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?

$\begingroup$ It depends on the distribution on integers. $\endgroup$ – Fedor Petrov Dec 23 '15 at 18:24

2$\begingroup$ @fedor The standard way to order is as a six tuple of of rational numbers treated as a point in projective space ordered by height. $\endgroup$ – Joe Silverman Dec 23 '15 at 19:49
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 $\frac{p1}{p^2} \cdot \frac{p1}{2p} \ge \frac{2}{9p}$.
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 nonexistence of a smooth ${\mathbb F}_p$point on the reduction mod $p$ of the conic is a necessary condition for the nonexistence 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 $$\frac{1}{2} \frac{\#{\mathbb P}^2({\mathbb F}_{p^2})  \#{\mathbb P}^2({\mathbb F}_p)}{\#{\mathbb P}^5({\mathbb F}_p)} \frac{p1}{p} = \frac{(p1)^2}{2(p^3+1)} \ge \frac{3}{14p} .$$ (The factor $(p1)/p$ 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 $p/(2(p+1)^2)$.

$\begingroup$ Thanks Michael Stoll. Why the constant $c$ is independent of $p$? $\endgroup$ – user84475 Dec 23 '15 at 18:20

$\begingroup$ It is more than $c/p$ for the natural measure on $p$adics. But there is no probabilistic measure on integers which is uniform modulo any prime. $\endgroup$ – Fedor Petrov Dec 23 '15 at 18:25

1$\begingroup$ @MichaelJoyce : thanks, I've fixed the omission. $\endgroup$ – Michael Stoll Dec 23 '15 at 20:05

$\begingroup$ It would seem to be more natural to restrict to 6tuples with $\gcd(A,B,\ldots,F)=1$, since the quadratic form is determined by the homogeneous coordinates $[A,B,\ldots,F]$. Certainly this is the right thing to do if one wishes to generalize from $\mathbb Q$ to a number field. Of course, this won't affect your conclusion that the density is 0. $\endgroup$ – Joe Silverman Dec 23 '15 at 22:43
Problems of this type are considered by Serre in the paper:
Serre  Spécialisation des éléments de $\mathrm{Br}_2(\mathbb{Q}(T_1,\ldots, T_n))$
The case relevant to you is Exemple 4. Here Serre shows that
\begin{align*} &\#\{A,B,C,D,E,F \leq N : \\ & Ax^2+Bxy+Cy^2+Dx+Ey+F=0 \text{ has a rational point} \} \ll \frac{N^6}{(\log N)^{1/2}}. \end{align*}
This gives a more precise quantitative version of "probability $0$" mentioned by Michael Stoll.
Hooley has in fact shown that Serre's bound is sharp. This result is the object of the paper:
Hooley  On ternary quadratic forms that represent zero II.
There is a paper, that answers this in more generality.
https://www.dpmms.cam.ac.uk/~taf1000/papers/isotropic.html
In your case, the probability is 0 (asymptotically) under conditions that the distribution is piecewise smooth and rapidly decaying, though likely can be generalized, in your case.

$\begingroup$ The linked paper is definitely helpful. $\endgroup$ – Sebastian Goette Dec 24 '15 at 10:07