[fnord_too]

I think this may be solvable the following way:
At some point you will will be 1 or 2 away from every number on the low side (i.e. before you pass or hit any number you will be exactly one or two less than it). For each of these two equally likely occurences there are sequences that land you on the exact number. So you should be able to take both of these starting conditions for an arbitrary number and determine the chance of hitting that number.

For instance, consider 2. At some point you are on 0 or 1 (each equally likely, I can prove that if necessary).

From zero, there are several ways to hit 2, but they are grouped around the number of actions, so:
1 action F - 50%
4 Actions BBFF, BFFB, BFBF 18.75%
8 Actions I think is the next
and so on.

Intuitively, I think this forms a nice managable infinite series, and there is a similar one for starting 1 away. Solve both of those and average the P's and that is the answer. It would take me a long time to get into the right mindset to crunch through that, but I think it is a good approach.