The point is that your probability for the “first” integers will not be infinitesimal. If you think that drops off too quickly, instead of 2 use 1+e or something. p(n) = e/(e+2) * (1+e)^(-|n|). And replace n with s(n) if you don’t like that ordering of integers. But regardless, there’s some N for which there is an n with |n|N such that p(n)/p(m) >> 1.
The point is that your probability for the “first” integers will not be infinitesimal. If you think that drops off too quickly, instead of 2 use 1+e or something. p(n) = e/(e+2) * (1+e)^(-|n|). And replace n with s(n) if you don’t like that ordering of integers. But regardless, there’s some N for which there is an n with |n|N such that p(n)/p(m) >> 1.