Hole in One

What are the approximate chances that Tiger Woods shoots a hole in one on a 150 yard hole. Use thought, not statistics.

My guess is that if he gets the angle right, he will get a hole in one 1/2 the time. I also think that it wouldn't be too unreasonable to assume a uniform distribution of left right error of -10 yards to +10 yards. Assuming his angle needs to be within a 2 inch region to hit the hole, I would get a probability of


2 in / (20yd *3*12) * 1/2 = 1/720.


PS: I don't play or watch golf /images/smile.gif

None of the courses he plays have a par 3 that's only 150 yards.


To arrive at a number (assuming that he tries the shot to begin with), you have to account for too many free variables. Wind, tiny objects on the ground that may make a hole in one impossible, the probability of the ball bouncing out of the cup (if it landed directly in it). Then you have to ask whether or not he'd be trying to hole the shot anyhow (maybe bunker placement makes it more advantageous to miss the cup by a fair amount, to land at a position where there is more green to work with.


All in all, I'd have to say that the probablility of him getting a hole in one is about the same as anyone else getting a hole in one, EVEN THOUGH he'll get closer to where he wants to be more often than someone else. It might even be less.


Look at poker. Terrible players drag more huge pots than professionals, as they find themselves contending over monster pots more often, as they play more hands. Bad golfers will always be aiming at the pin, even though they'll end up in a bunker more often than Tiger. So they'll hit hole-in-one's more often.


So I can't offer a number, but I'd lean (after some rambling thought that I verbalized) toward saying that an average golfer will make more of them as a function of number of holes of that distance played (not attempts at a hole in one) than he.


~D

Pros make more holes in one than hacks, percentage wise. I have seen stats on this, but will avoid them since Sklansky said not to. But do you see why your comparison to bad poker players winning more pots is flawed?


Some thoughts: Every poker pot must be awarded to somebody. Nobody need make a hole in one. The outcome of each poker hand is affected by random distribution than golf shots. For instance, 9-2 can beat AA a lot easier than a shank can go into the hole in golf. Sure, some holes in one are really lucky, like very poor shots that hit a fence or sprinkler or something. But many are pretty decent shots that happen to go in. Likewise, most excellent shots do not go in. But clearly, the better you hit a shot the more likely it is to go in. Maybe the more aggressive player who has decent skills will make more of them than a conservative skilled player, but no way does an aggressive bad player expect to make more of them than a good player.


One thing to think about is why some players make a lot of holes in one. Why did Art Wall make so many more than Ben Hogan? Wall was an excellent player, sure, but why so many? There are a few players out there who have made a lot more than their share. This may be luck.

I would estimate about 2000-1. Reasoning- if Tiger plays 250 rounds per year (He might play more, but I doubt he plays more than 275-300) he will play 4500 holes per year. Out of that, he can expect to see about 1000 par 3's.` No golfer expects to make more than 20 holes-in-one in a lifetime. 20 is a lot. For Tiger holes in one can occur on any par 3 and some par 4's. So I would say Tiger might expect to make a hole in one every two or three years, in all his play including practice rounds and rounds at his home course. So this would make his odds on any par 3 around 2000 or 3000 to 1. 150 yards is easier than his average par 3, so the chances would be better. But I don't think the chances on a 150 yard hole are so much greater even for a player like Tiger than they would be on any average par 3. Much better than on a brutal par 3 sure, but I doubt there is a huge difference between 150 and 200 yards in terms of the odds. So I would say around 2000-1, which might be worse odds than some expect. 2000-1 might actually be low.

I don't have alot of time right now, so instead of an answer I'll try to get inside your head and say what I think this is about. It's similar to estimating EVTH in that you have to decide how one variable depends on another, and once you do, you can make a very close estimate. In this case you would probably consider the relationship of probability to distance from the hole using the realization that area goes down as the square of distance, etc.


