Okay, here is a more detailed answer: let P[win] be the probability that you bust your opponent. I will refer to that as P[win], and your expected value is 75 P[win]-100(1-P[win]). We need to estimate P[win].

Here is one approach: let X_n represent your profit after n bets. Since each bet is in your favor, studying X_n directly isn't so nice, so a standard trick is to consider Y_n=a^{X_n}.

We start with X_0=0, and Y_0=1. Computing the probability that you win the first hand, we get

E Y_1=a^3 (378/1081) + a^{-1} (703/1081).

If we choose a so that this is 1, then the E Y_1=1. More importantly, for this value of a, each bet doesn't change the expected value of Y_n. In particular, the expected value of Y at the end of the game is still 1. Let T denote the time when the game ends.

At the end of the game, either you won, and X_T=100 (so Y_T=a^100), or you lost and X_T=-100 and Y_T=a^{-100}. Since EY_T=1, this gives us

1=E Y_T=a^100 P[win] + a^{-100} P[lose]

(Note: this isn't quite true--I'll comment on that later).
Solving numerically for a, we get roughly .3607, so the probability that you win is, to over 40 decimal places, 1.

The problem with this analysis is that since your profit is going up by $3 at a time, you probably won't hit $100 on the dot. We actually need to modify the last equation slightly to take that into account, but the modification involves terms that are less than a^100, which is basically 0.

It is surprising to many people that the chance of you winning is so high. When I presented a similar fact at a grad student seminar a few years ago, one of the students flat out told me that I was wrong and that he refused to believe that the arithmetic was correct. It might help to realize that after 100 bets, your expected profit is over $40, so by the earliest time your opponent could possibly have won, the expected value of all of the bets is almost half of his original bankroll.