What's the probability of multiple holes-in-one?
September 21, 2009 11:44 AM Subscribe
Some guy recently got 3 holes in one on the same course (different holes) within 5 days. My cousin contends that the odds against this are 5 billion to one, based on something he read. He's willing to have me prove him wrong, but this is well beyond my ability to calculate. I'm good with any time period -- the likelihood of someone doing this in a year, or 10 years, or whatever.
Part of the problem is that he is convinced that the likelihood of a golfer getting even a single hole-in-one is minute. I found a statistic saying that it's around 12,000 to 1 for a round of golf, or possibly for each hole. But even given that, I have no idea how to estimate how many people play at least 3 games on the same course in any given 5 day period, especially since these periods can overlap.
I'd prefer to use stats for amateur only golfers, but really anything vaguely reasonable is fine -- if I understand the caculations I can sub in any other numbers that my cousin prefers.
Part of the problem is that he is convinced that the likelihood of a golfer getting even a single hole-in-one is minute. I found a statistic saying that it's around 12,000 to 1 for a round of golf, or possibly for each hole. But even given that, I have no idea how to estimate how many people play at least 3 games on the same course in any given 5 day period, especially since these periods can overlap.
I'd prefer to use stats for amateur only golfers, but really anything vaguely reasonable is fine -- if I understand the caculations I can sub in any other numbers that my cousin prefers.
A big problem is going to be agreeing on how to calculate the odds.
This article has a few different sources for HIO stats.
posted by edgeways at 12:01 PM on September 21, 2009
This article has a few different sources for HIO stats.
posted by edgeways at 12:01 PM on September 21, 2009
Basically, you need to know how many par-3 holes the golfer played over the five days. The odds of getting 3 holes-in-one will then be given by the binomial distribution. This assumes that all holes-in-one are independent events with equal probability.
Let's say your golfer played 18 holes on each of the 5 days, and that the course has 4 par-3s. This gives us 3 successes in 20 trials, with the probability of each success 1/12,500. The total probability is then about 5.83 E -10, or about 1 in 2 billion.
posted by mr_roboto at 12:12 PM on September 21, 2009
Let's say your golfer played 18 holes on each of the 5 days, and that the course has 4 par-3s. This gives us 3 successes in 20 trials, with the probability of each success 1/12,500. The total probability is then about 5.83 E -10, or about 1 in 2 billion.
posted by mr_roboto at 12:12 PM on September 21, 2009
Response by poster: Mr. Roboto, I don't know how many games he played over that time -- I cannot find a story about this. But the relevant thing is really how many people play at least 3 rounds in any given 5 day period and get at least 1 hole-in-one per round. My feeling is that there are sufficient people playing that much golf that even if we agree that 1 in 100,000 shots gets a hole in one, we get a higher likelihood that over the course of a year or 5 years or something the probability of *someone* doing it -- certainly significantly more likely than 1 in 5 billion. But my gut isn't going to convince my cousin.
Estimating the numbers is part of the problem here: I have no idea where to start with figuring out how many amateurs play golf at least 3 times in a week. I have no clue how to deal with the overlapping time periods either.
posted by jeather at 12:14 PM on September 21, 2009
Estimating the numbers is part of the problem here: I have no idea where to start with figuring out how many amateurs play golf at least 3 times in a week. I have no clue how to deal with the overlapping time periods either.
posted by jeather at 12:14 PM on September 21, 2009
Response by poster: Sure, 1 in 2 billion for this specific guy to get 3 holes in one on those specific 5 days. That's entirely reasonable as a probability, but it's not what I'm looking for. I am trying to find out what the probability of any random golfer getting 3 holes in ones during any random 5 day period within the course of a year, say -- or even during any given 5 day period, what's the likelihood that someone, somewhere, has gotten 3 holes in one?
posted by jeather at 12:22 PM on September 21, 2009
posted by jeather at 12:22 PM on September 21, 2009
Well, it looks like there are about 500 million rounds of golf played in the US each year, by about 26 million golfers. This gives us about 20 (to make it a nice round number) rounds per person per year. If each round is an 18-hole round at a course with 4 par-3s, that means an average of 80 hole-in-one attempts per person per year. Plugging this into the binomial distribution, the chance of an average golfer getting three holes-in-one in a year is about 4.2 E -8, or about one in 20 million.
posted by mr_roboto at 12:25 PM on September 21, 2009
posted by mr_roboto at 12:25 PM on September 21, 2009
Sure, 1 in 2 billion for this specific guy to get 3 holes in one on those specific 5 days. That's entirely reasonable as a probability, but it's not what I'm looking for.
