He was computing the probability of a Hamming distance of exactly D instead of at most D. To get the "at most" version, you could just add up the exact ones, b

He was computing the probability of a Hamming distance of exactly D instead of at most D. To get the "at most" version, you could just add up the exact ones, but it's better to just use the central limit theorem instead. As for your math, I think you are over counting a little :). The problem is, 100...0 is counted by both ??0...0 and ?0?00...0, where ? marks a digit that you've allowed to be free, and 0 marks a "chosen" 0.