Let $G$ be a random graph, $G(n,p)$, with $n$ vertices, and each edge is included in the graph with probability $p$ independent from every other edge.
Suppose there is at least one path between $v$ and $u$. What is the probability of having a path of length $l$ between $v$ and $u$ in $G$?
This question was asked for many times on this site, but none of has an exact and straightforward answer. In the other hand, I read numerous papers, but I can't find the solution for it. Do you have any suggestion?
To have a path of length $l$, the graph must contain edges from $v$ to $w_1$, $w_1$ to $w_2$, ..., $w_{l-2}$ to $w_{l-1}$ and $w_{l-1}$ to $u$. The vertices $w_i$ may be chosen arbitrarily. The number of possible sequences $w_i$ is easily seen to be $\prod_{i=0}^{l-2} (n-2-i)$, and then for any fixed choice, the probability that all of the necessary edges are in the graph is just $p^l$. So if I am just naive and don't worry about double counting, I get a probability of
$$\left ( \prod_{i=0}^{l-2} (n-2-i) \right ) p^l.$$
Now fix the double counting issues, if there are any. Finally divide by the summation of these probabilities over $l$, to take into account the assumption that at least one path exists.