Q: When a binary number is incremented by 1, there may be a carry that may carry over several bit positions i.e. 1010111 is incremented and the carry distance is 3. Given a natural number n, find the average carry distance when incrementing a binary number b in the inverval 2n <=b <= 2n+1, assuming all 2n numbers in this interval occur equally often. So far, I have gotten a pattern such that 2n / 2n. However, this is just equal to 1. I am really stuck here. Any help, hints or clarification would be great, thanks.