If you had average score & standard deviation for players scores could you calculate for each player the odds of winning a 3 day club championship golf tournament with no handicaps?

Sure. Details will depend on assumptions made.

I assumed a normal distribution of independent scores . Since the sum of normal variables is a normal, each player over 3 days has a normal mean score of 3*A and variance of 3*V, where A and V are his average and variance for one round. This assumes score independence within a player (day to day) and between players, which may not be true.


The following link addresses this problem of a winner among n players.

http://www.untruth.org/~josh/math/normal-min.pdf

It suggests simulation but gives the analytic method also. I programmed a simulation in Excel VBA. Took about 25 lines. I bet in a program like R it can be done in one or two lines after data are entered or read in.

It looks like the simulation correlated very well with the calc. I would be interested in running a simulation for the 10 players, how would I go about that if I am not literate in programming?

Feel free to post the 10 players' single round averages and standard deviations.

I am very confident that at least one person will run a simulation of millions of 3-day tournaments to derive good estimates of the probabilities that each of the 10 players will win the tournament.

(If nobody else volunteers in a timely manner, I'd be happy to step up.)

Thank you! Below is the average index & St Dev for each, the winner would be the player with lowest cumulative index over 3 rounds.

Avg St Dev
A 1.92 1.98
B 1.71 3.41
C 2.46 2.57
D 2.43 2.13
E 3.43 1.84
F 3.27 2.63
G 2.17 2.66
H 4.11 2.86
I 4.98 3.16
J 2.20 2.83

Based upon the player information you provided, here are the results of the simulation of 10 million 3-day tournaments.

Player	Win Pct
A	13.9
B	28.8
C	10.5
D	7.8
E	0.9
F	4.2
G	15.1
H	1.9
I	0.9
J	16.0

Hope that helps.

That is really cool, thank you so much!

Just to confirm, I got essentially the same results. The question I have is whether a symmetrical distribution like the normal appropriate for golf scores.

Would an average 85 golfer be more likely to shoot 80 or 90 occasionally?

This is a great question, could this be determined by studying lots of data and see the results or is this an assumption that must be made? My personal feeling is variance is more likely towards higher scores than lower, but they may be a personal bias or tendency.

Well, going a little theoretical here. A golf score is essentially a sum of random variables, namely the score on each hole The central limit theorem states that the sum of independent random variables is approximately normally distributed under some very general conditions. This suggests to me that for professional golfers we can assume normality. It should also work for amateurs but with the same bias as you had, I suspect a right-skewed distribution is more likely based on my own duffer experience.

Certainly, if you had the data, simply put it in a frequency distribution (histogram) and plot it to see visually if there is skewness and its direction. There are also tests for skewness that can be applied.(Google "tests for skewness")

Something has puzzled me about golf scores for a while and I wonder if anyone has an explanation.

Bookmakers don't differentiate between close matches and one-sided matches when pricing up the odds of both players scoring the same number. My own sampling - comparing 100 matches with one player greater than a 6/4 dog and 100 matches with both players less than 5/4, shows that the bookmakers appear to be correct as both samples had 11 ties.

This would indicate to me that there is an equal probability of scoring over quite a wide range rather than a normal distribution?

One reason for ties being a bit common, is the discrete nature of a golf score over what I would consider a relatively narrow range for professionals, say 62 to 80. However, for OP’s question, we are dealing with averages or sums, which makes the underlying random variable more applicable for a normal.

In fact, if you are correct is assuming a uniform distribution for a particular round, you can show that the sum of 12 scores can well be represented by a normal. So, if OP’s sample averages were based on 12 or more rounds, the normal distribution would be a good model to use. I used this approximation to generate normal variates in the simulation program I ran. Not sure how whosnext generated normal rv's.

Yeah I suppose a more uniform distribution of single rounds does make sense. I am still puzzled that oddsmakers appear to ignore the different skill level in their tie prob calc. You would think that the reason that one player is favourite over another is that he has more low scores in his distribution and these are highly unlikely to lead to ties (as are the high scores of the inferior player). Perhaps it is just that even with this case, almost all the actual scores are within both players distributions but the more skilled player is always slightly more likely to have a lower score.

Does this work? On a par 5 for examples a golfer can get a 4 or a 6. A 3 may be possible, as is a 7. Anything less though it essentially impossible, though higher than 7 will occur.

I'd think a player's average golf score would have a positive skew. Another way to think about it: there is certainly some best case round score for the golfer, but there is not as neat of a worst case score.

Does this make sense?

Yes, that's correct, but the central limit theorem says that the underlying random variables don't need to be normal, but the sum of them still is.

The part left out about the central limit theorem is that it needs a sufficiently large sample - I'm not sure whether 54 counts as sufficiently large.