This answer only examines the betting strategy the OP advocates (always bet half of what you have) and some of its variants. With this strategy:
One never gets broke (one's fortune is always positive). One doubles up almost surely if $p$ is large enough but with a probability which is less than $1$ if $p$ is small enough.
Note that there is no theorem saying that every positive submartingale reaches every level almost surely, hence arguments based on averages of increments only, cannot suffice to reach a conclusion.
In the strategy the OP advocates, the fortune performs a multiplicative random walk whose steps from $x$ are to go to $\frac32x$ and to $\frac12x$ with probability $p$ and $1-p$ respectively. Thus, the logarithm of the fortune performs a usual random walk with steps $\log\frac32$ and $\log\frac12$, whose constant drift is
$m(p)=p\log\frac32+(1-p)\log\frac12$ hence $m(p)=p\log3-\log2$ has the sign of $p-p^*$ where $p^*$ solves the equation $3^{p^*}=2$, hence $p^*=\frac{\log2}{\log3}=.6309...$
If $p\geqslant p^*$, $m(p)$ is nonnegative hence the logarithm of the fortune reaches the set $[C,+\infty)$ almost surely for every $C$. In particular, one doubles up almost surely.
If $p<p^*$, $m(p)$ is negative hence the logarithm of the fortune has a positive probability $q$ to never visit the set $[\log2,+\infty)$, for example. This means that with probability $q$, one will bet an infinite number of times without ever getting broke nor doubling up. In effect the fortune at time $k$ will go to zero like $a(p)^k$ when $k\to+\infty$, with $a(p)=\frac123^p<1$.
There is no easy formula for the probability $q$ to never double up but the probability to double up $n$ times decreases exponentially like $\exp(-b(p)n)$ when $n\to\infty$, where $b(p)$ is the unique positive solution of the equation $p(3^b-1)=2^b-1$.
If $p<p^*$, the strategy where one always bets half of what one has fails. What happens with the strategy where one always bets a proportion $r$ of what one has? The same analysis applies and shows that one doubles up almost surely, for every $p\geqslant\wp(r)$, where $p=\wp(r)$ solves the equation $(1+r)^{p}(1-r)^{1-p}=1$ (note that $\wp(\frac12)=p^*$).
The other way round, for every given $p>\frac12$, a strategy which wins almost surely is to bet a proportion $r$ of what one has, provided $p\geqslant\wp(r)$. Since $\wp(r)\searrow\frac12$ when $r\searrow0$, one gets:
For each $p>\frac12$, the strategy where one always bets a proportion $r$ of what one has wins almost surely for every small enough positive value of $r$ (for example, $r\leqslant 2p-1$).
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
$$\frac{p}{1-p}<\frac1{-\ln p}<\frac1{1-p}=\frac{p}{1-p}+1\;.$$
Best Answer
If you count rolling a $1$ as $-35$, then your reference point is your current $35$, and you shouldn't include $35$ in the chance of further winnings.
On the other hand, you can certainly keep rolling while your score is below $43$.
So I would take the value of the next roll to be at least $(1/6)(-35)+(5/6)8$ which is positive.