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 $p_{39,62}\approx0.1842$.
In fact your starting capital is multiplied by $(1/2)^i(3/2)^{n-i}$ in $n$ rounds by probability of $\binom{n}{i}/2^n$,
disregarding the order of loses and wins.
Since it can never become $1$, you can never return to any of your previous positions, including
"quit with no loss or profit".
Suppose $j = \underset{(1/2)^i(3/2)^{n-i}>1}{\max}(i)$, so you profit by chance of $p_j = \sum_{i=0}^j \binom{n}{i}/2^n$, and lose by chance of $1-p_j = \sum_{i=j+1}^n \binom{n}{i}/2^n$.
Also you can use $\min$ function similarly, if you like:
Suppose $k = \underset{(1/2)^i(3/2)^{n-i}<1}{\min}(i)$, so you lose by chance of $p_k = \sum_{i=k}^n \binom{n}{i}/2^n$, and profit by chance of $1-p_k = \sum_{i=0}^{k-1} \binom{n}{i}/2^n$.
Best Answer
We'll call $X_i$ profit from round $i$, which has moments $$E[X_i]=-0.5+10(0.1)-1(0.7)-2(0.2)=:\mu\\ \text{Var}(X_i)=E[X_i^2]-(E[X_i])^2=10^2(0.1)+(-1)^2(0.7)+(-2)^2(0.2)-\mu^2=:\sigma^2.$$
Since the $X_i$ are independent, CLT tells us $$ \frac{\frac{1}{n}\sum_{i=1}^n X_i -\mu}{\sigma/\sqrt n}\overset{d}{\rightarrow}N(0,1),$$
which gives an asymptotic approximation for the probability that the profit after 100 rounds is $\leq5$:
$$P\left(\sum_{i=1}^{100} X_i\leq 5\right)=P\left(\frac{\frac{1}{100}\sum_{i=1}^{100} X_i -\mu}{\sigma/\sqrt {100}}\leq \frac{\frac{5}{100}-\mu}{\sigma/\sqrt {100}}\right)\approx \Phi \left(\frac{\frac{5}{100}-\mu}{\sigma/\sqrt {100}}\right)=\Phi \left(\frac{0.5-10\mu}{\sigma}\right),$$ where $\Phi$ is the CDF of a standard normal.