I am interested to find out what’s the probability that an Erdos-Renyi graph $\mathcal{G}(n,p)$ is a regular graph?

I believe this is a really hard question whose non-asymptotic results are probably impossible to find. Therefore, I would probably settle for any asymptotic large deviation result or may be some upper bound of the probability.

I think there is no closed form of joint degree distribution of Erdos Renyi graph, which makes this a hard problem.

Any result in this direction will be very much appreciated.


I don't think this counts as a full answer but I believe it is too long for the comments. It may give insight on how to approach this

  1. Let $G$ and $G'$ be two labelled graphs on $\{1,\ldots, n\}$ [i.e., $G=G'$ iff $ij \in E(G)$ implies $ij \in E(G')$ and vice versa mere isomorphism doesn't count] with the same number $m$ of edges. If a graph is drawn according to ER$(n,p)$, the $G$ and $G'$ have the same probability $P_n(m,p)$ of being chosen (which is a function of $n,p$ and $m=|E(G)| = |E(G')|$). This I know can be calculated explicitly.

  2. There is a formula (I think) that approximates the number $N(k)$ of $k$-regular graphs on $\{1,\ldots, n\}$.

So the probability $P$ of chossing a $k$-regular graph if drawing according to ER$(n,p)$ would come to

$$P= \sum_{k=1}^{n-1} N(k)P_n\left(\frac{kn}{2},p\right)$$


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