A Curious Probability
What is the probability of selecting 4 setwise coprime positive integers?
The answer is 0.9239.
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1 solution
the exact value is π 4 9 0
You need to be careful with the phrasing of your problem.
It is impossible to "randomly select" a (positive) integer. Instead, what you are looking for is the limiting probability.
The known probability of 3 integers chosen at random being pairwise coprime is 0.285747..., which is not the reciprocal of Zeta(3). I'e like to see a citation of such a claim, as my understanding is that for more than 3 integers, the expression for the probability is not proven.
To paraphrase Carl Sagan, extraordinary problems require extraordinary care in stating the problem precisely.
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I believe the probability of selecting 3 coprime integers is 0.8319. Here's a link with that number. http://mathworld.wolfram.com/RelativelyPrime.html
This can be proven by showing that the probability of selecting n integers that share a prime factor p is 1 − p n 1 . The infinite product for all primes p is then i = 1 ∏ ∞ 1 − p i n 1 . With a little algebra, this product becomes ( i = 1 ∏ ∞ p i n − 1 p i n ) − 1 which is the reciprocal of the zeta function expressed as a product of primes.
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Okay, this is a perfect example of why wording of the problem is so important. Your problem, as stated, asks for the probability that no pair of integers share a common factor. This is not the same as none of them having a common factor. 0.83190...is the probability that 3 random numbers don't have a factor common to all. But 0.285747.., is the probability that no pair of numbers share a common factor. To the ear, it almost sounds the same, but they're different problems.
I gave the answer to your problem of 4 pairwise coprrime numbers as 0.114884...., which I think is correct. Maybe I should run random trials to see if this is approximately right.
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@Michael Mendrin – I think I will remove this problem. It is too easy to misinterpret the question.
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@Steven Zheng – Just repost it with more careful wording. I've had to do that sometimes.
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@Michael Mendrin – I noticed you solved my problem Hein's Egg. My first answer was wrong, and I think you solved it correctly once the new answer was updated. I was wondering how you solved it?
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@Steven Zheng – Actually, Steven, the first answer I gave to that one, 56.5508, is the correct one. It's in my notes, which I had to go find. I integrated circular disks from bottom to top of of Piet Hein's superellipse. When that didn't work, I tried a couple of other interpretations of the problem, the last one being that of a superellipsoid, which I looked up in Wolfram, and used the volume formula provided there. Working out that volume formula by double integration was simply too hard for me to do. I think your original answer was the case where a = 2 and b =3, so that it would have looked more like a curling stone, or an oversized round pill, which would have looked nothing like as in the picture. In retrospect, that's what threw me off, I kept thinking of a solid that looks like what's in the picture.
@Michael Mendrin – How did you calculate the probability of 4 coprime positive integers? How is the formula derived?
It can be shown that the probability of selecting n pairwise coprime positive integers is the reciprocal of zeta(n), the zeta function.
This can be proven by showing that the probability of selecting n integers that share a prime factor p is 1 − p n 1 . The infinite product for all primes p is then i = 1 ∏ ∞ 1 − p i n 1 . With a little algebra, this product becomes ( i = 1 ∏ ∞ p i n − 1 p i n ) − 1 which is the reciprocal of the zeta function expressed as a product of primes.