No. One in two billion for any golfer playing 20 par-3s over any period of time.
posted by mr_roboto at 12:26 PM on September 21, 2009
No. One in two billion for any golfer playing 20 par-3s over any period of time.
posted by mr_roboto at 12:26 PM on September 21, 2009
This seems to say that there's a 1 in 5000 chance of a player making a hole in one in a given round. The source is a Ph.D. mathematician, and the information is from Golf Digest, as reported by the NYTimes, so it's about as definitive as you're likely to get. It assumes that there are 4 ace-able holes in a given 18 hole round. It also sez that the likelihood of three aces in a single round is 1 in 2 trillion.
The thing to remember is that previous events do not effect, statistically, the likelihood of future events. If you flip a perfect coin 9 times, and it is a head each time, it doesn't mean that you're more likely to get a tail on the next flip. However, before you start flipping, the likelihood of 10 heads in a row is exceedingly small.
posted by Damn That Television at 12:30 PM on September 21, 2009
The thing to remember is that previous events do not effect, statistically, the likelihood of future events. If you flip a perfect coin 9 times, and it is a head each time, it doesn't mean that you're more likely to get a tail on the next flip. However, before you start flipping, the likelihood of 10 heads in a row is exceedingly small.
posted by Damn That Television at 12:30 PM on September 21, 2009
Plugging this into the binomial distribution, the chance of an average golfer getting three holes-in-one in a year is about 4.2 E -8, or about one in 20 million.
To take this a step further, since there are over 20 million golfers in the US, odds are better than even that someone gets 3 holes-in-one every year. For 26 million golfers, the chances of someone getting 3 holes-in-one in a single year is about 66%.
posted by mr_roboto at 12:31 PM on September 21, 2009
To take this a step further, since there are over 20 million golfers in the US, odds are better than even that someone gets 3 holes-in-one every year. For 26 million golfers, the chances of someone getting 3 holes-in-one in a single year is about 66%.
posted by mr_roboto at 12:31 PM on September 21, 2009
The thing to remember is that previous events do not effect, statistically, the likelihood of future events. If you flip a perfect coin 9 times, and it is a head each time, it doesn't mean that you're more likely to get a tail on the next flip.
Playing golf isn't exactly flipping a coin though. The fact that this guy has a sufficiently long and accurate drive to make a hole-in-one at all puts him well ahead of the vast majority of other golfers.
That's the problem with trying to calculate probability for non-random events. You can create some high level statistics, but they're meaningless in the face of the individuals actually performing the task. There's a significant amount of luck in getting a hole-in-one, but there's a big skill element as well, so the fact that this guy was capable of making one put him in a lot better position to make two more.
posted by jacquilynne at 12:47 PM on September 21, 2009
Playing golf isn't exactly flipping a coin though. The fact that this guy has a sufficiently long and accurate drive to make a hole-in-one at all puts him well ahead of the vast majority of other golfers.
That's the problem with trying to calculate probability for non-random events. You can create some high level statistics, but they're meaningless in the face of the individuals actually performing the task. There's a significant amount of luck in getting a hole-in-one, but there's a big skill element as well, so the fact that this guy was capable of making one put him in a lot better position to make two more.
posted by jacquilynne at 12:47 PM on September 21, 2009
The odds vary from person to person. This isn't the lottery. Playing golf takes skill and talent.
Tiger Woods' or Phil Mickelson's odds of doing this are much smaller than the average golfer.
So you cannot make a blanket statement and simply say: "The odds of getting 3 holes in one are x."
posted by Zambrano at 12:59 PM on September 21, 2009 [1 favorite]
Tiger Woods' or Phil Mickelson's odds of doing this are much smaller than the average golfer.
So you cannot make a blanket statement and simply say: "The odds of getting 3 holes in one are x."
posted by Zambrano at 12:59 PM on September 21, 2009 [1 favorite]
The odds vary from person to person. This isn't the lottery. Playing golf takes skill and talent.
Tiger Woods' or Phil Mickelson's odds of doing this are much smaller than the average golfer.
Which is why the mathematical construct you referred to as "the average golfer" is useful for addressing questions like this. It is certainly possible to draw conclusions about how this theoretical average golfer is expected to perform.
posted by mr_roboto at 1:04 PM on September 21, 2009
Tiger Woods' or Phil Mickelson's odds of doing this are much smaller than the average golfer.
