Probability of prime numbers

Question

Say we use the Euclidean construction for prime numbers and take a set $S$ solely containing prime numbers, so that $p_n$ is the greatest prime within S. What is the probability that $1+p_1 \cdots p_n$ is a prime number?

Answer 1

Heuristically, it might be reasonable to expect that $1 + p_1\cdots p_n$ behaves like any other number of its size that is known to be coprime to $p_1, p_2, \ldots, p_n$. The magnitude of $1 + p_1\cdots p_n$ is roughly $e^{p_n}$, so the density of primes in that size range is about $1/p_n \sim 1/(n\log n)$. To account for the avoidance of the first $n$ primes, we can apply the correction factor $\prod_{k=1}^n \frac{p_k}{p_k - 1}$, which by Mertens's theorem is about $e^{\gamma} \ln p_n \sim e^{\gamma} \ln n$.

Thus one might reasonably expect (as a heuristic) to find primes of the form $1 + p_1\cdots p_n$ with frequency about $e^\gamma / n$, or about $e^\gamma \log n$ occurrences up to $n$. Of course, since this isn't actually a random process, the true number of occurrences will be primarily dominated by the actual results for small $n$. But heuristically, yeah I'd expect there to be infinitely many primes in the sequence (but good luck computing high enough to find more than a few dozen of them).