Given a binary string of length n with i number of 1s and each digit is changed to the other, independently, with probability $\frac{1}{n}$. What is the probability that more 0s will be changed than 1s?
My approach so far: a - number of 0s flipped, b - number of 1s flipped, and $p = \frac{1}{n}$.
$$P(a>b) = \sum_{k=0}^{i} (P(a>k)P(b=k))$$ $$ =\sum_{k=0}^{i} ((\sum_{w=k+1}^{n-i} P(a=w))P(b=k)) $$ $$ = \sum_{k=0}^{i} ((\sum_{w=k+1}^{n-i} {(n-i) \choose w} p^{w}(1-p)^{n-i-w}) {i \choose k}p^{k} (1-p)^{i-k}) $$
At this point, I feel it is too complicated and I cannot seem to simplify it any further. Did I go wrong somewhere? Is there an easier method? Can it be further simplified? If not is there any good lower bound we can deduce? Any help would be appreciated. Thank You.