The problem simply does not make sense in the case that the gambler has less than $1. If the cost is less than or equal to the gambler's asset's, the gambler plays, and each time he gets more than he loses, so he wins automatically. If the cost is greater than the gambler's assets, then he wins automatically. The same argument works for when the amblers assets is exactly $1, the gambler simply cannot lose money, unless the gambling cost is higher, in which case the gambler loses, so no solution is valid or A<=1. For A close to 1, the solution is also hard. If the cost to enter the game, C, is bounded by (1+A)/2<C<A, then if the gambler enters and only wins $1 (50% likely) then he loses. Also there is a slight chance that the gambler will lose even if he gets $2. This means that the probability of the gambler losing is greater than 50%, which means it is not a solution to the original problem. Therefore 1<C<(1+A)/2 for all A>1.
Edit: I think that an exact solution to the original problem does not exist for most values of A and B (B=casino assets). I realized that, using the above argument, assuming that a solution exists for a given A and B implies that the the probability of the gambler breaking the bank with C=(1+A)/2 (I'll call it P(A,B), as it is a function of the casino's assets), is bounded by P(A,B)<=0.5, as the cost cannot increase, as that would make the bank more likely to win, so it can only decrease, which increases the chance of the gambler winning. Actually, extending this argument, assuming that a solution exists for a given A implies that the probability of the gambler never running out of money (assuming the bank has unbounded assets) with C=(1+A)/2 (I'll call it P(A)) is bounded by P(A)<=0.5. This is true because this is the worst case that the gambler can still win with, so the probability of the gambler winning with any other fair scenario must be >P(A).
This is useful because it gives the ability to give tighter bounds on the solvable values of A.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.