Golf & Probability Puzzle

• Jul 25th 2005, 07:20 AM
TurkeyRun
Golf & Probability Puzzle
Can someone help me with the following question:

In the 1989 US Open Golf Championship, 4 golfers made hole-in-one on the 6th hole.The probability of a professional golfer making a hole in one (on any given hole) is estimated to be 1/3709.

There were 155 golfers participating in the second round. Assuming the estimate given above, what is the probability that exactly 4 of the 155 golfers would get a hole-in-one on the sixth hole?
• Jul 27th 2005, 11:53 AM
mactev
Assuming each hole is independent from the next and that the probability of getting a hole in one remains constant from hole to hole, this question is simply a case of using the Binomial Distribution.

So the probability that exactly 4 of the 155 golfers got a hole-in-one on hole 6 is:

p(x)=(n/x) p^x (1-p)^n-x

where

(n/x)= n!/x!(n-x)!

so p(4)=0.0000001

It was basically a fluke!
• Jul 27th 2005, 12:43 PM
Rebesques
e?
Quote:

the probability of getting a hole in one remains constant from hole to hole
That never happens. If there's no luck at start, there will be none later. ..
• May 24th 2008, 01:09 PM
jessica44
I think this is just a sure luck for these 4 golfer. I mean they have to be like math geniuses to calculate their probability on that golf course. I really thinks this is just a mere coincidence.

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