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.]
You are correct $($assuming you are talking about French Roulette, which has no $00)$ that $X\sim\operatorname{Binomial}(10,\frac{18}{37})$.
However, you've miscalculated. If you win, then you are given twice your bet including your bet, not in addition to your bet. That is, if you bet on red and win, then you only double your bet, not triple it. Your net gain for each win, then, is only a dollar, since you're betting a dollar each time.
You'd be better off starting by computing your expected number of wins, which is just $10\cdot\frac{18}{37}$. At that point, the answer will depend on what is meant by "expected winnings."
If it means "expected net gain," then note that you will gain a dollar for each win and lose a dollar for each loss, so since your expected losses are $10\cdot\frac{19}{37},$ then your expected net gain is $10\cdot\frac{18}{37}-10\cdot\frac{19}{37}=-\frac{10}{37}$ dollars.
If it means "expected gain," then the answer is simply $10\cdot\frac{18}{37}$ dollars.
If it means "expected total amount that you will be given by the croupier," then the answer will be $2\cdot10\cdot\frac{18}{37}$ dollars. (Note: If you simply subtract the $10$ dollars that you bet from this, then you will find the expected net gain in another way.)
Addendum: More generally, suppose that you play $n$ independent games with win probability $\theta.$ Let $X$ be the number of wins, so that $X\sim\operatorname{Binomial}(n,\theta).$ Then your expected number of wins is $$\Bbb E[X] = \sum_{x=0}^n\binom{n}{x}\theta^x(1-\theta)^{n-x},$$ but that's a bear to actually calculate. Instead, we will let $X_k$ be the number of wins on the $k$th trial for $1\le k\le n,$ so that $X\sim\operatorname{Bernoulli}(\theta)$ for each $k,$ and $$X=\sum_{k=1}^nX_k.$$ Since expectation is linear over sums of random variables, then $$\Bbb E[X]=\Bbb E\left[\sum_{k=1}^nX_k\right]=\sum_{k=1}^n\Bbb E[X_k]=\sum_{k=1}^n\sum_{x=0}^1x\theta^x(1-\theta)^{1-x}=\sum_{k=1}^n1\cdot\theta^1(1-\theta)^{1-1}=\sum_{k=1}^n\theta=n\theta.$$
So, our expected number of wins is $n\theta,$ which in your particular example is $10\cdot\frac{18}{37}.$
Best Answer
Guide:
Number the rounds $1,2,3$ and write: $$X=X_1+X_2+X_3$$ where $X_i$ denotes the winning in round $i$ for $i=1,2,3$.
Apply linearity of expectation to find $\mathbb EX$.
Realize that the $X_i$ are independent and use the rule: $$\mathsf{Var}(U+V)=\mathsf{Var}(U)+\mathsf{Var}(V)$$ for independent random variables $U,V$ for which variance is defined.