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#1
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Golf-Probability Question
I alluded to this on another thread. In order to answer it accurately, one would have to have a pretty good idea about both the subject of golf and the subject of probability and statistics. The answer will get screwed up if you don't understand both. JUST LIKE SO MANY OTHER THINGS. AND GETTING THAT THROUGH ALL YOUR THICK SKULLS IS WHY I HAVE BEEN PUT ON THIS EARTH.
Where was I? Oh yeah, golf. Assume American golfers only. Rank them from one to ten million. Now assume a typical PGA men's tournament comprised of the top 150 golfers, all healthy, playing their best and trying to win. Not trying to get a minor check. 72 holes. You are given the opportunity to nominate a golfer to be the 151st player. You are asked to pick someone who is pretty steady in his performance. Not someone whos scores are all over the place. You are also asked to pick someone who is at least 50-50 in your mind to come in at least 140th. Say you have to bet on it. In order to have a good bet, the golfer you pick should be no worse than the xth best golfer in the country. What, approximately is x? |
#2
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Re: Golf-Probability Question
x = probably somewhere around 1000. There are lots of really good golfers, and the difference between them flattens out quite rapidly after the ultra-elite echelons. And golf is high variance; performance, even for those at the top, varies significantly from day to day, so it's not that hard for someone who is ranked far below you to beat you on any given day.
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#3
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Re: Golf-Probability Question
Some random thoughts I have about this topic:
The standard deviation from mean score among professional golfers seems pretty high. In a field of the 150 top American golfers over 72 holes on a moderately difficult championship course, I would expect (based on a typical week on the PGA tour) 140th place to be something like +18 or +20. That means we are looking for a golfer who is 50/50 to be better than +18 over 72 championship holes. I believe this is less people than one might think. The U.S. Open district qualifiers are a nice point of reference here. In order to participate, golfers must either be professionals or have a USGA handicap better than 1.4 . This means - for all intents and purposes people who are consistently scoring in the mid-70's and at their best scoring just over par. (Remember, handicaps aren't average scores, they are 'average bests' that golfers will fall short of 80% of the time in practice. Now if you go to a U.S. Open district quailfier (I've been to many), the people who qualify for the regionals are typically right around scratch during the district qualifiers. The rest of the field can drift WAY back. My estimate is that less than 25% of the field at a distrct qualifier breaks +10 over the 36 holes. Bottom line, I think we have to start by limiting our possible candidates to people who consider themselves better than scratch golfers. Lots of people who claim to be 2 or 3 handicaps could, in theory, finish better than +18 over 72 championship holes, but not at a 50% rate. No way. Another issue is the standard deviation. Because of the way handicaps are figured (average of 10 best rounds in last 20 rounds), they badly mask high variance. Among self-identified "plus golfers," you still have many people who could not turn in a +18 on a championship course over 72 at .5 probability. So I would shrink it to less than that group. I think we're looking for someone who is a solid plus 2 handicap, maybe a plus 3. I would suspect they would fit the bill. I have no idea how many plus 3 golfers there are in the U.S., but it can't be that many. Maybe five or six hundred. I know that is not the elegant or parsimonious answer you are looking for, DL, but i'll keep thinking on it. mg |
#4
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Re: Golf-Probability Question
Also, one note: I think I once read somewhere that the average PGA pro is likely to be about a plus 8 handicap on a typical moderately tough country club course from the men's tees. And since that's the frame of reference where almost all amateur golfers rate themselves, it's worth considering. I would estimate that "scratch" from the standard rating is something like a 5 or 6 from the championship tees at a tough course.
mg |
#5
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Re: Golf-Probability Question
[ QUOTE ]
The standard deviation from mean score among professional golfers seems pretty high. In a field of the 150 top American golfers over 72 holes on a moderately difficult championship course, I would expect (based on a typical week on the PGA tour) 140th place to be something like +18 or +20. That means we are looking for a golfer who is 50/50 to be better than +18 over 72 championship holes [/ QUOTE ] I agree. I also believe US Open or equally tough courses will be harder for the worse players. I think that on the easiest course 2-3 times as many players would make it. |
#6
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Re: Golf-Probability Question
So basically we just need a guy who can come in the top 93% of the field half of the time against the World's best 150 golfers.
Not knowing how to apply any specific mathematical formulas to this problem, but with what I consider to be a decent understanding of golf, I'd say that you can dip pretty far below the 150 mark to satify that condition. As someone already mentioned, the relative skill level of golfers once we're out of the extreme top really begins to smooth out. There's probably very little difference between the guy ranked 100th and the guy ranked 150th, and even less difference between 150 and 200. Just off the top of my head I'd say you could safely go down as far as 1500 and still have a 50% chance of him NOT finishing in the bottom 6.67% of the field, but I wouldn't be surprised if the number were much, much higher. I can't wait for Sklansky to break out some bell curves on this [censored]. |
#7
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Re: Golf-Probability Question
1000ish was my gut instinct, but it wouldn't completely shock me if the number was as low as 5000 or so.
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#8
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Re: Golf-Probability Question
I think there are 4 steps to answering this problem rigorously:
1)Estimate the necessary score required to finish 140th out of the 150 bets golfers in the U.S. over 72 holes on course X. 2)Convert this score to a handicap specific to course X, remembering that handicaps are not mean scores and do not handle variance the same way as we would like in this problem. 3)Translate that handicap to a handicap applicable on the average courses used by amateur golfers. 4)Determine how many amateur golfers can attain that hadicap. As I rambled through in my above post, I'm thinking we're talking about something below scratch here, maybe a plus 2 or plus 3 handicap. No idea how many people have that hadicap. mg |
#9
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Re: Golf-Probability Question
So far it seems that if errors have been made they are more along the lines of golf errors than math errors. That makes me happy.
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#10
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Re: Golf-Probability Question
[ QUOTE ]
all healthy, playing their best and trying to win. Not trying to get a minor check. [/ QUOTE ] that's not a reasonable assumption. if it were, the variance in scores would be much bigger than in your typical tournament. |
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