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It's well known that the following $0$-$1$ holds for random graphs:

If $P$ is a first order property of graphs, then for any constant $p$, the probability that $P$ holds in $G(n,p)$ has a limit of zero or one as $n \to \infty$

My question is for an example of a higher order property (preferably just second order) for which the $0$-$1$ law doesn't hold, i.e. property $P$ so that the probability that $G(n,p)$ has property $P$ doesn't converge to $0$ or $1$. Most properties of random graphs that I can think of either do or don't hold for $G(n,p)$ asymptotically for fixed $p$ (for instance whether or not it has a non-trivial automorphism). Any help is appreciated.

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"There are an even number of vertices."

This is second order: say that there is a unary function $f $ with no fixed points such that for all $x $, we have $f (f (x))=x $.

EDIT: here's another example. I'll assume $p={1\over 2}$ here, and leave it as an exercise to generalize it to other $p$.

Consider the sentence "No more than than half the pairs of vertices are connected by an edge." Clearly if $p={1\over 2}$, the asymptotic probability of this sentence is ${1\over 2}$, so this is a counterexample to the $0-1$ law. Perhaps surprisingly, the sentence is expressible in a second-order way! HINT: we want to say "there is a surjection from $\{$pairs not corresponding to an edge$\}$ to $\{$pairs corresponding to an edge$\}$. Quantifying over functions and expressing surjectivity is easy in second-order logic, but these functions have domain and range the set of ordered pairs, not the domain of the graph itself! Do you see a way around this? (HINT: a function can be thought of as a special kind of binary relation. Think about 4-ary relations . . . )

Note that these two examples represent the two different ways the $0-1$ law can fail: either the limit of the probabilities doesn't exist (first case), or it exists and doesn't equal $0$ or $1$ (second case).

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    I'm not following your second example: I understand that your statement is second order, but how is that related to $G(n,p)$? – 2017-01-25
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    @MarcusM Regardless of what $p$ is, what is the probability that $G(n, p)$ satisfies the sentence in my answer if $n$ is even? (HINT: what does "$n$" mean in "$G(n, p)$"?) If $n$ is odd? Now think about $n\rightarrow \infty$ . . . – 2017-01-25
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    Okay, I understand that; I get that that's a valid example, but I find a bit disappointing, because it has nothing to do with the probability at play. – 2017-01-25
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    @MarcusM That's sort of the point - second-order logic lets you say "inherently unprobabilistic" things. But I'll add another example. – 2017-01-25
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    Great, thanks! That's a good way of thinking about it. – 2017-01-25