[shcisamax]

My husband made a hole in one yesterday on the third hole. He thought it was funny because it was the third hole in one he has made in his lifetime....(last time was 25 years ago). When we got home, he realized all three of the holes in one have been on the third hole. What are those odds? AND...all three balls are #3 balls. What are those odds?

[Taltarzac725]

Originally Posted by shcisamax

My husband made a hole in one yesterday on the third hole. He thought it was funny because it was the third hole in one he has made in his lifetime....(last time was 25 years ago). When we got home, he realized all three of the holes in one have been on the third hole. What are those odds? AND...all three balls are #3 balls. What are those odds?

18 hole or 9 hole courses? How high do the ball numbers go? How often does your husband golf? http://golf.about.com/cs/beginnersgu...numbersonb.htm

[shcisamax]

2 were 9 hole courses and one was an 18. I think the balls go to 5....leave that part out of the equation. Although he golfs sporadically but figure once or twice a week.

[Dr Winston O Boogie jr]

Ball numbers go up to 8 and recently there have been some weird numbers produced by some companies. I would guess that there really aren't enough of these weird numbers to make an impact on any calculations.

[shcisamax]

What is the chance of getting three holes in one on the third hole. I guess you need to know how many times he has played....hmmmn. I guess the answer to the question would be really slim?

[gmcneill]

For your husband, the odds are 100% because he did it!!

For the rest of us, the odds are not as good.

My congrats to him! And may it be only 25 more rounds, not 25 years, before his next one.

[shcisamax]

Haha. So true. Actually, the first one was in '70, the second was 15 years later in '85, and now more than 25 years later, so at that progression, this may not work out for another.

[shcisamax]

Just looked up on NationalHoleInOneRegistry.com
An average player making an ace is 12,000:1
A hole in one is scored once every 3,500 golf rounds

More interesting facts if you are interested.

[blueash]

The odds of making a hole in one on a particular hole is one in how ever many holes are on that course. So for a nine hole course, it is one in nine (assuming you are going to make a hole in one on one of them, which is a given for the problem). Then the odds of the next one being on the same hole is one out of the same. We have to be careful because we have both 9 and 18 hole courses. So as long as his first hole in one was on hole one thru nine we can ignore that. So, he gets a hole in one on a hole 3. The chance it would be #3 is 1/9 The next time he gets a hole in one it is on an 18 hole course, so it is 1/18th likelihood to be on the same number. So that is 1/18 times 1/9 or 1/162 The next is on a nine hole course, which is 1/9 likelihood. So the odds of all three holes in one being the same 3 number hole is 1/1458 given your facts. Now, that it was a #3 ball, you multiply 1/1458 times 1/n with n= total number of choices for numbers on his balls (his golf balls). If he only plays with #1,2,3,4 then the chances for having used a #3 ball is 1/4, and he used it 3 times, so 1/4 times 1/4 times 1/4 or 1/64. We must assume that he does not subconsciously or intentionally always pull out a #3 ball when he gets to the third hole. So to have used a #3 ball on the same hole for 3 holes in one is now 1/1458 times 1/64 or 1/92,302 This ignores any factor for what are the odds he would even get a hole in one as that was not your question. All math done on a napkin, so please someone get out a calculator and make corrections as needed. If he has more or fewer options for ball numbers, the calculation of course changes.

[shcisamax]

But you haven't factored in the national registry fact that the odds of a hole in one is 12,000 to 1.

[CalcTeacher]

This isn't related to math at all...and I'm not a golfer so maybe I don't really know what I'm talking about...but don't holes have "pars" and isn't it really impossible to make a hole-in-one on a par 5 and a par 4? So that would only leave the par 3 to consider. Are 9-hole courses all par 3s? Maybe that's why it matters whether it was 9holes or 18holes?

[shcisamax]

The second hole in one was 165 yards so I assume it was a par 4 but I do know it was an18 hole course. Will calc help with this?

