Let's first look at the first question:
If you bet $1 on black for 100 consecutive spins, how much money will
you end up with in expectation?
So you want to know what your final return will be, at the end of 100 spins. Call this $R$. That is just giving it a name, but what is your final return? You can see that it is the sum of the returns from each bet. So let the return on the $i$th bet be $R_i$, then note that $R = R_1 + R_2 + \dots + R_{100}$. So the expected return is $E[R] = E[R_1] + E[R_2] + \dots + E[R_{100}]$ by linearity of expectation.
So to calculate $E[R]$, we'll be done if we calculate each $E[R_i]$. Let's try to calculate a particular $E[R_i]$. You bet $1$ dollar, and you get back $2$ if the ball lands on black, and $0$ if it doesn't. In other words, you gain $1$ dollar if it lands on black, and lose $1$ dollar if it doesn't. The probability of the former is $18/38$, and that of the latter is $20/38$. In other words, $R_i$ is $1$ with probability $18/38$, and $-1$ with probability $20/38$, so the expected value of $R_i$ is $E[R_i] = \frac{18}{38}(1) + \frac{20}{38}(-1) = \frac{-2}{38}$. Now, as this is the same for each $R_i$, we have $E[R] = E[R_1] + E[R_2] + \dots + E[R_{100}] = \left(\frac{-2}{38}\right)100 \approx -5.26$.
For the second question, let the amount you walk away with be $W$. Let $p = 18/38$, the probability that your bet on black succeeds. There are $5$ possible outcomes:
- you win your first bet: probability $p$
- you lose your first bet, and win your second: probability $(1-p)p$
- you lose your first two bets, and win the third: probability $(1-p)^2p$
- you lose your first three bets, and win the fourth: probability $(1-p)^3p$
- you lose all four bets: probability $(1-p)^4$
In the first four outcomes, you walk away with $16$ dollars, so the probability of that happening (let's call it $q$) is $q = p + (1-p)p + (1-p)^2p + (1-p)^3p = 1 - (1-p)^4 = 1 - (20/38)^4 \approx 0.92$.
[More simply, you could think of it as just two outcomes: (a) that you win some bet, which has probability $q = 1 - (1-p)^4$, and (b) that you win no bet (lose all bets), which has probability $(1-p)^4$.]
In other words, $W$ is $16$ with probability $q$, and $0$ with probability $1-q$. So the expected amount of money you walk away with is therefore $E[W] = q(16) + (1-q)0 = (1-(1-p)^4)16 \approx 14.77$.
[Aside: Note that this is less than the $15$ you came in with. This shows that you can't win in expectation even with your clever betting strategy; a consequence of the optional stopping theorem.]
On a single play, your expectation is
$$\frac{1}{38}(35)+\frac{37}{38}(-1)=-\frac{2}{38}.$$
For multiple plays, your analysis is correct. Let random variable $X_i$ be your net gain in the $i$-th bet. Then your total gain $Y$ is $X_1+X_2+\cdots+X_n$.
By the linearity of expectation,
$$E(Y)=E(X_1+X_2+\cdots+X_n)=E(X_1)+E(X_2)+\cdots+E(X_n)=-\frac{2}{38}n.$$
In particular, if $n=38$ then $E(Y)=-2$.
Remark: Your calculation was basically correct. The small departure from the correct $-2$ is probably due to writing down an intermediate calculator result, and then reentering the number, perhaps not to full accuracy. It is a good thing to learn to use the memory feature of your calculator. It will save on roundoff errors, and on time. It will also help cut down on keying errors, which are a significant source of mistakes.
The independence of your bets plays no role in the calculation, as long as the roulette wheel itself is fair. We have precisely the same expectation if we bet on our favourite number, say $13$, all $38$ times. And remember, on average, the more you play, the more you lose.
Best Answer
The error in your calculation is that the chance of winning at least once in four tries is $1-(\frac {20}{38})^4 \approx 0.9233$. You can't keep adding the $\frac {18}{38}$s-the fact that the sum exceeds $1$ should tip you off. This is because you might win more than one-the events are not mutually exclusive.
Your expected win from one series is then $10 \cdot 0.9233 + (-150) \cdot 0.0767=9.233-11.51=-2.28$
The important truth is that no series of losing bets can be winning. You can have a high probability of profit, as here, but the expectation will be negative.