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

[jay_shark]

Ncray has mentioned that in his post already since 5000/5001 is about 1 anyway so it won't change the probability much at all .

Mine is the exact answer btw .

[mykey1961]

Jay, Jay, Jay.

Exact answers after you make assumptions... shame.

Anyone who has golfed knows that they are not independent events.

Watching the path of the ball, both in the air and on the green provide information to those following. Also knowing which club was used, and how far that player usually hits that club helps.

[ncray]

Here's another possibility. They have a distribution for golfers' ace abilities. For example, take 4 golfers, A-D. For simplicity, assume that they golf in the order A, B, C, D. Take A's ace probability to be 1/2249.73, B's to be 1/11189.7, C's to be 1/15148.5, D's to be 1/5001.

Then (1/2249.73 + 1/11189.7 + 1/15148.5 + 1/5001) / 4 = 0.00019996 = 1/5001. So their average ace getting ability is 1/5001.

The odds of back to back aces (AB ace, BC ace, CD ace) are =
1/2249.73*1/11189.7 + 1/11189.7*1/15148.5 + 1/15148.5*1/5001
= 5.882327667117519*10^-8 = 1/17000001

This is just one of an infinite number of solutions. I highly doubt the writers of the Yahoo story were able to find a distribution of golfers' ace abilities and figure out the odds in the article.

[BruceZ]

[ 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 ]

[ QUOTE ]
Mine is the exact answer btw .

[/ QUOTE ]

Exact for what? If you are excluding cases where more than 2 players make aces, then you would need to exclude ppqp and qppp which are now being counted. No other case of 3 or more is currently being counted.