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In this case, assuming $c\lt 1$ is independent of $n$, the number of cycles is asymptotically Poisson with constant mean $f(c)$. So asymptotically there is a constant nonzero probability $e^{-f(c)}$ t …

Take $2n$ vertices. Regular bipartite graphs have $n$ vertices of each colour, and most have only one colouring because they are connected (for degree $\ge 3$). So take the asymptotic number of regul …

First make a "random" quadrangulation, then take a random subgraph of that. Since every planar bipartite graph can be made into a quadrangulation by adding edges, at least you know that every outcome …

(To answer Anthony, I'm taking the two graphs to be on the same vertex set.) This is an argument, without details, that the answer is $\Omega(n^{3/2})$. Generate the graphs two vertices at a time. A …

Consider subgraphs consisting of two cycles with an edge in common (i.e. a theta-graph or something more complex). The number of such labelled graphs with $t$ vertices is at most $n^t$, and the probab …

I'm not going to do this calculation, but here is how it can be done. Any interval which contains the length of the longest cycle with probability tending to 1 will also contain the length of the long …

A random tournament is strongly connected with probability tending to 1 exponentially fast, and all strongly connected tournaments have hamiltonian cycles.

I'm assuming that the degree of a vertex is the number of edges incident with it (other definitions are possible). The degree of a vertex in $G_k(n,p)$ has a binomial distribution $$\operatorname{Bin} …

It's a nice question that I don't know a complete answer to. However it can be seen that the measures are not exactly the same. Take $p$ to be very small. The most likely connected $G$ is a tree, a …

Divide the vertex set into a fixed number of parts, in any way you like (such as randomly). Choose two probabilities $p_1,p_2$, where $p_1\le n^{-\varepsilon}$ for some fixed $\varepsilon\gt 0$ and $p …

I can't find a published proof of this known result, but here is a close miss. In this paper, page 256, is a short proof that a random undirected graph has a unique vertex of maximum degree almost su …

Second edition This is a partial answer to the question per the "Clarification Update", but first I'll generalize a little. Suppose that for each $n$ we have a degree sequence $n_0,n_1,n_2,\ldots$, …

There is a very efficient method. See Nicholas C. Wormald Generating Random Unlabelled Graphs

Make an intermediate graph $H$ with arbitrary random nonnegative edge lengths. Now define a graph $G$ with the same edges, with the length of each edge $vw$ in $G$ being the distance between $v$ and $ …

I don't know if this has been programmed. I'll describe one method based on counting. Let $c_{n,m}$ be the number of labelled connected graphs with $n$ vertices and $m$ edges. Choose a pair of dist …