspect, since it agrees with the current standings. Any score where A+2<B and A+4<C will win for A.

One such outcome, for example, is:

P(4,3,2) = 0.00364

Adding up all of these (143 outcomes) gives us 0.70275

You can repeat this for all the possible end-results to get:

P(A<B<C) = 0.70275

P(A<B=C) = 0.1026

P(A=B<C) = 0.094835

P(A<C<B) = 0.067447

P(B<A<C) = 0.028418

P(B<A=C) = 0.002017

P(A=B=C) = 0.000915

P(A=C<B) = 0.000768

P(C=B<A) = 0.000085

P(B<C<A) = 0.00008

P(C<A=B) = 0.00005

P(C<A<B) = 0.000035

P(C<B<A) = 0

And that is, at least roughly, a good idea of how things stack up. Keep in mind that these are all based on these distributions P1 and P2. However, you can see that by far, the most likely winner is A. The probability of A winning (without a tie for first place) is:

P(A<B=C) = 0.1026

P(A<C<B) = 0.067447

P(A<B<C) = 0.70275

That's 87.2797% probability of winning without a tie. A could also win with ties, which gives an additional contribution of:

P(A=B<C) = 0.094835

P(A=B=C) = 0.000915

P(A=C<B) = 0.000768

which brings the total to 0.872797 + 0.094835 + 0.000915 + 0.000768 = 0.969315

You can break it down as follows:

P(A wins, not tied) = 87.2797%

P(A wins, incl ties) = 96.9315%

P(B wins, not tied) = 3.0515%

P(B wins, incl ties) = 12.635%

P(C wins, not tied) = 0.0085%

P(C wins, incl ties) = 0.1853%

And that's probably the best way to look at it. The probability that C wins is an incredible longshot, roughly one in 500 (0.2% rounding generously in his favor) and it is even more unlikely that he win alone (not tied). B has a small chance (3%) to upset A and win (without tying) and relatively good chances (12.5% is one in eight) of winning by tie.

- Add to this Answer
- Ask a Question

Rating(1-10) | Knowledgeability = 10 | Clarity of Response = 10 | Politeness = 10 |

Comment | No Comment |

I can answer all questions up to, and including, graduate level mathematics. I do not have expertise in statistics (I can answer questions about the mathematical foundations of statistics). I am very much proficient in probability. I am not inclined to answer questions that appear to be homework, nor questions that are not meaningful or advanced in any way.

I am a PhD educated mathematician working in research at a major university.**Organizations**

AMS**Publications**

Various research journals of mathematics. Various talks & presentations (some short, some long), about either interesting classical material or about research work.**Education/Credentials**

BA mathematics & physics, PhD mathematics from a top 20 US school.**Awards and Honors**

Various honors related to grades, various fellowships & scholarships, awards for contributions to mathematics and education at my schools, etc.**Past/Present Clients**

In the past, and as my career progresses, I have worked and continue to work as an educator and mentor to students of varying age levels, skill levels, and educational levels.