In offering the above statement I look great if I'm right, and lose little if I'm off, so it's positive EV to make this post as I see it /images/smile.gif

let's say he can pepper a five yards by five yards green at that distance


that's a 32,400 sq inches area


a hole is about 12.5 sq inches


so he will he will hole in one once in every 2600 shots


you might even say if he hits parts of a line one inch wide leading one yard towards him from the hole it will bounce in, so that gives him say another 4 square inches to aim at


so he might hole once in every 2000 shots

The first couple days of a tournament, I would say approximately zero, as Tiger looks to make a solid score and advance past the cut. I assume this style of play would have him making very solid, plodding shots, to maximize the chances of birdie/par, and minimize the chances of an extreme(hole in one/bogey). This style generally sees Tiger in the lead or near the lead at the end of a tournament, so continuing this style of play has a very high expected earn for him. So I suppose this means in general, the more skilled the player is, the greater benefit he gains from reducing randomness and high variation plays in his game. So Tiger should see fewer holes in one than a less consistent, more random golfer.


BQ

None of the courses he plays have a par 3 that's only 150 yards.


And if they did, I doubt they would put the pin where he could shoot at it.


If there was a 150 yard hole with the pin in the middle of the green, I like Mike Haven's solution.

Assume if the angle is right, and he hits the green after some point, he will miss. This point would be some distance before the hole, or else it would bounce over or roll over too fast to go in. Assume hitting anywhere from this point back to the edge of the green is OK, and if he misses the green it won't bounce onto the green and go in. Then the question is how often will he touch down between the edge of the green and this point? He would aim for a point half way from the edge to this point. Since he's an excellent golfer, this halfway point would be more likely than any other point. Assume the probability decreases symmetrically in either direction. Note it doesn't matter the shape of the probability distribution if we assume it's symetrical. All that matters is the chance he gets outside the ends. Let's assume there is a 10% chance he hits before the edge of the green, so then there would then be a 10% chance that he hits after the halfway point. Someone with more experience could give me a better estimate of how often he would hit the green. Then there would be an 80% chance that he would make a hole-in-one if the angle was right.


Now for the angle error. At this distance, angle error is essentially the same as left/right distance error. Hitting dead center would be more likely than anything else. I would think the shape of the probability distribution would be gaussian, but we we're not supposed to use statistics (or did he mean published statistics?). I'm going to assume it's gaussian anyway because that is easiest for me to think about; why should I turn my brain off just to get a less accurate answer? If you want, you can assume some simpler distribution. Anyway, let's say being off 10 yards is 2 standard deviations down which would mean there is a 4.6% chance of being off more than 10 yards in either direction. A standard deviation then would be 5 yards. I'll guess the hole is twice the diameter of a golf ball (I haven't looked at either for 20 years). Then we could miss by 1/2 golf ball diameter in either direction. I'll guess that's 1 inch. That would be +/- 1 inch or 1/36 of a yard or 1/(5*36) = 1/180 standard deviations. From a table of the gaussian distribution and some interpolation, the probability of this is about 0.2%.


Assuming distance error and angle error are independent, the probability of a hole-in-one is 80% * 0.2% = 0.18% or 18/1000. Of course I may only have a 50% confidence that this is 90% correct /images/smile.gif

if he made this hole in 1 every 500 shots he should give up golf and start carnival-work


your story reminds me of a joke


so there were this girl and guy looking over the side of the cruise-ship at sharks swimming below


she asked, "if i fell in there, would they eat me whole?"


he answered, "no - they'd leave that bit"

There are holes of approximately that distance that Tiger plays. And to say that his chance of making an ace on Thur. or Fri. is zero- well that is just ridiculous. I think the 2000 to 1 odds is overlooking 2 things. First, some shots on line but long will hit the pin and go in. Second, the ball rolling through those sq. inches at a slow enough speed will also fall in. Add to that the fact that he IS Tiger, I think about 900 to 1 is ball park. To fine tune it any further, other factors need to be known. ( firmness and contour of green, pin placement, wind speed and direction, etc.)

Speaking of carnival work, I think I was cheated at a recent state fair (big surprise). You had to shoot a star out of a piece of paper completely using a machine gun with limited bullets. As far as I could see I did it and left a huge hole, but when the guy hauled the card back in, there was a small piece of star left. Others have had the same experience. Last year I left that exact same little piece after thinking I shot the whole thing. The next day I remembered something. When the guy hauled the card in, he held up a new card in front of the old one so I couldn't see my target right away. I think he used slight of hand and swapped my card with a standard one with a little piece left.