Which is why the mathematical construct you referred to as "the average golfer" is useful for addressing questions like this. It is certainly possible to draw conclusions about how this theoretical average golfer is expected to perform.
posted by mr_roboto at 1:04 PM on September 21, 2009
If you want to assume that there's a 1/2B (:=alpha) chance of a golfer hitting those 3/3, and that all 26M (:= beta) golfers (who average 20 games a year) have one eligible attempt, then the chance that nobody hits a 3/3 is (1-alpha)^beta. I'll ignore terms smaller than (alpha*beta)^2, and approximate the result with a taylor expansion around alpha=0. You get the probability of at least one success out of 26M eligible attempts at about 1.27%
Substitute any number of eligible attempts you want, as long as it's less than 100M the same result will hold, number/2B -2(number/2B)^2. As the number gets closer to 1/2B you need to include more terms.
posted by a robot made out of meat at 1:28 PM on September 21, 2009
Substitute any number of eligible attempts you want, as long as it's less than 100M the same result will hold, number/2B -2(number/2B)^2. As the number gets closer to 1/2B you need to include more terms.
posted by a robot made out of meat at 1:28 PM on September 21, 2009
Which is why the mathematical construct you referred to as "the average golfer" is useful for addressing questions like this.
Not really. Not without knowing more about the distribution of holes-in-one among all golfers; knowing the average figure alone is not sufficient to calculate the overall probability.
Demonstration: consider two countries, X and Y, each of which has only two golfers, X1 and X2; and Y1 and Y2.
Let's say X1 and X2 each have a 1/100 chance of getting a hole-in-one in a round of golf (and they never get more than one in a single round; perhaps the only course in the country has only one par-three hole). Thus, the "average golfer" in country X has a 1/100 chance of getting a hole-in-one in a round of golf. It is not difficult to calculate that X1 and X2 each have a .0000098506 chance of getting three or more holes-in-one in five rounds of golf, and one can say that the average golfer in country X has a .0000098506 chance of getting three or more holes-in-one in five rounds of golf.
Now, let's say Y1 has a 1/50 chance of getting a hole-in-one in a round of golf (and again, never gets more than one in a single round), while Y2 never gets a hole-in-one. Again, the average golfer in country Y has a 1/100 chance of getting a hole-in-one in a round of golf. It turns out Y1's chance of getting three or more holes-in-one in five rounds of golf is .0000776192, and of course Y2's chance of the same is zero. So the average golfer in country Y has a .0000388096 chance of getting three or more holes-in-one in five rounds, nearly four times the probability of the average golfer in country X doing the same.
So if you are told only that an "average golfer" has a certain probability of getting a hole-in-one, that is insufficient information to determine the probability, averaged over all golfers, of getting a series of holes-in-one across multiple rounds. Simply assuming that every golfer has the same equal probability of making a hole-in-one will result in a probability of a series of holes-in-one that is too low.
posted by DevilsAdvocate at 1:37 PM on September 21, 2009
Not really. Not without knowing more about the distribution of holes-in-one among all golfers; knowing the average figure alone is not sufficient to calculate the overall probability.
Demonstration: consider two countries, X and Y, each of which has only two golfers, X1 and X2; and Y1 and Y2.
Let's say X1 and X2 each have a 1/100 chance of getting a hole-in-one in a round of golf (and they never get more than one in a single round; perhaps the only course in the country has only one par-three hole). Thus, the "average golfer" in country X has a 1/100 chance of getting a hole-in-one in a round of golf. It is not difficult to calculate that X1 and X2 each have a .0000098506 chance of getting three or more holes-in-one in five rounds of golf, and one can say that the average golfer in country X has a .0000098506 chance of getting three or more holes-in-one in five rounds of golf.
Now, let's say Y1 has a 1/50 chance of getting a hole-in-one in a round of golf (and again, never gets more than one in a single round), while Y2 never gets a hole-in-one. Again, the average golfer in country Y has a 1/100 chance of getting a hole-in-one in a round of golf. It turns out Y1's chance of getting three or more holes-in-one in five rounds of golf is .0000776192, and of course Y2's chance of the same is zero. So the average golfer in country Y has a .0000388096 chance of getting three or more holes-in-one in five rounds, nearly four times the probability of the average golfer in country X doing the same.
So if you are told only that an "average golfer" has a certain probability of getting a hole-in-one, that is insufficient information to determine the probability, averaged over all golfers, of getting a series of holes-in-one across multiple rounds. Simply assuming that every golfer has the same equal probability of making a hole-in-one will result in a probability of a series of holes-in-one that is too low.
posted by DevilsAdvocate at 1:37 PM on September 21, 2009
Yeah, you can't say "these are the odds". Period. Everyone is at a different skill-level.
It's a sport. It's not picking balls out of a cylinder for the lottery.
