Hi @bpet, thanks for the reply.
Actually, the total capacity of all urns is simply the sum of their individual capacities, i.e., C(U_1, ..., U_m) = Sum_i=1^m C(U_i) = m + (m-1) + ... + 1 = m*(m + 1)/2. Thus, adding the (m+1)th ball would not be an issue. I'm generally interested in the specific...