Let $f \in \mathbb{Z}[X]$ be an irreducible non-constant polynomial, and consider this polynomial modulo $p$ for each prime $p$. On average, how many roots does $f$ have modulo $p$? I.e., if $r(p)$ denotes the number of roots of $f$ modulo $p$, what is \begin{equation} \lim_{X \rightarrow \infty} \frac{1}{\pi(X)}\sum_{p \leq X}r(p) \ ? \end{equation} where $\pi(X)$ is the prime counting function. If the degree of $f$ is $1$, then the answer is trivially $1$. If the degree of $f$ is $2$, then by Quadratic Reciprocity and Dirichlet's theorem on primes in arithmetic progressions, you can again show the answer is $1$.
Does this quantity always exist? Is it equal to 1? Even information on other specific cases, such as cubic or quartic irreducible polynomials, would be helpful.
This quantity always exists and is always equal to $1$. By Chebotarev's density theorem, the density of primes $p$ such that $f$ has $k$ roots $\bmod p$ is equal to the proportion of elements in the Galois group of $f$ which fix $k$ roots when acting on the roots. Since the Galois group acts transitively on the roots, the average number of fixed points is $1$ by Burnside's lemma.