TITLE: Odds of ellipse from five random points
QUESTION [8 upvotes]: 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.

REPLY [2 votes]: 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.