I noticed that many games involve stars. Shoot out a star, hit a star with a dart, throw a football through a star shaped hole. Mathematically I believe stars maximize perimitter for a given area, or something similar to that. I suppose that makes the games look easier than they are.


In another game you had to place 5 circles so they completely covered a big circle. I was barred from this game when they caught me making some relative measurements.


Anyone know anything about beating any of these games and cheating methods? I know there isn't alot of money to be made in analyzing carnival games, but I would have liked to have won the giant dog. I had to settle for a smaller one, although I did win a giant snake in a pool game.

How often do you think he'll hit the green, and how tight will his shots be left to right 95% of the time?

I think 15 ft. left or right is app. correct. It will be a tighter pattern front to back.

From Scarne's complete guide to gambling..


..If there ever was a game of skill or science, Spot the Spot is it.......


...Experiments I have made show that chance and skill both play a role in Spot the Spot. The probability that a skilled player can drop a disk from a six-inch height onto the exact position on the red circle is about 1 in 3......This means that the probability that a player will make five perfect drops is 1 in 243......


...The operator is able to succeed in his demonstration because he has practiced a smooth,deceptive move. His hand sweeps quickly over he red spot in a fast back-and-forth movement which is confusing and hides the fact that the disk, at the moment he drops it, is only an inch or so above the layout........


...some operators also carry a slightly larger set of disks....

Assume that he can hit center of green about 40% of the time, center is a 20 ft circle around cup, and that any where in center is equally likely. Area of center is the square of (20*12/2)*pi. Assume cup is 3 inches in diameter, and has area 1.5*1.5*pi. Then probability of hole in one is area of cup divided by area of center, times the .40 chance that he is in the center. I get about 0.0000625.

this is a good prediction but i think your maths is slightly off


area of centre is 45257 sq ins and area of hole is 7.07 sq ins


so he holes 1 in 6401 hits of the green with probability of 0.015625 x 0.40 = 0.00625 or once in every 16,000 shots

The cup is 4 1/4 in. in diameter. I think your 40% estimate is low. And, a ball will finish in the hole more often than just the number of times it should stop within that diameter. If the hole were just a painted circle on the green, there would be many, many fewer holes-in-one than what there actually are. Balls hit the pin and fall in; balls rolling fast, medium and slow speeds sometimes find the bottom of the cup. Occasionally, a bouncing ball will land and stay in the hole.

Between 1987 and 1996 there were 323 aces on the PGA Tour. There were 443 tournaments. One ace every 1.37 tournaments. There are approximately 2,367 par 3s played in 1.37 events. (144 players x 8 before cut, + 72 x 8 after the cut= 1,728 x 1.37). So the odds of a PGA pro making a hole-in-one on a single shot were roughly 1 in 2,367. Add to this that the ave. par 3 is much longer than 150, and many (maybe most) of these shots were played having somewhere other than the hole as the target, and all of these shots were played by players of lesser talent than Tiger Woods, and you begin to get an idea of his chances of making it from 150.

i am not a real good golfer and i would bet i can make one at 100 yards in 500 trys.

The probability of you making the 1st try I would think are worse than 500 to 1. The probability of you making the 10th try may be better than 500 to 1.

[BruceZ]

I just noticed an error in my 500-1 estimate below. The probability of getting the angle right must be doubled to 0.4% (actually 0.44%) since the 0.2% was for being within an inch on only one side. This would make my final result 80% * 0.44% = 0.35% or 1 in 284.

Again this assumes we miss by over 10 yards in either direction 5% of the time, and we miss the green 10% of the time. Changing the percentage for hitting the green or changing the left-right error yardage (not percentage) will change the result by approximately a proportional amount. What would these numbers be for you Ray?

It seems to me, you could do well to make guestmates based on:

How often you assume the player will hit the ball onto the green.

How much surface area the ball will "test" before comming to a stop.

How much surface area the entire green is.

How big the hole is.

Tiger would have one chance to get a hole-in-one on each hole that he plays.