# Picking random natural number

It is not possible to pick a random natural number (out of all natural numbers), such that each number has the same probability.

But is it possible to define a probability distribution over the natural numbers, which is such that larger numbers get a higher probability of being picked?

• This is vague. If you really mean that $n>m\implies p_n>p_m$, then clearly not, since that would make $\sum p_n$ diverge. Knowing that $\sum p_n=1$ implies that $\lim_{n\to \infty}p_n=0$.
– lulu
Commented May 12 at 21:11
• Can you write an answer @lulu ? Commented May 12 at 21:31

Suppose $$p(2) =1/1000$$. Then $$p(x) \gt 1/1000$$ for $$x=3,4,5,...,1003$$. But those probabilities would then add up to more than $$1.0$$, which is impossible for probabilities.
You should be able to easily write this in a more general form for $$p(2)= \textrm { any }\epsilon \gt 0$$, or if you allow some initial batch of smaller numbers to have a probability of $$0$$, then by picking some starting number other than $$2$$ that does have positive probability.
• @K.Jiang - Well, it's pretty loose, but when you assume equal probabilities, you can still make the same argument, and take enough terms so that the sum exceeds 1. But in our case here, the individual probabilities are larger, so we're going to get an even larger sum - or if you imagine yourself adding together consecutive $p(n)$'s, we might exceed 1 "earlier" than in the equal probability case. Commented May 13 at 3:07