Golfers defy 17 million-to-one odds with back-to-back holes in one

Two golfing friends have defied odds of 17 million to one by achieving the sport’s holy grail of a hole in one in consecutive shots.

Steve Wilmshurst, who is 58, and his 70-year-old playing partner Liam Nairn achieved the extraordinary feat on Monday while playing the 16th hole at the Studley Wood Golf Club in Oxfordshire.

Wilmshurst and Nairn were playing as part of a foursome and the National Hole-In-One Registry quotes the odds of two players achieving that feat on the same hole as 17 million to 1. An individual player is quoted at 12,000 to one to make a hole in one.

The odds, at the start, of two consecutive shots being a hole in one would be 12,000×12,000. One in 144 million.

But, well, some holes might get a hole in one, others definitely won’t. So, conditional one the first one going in the second is more likely – it’s possible to do a hole in one on this hole that is.

17 million to one sounds weird though.

And of course, once the first one went in then the odds of the second doing so are one in 12,000.

If only we still had that yacht floating around to explain this to us…..

Quoting a single odds for a hole-in-one on a par 3 or any hole is utter nonsense.

The odds of getting a hole-in-one on a 100 yd par 3 is going to be a lot lower lower than on a 250 yd par 3. And even then the odds of getting a hole-in-one on a 100 yd par 3 that is up a steep hill with a big bunker in front of the green are going to be a lot higher than on a 100 yd hole that slopes gently downhill and there are no obstructions between tee and green.

And then there’s the ability of the golfer. The odds on a professional or low handicap golfer getting a hole-in-one on any hole is going to be a lot lower than, say, a 36 handicap golfer.

And that’s before we start considering the weather and pin position on any given day.

Depends on many things, like the skill of the players. My odds would be considerably longer as I’d be likely to hit the ball in completely the wrong direction. I’d be reliant on a seagull catching it mid-air and dropping it in the hole upon realising it wasn’t a chip.

Ah.

Minor bit of Googling gives odds of 12,500-1 for an amateur, but 2500-1 for a pro.

I assume they’ve adjusted those “base” odds according to the handicaps of both Mr Wilmshurst and Mr Nairn.

Although I think it still looks a bit odd, as same hole, different players, each hole-in-one is still an independent event, shirley?

Without googling, what do you call a hole in one on a par 5 ie four under par?

It’s a real thing, apparently, though for obvious reasons incredibly rare.

BiG – we would be pretty evenly matched and I’d suggest we gave it a try except that it would be unreasonable to rely on so much avian intervention.

Tim – “once the first one went in then the odds of the second doing so are one in 12,000” but only if you weren’t looking at the odds of two holes in one in the first place which you were. This is what is technically known as pulling a fast one.

BiG, must be pretty brainy seagulls on your course. Not many can tell the difference between a chip and a drive…

How many times have two or more golfers in a group played a whole together?

17 million times wouldn’t seem a lot, so why hasn’t this happened before?

“ Although I think it still looks a bit odd, as same hole, different players, each hole-in-one is still an independent event, shirley?”

It’s like coin tossing.

Tossing a coin twice and the chance of each being a head is 0.5 but the chance both coming up heads is 0.25.

Ducky: not quite independent, as the length etc as BiND articulated, plus conditions on the day, will affect both players more or less equally. However I expect those things are less important than the individual strokes played. Sqrt(17M) is about 4123 so that doesn’t seem unreasonable for individual odds.

@interested – it’s a Condor (without googling).

@ Ducky – I recall a few years ago someone making a lot of money betting on there being a hole in one during a particular major tournament. The bookies hadn’t taken into account there being 140 of the best golfers in the world each having 16 (if they made the cut) or 8 chances to get a hole in one over 4 days. That represented around 1,700 attempts. At one in 2,500, The odds should have been less than 2-1 and the bookie had given something silly like 50-1.

