One method I did not see listed in that linked MathWorld article is to choose the points from a standard normal distribution in the plane, i.e., with density given by p(x,y) = (2π)^(-1) exp(-(x^2 + y^2)/2). This is of course circularly symmetric, but it's not at all clear to me that four points picked from this distribution will have the same chance of their convex hull being a quadrilateral as four points picked from the uniform distribution on a disk. Now I'm really curious: What *is* the probability that the convex hull of four points chosen from the standard normal distribution in R^2 will be a quadrilateral??? —Dan ----- Keith, Very interesting! I will have to study your message carefully, but I just wanted to mention something Ed Pegg pointed out. Sylvester looked at this question, see https://mathworld.wolfram.com/SylvestersFour-PointProblem.html For a square, his theorem is that the prob. of getting a convex quadrilateral is 25/36 , which is exactly the value you found! -----