Any mathematicians want to try this?

[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]

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/beginnersguide/a/bfaq_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]

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]

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.

[getdul981]

Somebody has waaaayyyy too much time on their hands.

[shcisamax]

[QUOTE=
If I'm not mistaken, calculus is a geometric function.[/QUOTE]

Referring to CalcTeacher who is a math whiz.

[CalcTeacher]

Calc Teacher is not a math "whiz"! I'm very good at what I do, but on any given day I too can't add two numbers together or second guess a calculated tip at a restaurant...hahaha! But, as stated earlier, calculus is more geometric and this problem is definitely rooted in probability and statistics. What CalcTeacher IS is very well connected...I know people that can help me get the right answer. The problem is they're going to look at me like I'm nuts for asking...a.k.a. I have too much time on my hands!

[kittygilchrist]

Jeepers, I have to know which course you were on. My house is on the third hole of Destin, so I want some credit too!

[shcisamax]

I agree. This thread is only for those who have too much time on their hands or are really interested in odds. Guess we have come to the conclusion that the odds are remarkably low.

[skyguy79]

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?The odds? Astronomical, simply astronomical.

[Buckeye Guy]

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.
Unknown information is:
1. How many hold does the golfer play: 9, 18 or 27?
2. How many numbers can there be for golf balls: 1 to 9 or 1 to 99 or something else.

Assumed answers to unknown information:
Golfer plays 18 holes,
The only number of the balls are 1 through 9.

The answer is:
1 in 162 for each time he gets a hold in one (18 * 9). This is 1 in 4,251,528 for 3 holds in ones (162 ^ 3).

However if her husband only used #1, #2 and #3 balls, the answer is 1 in 54 for each hold in one. This is one in 157,464

NOTE: There was only one undefeated football team last season (college or pro)

Buckeye Guy

[shcisamax]

Unknown information is:
1. How many hold does the golfer play: 9, 18 or 27?
2. How many numbers can there be for golf balls: 1 to 9 or 1 to 99 or something else.

Assumed answers to unknown information:
Golfer plays 18 holes,
The only number of the balls are 1 through 9.

The answer is:
1 in 162 for each time he gets a hold ion one (18 * 9). This is 1 in 4,251,528 for 3 holds in ones (162 ^ 3).

However if her husband only used #1, #2 and #3 balls, the answer is 1 in 54 for each hold in one. This is one in 157,464


Now there is someone who takes figuring out odds seriously.

[SALYBOW]

My answer would be that the odds are astronomical!!! But three holes in one are even more astronomical. I am still working on my first one.

[DougB]

I would think you would also have to take under consideration the tee used each time, the weather conditions on each day, what order he was golfing in each time, etc.

[blueash]

You have to start with certain assumptions. You take as a given that the golfer is going to get a hole in one. Of course he is not likely to get it on a par 4 or 5 hole. However, I will assume that whether or not the third hole is a par three is randomly distributed. You simply need to answer the question as posed. What are the odds that all 3 of his holes in one would be on the same number hole. Obviously if you are playing with no holes in one in your life, the chance your first will be on a number 3 hole is 1/9 for a nine hole course, and 1/18 for an 18 hole course. So now your life history has one hole in one for every round you play in the future. As you start your round on a 9 hole course I ask you, "What is the chance that if you get a hole in one today, it will be on the same number hole as the previous ace you made?" The answer is 1/9. It is your lucky day and you do get an ace. The same calculation applies for all future rounds when going for your 3rd ace. You take the odds of each individual event expressed as a fraction or decimal and multiply them to get a result. Easier example.. What are the chances of rolling a single die and getting 6 three times in a row. For each individual roll it is 1/6. So the result is 1/6 times 1/6 times 1/6 or 1/216. It is a different question than asking what are the odds of rolling the same number three times in a row. Because the odds of rolling a number on the first roll is 6/6 because you are going to get some number, the question there changes to what is the chance of matching your second roll to the first, which is 1/6. So to say it another way, "the odds of rolling doubles is 1/6." And of course the odds of rolling three dice at once all having the same number is 1/36. But the odds of rolling 3 dice and getting all 6's is 1/216. And yes I have time on my hands.
If you want to factor in the odds of making a hole in one at all, you can multiple the previous result by the odds of a hole in one on a round. You will need to know how many par 3's are in the average course if you want to know the odds of an ace in a round to do the correction for a course with excess par 3's.

[Arctic Fox]

The biggest unknown is "how good a golfer is your husband?"

If he consistently lands his drive on the green, then his chances of a hole-in-one are considerably better then for those who consistently hook, slice, leave the ball short, over-hit etc. etc.

My only chance of a hole-in-one would be if landing on the wrong green counted.

[Arctic Fox]

as an aside, since we don't have enough information to fully answer your original question:

the odds of your husband getting his first hole-in-one on the third hole (of a nine-hole course) using a #3 ball (assuming there are eight numbers) are 1 in 72

the odds of him getting his first and second holes-in-one 3/3 are 1 in 5,184

and the odds of him getting his first, second and third holes-in-one 3/3 are 1 in 373,248

and these odds are without factoring in the chances of him getting each of those holes-in-one in the first place