This recent question asks whether it is possible for $n!-1, n!+1$ to both be prime when $n > 3$. According to the answers, this is an open problem.
I am trying to figure out how you would justify this conjecture probabilistically. Since $n$ informally has a probability of about $1/\log(n)$ of being prime, and $\log(n! \pm 1) \approx n \log(n)$, a naive estimate suggests that we should expect about $$ \sum_{n \geq 2} \frac1{n^2 \log^2(n)} \approx 0.69 $$ examples (of course $n=3$ itself is an example). The answers to the question linked above contain checks up to 2000, at which point the remainder of the sum is quite small (in fact WA has a weird bug in evaluating this number...).
However, this is not a fair computation: because the numbers are $n! \pm 1$, they have a significantly larger "chance" of being prime: they are not divisible by any primes below $n$.
My question is:
If we incorporate this fact into our probabilistic estimate, what kind of conclusion do we reach about the probability of the existence of a twin prime pair of the form $n!\pm 1$ for $n > 3$? Can we make an even more sophisticated argument?
The probability that a number is not divisible by any primes $\le n$ is essentially the probability that a number around the size of $n^2$ is prime, which is around $\frac1{\log n^2} = \frac1{2 \log n}$. Since we can assume this about $n! \pm 1$ for free, we should get to multiply the probability by $(2 \log n)^2$, which gives us a convergent sum of $\sum_{n>2000}\frac4{n^2}$ by the same logic.
Of course this is still just a heuristic argument...