"Steve Gray" <stevebg@adelphia.net> asks:
In the unit square pick 4 points A,B,C,D at random (x,y each uniformly distributed). Draw lines AB, BC, CD, DA. What is the probability that two of these lines will cross (that is, we get a reflexive quadrilateral)?
It's easy to see that the answer is two thirds of the probability that the convex hull of {A,B,C,D} has four sides. That holds for any i.i.d. selection of the points (consider the points drawn from a circle).
With 5 points there are several possible topologies: no crossings, one, two, three, or five. Is 4 crossings possible?
No, but I forget the good proof, and hesitate to give you a tedious one. For N points, just characterizing the topologies gets interesting. Do we even have a closed form for the maximum number of self-intersections of the tour {p1, p2, ..., pn, p1}? Dan Hoey@AIC.NRL.Navy.Mil