Well this reminds me of the martingale betting strategy which goes like this :

- You bet one dollar initially

- If you win, you quit

- If you lose, you bet 2 dollar next time

The winnings in this gam can be written as

\(W_n = \sum_{j=1}^{n}B_j X_j\)

Where \(B_j\) is the bet where \(B_1=1\) and for \(j>1\),\(B_j=2^{j-1}\) if we lost j-1 times consecutively.

Therefore our expected winning will still be zero since it's a fair game.

\(E[W_n]=1\times [1-2^{-n}]-[2^n-1]\times 2^{-n}=0\)

However, we'll eventually win which means that with probability one

\(W_{\infty}=\lim_{n->\infty}W_n=1\)

Well in your case, we can assume we bet x dollars initially. Then I think the problem will be choosing an optimal x, such that \(x \times [2^n-1] < 100\) and n is big enough since our probability of eventually wining x will be \(1-2^{-n}\)