I'm trying to evaluate the probability that some even integer $2n$ has no Goldbach partitions using the following approach...
First, visualize the distribution of primes from $1$ to $n$ as a binary string representing odd integers, with $1$ for primes and $0$ for composites:
$$ 011101101101001... $$
Now in order for $2n$ to have no Goldbach partitions, the following binary sequence (from $n$ to $2n$), spelled backward, would have to be of the form:
$$ x000x00x00x0xx0... $$
This means that prime numbers from $n$ to $2n$ would have to be distributed within the $x$'s.
In order to get the probability that $2n$ has no Goldbach partitions, we can calculate the ratio of the number of possible distribution of primes from $n$ to $2n$ within the $x$'s, over the total number of possible distribution of primes from $n$ to $2n$.
Number of primes from $n$ to $2n$: $$ \pi\left(2n\right)-\pi\left(n\right) $$
Number of odd integers from $n$ to $2n$: $$ \frac{n+1}{2} $$
Number of odd composites from $1$ to $n$ (number of $x$'s): $$ \frac{n+3}{2}-\pi\left(n\right) $$
Number of possible distributions of primes from $n$ to $2n$: $$ \frac{\left(\frac{n+1}{2}\right)!}{\left(\pi\left(2n\right)-\pi\left(n\right)\right)!\left(\frac{n+1}{2}-\left(\pi\left(2n\right)-\pi\left(n\right)\right)\right)!} $$
Number of possible distribution of primes from $n$ to $2n$ within the $x$'s: $$ \frac{\left(\frac{n+3}{2}-\pi\left(n\right)\right)!}{\left(\pi\left(2n\right)-\pi\left(n\right)\right)!\left(\left(\frac{n+3}{2}-\pi\left(n\right)\right)-\left(\pi\left(2n\right)-\pi\left(n\right)\right)\right)!} $$
Probability $P$ that $2n$ has no Goldbach partitions:
$$ P\approx\frac{\left(\frac{n+3}{2}-\pi\left(n\right)\right)!\left(\frac{n+1}{2}-\pi\left(2n\right)+\pi\left(n\right)\right)!}{\left(\frac{n+1}{2}\right)!\left(\frac{n+3}{2}-\pi\left(2n\right)\right)!}$$
Questions:
1: This approach does not take into account the fact that $3$ consecutive odd integers cannot be all primes. I need to show whether this fact decrease or increase the value of $P$.
2: A quick search about this subject shows some other formulas to approximate $P$, like for example:
$$ \prod_{a=2}^{n/2}\left(1-\frac 1{\log a\cdot \log (n-a)}\right) $$
but the values for $P$ obtained from those formulas are incredibly higher than what i get with mine. Could it be only because of the fact from question 1, or is there a problem with the whole approach ?
Some results for $P$ using my formula:
Few things need accounting for:
$$n^2\bmod 3\to p\equiv q\bmod 6\\n^2\bmod 2\to p\equiv q \bmod 4\\p\equiv q\bmod 4\land p\equiv q\bmod 6\implies p\equiv q\bmod 12\\p\not\equiv q\bmod 4\land p\not\equiv q\bmod 6\implies p\not\equiv q\bmod 12\\ \gcd(n,p)=gcd(n,q)=1$$
and a few others for a start. There are also gaps in this new arithmetic progression, lining up with other gaps in another ( possibly different) arithmetic progression, to consider. These eliminate gaps, the relevant primes could possibly hide in.
It's literally pigeonhole principle, if the number of primes of a specific form above n, outnumber the gaps not covered by gaps in another ( possibly different) arithmetic progression of primes below n, then it's a shut case, it will cause a partition.
The exception to your first rule is of course 3,5,7.