Theory with the independence property, but not in two variables It's known that a theory $T$ has the independence property if and only if there is a model $M$ of $T$ and a formula $\phi(x,\overline{y})$ which has the independence property with respect to $M$. Here $x$ is a single variable and $\overline{y}$ is a finite tuple of variables. Are there examples of theories $T$ with the independence property such that for any model $M$, there is no formula $\phi(x,y)$ where both $x$ and $y$ are single variables which has the independence property? 
I've tried to construct examples, say by taking a bijection $f: \omega \times \omega \to [\omega]^{<\omega}$ and then forming a theory $T$ in a language with a single ternary relation $R(x,y,z)$ to be the complete theory of the structure $<\omega, R>$, with $R$ interpreted to satisfy $R(x,y,z) \iff x = f(x,y)$. In the resulting theory, the formula $R(x,y,z)$ will clearly have the independence property, but it's not so clear how to analyze the definable sets in two variables for this theory. I also realize that the resulting theory is highly dependent on choice of $f$. 
The same question could also be asked with NIP replaced by other combinatorial properties such as stability. 
 A: Let $L$ be the language with a single ternary relation $R$. Let $T$ be the theory containing the single sentence $\forall x\forall y\forall z(R(x,y,z)\rightarrow (x\neq y\land y\neq z \land x\neq z))$, which says that whenever $R(a,b,c)$ holds, the elements $a$, $b$, and $c$ must be distinct.
Let $K$ be the class of finite models of $T$. You can check that $K$ is a Fraïssé class, so it has a Fraïssé limit $M$. Let $T^* = \mathrm{Th}(M)$. Then $T^*$ is countably categorical and has quantifier elimination. 
Now $T^*$ does not witness the independence property (or the order property, or any other bad property) with a formula in two variables, since there are no interesting formulas in two variables: Every formula $\varphi(x,y)$ is equivalent to $\top(x,y)$, $\bot(x,y)$, $x = y$, or $x\neq y$. Indeed, any pair of two element substructures of $M$ are isomorphic, and hence conjugate by an automorphism of $M$. 
But the formula $\varphi(x,y;z)$ has the independence property. Indeed, for any $n$, there is an element of $K$ with domain $\{a_i,b_i\mid i\in [n] \}\cup \{c_X\mid X\subseteq [n]\}$ and $R = \{(x,y,z)\mid x = a_i, y = b_i, z = c_X \text{ and }i\in X\}$, and hence we can find this configuration in $M$. 
Closing comments:


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*Everything I said above also works if we add an axiom to $T$ asserting that $R$ is symmetric, so that we're dealing with $3$-hypergraphs, as suggested by Kyle in the comments (though considering forbidden configurations is an unnecessary complication). 

*Incidentally, $T^*$ is simple (in both the symmetric and general case). 

*You can increase the arity from $3$ to $n$ to get examples where the minimum number of free variables in a formula with the independence property is arbitrarily large.
