Coin Tossing Game Optimal Strategy

It really is as simple as "the bet is in your favor-take it." $S=x$. You win $100(3^{100}-1)$ with probability $2^{-100}$ and lose $100$ with almost certainty. This presumes somebody can pay you that much. Then the expected win is $$\frac{1}{2^{100}} \cdot 100(3^{100}-1) -(1-\frac 1{2^{100}}) \cdot 100\approx 4\cdot 10^{19}.$$

To maybe make this less unbelievable, imagine a two round game. Clearly on the last throw, you want to bet all you have, increasing your expected fortune by $50\%$. On the first throw, your expected balance is $\frac {x-S}{2}+\frac {x+2S}{2}=x+\frac{S}{2}$ which (given the rules) is maximized when $S=x$. If you won on the first flip, you now have 3S, so you can now bet $S_{2}$ with the condition that $S_{2} \leq 3S$. Your expected balance after the second flip is then $\frac{3S-S_{2}}{2} + \frac{3S+2S_{2}}{2} = \frac{3S+S_{2}}{2}$, which is again maximized if $S_{2}=3S$, so that your expected balance will be $\frac{6S}{2} = 3S$, and your expected profit (assuming you won on the first flip) will now be $3S-S=2S=2x$.

Alternately, your result is the same if you interchange the two flips. Since you should be all on the last flip, you should on the first as well.

In my opinion this calls for the Kelly Criterion (http://en.wikipedia.org/wiki/Kelly_criterion). In this case, the fraction of your wealth that should be bet is $\frac{0.5 \times 3-1}{3-1}$.