Not necessarily anything. Attempts to read deeper meanings into equations and theorems often lead to mystical claptrap.
I think the problem here is that the example comes from a discrete distribution, Normal distributions are continuous distributions.
What problem?
From the OP: “What has the geometry of a cirlce got to do with pulling colored balls from a bag?”.
I’m not sure Pi has anything to do with the analysis of problems of this nature. Well, at least not to calculate simple probabilities of a discrete distribution.
You went to all that trouble to format your posts, but this character shows up as a question mark on my system (Win2K, Firefox).
Actually, the probability that any two numbers picked at random, not necessarily large, are relatively prime is 6/(pi^2).
See this page for a cite:
http://mathworld.wolfram.com/RelativelyPrime.html
That’s still one of the most remarkably unexpected appearances of pi, along with the fact that the sum of the reciprocals of the squares of the integers is (pi^2)/6.
Hey, those two values are reciprocals themselves! I wonder if there’s a connection.
Ed
What does “at random” mean in that context?
I was about to ask the same thing. It’s meaningless to give the probability of two randomly picked integers being relatively prime if we don’t know what probability distribution on the integers is being used in the first place. I’m sure it’s probably an interesting result, but it’s kind of misleading when it’s so vaguely stated.
I think that what it means is that if you take an upper bound U and pick the numbers from a uniform distribution on [1,U], you can determine the probability P(U) that the numbers will be relatively prime, and that in the limit U --> infinity, P(U) --> 6/pi[sup]2[/sup] (of course, for any finite U, P(U) is rational). Amusingly, I once saw a numerical experiment which used the digits of pi themselves as the source of the “random” numbers, and thereby managed to use a million digits of pi to derive an estimate of pi accurate to three decimal places.
Unfortunately, the mathworld page I cited doesn’t have any more information about this.
Ed
That would be my guess as well. It’s a pretty natural way of doing it, but such a probability distribution wouldn’t atually satisfy the Kolmogorov probability axioms (it’s not countably additive).
Are you saying that the limit of the uniform distribution on {1,…,U} (as U → infinity) is not a probability distribution? In that case, I agree, but the limit that’s being taken is not of the distribution but of the probability of relative primality of two numbers chosen according to that distribution. I don’t see anything problematic about that limit (except for the somewhat loose way it was originally given).
The result is basically just an interpretation of the identity
6 / pi[sup]2[/sup] = [product over primes p](1 - 1/p[sup]2[/sup]),
with each factor of (1 - 1/p[sup]2[/sup]) representing the probability that two “randomly chosen” numbers do not both contain a factor of p. This identity is in turn easily reducible to the more common identity
pi[sup]2[/sup] / 6 = [sum over n]1/n[sup]2[/sup]
already mentioned.
should be a Unicode “Greek small letter pi”.
I just found out about another place where pi makes an unexpected (at first glance) appearance:
The average number of ways in which a non-negative integer can be expressed as the sum of the squares of two integers is pi!
To be precise:
Let r(n) be the number of ways that n, a non-negative integer, can be expressed as the sum of squares of two integers. For example r(5) = 8, because
5 = (2^2) + (1^2) = (1^2)+(2^2)
= (-2^2) + (1^2) = (1^2)+(-2^2)
= (2^2) + (-1^2) = (-1^2)+(2^2)
= (-2^2) + (-1^2) = (-1^2)+(-2^2)
Now let R(z) = r(0) + r(1) + . . . + r(z-1),
Then A(z), the average number of ways for the first z integers is:
A(z) = R(z)/z
I saw a proof yesterday that the limit of A(z) as z goes to infinity is pi.
When you realize that R(z) is equal to the number of lattice points inside a circle centered at the origin, with radius sqrt(z), the connection with pi starts to make sense.
Ed