Hole in one odds from Yahoo news story

[SheetWise]

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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.

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So what? An unnecessary step introduces a small error so it's OK?

Then pick a situation where it introduces large error.

Let's say the foursome were shooting a six-sided die with the expecation of it landing on one. What's the probability that there will be back-to-back aces after each in the foursome roll the die once?

[jay_shark]

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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.

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I think that claiming that the odds is about 5000:1 to hit a hole in one is just as absurd as claiming that the events are independent .

[mykey1961]

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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.

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I think that claiming that the odds is about 5000:1 to hit a hole in one is just as absurd as claiming that the events are independent .

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You doubt Golf Digest?

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

[jay_shark]

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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

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Mine is the exact answer btw .

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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.

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I was excluding cases where three consecutive players hit aces. 3 players may hit aces , it just can't happen consecutively .

P(exactly 2)=P(ppqq,ppqp,qppq,qqpp,pqpp)=P(ppq,qppq,qpp)

If we were interested in the probability of at least two consecutive aces then you would include

P(exactly 3 ) =pppq,qppp
P(exactly 4) = pppp

But you know all this [img]/images/graemlins/smile.gif[/img]

[mykey1961]

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... 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

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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.

[pzhon]

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I think that claiming that the odds is about 5000:1 to hit a hole in one is just as absurd as claiming that the events are independent .

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You doubt Golf Digest?

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


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That link is quite a contrast to what we have seen on this thread.

[img]/images/graemlins/diamond.gif[/img] The calculation was not made by a journalist at Golf Digest. Golf Digest hired a mathematician. Mathematicians are not always right, but when we err, our errors usually make sense anyway. I expect that the mathematician either analyzed reams of data or came up with a mathematical model, perhaps determining the effective area of the resting point distribution and divided by the area swept out by a ball as it bounces and rolls.
[img]/images/graemlins/diamond.gif[/img] 5000:1 was for a round, not a hole. A round usually includes many par 3 holes.
[img]/images/graemlins/diamond.gif[/img] Not all par 3 holes are equally difficult. This seems to have been used by the mathematician if his probability of getting two holes in 1 is to be consistent with his probability of getting one, unless there are far fewer par 3 holes on average than I would guess.

[mykey1961]

From my experience a normal course 18 holes of par 3, 4 and 5's has either 2, or 3 par 3s, and the same number of par 5's

There are some 9 hole courses, and some of those that are all par 3's.

[DrVanNostrin]

If hole in ones are independant events (which they're not). The number of hole in ones per group, per par 3 should follow a Binomial(4, 1/5001) distribution. The probability of getting back to back aces would be:

P(4) + P(3) + 0.5*P(2)

(P(2) is multiplied by 0.5 because of the 6 ways to choose 2 from 4 only 3 would result in back to back hole in ones.)

=1.19936E-07 (which the same thing jay got to 4+ digits)

This results in being an ~8.3 million:1 dog.

The number 17 million:1 makes no sense. Since if one player gets an ace other players in that group should be more likely to get one. Among many other things already listed that makes it easier, getting a hole in one implies that the hole is easier to ace on.

[mykey1961]

17 million to 1 makes sense when you consider it's filtered through journalists.

The "raw data" showed approximately 528 million rounds, and 100 thousand holes in one.

The math guy estimated 4 potential chances for a hole in one per round

If we assume the "raw data" was reported correctly, the chance of a player making a hole in 1 would be 1 in 5280. As "journalists" we'll use 1 in 5000.

With the assumption that there are 4 chances per round to make a hole in one, each chance would be a 1 in 20,000 shot.

The chance that 2 out of 4 people would make a hole in one on a specific hole would be 1/20,000 * 1/20,000 * 19,999/20,000 * 19,999/20,000 * 6 = 1/66,673,334

Since there are 4 chances per round, the odds of it ocurring are 1/16,668,334

As "journalists" we'll report that as 17 million to one.

[Copernicus]

I dont know how many rounds are played in a year, but I do know that there are 12 to 18 "back to back" holes in one witnessed and reported every year, so somehow I think the odds above are way off, probably because they arent totally random events, but depend a lot on hole placement and conditions.

I know that little factoid because I was involved in one...second one in on top of my playing partners ball. We wrote to Golf Digest about it and that was their response.