Don’t see your logic there BiND. Since it’s a Hx1 what’s in front of the green’s irrelevant. May even be an incentive to try to carry to the apron. And up hill course actually makes it easier. The ball’s carrying less energy when it lands.. Doesn’t bounce through.

Some decades ago in fact. I recall talking to a bookie about it just after it happened. He said “Good Luck to ’em. Of course, it’ll never work again…..”

Could the 1/12000 be the probability of getting a hole in one in a game?

Then the probability of getting two consecutive holes in one would be greater* than (1/12000)^2 because there’s more than one chance you’ll get consecutive holes in one. E.g., could happen on hole 1 then hole 2, or could happen on hole 2 then 3, etc.

*working out how much greater is left as an exercise for the reader. It’s a long time since I last used the Binomial Theorem and I cannae be ersed to look up the formula.

The golf punters were known as the hole on the wall gang. They concentrated on small independent bookmakers placing apparently innocuous stakes which became in their own words “cluster bombs”when 2 or more of the 33/1 shots came off. Many refused to pay, some went out of business. It was a well-planned and well executed completely legal strategy.

2 holes in one? Hold my beer says Kim Jong Il.

https://golf.com/news/behind-kim-jong-ils-famous-round-of-golf/

The important statistic that’s been neglected is the record of those getting Hx1’s on this hole. And more than anything else, that’s the thing will determine the odds.

The 1 in 17M suggests this should have happened about 1,000 times in the last 50 years based on some pretty basic assumptions about the number of golfers & rounds of golf. 1 in 144M seems much more reasonable odds given actual reports of this happening.

As noted by others, 1 in every 144M shots should result in consecutive holes in one; very different to the odds of any given shot being part of consecutive holes in one.

Tossing a coin twice and the chance of each being a head is 0.5 but the chance both coming up heads is 0.25.That’s mathematically. Theory world. It assumes a coin toss is a random number generator. In the real world, it coming up heads in the first toss increases the chances of heads in the second. Because the coin may not be a random number generator.

If only we still had that yacht floating around to explain this to us…..Maybe it was banned after they followed Murphy’s recommendations that ‘noone needs a yacht’?

@ bloke in Spain.

No!

The chance of it coming up heads the second time is exactly the same as it coming up heads the first time. That may not be a probability of 50%, if the coin is not ‘fir’, but the coin does not have a memory, so what happened on the first toss cannot freer what happens on the second toss.

@Stuart Caldwell

Mathematics is not reality. It’s theory. It would be possible to make a coin came up heads more than 50% of the time. So you can’t ignore the possibility. The more important the outcome of the coin toss, the more the possibility. Because there’s more incentive to fix the coin.

One should never ignore that the world is not a machine. The people in it have agency.

That’s why I said the important aspect of the golf thing is to look at the history of the hole. The course designer may have designed a hole that favours Hx1s. So all your clever statistics go out the window.

@bis

Like the 16th at Augusta. I think it’s on the final day at the Masters that they always put the hole in a ‘bowl’ like depression toward the back of the green. The ball feeds naturally towards it as it runs from a number of areas on the green. Still f’ing difficult but easier than the hole was in an area where the ball ran naturally away from it.

@BiS, the coin-tossing example in statistics always specifies (or should always specify) ‘a fair coin.’ In practice, also, it should be tossed by a fair machine, not a human being.

In related matters, what are the odds in the true story of someone I knew, playing golf for the first time, who teed off on the first and got a hole in one on the eighteenth? He gave up golf at once, figuring that he was never going to beat 67 under par.

That’s nothing, how many times have you seen somebody pot the black from the break in a game of pool?

bis,

Don’t see your logic there BiND. Since it’s a Hx1 what’s in front of the green’s irrelevant. May even be an incentive to try to carry to the apron.Higher handicap players are much more likely to be getting Hx1 or even on to the green by running the ball on to the green rather than pitching it all the way to the green and landing just short of the flag and the ball rolling in to the hole.

