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 [img]/images/smile.gif[/img]

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 [img]/images/smile.gif[/img]

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 [img]/images/smile.gif[/img]