Re: Probability of picking a positive rational number at random
- From: S_Paske@xxxxxxxxxxx
- Date: Thu, 13 Mar 2008 15:03:57 -0700 (PDT)
On Mar 13, 3:16 pm, Randy Poe <poespam-t...@xxxxxxxxx> wrote:
On Mar 13, 4:10 pm, S_Pa...@xxxxxxxxxxx wrote:
From my understanding, the probability of picking a positive rational
number at random which is <1 is 50%, and also the probability that
this number >1 is also 50%.
Any thoughts are appreciated.
Thanks
Depends on your distribution for picking "at random".
I'm guessing that you mean the numerator n and
denominator d are independent random variables
chosen from the same distribution. In that case,
regardless of the distribution, p(n < d) = p(d < n).
However, if there is a nonzero probability that
n = d, then neither p(n<d) nor p(d<n) is 50%.
One more general comment: There is no such thing
as a uniform distribution over all the natural
numbers. However, you might choose n and d as
uniformly distributed on some finite set, say
{1, 2, ..., 10000}.
What is the probability that the number is <1/2. What about <2?
How about <1/3 or < 3?
Depends on the details of the distribution.
- Randy
I am thinking that if i picked a random integer n>=1 i could use this
distribution:
The probability of the exponent of its 2 prime factor=0 is 1/2
The probability of the exponent of its 2 prime factor=1 is 1/4
...
The probability of the exponent of its 3 prime factor=0 is 2/3
The probability of the exponent of its 3 prime factor=1 is 2/9
The probability of the exponent of its 3 prime factor=2 is 2/27
...
The probability of the exponent of its n'th prime factor=k is (Pn-1)/
Pn^k
Each prime factor would be independent of the others.
To extend this to the rationals, if k>0 then split each probability
into equal parts 1 for negative exponent, 1 for positive exponent.
I am pretty sure with this model that picking a positive rational at
random would yield 50% between 0 and 1 and 50% > 1.
But what is the probability of between 0 and 1/2 or between any
interval in general?
.
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