So, as Joey on Friends says: "The question is moo."
posted by Zambrano at 3:10 PM on September 21, 2009
It's a sport. It's not picking balls out of a cylinder for the lottery.
So, as Joey on Friends says: "The question is moo."
posted by Zambrano at 3:10 PM on September 21, 2009
Dudes. With 26 million people playing golf in the US, and tons of statistics gathered, it's totally possible to predict (with some quantifiable level of confidence) the frequency of holes-in-one. This is the nature of statistics. The question is about the likelihood that any golfer will hit a series of holes-in-one in a given amount of time. The tools we have at hand can answer this question!
Do you doubt that I could predict the number of grand slams that will be hit during a season of MLB? I mean, of course I won't hit the number right on the head, but I could give you a range (100-200 or so), that I'd be pretty damn confident about. This is despite the varying skill levels of professional hitters and pitchers.
posted by mr_roboto at 3:23 PM on September 21, 2009
Do you doubt that I could predict the number of grand slams that will be hit during a season of MLB? I mean, of course I won't hit the number right on the head, but I could give you a range (100-200 or so), that I'd be pretty damn confident about. This is despite the varying skill levels of professional hitters and pitchers.
posted by mr_roboto at 3:23 PM on September 21, 2009
There's no argument that you can predict the frequency of holes-in-one, or that it would help to solve the clarified question: "I am trying to find out what the probability of any random golfer getting 3 holes in one during any random 5 day period within the course of a year, say -- or even during any given 5 day period, what's the likelihood that someone, somewhere, has gotten 3 holes in one?"
The argument is whether he and his cousin can truly resolve the original question, which is not the same. Can you use statistics taken from a pool of golfers ranging from those who are lucky to ever hit a green to those who routinely knock it stiff to get a meaningful idea of how much this one particular guy of unknown skill level "defied the odds" when he shot three holes-in-one in 5 days? The odds are nowhere near the same for a guy who hits less than 50% of his par-three greens as they are for a scratch golfer (which I bet this guy was).
posted by gimli at 6:15 PM on September 21, 2009
The argument is whether he and his cousin can truly resolve the original question, which is not the same. Can you use statistics taken from a pool of golfers ranging from those who are lucky to ever hit a green to those who routinely knock it stiff to get a meaningful idea of how much this one particular guy of unknown skill level "defied the odds" when he shot three holes-in-one in 5 days? The odds are nowhere near the same for a guy who hits less than 50% of his par-three greens as they are for a scratch golfer (which I bet this guy was).
posted by gimli at 6:15 PM on September 21, 2009
With 26 million people playing golf in the US, and tons of statistics gathered, it's totally possible to predict (with some quantifiable level of confidence) the frequency of holes-in-one.
No one here is disputing that.
The question is about the likelihood that any golfer will hit a series of holes-in-one in a given amount of time. The tools we have at hand can answer this question.
If the "tons of statistics" includes not merely the overall average hole-in-one likelihood, but the distribution over all golfers, then yes, we can. If not, then we can't.
Do you doubt that I could predict the number of grand slams that will be hit during a season of MLB? I mean, of course I won't hit the number right on the head, but I could give you a range (100-200 or so), that I'd be pretty damn confident about.
No, I do not doubt that. That would be akin to predicting the total number of holes-in-one among all players in a given number of rounds played, which, I agree, you could do knowing nothing more than the overall average hole-in-one rate among all golfers. But that is not the question at hand; the question at hand is akin to predicting how many players will hit three or more grand slams in a given season, and merely assuming that all baseball players have an identical probability of hitting a grand slam in any single at bat, equal to the overall average, underestimates the true likelihood due to varying skill levels.
posted by DevilsAdvocate at 6:28 PM on September 21, 2009
No one here is disputing that.
The question is about the likelihood that any golfer will hit a series of holes-in-one in a given amount of time. The tools we have at hand can answer this question.
If the "tons of statistics" includes not merely the overall average hole-in-one likelihood, but the distribution over all golfers, then yes, we can. If not, then we can't.
Do you doubt that I could predict the number of grand slams that will be hit during a season of MLB? I mean, of course I won't hit the number right on the head, but I could give you a range (100-200 or so), that I'd be pretty damn confident about.
No, I do not doubt that. That would be akin to predicting the total number of holes-in-one among all players in a given number of rounds played, which, I agree, you could do knowing nothing more than the overall average hole-in-one rate among all golfers. But that is not the question at hand; the question at hand is akin to predicting how many players will hit three or more grand slams in a given season, and merely assuming that all baseball players have an identical probability of hitting a grand slam in any single at bat, equal to the overall average, underestimates the true likelihood due to varying skill levels.
posted by DevilsAdvocate at 6:28 PM on September 21, 2009
But that is not the question at hand; the question at hand is akin to predicting how many players will hit three or more grand slams in a given season, and merely assuming that all baseball players have an identical probability of hitting a grand slam in any single at bat, equal to the overall average, underestimates the true likelihood due to varying skill levels.
