Let say, I have two binary strings with length N, chosen from a set where there are $2^N-K,(K \ge 0)$ independent strings. What would be the expected number of Ones at the same index from two randomly picked strings from the set?

For example, 0010 and 1010, the number of ones at the same index is 1. Can it be by somehow related to the expected hamming distance between two binary strings?

------my own guess, some one could please verify---------

**sorry for the mess, unintentionally, the new problem was posted with this part..

having 1 at the i-th position is an independent event. So, let P(c_i=1) is the probability of having a common 1 at i-th position. Then, the expected number of shared 1s will be $\sum_{0..N-1} P(c_i)$. From the $2^N-K$ set, for ith position, count the number of 1s (denote $N^1_i$), and the number of 0s (denote $N^0_i$). Then $P(c_i)=\frac{N^1_i C 2}{(N^1_i+N^0_i) C 2}$. (C is combinations) when $N^1_i>=2$, otherwise $P(c_i)=0$.

For an example of {11110,1111,01110}, it gives me 3.66666, which sounds correct.

`$2^N-K$`

of the`$2^N$`

binary strings of length $N$, and then, from those`$2^N-K$`

strings, you randomly choose two and check in how many places both have 1's. But that seems equivalent to just choosing 2 among all`$2^N$`

strings, so it's probably not what was intended. I'll vote to close, as "not a real question", with the hope that the OP will clarify what he intended to ask. – Andreas Blass Oct 2 '12 at 8:37