Yes, and it's simpler than that:  every bet has an expected return of 99%, so the sum of all bets has an expected return of 99%, however you bet.

I'm not suggesting that Rick has found a winning strategy; I don't believe there is one.

I was just asking whether anyone can find the probability of his strategy busting, given that he can afford to double N times.

To make it more abstract:

a. start with x=0
b. add 1 to x (with probability p) or subtract 1 from x (with probability (1-p))
c. if x is negative, stop - you've won
d. if x is N, stop - you've lost
e. go to b.

You will eventually stop at c. (won) or d. (lose).  What is the probability of stopping at d?  Show your working!

It's easy enough to run a simulation and find an approximation for a particular p and N.  I'm looking for an analytical answer.