nt.number theory - Probabilistic interpretation of prime number theorem

First of all, I assume you understand that this is meant to be a nonrigorous argument, so there will be a limit to how rigorous I can make my answer.

The intuition here is that $n$ is prime if and only if it is not divisible by any prime $<n$. So we "should" have
$$f(n) approx prod_{p < n} left( 1-1/p right).$$
Similarly
$$f(n+1) approx prod_{p<n+1} left( 1-1/p right) = prod_{p < n} left( 1-1/p right) cdot left{ begin{matrix} left( 1-1/n right) mbox{if} n mbox{is prime} \ 1 mbox{if} n mbox{is not prime} end{matrix} right. approx f(n) cdot left{ begin{matrix} left( 1-1/n right) mbox{if} n mbox{is prime} \ 1 mbox{if} n mbox{is not prime} end{matrix} right. .$$
Since $n$ is prime with "probability $f(n)$", we interpolate between the two cases next to the brace by writing:
$$f(n+1) approx f(n) left( 1-f(n)/n right).$$
One might argue that it would be better to interpolate with a factor of $(1-1/n)^{f(n)}$, but this will make no difference in the asymptopics as $(1-1/n)^{f(n)} = 1-f(n)/n+O(1/n^2)$.

This argument is famously fishy, because it gives the right answer, but the intermediate point is wrong! The actual asymptopics of $prod_{p<n} left( 1-1/p right)$ do not look like $1/log n$, but like $e^{-gamma} /log n$. I've never seen a good intuitive explanation for why we get the wrong estimate for $prod_{p<n} left( 1-1/p right)$, but the right estimate for the density of the primes.