[Dr Winston O Boogie jr]

Originally Posted by blueash

The odds of making a hole in one on a particular hole is one in how ever many holes are on that course. So for a nine hole course, it is one in nine (assuming you are going to make a hole in one on one of them, which is a given for the problem). Then the odds of the next one being on the same hole is one out of the same. We have to be careful because we have both 9 and 18 hole courses. So as long as his first hole in one was on hole one thru nine we can ignore that. So, he gets a hole in one on a hole 3. The chance it would be #3 is 1/9 The next time he gets a hole in one it is on an 18 hole course, so it is 1/18th likelihood to be on the same number. So that is 1/18 times 1/9 or 1/162 The next is on a nine hole course, which is 1/9 likelihood. So the odds of all three holes in one being the same 3 number hole is 1/1458 given your facts. Now, that it was a #3 ball, you multiply 1/1458 times 1/n with n= total number of choices for numbers on his balls (his golf balls). If he only plays with #1,2,3,4 then the chances for having used a #3 ball is 1/4, and he used it 3 times, so 1/4 times 1/4 times 1/4 or 1/64. We must assume that he does not subconsciously or intentionally always pull out a #3 ball when he gets to the third hole. So to have used a #3 ball on the same hole for 3 holes in one is now 1/1458 times 1/64 or 1/92,302 This ignores any factor for what are the odds he would even get a hole in one as that was not your question. All math done on a napkin, so please someone get out a calculator and make corrections as needed. If he has more or fewer options for ball numbers, the calculation of course changes.

You're assuming that all nine hole golf courses have nine par 3s or nine holes that are reachable in one shot. Most nine hole golf courses have two par threes. some have only one and a few may have three or more. In the Villages we get used to calling nine hole par three golf courses, "Executive Courses". An executive course can actually differ from a par three course.
Most executive course consist of a mixture of par 3s, 4s and sometimes even short par 5s. The difference is that they are usually shorter holes and may contain more par threes than the standard.
While there is no official definition of an executive course, I would say that a course where the par fours are in the 250-350 yard range, par threes under 200 yards and par fives under 475 could be defined as an executive course.
Executive courses and par three courses could be nine holes or eighteen holes. I have played several 18 hole par three courses.
So another factor that has to be looked at is how many courses have a par three (or a reachable par four for some people) as their third hole.
NGF gives all kinds of statistics for golf. I think that say that the odds of a hole in one on any par three hole by all players is 1:12000. There is a different ratio for professionals and still another for tour players. I think that they also have the odds of a player making more than one hole in one in a lifetime. Your husband's feat is incredibly rare. The only thing I have ever heard close to that was a player playing eighteen holes on a par 36 nine hole course made a hole in on the sixth hole twice in one round. That was in the millions to one range and I believe that three in a lifetime is in the millions to one and your husbands feat may be in the billions to one.

[Dr Winston O Boogie jr]

Originally Posted by shcisamax

The second hole in one was 165 yards so I assume it was a par 4 but I do know it was an18 hole course. Will calc help with this?

165 yards would be a par 3 on any course. True most holes in one come on par threes. There are however, some par fours that can be reached in one shot by a minority of golfers. It is possible for this small group of long hitters to make a hole in one on these types of par fours.
If I'm not mistaken, calculus is a geometric function.

[Dr Winston O Boogie jr]

I am not a mathematician, but I think that the factoring would go something like this.

Odds on a hole in one by any player.
Odds of a player making three holes in one in a life time.
Odds of a player making three holes in one on the same hole in a lifetime.
Odds of a player making three holes in one on the same numbered hole on different golf courses in a life time.

If the first one is 1:12,000, you can only imagine what the number is for the last one.

There are also statistic for making a hole in one depending on the length of the hole. There is a much greater chance of a hole in one being made on a hole that is 100 yards compared to a hole that it 250 yards. Most players cannot reach a 250 yard par 3. Almost all can reach a 100 yard hole so there will be many more chances of a hole in one on a shorter hole.
Then there are the odds of any of these by players age, players gender and players skill level.