I found a math discussion site on the web where someone calling themself Poliwrath
posed and solved the question of what is the probability that if A,B,C,D are random points in a square, then AB and CD intersect?

Poliwrath's clever reasoning is as follows: If X,Y,Z are any three random points in the unit square, let T be the expected area of the triangle XYZ. (This just happens to be 11/144.) Hence the probability of (A lying in BCD) = T, and same for B lying in ACD, C lying in ABD and D lying in ABD. Since these 4 events are disjoint (!), the probability that one of them occurs is 4*T, and so the probability that none of them occurs, i.e. the probability that the convex hull of A,B,C,D is a quadrilateral, is 1-4T.
Among the 24 equally likely ways that A,B,C,D can be assigned to the corners of any given quadrilateral, just 1/3 of them involve AB interesecting CD (since wherever A is, B can be adjacent with prob. = 2/3, or opposite, with prob. = 1/3). This shows that P = [the probability AB and CD intersect] is (1-4T)/3. Assuming T = 11/144, this gives P = 25/108. (I don't know the calculation giving T = 11/144, but both this figure and the 25/108 are well approximated by my computer simulations, so I believe them.)

--Dan