A second order property that is a counter-example to the $0$-$1$ law for random graphs 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.
 A: "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).
