my original solution for d4 was to take the top two numbers and subtract. It almost works; it's just that if you see a 2 and no 6, then you have to pretend the 2 is a 0 in order for the odds to work out. Which then has to be explained and bleah.

I guess this is the winner, then

                a
      4        /
        .__d__.
       / \     \
      c   b  1  c
     /     \     \
__b_.   2   .__a__. 5
     \     /     /
      a   d  3  b
       \ /     /
        .__c__.
     6         \
                d

(upper faces hi-lo/lo-hi or upper edge ac|bd (or ab|cd) for d2, smallest side face for d3, upper edge for d4, left side for d6, upper edge + hi-lo for d8, upper edge + next clockwise for 4-choose-2,... whee)