Let $p$ be the probability that the coin comes up heads. Suppose that you’ve tossed a head $n$ times. If you play one more time, your expected payoff is $100(n+1)p$; if this is less than the $100n$ that you’ve won so far, you shouldn’t play again. The inequality $100(n+1)p<100n$ boils down to $n(1-p)>p$, or $n>\frac{p}{1-p}$; if $n=\frac{p}{1-p}$, your expected payoff is equal to your current winnings. Thus, if $n\ge\left\lfloor\frac{p}{1-p}\right\rfloor+1$, you should not play again. If $n=\frac{p}{1-p}$, you can play again or not. And if $n<\frac{p}{1-p}$, you should play again. We can get a slightly nicer expression for the cutoff: you should stop when $n$ reaches
$$\left\lfloor\frac{p}{1-p}\right\rfloor+1=\left\lfloor\frac1{1-p}-1\right\rfloor+1=\left\lfloor\frac1{1-p}\right\rfloor\;.$$
For a fair coin, with $p=\frac12$, this says that if $n>1$, you shouldn’t play again: if you toss heads once, you can go ahead and try again or not, as you please, but if you toss heads twice, you should quit. If $p=4/5$, the cutoff is $n>4$: if you’ve managed to toss four heads in a row, you can try again or not, but if you’ve managed five, quit.
To see how this relates to Sam’s calculus-based answer, start with the Maclaurin series for $\ln(1-x)$:
$$\ln p=-\sum_{n\ge 1}\frac{(1-p)^n}n\;,$$
so
$$\frac1{-\ln p}=\frac1{\sum_{n\ge 1}\frac{(1-p)^n}n}=\frac1{1-p}\cdot\frac1{1+\frac12(1-p)+\frac13(1-p)^2+\ldots}\;.$$
Clearly $1<1+\frac12(1-p)+\frac13(1-p)^2+\ldots<\sum_{k\ge 0}(1-p)^k=\frac1p$, so
$$\frac{p}{1-p}<\frac1{-\ln p}<\frac1{1-p}=\frac{p}{1-p}+1\;.$$
You are doing fine, but you can do less cases if you use the linearity of expectation. You can consider the two flips indepently, as the return of each does not depend on the other flip. You can just add the expected value of each flip. Then the expectation of the first filp is $-\frac 12$ and the expectation of the second is $+1$, for a total $+\frac 12$. You can complain that I went through four cases as did you, but my cases were simpler. If there are $n$ flips, you would have to do $2^n$ cases, while I would have to do $2n$. If $n=10$, I am way ahead.
Best Answer
The game may last very long, but the probability that it does not end is zero.
Since the coin is assumed to be fair, the expectation value of the game is zero. Now the expectation value is given by $E = P*70 + (1-P)*-30$, where $P$ is the probability of winning the game. Setting $E$ equal to zero, we can solve for $P$ with the result: $P = 0.3$.