Is this a valid method to calculate a lower bound on the number of twin prime pairs that occur over $(6n$, $36n^2]$?
Divide the number line into groups of six, each of which contains a potential twin prime pair at $6i\pm1$ for $i = 1, 2, 3...\infty$. For any prime number $p$, the fraction of groups that contain multiples of $p$ landing on a potential prime (I will refer to this as a collision) is $2/p$. The fraction of groups having no collision is $\frac{p-2}{p}$. Over a finite number of groups $N$, the maximum number of groups containing a collision is $\lceil N\frac{2}{p} \rceil$. The minimum number of groups containing no collision is $\lfloor N\frac{p-2}{p} \rfloor$.
To enable the calculation for the groups over $(6n$, $36n^2]$, consider the set of all numbers equal to $6i\pm1$ for $i = 1,2,3...n$, which is a superset of the primes over $[5, 6n]$. Using a superset of the primes will eliminate more groups than necessary, yielding a lower bound on the number of surviving groups containing twin prime pairs. For a given $i$, the minimum number of groups surviving collisions with $6i-1$ is
$$N\frac{6i-3}{6i-1}-1$$
The minimum number of groups that survive collisions with $6i+1$ is
$$N\frac{6i-1}{6i+1}-1$$
Considering $6i\pm1$ together, the minimum number of groups that survive is
$$N\frac{6i-3}{6i-1} \frac{6i-1}{6i+1} - 2$$
Over $(6n$, $36n^2]$, $N=6n^2-n$. Considering $i = 1,2,3...n$, the minimum number of groups containing twin prime pairs is
$$(6n^2-n)\prod_{i=1}^{n}{\frac{6i-3}{6i+1}} - 2n$$