Probability that p and q are both prime provided q-p=2r Hello,
I would like to know whether there is a way, thanks to the prime number theorem, to give some kind of an equivalent of the probability that two positive integers $p$ and $q$ less than a given positive integer $n$ are both prime if one knows that $q-p=2r$, where $r$ is a given positive integer.
Thank you in advance.
 A: A good estimate (in some sense) is $$\frac{\alpha_{2r}C_2}{\ln^2N}$$ where $$C_2 = \prod_{p\ge 3} \frac{p(p-2)}{(p-1)^2} \approx 0.66016.$$   and $$\alpha_{2r}=\prod_{p \mid 2r}\frac{p-1}{p-2}$$ is the product is over all odd primes $p$ which divide $2r.$ 
Notice that $\alpha_2=\alpha_4=\alpha_8=\dots=1$ and $\alpha_6=\alpha_{12}=\alpha_{18}=\alpha_{24}=\dots=2$ 
More careful statement: Let $\pi_{2r}(N)$ be the number of primes $p$ up to $N$ with $p+2r$ also prime and   $\alpha_{2r}=\lim_{N \to \infty}\frac{\pi_{2r}(N)}{\pi_2(N)}.$ The twin prime conjecture is indeed open: there is no proof that $\lim_{N \to \infty} \pi_2(N)=\infty.$ However there is every reason to expect that 


*

*$\pi_2(N) \sim C_2\frac{N}{(\ln{N})^2}.$

*$\alpha_{2r}=\prod_{p \mid 2r}\frac{p-1}{p-2}$ where the product is over all the odd primes which divide $r.$


The great article Heuristic Reasoning in the Theory of Numbers (read it!) discuss the heuristics and some computational (circa 1959) evidence behind this.
