Re: Probability of picking a positive rational number at random

On Mar 16, 1:19 am, quasi <qu...@xxxxxxxx> wrote:
On Sat, 15 Mar 2008 19:59:32 -0700 (PDT), S_Pa...@xxxxxxxxxxx wrote:
On Mar 15, 9:04 pm, Tim Little <t...@xxxxxxxxxxxxxxxxxxxxxxxxxx>
wrote:
On 2008-03-15, Ross A. Finlayson <r...@xxxxxxxxxxxxxxx> wrote:

So now you're going from a probability distribution over [0,1] defined
by Lebesgue measure to a completely undefined probability distribution
over the set of all well-orderings of distributions of natural
numbers?  How do you know that one with the properties you want even
exists?

Can't you just use classical probabilities to show that?

Classical probability is more limited than measure-theoretic
probability, so it can't help you.  Can you even explain in terms of
mathematical properties what you mean by your desired uniform
distribution?

Simply asserting "for all x,y: P({x}) = P({y})" isn't enough, since
there are infinitely many distributions satisfying that relation.
Feel free to try to define a uniform distribution over all uniform
distributions over all well-orderings of distributions over natural
numbers.  That won't get you any closer, and I'll feel free to call it
ridiculous.

That iota is great among infinitesimals represents that summing
infinitely many of them yields one.

Originally you said "greatest".

In a theory like ZFC with well-orderable reals, and the ability to
"sample" an element of the population of real numbers by infinitely
many Bernoulli trials implying the ability to select an ordinal at
uniform random from their initial ordinal, because well-orderings of
the reals are so random, there is the ability to sample from the
naturals a value such that for any other value, the probability of
its selection is exactly the same, and the sum of their
probabilities, their whole, in that they are mutually exclusive
events one of which occurs, censored to countable, is one.

Apart from being a gross run-on sentence, this is rubbish.
Well-orderings aren't random.

Simply, sample a real by infinitely many Bernoulli trials and
discard it if a particular well-ordering has that real not mapping
to a finite ordinal.

So your "algorithm" is: Generate a countably infinite number of binary
digits.  Do this (on average) *uncountably* many times, until you get
a countable ordinal.  I don't know what you're on, but it's not
mathmatics.

- Tim

How would a mathematician describe the process of showing that two so
called randomly chosen integers

Anyone who understands probability would stop you right there.

In fact, many of us here in sci.math have already tried to make clear
to you that there is no such thing as a "randomly chosen integer",
unless you first specify a valid distribution.

You _can_ define a distribution on a countable set -- simply assign
probabilities to each element of the set, and make sure the
probabilities are nonnegative and sum to 1. But don't expect such a
distribution to be uniform -- it can't be. Also, don't expect such a
distribution to be unique. In fact, there are uncountably many such
distributions.

have a probability of no common factor = 6/pi^2?

This has already been explained to you. Once again, there's no such
thing as a random pair of positive integers.

What _is_ true is this ...

For each positive integer n,

   let A_n = {(x,y) in [1,n] x [1,n] | gcd(x,y) = 1}

   and let r_n = card(A_n) / n^2.

Then the limit, as n -> oo, of r_n, equals 6/pi^2.

Thus, 6/pi^2 is a limit of probabilities on finite, uniformly
distributed sample spaces, with a different sample space for each n,
(namely [1,n] x [1,n]), but the limiting value, 6/pi^2, is not itself
a probability -- it's a density.

quasi- Hide quoted text -

- Show quoted text -

For each positive integer n,
let A_2 = {(x) in [1,n] | largest prime factor exponent < 2}
and let r_2 = card(A_2) / n^2
Then the limit, as n -> oo, of r_2, = 6/pi^2
Also the limit, as n -> oo, of r_n, = 1/zeta(n)

I would conjecture that the limit of 1/zeta(n) as n->oo = 1 ?
.

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