Consider the set of all possible two-to-one functions that map inputs from $\{0, 1\}^{n}$ (domain) to outputs in $\{0, 1\}^{m}$ (co-domain) and let $m > n$.
If I pick a function randomly from this set, what is the probability that it can be computed by a Boolean circuit of size $\text{poly}(m)$?
Note that while the co-domain of the set of functions so defined is $\{0, 1\}^{m}$. However, the range of each individual function is a set of size $2^{n-1}$. What that set looks like depends on what function we randomly pick.
I was thinking of using a counting argument here. Let $M = 2^{m}$ and $N = 2^{n}$. Then, the number of two-to-one functions is given by $$\frac{N!}{2^{N/2}}\times {M \choose N}.$$The number of circuits (made of AND, OR, and NOT gates) of size $k$ is at most $$2^{k^{2}} \times 3^{k}.$$
In general, in the asymptotic limit, the second expression is larger than the first expression -- which would indicate that, perhaps, all two-to-one functions can have polynomial sized circuit description.
However, not every circuit computes a two-to-one function. So, this does not tell me what fraction of two-to-one functions has polynomial sized circuits.