Do the functions of the collatz conjecture and their inverses model a Trapdoor Function? 

If given a, b, a^-1, b^-1 and your choice of f(x), is it “hard in the average case” to find some secret x?

I propose that the probability: 
Pr[f(F(f(x)))=f(x)] < n^-c
=Pr[(2^-n)^2] < (n^-(n-1))
Simplifies to:
1 < (log(2)) \* ((log(n))^-1) \* (e^(n+1))

Looking at the picture from Wikipedia that others are referencing.
Let D be the domain of all natural numbers that I can choose x from.
Let R be the domain of all natural numbers that you can choose f(x) from.
The arrow marked “easy” is the promise that I am choosing a natural number that can be calculated by applying functions A and B to 1.(c(x) exists)
It is hard on average for you to calculate how to apply given functions A,B A^-1 and B^-1 to f(x) to return x.(Sometimes this time is less than the verification process)
However, I can quickly apply A^-1 and B^-1 to x until 1 is returned, retain and reverse the order the operations applied and pass you this information (t) making finding x trivial by finding the shortest path to collision.