Post Deleted by Timothy Foo
Now, thinking about this a bit, let's say $f$ is the function that reverses the digits, so that $f(n)$ is the number that has the digits of $n$ in base 10 reversed. I think that when estimating $$|\{n \leq x: n,f(n) \mbox{ simultaneously prime}\}|,$$
then for each prime $p$, maybe you'll have to estimate the number of solutions to $$nf(n) \equiv 0 \bmod p,$$ where $n \in \mathbb{Z}/p\mathbb{Z}$. This is similar to like when estimating the twin prime constant (but I'm not claiming that the whole thing goes through the same way). The problem is that $f(n)$ is not so straightforward like $n+2$ is. For $p=3$, at least $f(n) \equiv n \bmod 3$, so that is ok. $p=11$ is also not too bad. For other $p$, it doesn't seem so straightforward.
In fact, it's not even as straightforward as this, because for $n,f(n)\in \mathbb{Z}$, when one fixes $n \bmod p$, $f(n) \bmod p$ is not fixed in general. The point is just that I think one might have to estimate the probability that $p\nmid f(n)\in \mathbb{Z}$, given that $p\nmid n\in \mathbb{Z}$.
Now, thinking about this a bit, let's say $f$ is the function that reverses the digits, so that $f(n)$ is the number that has the digits of $n$ in base 10 reversed. I think that when estimating $$|\{n \leq x: n,f(n) \mbox{ simultaneously prime}\}|,$$
then for each prime $p$, maybe you'll have to estimate the number of solutions to $$nf(n) \equiv 0 \bmod p,$$ where $n \in \mathbb{Z}/p\mathbb{Z}$. This is similar to like when estimating the twin prime constant (but I'm not claiming that the whole thing goes through the same way). The problem is that $f(n)$ is not so straightforward like $n+2$ is. For $p=3$, at least $f(n) \equiv n \bmod 3$, so that is ok. $p=11$ is also not too bad. For other $p$, it doesn't seem so straightforward.