By mustang6560 on 3/25/11

I'm starting to think I'm the only golfer in the world without an ace.

We've heard about two brothers acing the same hole back-to-back and we've heard about a husband and wife both record an ace during a round. But, how about three golfers acing the same hole on the same day and two of them were playing in the same group? Check.

It all happened on the 137-metre par-3 14th hole. Dragan Milosevic was the first to achieve the feat, playing in the morning group. Marking a triumphant "1″ on the card, Dragan was, needless to say, extremely confident he had sewn up the "Nearest the Pin" prize for the day.

But two afternoon players, Lance Robinson and Lindsay Howard, quickly dashed Dragan’s hopes, as they each scored aces of their own on the hole, achieving every golfer’s dream and giving the club an unprecedented three aces in one day.

As an expert statistician, I have calculated the odds of this happening. If the odds of making a hole in one for an average handicapper is 12k to 1, according to the National Hole-In-One Registry, and the odds of two players from the same foursome acing the same hole is 17 million to 1, then all you have to do is multiply these together and voila! The odds of three players acing the same hole on the same day and two of them are in the same group are 2.04e+11.

OK, kidding aside. Interestingly enough, the odds of a Tour player (3k to 1) acing a hole are only four times better than me acing a hole (12k to 1). Considering how much better they are, I'm surprised the odds aren't higher for me. One of these days, I'm going to get that allusive ace!



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