Probability that a Mersenne number is prime

Question

Let $p$ be a prime and let $M_p = 2^p-1$ be a (Mersenne) number. Is there any known result on the probability that $M_p$ is prime? In particular is it known whether the probability tends to $1$ as $p \to \infty$? If so, is there any known lower bound for the asymptotic rate with which this probability tends $1$? Thank you.

Answer 1

I'm pretty sure there is no such result: it is not even known that infinitely many $M_p$ are primes, or that infinitely many are composite.

Answer 2

For the already mentioned reason, this probability question can't be definitively solved right now. But for fun you could assume one of the conjectures about the distribution of Mersenne primes (I think Wagstaff conjectured an asymptote of C*loglog(x) for some funky constant C?) and see what that brings.

Answer 3

Wagstaff Mersenne Conjecture:

For a large prime $p$, the probability of $2^p−1$ being prime is about $\frac{\log{p}}{p}$.

Answer 4

There was a quotation of the "Wagstaff Mersenne Conjecture", with the claim that for large prime, the probability that $2^p - 1$ is prime is supposed to be about $\frac{\log{p}}{p}$. Since the probability that n is prime is about $\frac{1}{\log{n}}$ this would make the probability that $2^n - 1$ is prime for integer n about $1/n$. The number of Mersenne primes $2^p - 1$ with p ≤ M would be about log M.

$\log M (2 / \log{2})$, which is about 2.9 times bigger, seems to be a much better estimate for the number of Mersenne primes with exponents ≤ M. It overestimates the number of Mersenne primes by about 3 to 8 in the range where they are known. Some deviation is to be expected. If we picked an odd number near $2^n$ for every n, we would expect the same number of primes.

So with the known data, $\frac{2\log_2{p}}{p}$ seems to be a better estimate for the probability that $2^p - 1$ is prime if p is a prime.