And up hill course actually makes it easier. The ball’s carrying less energy when it lands.. Doesn’t bounce through.No, its much harder to judge club selection. We are allowed to use mearing devices that are quite accurate but not ones that tell you difference in height. On one of our holes depending on where drive lands I will be taking either 1, 2, or 3 extra clubs (that means instead of say a 9-iron I might have to take either and 8,7 or 6 iron depending on the elevation change). This makes it much harder to judge distance and control the ball and invariably I’ll be long or short. Same applies to a drive. Obviously the more you play the hole better the judgement.

Tossing a coin twice and the chance of each being a head is 0.5 but the chance both coming up heads is 0.25.In the real world, it coming up heads in the first toss increases the chances of heads in the second. Because the coin may not be a random number generator.That’s mathematically. Theory world. It assumes a coin toss is a random number generator.

Oh come on, you know that’s a theoretical example of probabilities given to explain the problem rather than claim that you’ll get 2 perfect coins and the toss will always be random.

Like Tim, I knew one of the betting shop owners who got taken to the cleaners by the Hole in One Gang. He took the bet because he was skint and needed cash in the till. When the bet was landed, he did a runner and never paid out. He took to living in an L-reg Peugeot in various Midlands lay-bys. It wasn’t even his car, which is about as low as a man can get, when you think about it.

I did see him once more, years later, when he was booking a hotel suite for a sale of Oriental carpets. You just can’t keep men like that down.

Oh come on, you know that’s a theoretical example of probabilities given to explain the problemRule of life that’s done me very well. Ignore the theories & deal with reality. The amount expensive disasters I’ve seen come out of what was theoretically correct are beyond number.

It’s like your Hx1 problem. You start with the history of the hole. Not try to statistically apply the history of all holes. This the one you’re interested in.

As for the difficulty of the hole, I bow to your superior knowledge. Personally I find golf pointless. Much exertion to achieve nothing. Game shooting I could go for. You get to eat the result.

My impression that an elevated green would be easier was from trying to apply a bit of physics. We start by presuming that the golfer can actually appreciate the green is elevated & allow for it. The amount of energy put into the shot will be subject to error. That translates into the velocity the ball is travelling when it arrives at the green. It’s going to be the same margin of error whatever the arrival velocity. Velocity is expressed as distance. In the trajectory, the ball loses energy to gravitational potential in the up ward arc & gains energy from gravitational potential in the downward arc. Reducing energy gained in the downward arc because it’s shorter means lower velocity. Since the margin of error is constant, the resultant distance from the error at the lower velocity is less.

@Steve, I see that your grasp of probability theory is as good as your grasp of geopolitics.

I believe the 1/12k odds are for a single shot and based on all amateurs/all par 3’s. So, if this was a short & easy par 3 that # could be halved (WAG). Also, could be lowered by their skill level & familiarity with the hole. If you were betting on both of them getting a hole in one prior to either taking a shot it’d be 1/144M (assuming average golfers & par 3). But, once the first golfer aces it’s down to 1/12k.

Bayes’ theorem applies here – once the first ace happens the odds of two in a row drops to 1/12k. (you’ve already got the first).

The difference between:

A – odds the next two players make a hole in one

B – odds that player 2 makes a hole in one (since player 1 already did, we don’t have to factor his odds into this)

Esteban, that’s for the individual second shot.. The odds at two Hi1 happening consecutively on the same hole is still P^2.

With P being the chance based on all shots

on that particular hole, by amateurs, over time, under varying circumstances.You need large numbers of attempts to get a significant answer on Student’s T….. And details in collating your data matter, hence the emphasis and the excessive amount of commas..

