What does a random countable non-rado graph look like? We can define the countable random graph $G$ as a graph whose nodes $\mathbb N$ and such that for any $n,m$, the probability that $n$ and $m$ are connected is $\frac 12$ (or $0 \lt p \lt 1$ if you want to generalize). $P(G \cong R) = 1$, where $R$ is the Rado graph. This makes $G$ pretty boring, since the probability that $G$ satisfies any given property is either $0$ or $1$, according to whether $R$ satisfies it.
My question is, what does the probability distribution of $G | G \ncong R$ look like? (For those unfamiliar with that notation, it denotes a conditional random variable. For a random variable $X$ and event $E$, $Pr((X|E) \in S) = Pr(X \in S | E)$, for any given non-random set $S$.) Intuitively, $G | G \ncong R$ is the random countable graph, subject to the condition that is not the Rado graph.  In particular, what is the absolute mode(s) of $G | G \ncong R$?
Note that is certain that $(G | G \ncong R) \ncong R$.
 A: There is an obvious problem even setting up this question: in the usual distribution, the set of non-Rado graphs has measure zero, and you can't in general condition against a measure-zero set.
However, there is a natural (family of) attempts to get a reasonable version:
Remember that the Rado graph is (up to isomorphism) the unique countable graph with the extension property: for any disjoint finite sets $U, V$ of vertices, there is a vertex $x$ connected to every $u\in U$ and not connected to any $v\in V$. So if you want to build a non-Rado graph, it's enough to ensure that the extension property fails, and moreover this is the only way to build a non-Rado graph in a sense.
How do we do that? Here's a natural attempt:


*

*Fix natural numbers $a<b$. Our graph will have natural numbers as vertices. Setting $U=\{x: x\le a\}$ and $V=\{y: a<y\le b\}$, we will guarantee that no $x$ is connected to each element of $U$ and no element of $V$, and otherwise things will be random.

*At stage $s$, we look at vertex $s$ and decide which vertices $t<s$ this vertex is connected to. There are $2^s$-many options here, and some of them are illegal: that is, some of them make $t$ connected to every element of $U$ and no element of $V$. So - using the standard distribution - we pick one of the legal options, at random.
It's now not hard to show that with probability one, we get the same graph up to isomorphism from this process; call this graph "$G_{a, b}$." However, it's also not hard to show that $G_{a, b}$ depends heavily on $a, b$ - in fact, $G_{a, b}$ is the unique graph (up to isomorphism) satisfying the limited extension property - there are finite $U, V$ with cardinalities $a, b-a$ respectively such that whenever $X, Y$ are disjoint finite sets of vertices, exactly one of the following holds:


*

*$U\subseteq X, V\subseteq Y$.

*There is some vertex $v$ which is connected to every element of $X$ and no element of $Y$.
So ultimately you don't get anything from this process which is much different from the Rado graph itself. This makes the outline above not very satisfying; however, it's not clear to me what a better option is.