You're totally right about this, of course. Can we make any reasonable assumptions about the shape of the distribution? Can we estimate how much a skew in the distribution will affect our probability predictions? The case you sketched out above, with two populations of two golfers each, is clearly an extreme case... the real distribution of hole-in-one likelihood will be much more uniform. How uniform does it need to be to make the assumption that all golfers have an equal likelihood of hitting a hole-in-one at each attempt "reasonable"; i.e. to get the answer right to within an order of magnitude?
My intuition is that it's probably a fine order-of-magnitude assumption, but we all know what intuition is worth in arguments about statistics....
posted by mr_roboto at 6:48 PM on September 21, 2009
You're totally right about this, of course. Can we make any reasonable assumptions about the shape of the distribution? Can we estimate how much a skew in the distribution will affect our probability predictions? The case you sketched out above, with two populations of two golfers each, is clearly an extreme case... the real distribution of hole-in-one likelihood will be much more uniform. How uniform does it need to be to make the assumption that all golfers have an equal likelihood of hitting a hole-in-one at each attempt "reasonable"; i.e. to get the answer right to within an order of magnitude?
My intuition is that it's probably a fine order-of-magnitude assumption, but we all know what intuition is worth in arguments about statistics....
posted by mr_roboto at 6:48 PM on September 21, 2009
Here's what I'd use as a starting point to get the shape of the distribution: handicap data. I'm having a hard time finding well-sourced data, but most of what I find looks very normal. So I'd start with the assumption that likelihood of hitting a hole-of-one scales (as a linear inverse of?) handicap, model handicap as a Gaussian distribution across all golfers (probably just drop off the hacks at > +35 completely), and see how this alters the probability predictions. Lots of assumptions, now, but we can work with ranges to see how sensitive our estimates are to the numbers we use.... And this officially just became too much work for me.
What the uniformity assumption does give us is a floor to the estimate. So over a year, the absolute minimum probability of some golfer in the US hitting 3 holes-in-one is at least 66%, right? The probability only gets higher once we take varying skill levels into account. I think this is already a surprisingly large number, and might support the point jeather is trying to make with his cousin.
posted by mr_roboto at 7:03 PM on September 21, 2009
What the uniformity assumption does give us is a floor to the estimate. So over a year, the absolute minimum probability of some golfer in the US hitting 3 holes-in-one is at least 66%, right? The probability only gets higher once we take varying skill levels into account. I think this is already a surprisingly large number, and might support the point jeather is trying to make with his cousin.
posted by mr_roboto at 7:03 PM on September 21, 2009
And one more conclusion: if we assume that all golfers have equal probability of getting a hole-in-one (I know), and the 26 million golfers in the US play at a constant rate across all days of the year (I know, I know), the probability of some golfer hitting three holes-in-one over a five day period is about 0.018%, or about 1 in 6000. This is a minimum number, and the probability will increase once varying skill levels are taken into account and clustering of games in summer and over the weekends is considered.
I think this is the answer to the question as asked.
posted by mr_roboto at 7:25 PM on September 21, 2009
I think this is the answer to the question as asked.
posted by mr_roboto at 7:25 PM on September 21, 2009
Eh. That assumes that everyone only ever plays at the same course, though.
posted by mr_roboto at 7:27 PM on September 21, 2009
posted by mr_roboto at 7:27 PM on September 21, 2009
Some guy recently got 3 holes in one on the same course (different holes) within 5 days. My cousin contends that the odds against this are 5 billion to one
It's already happened, so the probability is 1.
posted by pompomtom at 9:16 PM on September 21, 2009
It's already happened, so the probability is 1.
posted by pompomtom at 9:16 PM on September 21, 2009
Besides all the other debate swirling around; golf being a game of skill it would seem likely that the probability of the 3rd hole in one on the same course with three days is more likely than the first was.
IE: a driving range with a single static hole is going to see a lot more holes in one (if you take the sample from the people actually going for it) than a course with a hole at the same range.
posted by Mitheral at 6:39 AM on September 22, 2009
IE: a driving range with a single static hole is going to see a lot more holes in one (if you take the sample from the people actually going for it) than a course with a hole at the same range.
posted by Mitheral at 6:39 AM on September 22, 2009
This thread is closed to new comments.
posted by mr_roboto at 11:57 AM on September 21, 2009