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.