Given a conic $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ with integers and random coefficients, what is more probable? To find a rational point on the conic or not?



Conics, rational points and probability
For any given prime number $p > 2$ the probability that there is no $p$-adic point is $\ge c/p$ for some constant $c > 0$ indpendent of $p$. Since the sum over $1/p$ diverges, this implies that the `probability' (or rather, density) of conics with a rational point is zero.


EDIT: I assume that the `probability' is meant in the sense of density, i.e.
$$ \lim\_{N \to \infty} \frac{\#\{(A,B,C,D,E,F) \in ({\mathbb Z} \cap [-N,N])^6 : \exists \text{ rational point}\}}{(2N+1)^6}. $$
The probability of $p$-adic solubility (with respect to the standard measure on ${\mathbb Z}\_p^6$) comes down to the $p$-adic density of pairs of $p$-adic integers $a, b$ such that the Hilbert symbol $(a,b)\_p = -1$.
This is the case (e.g.) when $v\_p(a) = 1$, $v\_p(b) = 0$ and $b$ is a nonsquare mod $p$; the probability for this is $\frac{p-1}{p^2} \cdot \frac{p-1}{2p} \ge \frac{2}{9p}$.


EDIT 2: I realized that the map that produces $a,b$ from $A,B,C,D,E,F$
very likely does not preserve $p$-adic measure. So here is an alternative
and more direct argument. We work with projective coordinates, as
suggested by Joe Silverman. By Hensel's Lemma, the non-existence of a
smooth ${\mathbb F}\_p$-point on the reduction mod $p$ of the conic
is a necessary condition for the non-existence of $p$-adic points,
and this is also sufficient when the conic is regular over ${\mathbb Z}\_p$.
So whenever the reduction mod $p$ is the product of two conjugate and
distinct linear forms with coefficients in ${\mathbb F}\_{p^2}$ *and*
the intersection point of the two corresponding lines is regular,
then there is no $p$-adic point on the conic. The $p$-adic measure
(as a subset of ${\mathbb P}^5({\mathbb Q}\_p)$) of the corresponding
set is
$$\frac{1}{2} \frac{\#{\mathbb P}^2({\mathbb F}\_{p^2}) - \#{\mathbb P}^2({\mathbb F}\_p)}{\#{\mathbb P}^5({\mathbb F}\_p)} \frac{p-1}{p}
 = \frac{(p-1)^2}{2(p^3+1)} \ge \frac{3}{14p} .$$
(The factor $(p-1)/p$ is the probability that the intersection point
is regular.) The paper by Bhargava, Cremona, Fisher, Jones and Keating
linked in post.as.a.guest's answer gives the precise value $p/(2(p+1)^2)$.


