Making an ace to halve a hole

[Shindig]

Not me;  I still haven't had my first eagle, including but not limited to aces.

That puts me at least two behind this twosome:

http://www.pga.com/news/golf-buzz/pair-scots-sink-aces-same-hole-in-match-play

[iacas]

Wow.

Jerks.

Quote

According to the National Hole-In-One Registry, the odds of an average golfer sinking an ace are 12,000 to 1. The odds of two players in a foursome acing the same hole? That’d be 17,000,000 to 1.

I'm probably brain farting - or maybe these are just what the empirical data has shown over the many years and it's not mathematical at all, really - but how does 12,000:1 for one player (one chance) become 17,000,000:1 when two golfers out of four have to make it?

 

[David in FL]

That's a tough skin to have cut!

[Shindig]

49 minutes ago, iacas said:

Wow.

Jerks.

I'm probably brain farting - or maybe these are just what the empirical data has shown over the many years and it's not mathematical at all, really - but how does 12,000:1 for one player (one chance) become 17,000,000:1 when two golfers out of four have to make it?

 

IF there's something mathematical to it -- and I don't know enough about the numbers to say if there is or isn't -- it might mean the events are not independent.  For example, I suppose an ace on a 150-yard par-3 is more likely than one on a 200-yard par-3 due to the tighter dispersion of shots hit into it.  But I don't know if that's true or not.  However, this might mean that "given that someone just aced this hole, it's more likely we're on a hole where such events are more likely, and thus the chances you also make an ace are increased."

[allenc]

1 hour ago, iacas said:

Wow.

Jerks.

I'm probably brain farting - or maybe these are just what the empirical data has shown over the many years and it's not mathematical at all, really - but how does 12,000:1 for one player (one chance) become 17,000,000:1 when two golfers out of four have to make it?

 

Assuming it is 12000:1 for a hole in one, I believe they did the math wrong.  It should be approximately 24,000,000:1 that two players make a hole in one.  Approximating the probability of NOT making a hole in one as 1, it's 1/12000 * 1/12000 * 1 * 1 * 6.  There are 6 combinations of two players who could do it.

[iacas]

Just now, allenc said:

Assuming it is 12000:1 for a hole in one, I believe they did the math wrong.  It should be approximately 24,000,000:1 that two players make a hole in one.  Approximating the probability of NOT making a hole in one as 1, it's 1/12000 * 1/12000 * 1 * 1 * 6.  There are 6 combinations of two players who could do it.

Right.

Or, the first player is 1 in 12,000. But the last three are 3 in 12,000 (because each has a 1 in 12,000 chance, but any of the three of them can count, so 3 in 12,000). So it's 3 in 144,000,000 or 1 in 48,000,000?

[dennyjones]

That is so cool.  

I played in a 4 person scramble at Hawk Hallow this year.   The unusual part of this was the hole was 8" and placed in very unusual positions on the green.   There was quite a bit of prize money available.   The group in front of us got a hole in one on 17 and thought they had a skin.   When the awards were passed out, they didn't get a skin, there was a second hole in one on the same hole.   

[David in FL]

11 minutes ago, dennyjones said:

That is so cool.  

I played in a 4 person scramble at Hawk Hallow this year.   The unusual part of this was the hole was 8" and placed in very unusual positions on the green.   There was quite a bit of prize money available.   The group in front of us got a hole in one on 17 and thought they had a skin.   When the awards were passed out, they didn't get a skin, there was a second hole in one on the same hole.   

Imagine that.  A scramble with an unusual, and unlikely occurrence...  

[allenc]

5 hours ago, iacas said:

Right.

Or, the first player is 1 in 12,000. But the last three are 3 in 12,000 (because each has a 1 in 12,000 chance, but any of the three of them can count, so 3 in 12,000). So it's 3 in 144,000,000 or 1 in 48,000,000?

Almost, except you're missing the possibilities of aces by players (2,3), (2,4), and (3,4).  So it's 6 in 144,000,000, not 3 in 144,000,000.  So 1 in 24,000,000.

[iacas]

8 hours ago, allenc said:

Almost, except you're missing the possibilities of aces by players (2,3), (2,4), and (3,4).  So it's 6 in 144,000,000, not 3 in 144,000,000.  So 1 in 24,000,000.

Yeah… any way I figure it, I can't get 1 in 17M, unless it's empirical or factors in something like "he had to aim at the hole…" or "perhaps this hole location is a bit more accessible" or something.

[allenc]

1 hour ago, iacas said:

Yeah… any way I figure it, I can't get 1 in 17M, unless it's empirical or factors in something like "he had to aim at the hole…" or "perhaps this hole location is a bit more accessible" or something.

Since there one ball in the hole it might be harder to make the second because it could cause it to bounce out if you slam dunk it.   Not sure if that's true.  But that would make the odds longer rather than shorter.

[Golfingdad]

Considering that it was a match, they should have calculated the odds of both players in a twosome making a hole in one, since the other two players in the foursome don't matter as much.  Also, the story doesn't mention that anybody else was playing with them either.

[allenc]

Here we go:

Quote

What is the chance that a golfer will make a hole in one?

• Tour player making an ace: 3,000 to 1
• Low-handicapper making an ace: 5,000 to 1
• Average player making an ace: 12,000 to 1
• Two players from the same foursome acing the same hole: 17 million to 1
• One player making two holes-in-one in the same round: 67 million to 1

Maybe the answer is all four player aren't average golfers.  Although 1 low handicap and 3 average player still doesn't come out to 1 in 17 million.  Perhaps the average foursome has 3.5 average players and 0.5 low handicappers or whatever it needs to be.  You know, like the average family has two and a half kids.

5 minutes ago, Golfingdad said:

Considering that it was a match, they should have calculated the odds of both players in a twosome making a hole in one, since the other two players in the foursome don't matter as much.  Also, the story doesn't mention that anybody else was playing with them either.

Definitely.  But they didn't calculate it.  They looked it up on a website and those were the odds the website provided.

[StefanUrkel]

Wasn't there a story of a guy making like 3 holes in one in a round?  Or like 2 double eagles?   Must be a squirrel having some fun with them

[Buckeyebowman]

Reminds me of the time I was invited to play Trumbull Country Club. There was a gang of high rollers, maybe 5 or 6 foursomes, going out who invited us to be part of their "game". I told my buddy it might not be a good idea since I had never laid eyes on the course before, so we begged off.

We caught up with them in the Grill Room, and as it turned out 3 of them had aced the same short par 3! I actually shot a half decent score, but I couldn't keep up with these guys!

[iacas]

10 hours ago, allenc said:

Since there one ball in the hole it might be harder to make the second because it could cause it to bounce out if you slam dunk it.   Not sure if that's true.  But that would make the odds longer rather than shorter.

It could also make it more likely to keep the ball in the hole by acting as a dampener.

[NCGolfer]

Came here to read a story but got a math lesson!  Bonus.

Closest thing I've ever seen to this is in a tournament I was playing...one group had a guy leave it on the lip for closest to the pin.  He had to mark the ball and leave the metal marker on the side of the green.  In the next group a guy one-hopped it in for an ace.