Let $x \in \mathbb{R}_+$ and $k \in \mathbb{N}^{*}$.
Let : $$\mathcal{A}(x)=\#\{(a_1, a_2, \ldots, a_k) \in \mathbb{P}^k \mid (a_1, a_2, \ldots, a_k \text{ verifying some properties}) \, , a_k \leqslant x\},$$
Example1: for $n$ even number, let : $\mathcal{A}(n)=\#\{ (p,n-p)\in\mathbb{P}^2 \, | \, p \leqslant n \}$.
Example2: $\mathcal{A}(x)=\#\{ p=n^2+1\in\mathbb{P} \mid p \leqslant x \}$
Example3: $\mathcal{A}(x)=\#\{ (p,p+2)\in\mathbb{P}^2 \, | \, p+2 \leqslant x \}$
The well known probabilistic model for prime numbers can give an heutistic argument to estimate $\mathcal{A}(x)$ for $x \to +\infty$.
I suggest an other model, let $q \in \mathbb{P}$ and consider : $$\mathcal{B}_q = \{ b \in \mathbb{N} \mid \gcd(b, { \prod\limits_{\substack{p \leq q \\ \text{p prime}}} {\normalsize p}})=1\}.$$ Consider : $$\mathcal{S}(x, q)=\#\{(a_1, a_2, \ldots, a_k) \in \mathcal{B}_{q}^k \mid (a_1, a_2, \ldots, a_k \text{ verifying some properties}) \, , a_k \leqslant x\},$$ Using chinese remainder theorem, let: $$\mathcal{S}\Big({ \prod\limits_{\substack{p \leq q \\ \text{p prime}}} {\normalsize p}}, q\Big)=R_q.$$ Let $q(x)$ be the largest prime number verify ${ \prod\limits_{\substack{p \leq q(x) \\ \text{p prime}}} {\normalsize p}} \leqslant x.$
Using Prime number theorem we have $q(x)=(1+o(1))\log(x)$.
Let $\mathcal{S}(x) = \mathcal{S}(x, q(x)).$
Theorem: I proved the following theorem
$$\frac{\mathcal{S}(x)}{x} \underset{x \to +\infty}\sim \dfrac{R_{q(x)}}{{ \prod\limits_{\substack{p \leq q(x) \\ \text{p prime}}} {\normalsize p}}}.$$
Conjecture:
$$\begin{array}{rcl} \mathcal{S}(x) & \underset{x \to +\infty}\sim & \mathcal{A}(x) \big( \pi(q(x)) e^{-\gamma} \big)^k \\ &\underset{x\to+\infty} \sim & \mathcal{A}(x) \dfrac{\log(x)^k}{\log(\log(x))^k}e^{- k \gamma} . \end{array}$$
This conjecture gives:
$$\mathcal{A}(x) \underset{x \to +\infty}\sim x \dfrac{R_{q(x)}}{{ \prod\limits_{\substack{p \leq q(x) \\ \text{p prime}}} {\normalsize p}}} \ \dfrac{\log(\log(x))^k}{\log(x)^k} \ e^{k \gamma}.$$
Let $q \in \mathbb{P}$
Example 1: Prime number theorem
Let $\mathcal{A}(x) = \#\{p \in \mathbb{P} \mid p \leqslant x\}$
We have $\mathcal{S}({ \prod\limits_{\substack{p \leq q \\ \text{p prime}}} {\normalsize p}}, q) = R_q = { \prod\limits_{\substack{p \leq q \\ \text{p prime}}} {\normalsize (p-1)}} .$
according to the conjecture: $$\begin{array}{rcl} \mathcal{A}(x) & \sim & \dfrac{x}{\displaystyle{\small \prod_{\substack{p \leq q(x) \\ \text{p prime}}} {\normalsize p}}} \ { \prod\limits_{\substack{p \leq q(x) \\ \text{p prime}}} {\normalsize (p-1)}} \dfrac{\log(\log(x))}{\log(x)}e^{\gamma} \\ & \sim & x \displaystyle{\small \prod_{\substack{p \leq q(x) \\ \text{p prime}}} {\normalsize \left(1-\frac{1}{p}\right)}} \dfrac{\log(\log(x))}{\log(x)}e^{\gamma} \\ & \sim & x \dfrac{e^{-\gamma}}{\log(q(x))} \dfrac{\log(\log(x))}{\log(x)}e^{\gamma} \\ & \sim & x \dfrac{e^{-\gamma}}{\log(\log(x))} \dfrac{\log(\log(x))}{\log(x)}e^{\gamma} \\ & \sim & \dfrac{x}{\log(x)}. \end{array}$$ We know that this result is true (prime number theorem).
