Suppose you have a cyclically ordered sequence of n dice, each of which has s sides, and each of which beats the (cyclically) next one with the same probability p > 1/2.
I believe I saw some reference that claims p <= 3/4 under all circumstances; maybe another that sharpened this to p <= sqrt(5 - 1)/2 = .618....
Focusing on the combinatorics rather than the numbers: for a given set of dice, consider the directed graph with the dice as vertices and an edge a->b iff a beats b. What graphs are possible? (Is it every directed graph that never has a->b->a, at least if you allow the dice to have more than 6 faces?) [after a little googling] Any tournament (i.e., directed graph where for any a,b exactly one of a->b and b->a is present) can be realized in this way if you have enough faces. The first step is to find a set of permutations of the vertices such that a->b iff a precedes b in more than half the permutations. (There's a paper by McGarvey that proves this is always possible. I haven't read it. And another by Erdos and Moser that proves you only need O(n/logn) permutations, which will lead to O(n/logn) faces on each die. I haven't read that either.) So, let those permutations be a11 a12 a13 ... a1n a21 a22 a23 ... a2n ... ak1 ak2 ak3 ... akn (k permutations of n elements; n, remember, is the number of dice). Give the dice as many faces as there are permutations; and suppose bij is the rank of j in the i'th permutation; label the dice b11 b21 ... bk1 <-- die 1 b12 b22 ... bk2 <-- die 2 ... b1n b2n ... bkn <-- die n. Then die p beats die q iff b*p > b*q usually; i.e., if the rank of p > the rank of q in most of the permutations; i.e., if p comes after q in most of the permutations. Hmm, we have our arrows the wrong way round, but that's boring :-). So, the answer is indeed that everything is possible (well, except that I don't know about which edges you can arrange to be absent because neither of two dice beats the other) if you have as many faces as you like. There seem to be some theorems about which graphs are realizable with a given number of permutations, but nothing much like a complete answer yet. -- g