I either need Sage help, or math help, in order to figure out whether all these fun facts about pairwise intersections extend to triple (and higher order) intersections. In one-dimension, I can show that the expected length of the intersection of k unit intervals, conditional upon the intersection being nonempty, is 1/k. It follows (via the additivity of expected values of random variables and the multiplicativity of expected values of independent random variables) that if you take the intersection of k a-by-b rectangles, conditional upon the intersection being nonempty, the intersection has expected perimeter (2a+2b)/k and expected area (ab)/k^2. This extends easily to parallelograms and parallelohedra, but where I get stuck is thinking about centrosymmetric hexagons. Can anyone see how to prove (or disprove) that if you take the intersection of three translates of a centrosymmetric hexagon, conditioned upon the intersection being nonempty, the expected perimeter is P/3 and the expected area is A/9, where P and A are the perimeter and area of the hexagon? Alternatively, can anyone tell me why the following Sage code (for computing the expected number of vertices of the triple intersection) doesn't work? v = [] h = 866/1000 for i in range(10): s=4*floor(1000000*random())/1000000-2 t=4*floor(1000000*random())/1000000-2 u=4*floor(1000000*random())/1000000-2 v=4*floor(1000000*random())/1000000-2 p1=Polyhedron(vertices=[[-1,0],[-1/2,h],[1/2,h],[1,0],[1/2,-h],[-1/2,-h]]) p2=Polyhedron(vertices=[[-1+s,0+t],[-1/2+s,h+t],[1/2+s,h+t],[1+s,0+t],[1/2+s,-h+t],[-1/2+s,-h+t]]) p3=Polyhedron(vertices=[[-1+u,0+v],[-1/2+u,h+v],[1/2+u,h+v],[1+u,0+v],[1/2+u,-h+v],[-1/2+u,-h+v]]) p12=p1.intersection(p2) if len(p12.f_vector())>1: p123=p12.intersection(p3) if len(p123.f_vector())>1: v.append((p123.f_vector())[1]) sum(v)/len(v) If I knew how to fix the bug, I think I'd be able to do numerical experiments regarding perimeter and area as well. Thanks, Jim Propp