Maybe this question is off topic or duplicate, but maybe there is some information which I am not aware of.

Let $G_n$ denote the set of non-isomorphic simple graphs and $|G_n|=g_n$. Also, let $U_{n,m}$ denote the set of non-isomorphic simple graphs with $n$ vertices and $m$ edges, and $|U_{n,m}|=u_{n,m}$. I am interested in the below limit:

$$\lim_{n\rightarrow \infty}{T_n=\frac{\sum\limits_{m=1}^{\binom{n}{2}}{\binom{u_{n,m}}{2}}}{\binom{g_n}{2}}}.$$

The sequence $T_n$ shows the probability that two graphs among graphs with $n$ vertices have the same number of edges. Is it possible that the above limit exists and is not equal to zero?

Maybe this observation helps to see something. We have $$u_{n,1}+\cdots+u_{n,\binom{n}{2}}=g_n,$$

so, $$\binom{u_{n,1}}{2}+\cdots+\binom{u_{n,\binom{n}{2}}}{2}=\binom{g_n}{2}-\sum\limits_{1\leq i<j\leq \binom{n}{2}}{u_{n,i}u_{n,j}}.$$

Therefore, we can write $T_n$ as follows

$$T_n=1-\frac{\sum\limits_{1\leq i<j\leq \binom{n}{2}}{u_{n,i}u_{n,j}}}{\binom{g_n}{2}}.$$

Some numerical results: $T_4=0.09090909091, T_5=0.1051693405, T_6=0.1055417701, T_7= 0.09997318375,$ $T_8=0.09345209613, T_9=0.08721068637, T_{10}=0.08077014830$.