Separating words with random DFAs One of the interesting open problems about DFAs listed in Are there any open problems left about DFAs? is the size of a DFA required to separate two strings of length $n$.  I am curious if there any results about the ability of a random DFA to separate two given (nonrandom) strings.

Clearly a random DFA with sufficiently many states separates strings with high probability.  Specifically, if $u,v \in \Sigma^n$, a random DFA with $O(n)$ states is unlikely to ever revisit the same state once it reaches the first place where $u$ and $v$ differ, and therefore separates $u$ and $v$.

Can we do better?  Ideally, what is the smallest $f(n)$ s.t. that a random DFA with $f(n)$ states separates strings of length $n$ with positive probability (or perhaps probability $\ge 1/2$)?  A brief search didn't turn up many results on the properties of random DFAs; all I could find was http://arxiv.org/abs/1311.6830.
[Edit: this answer does not work, see comments.]

This is just an informal idea and I don't know if it helps, but it's too long to be given as a comment. Also, I am not at all familiar with random DFAs, so maybe I have a wrong intuition of how you should reason about probabilities on them, but hopefully this is not entirely worthless.

I will suppose that your bounds should depend on how much $u$ and $v$ differ; if they don't, it seems clear to me that the worst case are strings differing only by their first character (strings differing at a set $X$ of positions have more chances of being told apart than strings differing at a set $Y \subset X$ of positions, I'd say, and putting the difference as early as possible gives you opportunity to resynchronize).

I will also look at the probability that the words are distinguished, namely, they reach different states. I guess you would then need to adapt for being accepted or rejected based on how your random DFAs allocate final states. If each state has a probability 1/2 of being final, then when the strings end up at the same state they are not distinguished, and when they end up at different states they have probability 1/2 of being distinguished.

Now I will consider the word $w$ obtained from $u$ and $v$ as follows: $w_i = 1$ if $u_i = v_i$, and $w_i = 0$ otherwise. I think it is clear that $w$ is the only interesting thing to consider about $u$ and $v$.

Now, define $p(i)$ the probability that we are at the same state after reading prefixes of length $i$ of $u$ and $v$, and $q(i) = 1 - p(i)$ the probability that we aren't.

I think we have $p(i+1) = p(i) + q(i)/n$ when $w_{i+1}$ is $1$. Intuitively, we are at the same state after reading $i+1$ letters either when we were at the same state after reading $i$, or when we were at two different (random) states, we drew two transitions to random states, and they happened to be the same one. Likewise, we have $p(i+1) = 1/n$ when $w_{i+1}$ is $0$: you are drawing two random states, no matter where you started from.

From this I think you could compute the probability of being at the same state after reading $u$ and $v$.