$\pi(n)$ and $\pi_2(n)$ represent the count of primes and count of twin primes $\leq n$ respectively.
Suppose we want to estimate $\frac{\pi(n)}{n}$. One way which obviously is not error-free is to note that we are essentially taking the fraction $\frac{p-1}{p}$ for every prime $p$ in $2 \leq p \leq \sqrt{n}$. Let us call this estimate $\alpha_1$ and the accuracy of this estimate as $acc_1$.
\begin{align} \alpha_1 &= \prod\limits_{\substack{p \text{ $\text{prime}$ } \\ 2 \leq p \leq \sqrt{n}}} \frac{p-1}{p} \\\\ acc_1 &= \frac{\alpha_1}{\left(\frac{\pi(n)}{n}\right)} \end{align}
In a similar fashion, suppose we want to estimate $\frac{\pi_2(n)}{\pi(n)}$. One way which obviously is not error-free is to note that we are essentially taking the fraction $\frac{p-2}{p-1}$ for every prime $p$ in $3 \leq p \leq \sqrt{n}$. Let us call this estimate $\alpha_2$ and the accuracy of this estimate as $acc_2$.
\begin{align} \alpha_2 &= \prod\limits_{\substack{p \text{ prime } \\ 3 \leq p \leq \sqrt{n}}} \frac{p-2}{p-1} \\\\ acc_2 &= \frac{\alpha_2}{\left(\frac{\pi_2(n)}{\pi(n)}\right)} \end{align}
Claim
I claim that the ratio of the accuracies $\lim\limits_{n\to\infty} \frac{acc_1}{acc_2} = 1$ since although the estimators $\alpha_1$ and $\alpha_2$ are not error-free, and since $\alpha_2 = 2C_2\alpha_1$ in the limit, the errors in $\alpha_1$ and $\alpha_2$ get cancelled when we take the ratio $\frac{\alpha_1}{\alpha_2}$. This leads us to the following conclusion:
\begin{align} \lim\limits_{n\to\infty} \pi_2(n) = 2C_2 \frac{\pi^2(n)}{n} \end{align} where $C_2$ is the twin prime constant.
I am struggling to understand what is wrong with my claim. Appreciate any guidance.
EDIT: I want to clarify what has $\alpha_2$ got to do with $\pi_2(n)$?
Consider primes $\leq n$. Of these primes, we want to find primes that are not congruent to $k \bmod p$ (for non-zero $k$) and for $p$ in $3 \leq p \leq \sqrt{n}$. We know the primes (possibly excluding one prime) get equally distributed (becomes more equal for large $n$) among the $p-1$ residue classes (for prime $3$). Of these $p-1$ residue classes, we choose $p-2$ of them. So we take a proportion $\frac{p-2}{p-1}$. We then distribute these primes among the $p-1$ residue classes for prime $5$ and so on for each primes $\leq \sqrt{n}$. And so we get an estimate for the proportion of primes that are $\not\equiv k \bmod p$ is $\alpha_2$. (This estimate is not error-free). For $k = -2$, we get the count of twin primes or $\pi_2(n)$. For $k = -4$, we get the count of cousin primes or $\pi_4(n)$. For $k = n$, (for even $n$), we get the count of Goldbach primes.