Can’t use any of the fancier methods in statistics either, since

allof them either assume a relatively consistent environment, or assume a method of compensation for variations in that environment.(PTSD flashbacks to my student years where we had to

proveour experimentsdidprovide that “consistent environment”andaccounted for any variables we might not have thought of if we dared to use those fancy statistical methods.“The Dragoon” (the faculty head, prof. dr. dr. Lubsen, adressed as Hetty ( and don’t you dare use anything else…)) was a true lesbo old-style feminist (proper fugly butch) , a brilliant scientist in her own right, and a

very stricttaskmaster. We loved and hated her in equal measure… A rare gem you only learn to appreciate later in life. 😉 )Let’s say that it really is 1 in 12,000 to get a hole in one, and that all the golfers are going to land their approach on the putting surface (simplistic but we’ve got to simplify).

The ball is 42.7mm wide, so will drop into any cavity 85mm wide either side that it rolls over slowly.

Just assume that balls travel up to 5 metres on landing – they will traverse an area of roughly 0.4 of square metre that they could drop in.

Let’s assume that 1 in 12,000 is about the correct odds, then the area of the green should be 12,000 * 0.4 sq m, or 5000 sq metres after rounding, or a circle with radius 39 metres.

Well no, that’s stupid, golf greens typically have radii of 15 metres or so, so I’m calling bullshit on that 1 in 12,000 number – a lower number, perhaps 1 in 2,000 is more realistic.

Apols if it’s already been mentioned but – 2 successive hole-in-ones generally requires two par threes in succession. which doesn’t happen so often (#). That shifts the odds considerably purely by itself.

[ # – unless it’s the local pitch and putt. 🙂 ]

Forget it – I misread it!

3:01 pm:

4:33 pm:

Not quite up to Murphy levels, at least those two sentences aren’t in the same paragraph!

Most comments here are about the

actualodds and the course. This is, I believe, irrelevant to the question at hand: how did the writer of the article get from 12k:1 for once to 17M:1 for twice…For 4 people playing, the chances of two of them getting a hole-in-one,

assuming the 12k:1 stat is correctwould be one in 12000^2 * (11999/12000)^2 / ( 4C2 ) — yay A-level statistics. The odds of the other players not getting the hole-in-one I’m going to ignore from now on as a rounding error. The important thing is the 4C2, i.e. there are 6 ways it could be done: AB, AC, AD, BC, BD, CD and any of those would fulfil that particular requirement so the chances are 6 times greater than just if two people out of two had both done it. This gives us odds of 24M:1 vs 17M:1 which is pretty close for journalists.The actual statement was that two successive players got holes in one. This is harder because there are only 3 ways it could be done: AB, BC, CD. This gives us odds of 48M:1.

Now, of course, I agree with all you blokes and others that such a broad statistic can’t be used to generate an actual probability, and the probability of a successive hole-in-one would not be statistically independent of the first (i.e. the easiness of the hole, likelihood of favourable wind, ground conditions, etc. are probably higher than normal, this ain’t no perfectly-fair coin flip).

Grikath – not sure anything you wrote differs from what I did.

BTW, I find it amusing that this topic is driving so much discussion, this feels like a combo of Seinfeld and Big Bang Theory.

So, to recap my understanding, the odds that Bob makes a HIO on this swing = 1/12k. The odds that Jerry also makes a HIO on the next swing = 1/12k, regardless of Bob’s result. So, if Bob already made a HIO and Jerry is up next the odds of back-to-back are 1/12k.

OTOH, the odds that Bob makes one immediately followed by Jerry = 1/144M.

All of the other caveats about difficulty of the hole, their skill levels and familiarity apply, of course.

@BiW

You can’t? I’ve done quite a bit of shooting where the difference between the trajectories of downhill, flat & uphill are marked & I don’t have any trouble. There’s also good old F=mv³/2 to be taken into consideration because of drag.

But bunny usually ends up in the pot.

Well, yes, except the purpose of the post itself was to be able to make the joke about the name of the yacht that just sank – Bayesian.

Bugger, Tim!

Bugger, bugger, bugger!

Too clever by half there!

Right over my head, even though, like Drax, I have excellent reflexes.

The physics from BiS is impressive, but the ability of a skilled player to impart spin goes a long way to slowing a ball down.