Is a linear factor more likely than a quadratic factor? I choose a reducible (over $\mathbb Z$) monic polynomial of degree four with integer coefficients,
at random. Is it more likely to have  a linear factor or a quadratic factor ?
Formal version of the question : let $N>0$, and
$$B_N=\lbrace (a,b,c,d) \in [|-N,N|]^4 | X^4+aX^3+bX^2+cX+d \text{ is reducible},\rbrace$$
(so that $|B_N|$ has at most $(2N+1)^4$ elements ; in fact, it is known to have
$o(N^4)$ elements, and even $O(N^3)$ if I'm not mistaken). Let
$$
\begin{array}{lcl}
B_{N,1} &=&\lbrace (a,b,c,d) \in B_N | X^4+aX^3+bX^2+cX+d \text{ has a linear factor}\rbrace, \\
B_{N,2}&=&\lbrace (a,b,c,d) \in B_N | X^4+aX^3+bX^2+cX+d \text{ has a quadratic factor}\rbrace,
\end{array}$$ (note that in $B_{N,2}$ the quadratic factor may itself be reducible ; also the intersection $B_{N,1}\cap B_{N,2}$ is nonempty) and
$$
p_n=\frac{|B_{N,1}|}{|B_N|}, q_n=\frac{|B_{N,2}|}{|B_N|} 
$$
Is anything known about the asymptotic behaviour of $p_n,q_n$ and $\frac{p_n}{q_n}$ ?
 A: Here is a heuristic answer to the corresponding question over a finite field $\mathbb{F}_q$. As described in this post, it turns out that for fixed $n$, as $q \to \infty$, the distribution of irreducible factors of degree $k$ in a random monic polynomial of degree $n$ asymptotically approaches the distribution of cycles of length $k$ in a random permutation in $S_n$. 
Hence the probability that a random monic polynomial over $\mathbb{F}_q$ of degree $4$ has a linear factor is asymptotically the probability that a random permutation in $S_4$ has a fixed point, and similarly the probability of an irreducible quadratic factor is asymptotically the probability that a random permutation in $S_4$ has a 2-cycle. These probabilities are, if I haven't miscomputed, 
$$\frac{15}{24} = 0.625, \frac{9}{24} = 0.375$$
respectively. So linear factors are more likely.
As described in this post, as $n \to \infty$ the number of cycles of length $k$ in a random permutation of $S_n$ is asymptotically Poisson with parameter $\frac{1}{k}$. In particular, the probability that there is at least one such cycle is asymptotically $1 - e^{-\frac{1}{k}} \sim \frac{1}{k}$. For $k = 1, 2$ this gives
$$1 - e^{-1} \sim 0.632, 1 - e^{-\frac{1}{2}} \sim 0.393.$$
But it's unclear what relevance this has to the original question. 
