It's known that five points determine a conic section. Five random points can go right into the $6\times6$ matrix, and then the $A x^2 + B xy + C y^2$ part can be looked at. If $B^2-4AC<0$, it's an ellipse. Five random points will almost never produce circles or parabolas, so the results will be ellipses and hyperbolas. What are the odds of an ellipse?
In a random run of 100000 trials, I got 27974 ellipses. "It's less than $e/10$," seems like a solid answer. Anyone have anything more specific?
EDIT: As Oscar points out, I should have said "It's more than $e/10$." In my trial, real-values points were randomly picked from a unit square. Square Triangle Picking methods might be applicable.
EDIT2: Aretino points out that odds of a convex pentagon are $49/144≈0.34$. So how can points making a convex pentagon give a non-ellipse? Here's a picture. With the red points fixed, the black points are outside of the convex hull yet still yield a non-ellipse.
EDIT3: That spray of points above goes back to Newton, Philosophiae naturalis principia mathematica, 1687, where he solved the 4 point parabola (another version). If a point is between one of the two parabolas and the degenerate lines, then it gives a hyperbola.
Let $x$ be the probability of getting an ellipse. For the five points to form a convex polygon, they must either lie on an ellipse (with probability $x$) or lie on the same branch of a hyperbola: if each point has probability $1/2$ of lying on either branch of the hyperbola this should happen with probability ${1\over16}(1-x)$.
As the probability that five points chosen at random inside a square are the vertices of a convex pentagon is $49/144$, we have the equation $$ x+{1\over16}(1-x)={49\over144}, \quad\hbox{whence:}\quad x={8\over27}\approx0.296. $$ This result is however a little higher than simulations suggest.