Let $G_{n,m}$ denote an uniform random graph with $n$ vertices and $m$ vertices. Take two edges $e,f$ in $G_{n,m}$. What is the probability that these two edges share exactly one vertex?
I was thinking: Say $e=\{e_1,e_2\}$ and $f=\{ f_1, f_2\}$. Then this probability is $$ P(e \cap f \neq \emptyset) =P(e_1 = f_1) + P(e_1 = f_2) + P(e_2 = f_1) + P(e_2 = f_2) = 4 P(e_1 = f_1) $$ and $P(e_1 =f_1) = 1/n $ as if $f_1$ is fixed there are exactly $1$ out $n$ good outcomes. Is this correct?
Are these undirected graphs with no self-loops? If so, imagine the first edge is drawn randomly among $\binom{n}{2}$ possible edges. There are $2(n-2)$ other edges that share a vertex with the first edge ($n-2$ for each vertex in the first edge). So the probability that the next edge shares a vertex with the first is $$\frac{2(n-2)}{\binom{n}{2}- 1}$$ Note that in the case $n = 3$ the probability is $1$ which makes sense because any two edges share a vertex.