So, in that case, H(x) = - sum from i = 1 to m of, well, p-sub-i is 1/m, times log base 2 of 1/m, and if we sum up this m-times, it's just going to be this plus this plus this m times---that's going to get rid of the 1/m term---so that's going to be equal to - log base 2 of 1/m---well, - log base 2 of 1/m is just equal to log base 2 of m, using our log rules, and that's the same that we had in our special case, when all the probabilities were equal. 