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.
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