Here is Warren's lovely approach: "OK, so here is a vague half-proof that for your problem, the exact expectation equals 4. Do not use just 1 cube. Instead cover all of 3-space with a grid of equal cubes, usual fashion. Now cut that with a random plane [or a random plane with prescribed orientation]. "Now, consider all the polygons your plane is thus divided into (restricting attention to a huge ball on your plane to keep things finite, in limit as radius large). "We want to prove the average number of edges is 4. Well, the line-segments bounding your polygonal cells form a planar graph each of whose valencies is (generically) 4. The dual planar graph also (by Euler's formula, edge-count to vertex-count ratio, mumble) therefore must have expected valency 4." As Warren pointed out in a later email to me: "My proof works (appropriately altered) for any polyhedron which tiles 3-space in a face-to-face manner with a constant number K of cells sharing each edge. For the cube, K=4, for the rhombic dodec, K=3, and for infinitely long eq-triangular prisms, K=6. "So a random plane cut thru a rhombic dodec should be an N-gon with mean(N)=6 exactly." Hopefully some of the emails I've sent to math-fun make more sense now! (Sorry for the odd font; I'd fix it if I knew how to do it on my iPhone.) Jim