[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]

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?

The odds? Astronomical, simply astronomical.

[Buckeye Guy]

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.

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]

Originally Posted by Buckeye Guy

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