It is difficult to give a precise answer to this question, but personally I am pretty sure that a random four dimensional manifold has no reason at all to have a symplectic structure. What is true for sure is that symplectic manifolds are clustered among all four-manifolds, namely if you look in some places then probability to find a symplectic manifold equals zero.
Let me give one example. Let us consider the following two questions:

**Question 1.** Is the probability for a 4-manifold to have negative Euler characteristic $>0$?

**Question 2.** Is the probability for a symplectic $4$-manifold to have negative Euler characteristic $>0$?

I am sure that one can give a relatively precise answer to Question 1 and my common sense tells me that since the probability for a number to be negative equals $\frac{1}{2}$, the answer to the question is morally yes.

What is very surprising is that the answer to the second question is without any doubt NO. The probability for a symplectic 4-manifold to have negative Euler characteristic is zero. One can classify such manifolds and there is just tiny amount of them. Namely all such manifolds are blow ups of $S^2 \times \Sigma_g$ in at most $4g-5$ points where $\Sigma_g$ is a surface of genus $g$.

**Added.** It seems to me that it would not be unfair to say that the majority of constructions of compact symplectic 4-manifolds come for the moment from algebro-geometric analogues. In particular this is the case for Gompf sum, which is the most used construction. So in some sense the cloud of symplectic manifolds in dimension four that we see now is centred around algebraic surfaces. In particular this fact about symplectic 4-manifolds with negative Euler characteristic is inherited from the theory of algebraic surfaces (though one needs SW theory to prove it for symplectic manifolds). I think no one doubts that algebraic surfaces are not distributed among $4$-manifolds evenly.

I want to add that there is a conjecture saying that four manifolds of constant sectional curvature $-1$ are never symplectic. It is not clear (for me) how well this conjecture is justified but still there are no ideas of how to construct symplectic structures on these manifolds...

orientable4-manifolds are symplectic (with some orientation)?" – Paul Oct 28 '12 at 16:41