Hole in one odds from Yahoo news story

[dansalmo]

Two N.J. golfers get back-to-back aces
http://news.yahoo.com/s/ap/20070920/ap_o...nKtsRta8g6ek3QF

This story today in Yahoo had the follow claim about odds:

... The odds of a golfer scoring an ace: 5,000-to-1. But the odds of two players in a foursome performing such a feat [back to back] are 17 million to 1, according to a Golf Digest article in 2000.

....snip....


Assuming the 5000:1 is true (who really knows), the possible combinations of back to back, when two players do one after another, not any two in the group, can only happen 3 ways.

I do not see how they come up with the 17M:1 odds for this. Is this right or wrong?

DanS

[ncray]

5000:1 means 1/5001. Assuming independence gives (1/5001)^2 * (5000/5001)^2 * 3 = 25000000/208500050006667 = 1.19904x10^-7
17mil:1 = 5.88235x10^-8
Edit: there's no stipulation that the other 2 players can't also ace, but that just changes the 5000/5001 to 5001/5001, so there's hardly a difference

So if they didn't screw up the math, they're saying that one person getting a hole in one is not independent from another in a foursome. According to them, if one person aces, the next is less likely to than if he were just shooting alone.

[rufus]

ncray - shouldn't that be:
(1/5001)^2 * (5000/5001)^2 * 6 (about 2.2 * 10^-7 so 1 in 4 million)
since 4 chose 2 = 6

This works out to the average chance of a golfer in a 4-some of holing as somewhere around 1 in 10,000 instead.

[SheetWise]

[ QUOTE ]
ncray - shouldn't that be:
(1/5001)^2 * (5000/5001)^2 * 6 (about 2.2 * 10^-7 so 1 in 4 million)
since 4 chose 2 = 6

This works out to the average chance of a golfer in a 4-some of holing as somewhere around 1 in 10,000 instead.

[/ QUOTE ]
They were calculating the probability of two players in a foursome getting back-to-back aces. So, only three of the pairings qualify.

[jay_shark]

If we assume independence it should be :

(1/5001)^2*5000/5001 + 5000/5001*(1/5001)^2*5000/5001 +
(5000/5001)*(1/5001)^2 = 0.0000001199

ie ppq,qppq,qpp

[SheetWise]

[ QUOTE ]
If we assume independence it should be :

(1/5001)^2*5000/5001 + 5000/5001*(1/5001)^2*5000/5001 +
(5000/5001)*(1/5001)^2 = 0.0000001199

ie ppq,qppq,qpp

[/ QUOTE ]

Since the statement and question don't mention anything about the other two players, I dont see anything wrong with using -

(1/5001)^2*3 = 0.00000011995201

[mykey1961]

[ QUOTE ]
... The odds of a golfer scoring an ace: 5,000-to-1. But the odds of two players in a foursome performing such a feat [back to back] are 17 million to 1, according to a Golf Digest article in 2000.

....snip....


Assuming the 5000:1 is true (who really knows), the possible combinations of back to back, when two players do one after another, not any two in the group, can only happen 3 ways.

I do not see how they come up with the 17M:1 odds for this. Is this right or wrong?

DanS

[/ QUOTE ]

Part of the problem is the article doesn't claim 17 million to 1 is for [back-to-back], just 2 in a foursome.

DanS seemed to add the back to back part to the statistic for some reason.

[dansalmo]

I see the where the reporter got the 17M to 1 odds from, and you are correct, it is any two golfers in a foursome making an ace on the same hole.

http://golf.about.com/od/faqs/f/holeinoneodds.htm

That being said, I would think the the back to back thing is what he should have quoted since that was what whole story was about. So can anyone work out the odds of back to back, if any 2 in a foursome can make it 17M to 1?

[Copernicus]

neglecting more than two in the foursome making aces there are 6 ways for two holes in one to be made, three of which are back to back. so about 34 million to one..if the 17 million is right.

[Copernicus]

Found a quote of about 500 million rounds of golf played in the US in a year. With an average of 15 back to backs, thats odds of 33.3 million to one...amazingly close to the estimates from above.