Example 2: Goldbach's conjecture
Let $n$ be an even number, and $\mathcal{A}(n) = \#\{ (p, n-p) \in \mathbb{P}^2 \, | \, p \leqslant n \}$
Using C.R.T We have :
$\mathcal{S}({\displaystyle\prod\limits_{\substack{p \leq q \\ \text{p prime}}} {\normalsize p}}, q)= R_q = \displaystyle\prod_{\substack{3 \leq p \leq q \\ \text{p prime, } p | n}} (p-1) \prod_{\substack{3 \leq p \leq q \\ \text{p prime, } p \nmid n}} {\normalsize (p-2)}=\displaystyle{\small \Big( \prod_{\substack{p | n \\ \text{p prime} \\ p \leq q}} {\normalsize \frac{p-1}{p-2}} \Big)} {\small \Big( \prod_{\substack{3 \leq p \leq q \\ \text{p prime}}} {\normalsize (p-2)} \Big)}.$
according to the conjecture: $$\begin{array}{rcl} \mathcal{A}(n) & \sim & \dfrac{n}{\displaystyle{\small \prod_{\substack{p \leq q(n) \\ \text{p prime}}} {\normalsize p}}} \ \displaystyle {\small \Big( \prod_{\substack{p | n \\ \text{p prime} \\ p \leq q(n)}} {\normalsize \dfrac{p-1}{p-2}} \Big)} {\small \prod_{\substack{3 \leq p \leq q(n) \\ \text{p prime}}} {\normalsize (p-2)}} \ \dfrac{\log(\log(n))^2}{\log(n)^2}e^{2 \gamma} \\ & \sim & \dfrac{n}{2} \displaystyle {\small \Big( \prod_{\substack{p | n \\ \text{p prime} \\ p \leq q(n)}} {\normalsize \dfrac{p-1}{p-2}} \Big)} \displaystyle {\small \prod_{\substack{3 \leqslant p \leqslant q(n) \\ \text{p prime}}} \Big({\normalsize 1-\dfrac{2}{p}}\Big)} \dfrac{\log(\log(n))^2}{\log(n)^2}e^{2 \gamma} \\ & \sim & \dfrac{n}{2} \displaystyle {\small \Big( \prod_{\substack{p | n \\ \text{p prime} \\ p \leq q(n)}} {\normalsize \dfrac{p-1}{p-2}} \Big)} \dfrac{4 C_2 e^{-2 \gamma}}{\log(\log(n))^2} \dfrac{\log(\log(n))^2}{\log(n)^2}e^{2 \gamma} \\ & \sim & 2 C_2 \displaystyle {\small \Big( \prod_{\substack{p | n \\ \text{p prime} \\ p \leq q(n)}} {\normalsize \dfrac{p-1}{p-2}} \Big)} \dfrac{n}{\log(n)^2}. \end{array}$$ With $C_2 = \displaystyle{\small \prod_{\substack{3 \leq p \\ \text{p prime}}} \left({\normalsize 1-\dfrac{1}{(p-1)^2}}\right)}$, and we prove that $\displaystyle {\small \Big( \prod_{\substack{p | n \\ \text{p prime} \\ p \leq q(n)}} {\normalsize \dfrac{p-1}{p-2}} \Big)} \sim \displaystyle {\small \Big( \prod_{\substack{p | n \\ \text{p prime}}} {\normalsize \dfrac{p-1}{p-2}} \Big)}$
Then: $$\mathcal{A}(n) \sim 2 C_2 \displaystyle {\small \Big( \prod_{\substack{p | n \\ \text{p prime}}} {\normalsize \dfrac{p-1}{p-2}} \Big)} \dfrac{n}{\log(n)^2}.$$
Example3: the k-tuple conjecture: Prime numbers and sieving up to $q(x)=\log(x)(1+o(1))$
Conclusion: I propose above a model to estimate $\mathcal{A}(x)$ using chinese remainder theorem and prime number theorem.
Question: Is there a connection between the probabilistic model for prime numbers and the conjecture above ?