It is known (citation) that, given an irreducible polynomial $f(x)\in\mathbb Q[x],$ the expected number of roots of $f(x)\pmod p$ is $1$ as $p$ varies over the primes.
What is known about the variance?
For example, I am not totally convinced that it will come out nicely.
One approach (that works to compute the expected value) is to translate everything into algebra. Namely, Dedekind-Kummer says that, for all but finitely many primes $p,$ the number of roots of $f(x)\pmod p$ is equal to the number of degree-$1$ primes in the prime factorization of $p\mathcal O_{\mathbb Q(\alpha)},$ where $\alpha\in\mathbb C$ is some root of $f(x).$
Now let $K=\mathbb Q(\alpha)$ and $L$ be the splitting field of $f(x)$ with $G:=\operatorname{Gal}(L/\mathbb Q)$ and $H:=\operatorname{Gal}(L/K).$ We can then count the number of roots of degree-$1$ primes dividing $p\mathcal O_K$ (for unramified primes) by choosing a Frobenius $\varphi_p\in\operatorname{Gal}(L/\mathbb Q)$ for $p$ and then counting the number of cosets $H\sigma\in H\backslash G$ such that $$H\sigma=H\sigma\varphi_p.$$ Putting everything together, the number of roots of $f(x)\pmod p$ is $$\#\{H\sigma\in H\backslash G:H\sigma=H\sigma\varphi_p\}.$$ Plugging this into the expected value computation and then using Chebotarev's theorem finishes the statement after an application of Burnside's lemma. However, I think something more sophisticated than Burnside's lemma is necessary to compute this variance.