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, $r\leqslant 2p-1$).
We either end as winners with 150 or 151 chips or as losers with between 0 and 50 chips (if we lose early, with between 0 and 33 chips; in general the upper limit is $\frac13$ of the last peak). In a fair game this would mean a winning probability between $\frac12$ and $\frac35$ (just by the relation of up and down movements $+50:-50$ to $+50:-67$). Even this rough estimate (that does not take into account the bank advantage) matches well with Sharkoe's simulation.
A more precise calculation for $p_{k,d}$, the probability to reach $150$ when starting with $k$ and trying to win at least $d$ in the first sequence
- $p_{k,d}=0$ if $2k<d$
- $p_{k,1}=1$ if $k\ge 150$
- $p_{k,d}=\frac{12}{37}p_{k+2b,1}+\frac{25}{37}p_{k-b,d+b}$ with $b=\lceil \frac d2\rceil$
This allows us to compute the probybility exactly as a fraction. However, numerator and denominator have about $785$ digits, so the numerical value from rounding that fraction
$$ p_{100,1}\approx0.56097114279613511032732301110367086534$$
should be good enough for us.
Code in PARI/GP (with memoization for values $p_{k,1}$, $100\le k\le 150$), edited to use less memory):
Start=100;Target=150
A=vector(Target-Start+1,n,-1);
getP(k,d)={local(pkd);
if(2*k<d, 0, if(k>=Target, 1,
pkd=if(d==1&&A[k-Start+1]>=0,
A[k-Start+1],
12/37*getP(k+2*ceil(d/2),1)+25/37*getP(k-ceil(d/2),d+ceil(d/2))
);
if(d==1,A[k-Start+1]=pkd);
pkd))
}
getP(Start,1) +.0
The new code after editing does not reflect it, but the "worst" case among the cases occuring is $p_{39,62}\approx0.1842$.
Best Answer
You have the right idea. Your expected loss on any bet is $\frac {1}{19}$ of your wager in an American casino and $\frac {1}{37}$ in Monte Carlo. Your betting strategy does not change this.
If you had an infinite endowment (and the casino had no maximum bet size) even if you are down a very large sum of money at some point, eventually your double-or-nothing bet will hit and almost surely you will eventually hit a win and be up one dollar. But you don't have an infinite endowment and the casino will not allow you to make bets of unlimited size.
Supposing you have a budget of $\$5,000$ for example, if you hit a streak of 12 consecutive losses, you can not afford to double your bet. We can calculate your expected loss with a strategy of play to the first win, or walk away after losing 12 in a row.
$(1 - \frac {20}{38}^{12}) - \frac {20}{38}^{12}(2^{12}-1)\approx - 0.85$ in Vegas.
$(1 - \frac {19}{37}^{12}) - \frac {19}{37}^{12}(2^{12}-1)\approx - 0.38$ in Europe.
Spending enough time employing this strategy, almost surely you will bust.