I am thinking of a quantitative (possibly based on random graph theory) or qualitative (say, based on topological ideas, e.g. Baire's theorem in the Gromov-Hausdorff metric space) information about how many finite graphs (perhaps: how many graphs on a given number of nodes $n$) are line graphs, or perhaps just *generalized line graphs*: these are those graphs whose adjacency matrix $A$ can be written as $A+2I=J^T J$, where $J^T J$ is the Gramian matrix of a vector system and $I$ is the identity (see the Godsil-Royle 2001 or Cvetkovic-Rowlinson-Simić 2004 for references). 


Ideally, I would expect an assertion like: as $n\to \infty$, the probability of a graph on $n$ nodes to be a line graph goes to $0$. Or, even better: the probability goes to $1$ :)



How many line graphs are there?
(Per request, post edited to discuss $G\_{n,p}$ for values other than $p=1/2.$ This is a routine line of reasoning in probabilisitic combinatorics. See the first few chapters of [Alon and Spencer](http://www.wiley.com/WileyCDA/WileyTitle/productCd-1119061954.html) for more.)


Line graphs are [claw-free](http://mathworld.wolfram.com/Claw-FreeGraph.html). And the [random graph](https://en.m.wikipedia.org/wiki/Erd%C5%91s%E2%80%93R%C3%A9nyi_model) $G\_{n, p}$ has in expectation $\Theta (n^4 p^3 (1-p)^3)$ claws (i.e., it's roughly some constant multiple of this). Having claws is not a monotone property, so we don't expect a classical threshold for when (as a function of $p$) we should think there ought to be claws. But nonetheless, a standard application of the [second-moment method](https://en.m.wikipedia.org/wiki/Second_moment_method) would tell you for what $p$ this property will hold almost surely (i.e., with probability tending to $1$ as $n$ tends to infinity). I did not check the following, but it's right up to some logs that I'm not writing:


The analysis splits into three regimes based on $p$.


* If $n^{-4/3} \ll p \ll 1-n^{-4/3}$, then the number of claws will be concentrated about its mean, and the graph will have a claw with probability tending to $1$. Hence, the graph will fail to be a line graph almost surely.
* If $p$ (resp $1-p$) is approximately of the form $C n^{-4/3}$, then the graph (or its complement) will be very sparse. I would guess that the number of claws would follow a Poison distribution, so it will be claw-free with probability tending to some $f(C)$. In this range, the graph (or its complement) will be extremely sparse, and in fact being claw-free would almost surely coincide with the property of being a disjoint union of paths (or its complement). Thus, in this range, $G\_{n,p}$ (or its complement) will be a line graph with probability $f(C)$.
* If $p \ll n^{-4/3}$ (resp. $1-p \ll n^{-4/3}$) then $G\_{n, p}$ (resp. its complement) will almost surely be a disjoint union of paths, and hence it will be a line graph with probability tending to $1.$


[I'm being sloppy with the above treatment in the "critical window" when $p \approx n^{-4/3}.$ If you wanted to do this better, that's certainly possible.]


So in summary, unless $p$ is extremely small or extremely large, $G\_{n,p}$ will almost surely fail to be a line graph. In particular, this holds for $p=1/2$, so almost all graphs fail to be line graphs.


