Lulu has given you a method for the probabilities of reaching $500$ before reaching $0$: dividing the values by $100$ to give you labels, you want to find the solution to $$\psi_i=\tfrac 12 \psi_{i+1} +\tfrac 14 \psi_i +\tfrac 14 \psi_{i-1} \\ \text{ i.e. } 2\psi_{i+1} -3 \psi_i + \psi_{i-1}=0$$ with the boundary conditions $\psi_0=0$ and $\psi_5=1$. In particular you are trying to find $\psi_{1}$. As Rushabh Mehta says, you would get the same answer ignoring ties by starting with $\psi_i=\frac 23 \psi_{i+1} +\frac 13 \psi_{i-1}$
For the expected time, you do something similar but this times have $$\tau_i=1+\frac 12 \tau_{i+1} +\frac 14 \tau_i +\tfrac 14 \tau_{i-1} \\ \text{ i.e. } 4+2\tau_{i+1} -3 \tau_i + \tau_{i-1}=0$$ with the boundary conditions $\tau_0=0$ and $\tau_5=0$, and in particular you are trying to find $\tau_{1}$. My comment was that if you ignore ties by changing this to $t_i=1+\frac 23 t_{i+1} +\frac 13 t_{i-1}$ then you would get $\tau_i = \frac43 t_i$
The expected value of a $1$ dollar bet on black on an American roulette wheel is $\frac{18}{19}-1 = -\frac{1}{19} \approx -0.05263$ dollars (a little better than $-0.06$).
If you were to bet $1$ dollar on each of $1000$ spins, your expected gain would be approximately $1000 \times -0.05263 = -52.63.$
But you are not betting $1$ dollar on each of $1000$ spins in either scenario; you are betting varying amounts of money on (usually) fewer than $1000$ spins.
What's more, the amount bet depends on the previous spins.
The chance of winning seems a little easier to compute, though still not easy.
In Scenario 1, since you start with an unlimited bankroll, we cannot measure your progress by subtracting your initial bankroll from your final bankroll.
So let's just say you start with zero net gain and you are allowed to let your net gain go as negative as you need in order to keep playing.
During a long run of red and green spins, your net gain may go very negative indeed. But every time black wins, your net gain becomes equal to the number of times black has won so far.
Moreover, your strategy is that if your net gain ever reaches $80$ you stop playing.
So the only way that you will not win $80$ dollars within $1000$ spins is if black wins fewer than $80$ times in $1000$ spins.
The probability that this happens is very, very small, but it is not zero.
The probability is so small, in fact, I did not manage to get Wolfram Alpha to give me an answer. But the probability of black winning exactly $79$ times is approximately
$2.020 \times 10^{-164},$
the probability of black winning exactly $78$ times is approximately
$1.923 \times 10^{-165},$
and the probability for each smaller number of wins is progressively much smaller even than that.
I think we can safely say that the probability of losing is between
$2.2\times 10^{-164}$ and $2.3\times 10^{-164}.$
You could make this a little more exact by adding in some of the other $78$ losing cases.
This is not exactly zero, so your probability of winning is not exactly $1.$
But the probability is so close to $1$ that you are almost certainly not going to simulate any losing case.
In Scenario 2, you have (almost) an asymmetric random walk with two absorbing states, also known as gambler's ruin.
The state is the size of your bankroll, the initial state is $256,$
and the absorbing stats are $0$ and $336.$
I say that this is almost that problem, because the usual random walk does not limit the number of steps that may be taken,
whereas you limit your problem to $1000$ steps.
Best Answer
Remember that each $X_i$ denotes the loss or gain of the person in the $i$-th game. You're denoting each loss by $(-10)$ , each gain by $(+10)$. So, as you have defined $Y = X_1 + X_2 + \cdots + X_{100}$ , the event that he gains some positive amount after $100$ games is $[Y > 0]$ , NOT $[Y > 1000]$.
Now, by CLT, $~Z :=\dfrac{Y + 50}{\sqrt{9475}} \stackrel{\text{approx}}{\sim} \text{N}(0,1)$ . Thus : \begin{align*} \mathbb{P}(Y > 0) &= \mathbb{P}\left(Z > \frac{50}{\sqrt{9475}}\right)\\ &\approx 1 - \Phi\left(\frac{50}{\sqrt{9475}}\right)\\ &\approx 0.3037 \end{align*}
NOTE : In case you want, you may assume that the person is using $\$ 10$ for playing a game, you may denote the total gain and loss by $(+20)$ and $0$ respectively, and keep the value of $X_i$ to be $(+10)$ in case of a draw.
Under that setup, the expectation and variance of $Y$ will change, but the event whose probability you want to drive will be $[Y > 1000]$. However, the final answer will be the same.