>>14317089You got one combination that gives you 2, two combinations that sum up to 3, 3 comb. that gives 4, and so on until 1000 comb. that give 1001, but, then 999 give 1002, 998 give 1003 and so on until 1 gives 2000.
So the probability is
[eq] \frac{1}{N^2}( \sum_{p \le N} (p-1) + \sum_{p \in ]N, 2N]} (2N+1 - p) ) = \frac{1}{N^2}( \sum_{p \le N} p - \sum_{p \in ]N, 2N] p + \pi(N) + (2N+1)( \pi(2N) - \pi(N) )} )[/eq]
You can compute it for N = 1000 and get the exact result of 14.51%.
Naturally, for N large it starts to converge to 0, but one interesting question is if the PNT gives you any asymptote. The problem being, of course, the term.
