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.]
Best Answer
We have : $$ P(S \geq m+1) = P(X_1 + ... + X_m \leq n) $$
where $X_i$ are iid uniformly distributed in $\{1,2,...,n\}$. Therefore, it is sufficient to find the RHS.
The answer is just $\sum_{1 \leq i_1,...,i_m \leq n} 1_{i_1+...+i_m \leq n} P(X_1 = i_1,X_2 = i_2,...,X_m = i_m)$, which just becomes $\frac {K}{n^m}$, where : $$ K = |\{(i_1,i_2,...,i_m) \in \{1,2,...,n\}^m : i_1+i_2+...+i_m \leq n\}| $$
We claim that the set given in the definition of $K$ is in bijection with the set $\{(i_1,i_2,...,i_m) \in \{0,1,...,m\}^m : i_1+i_2+...+i_m \leq n-m\}$. The bijection is given by taking a tuple and subtracting $1$ from each entry. It is easy to see, as $n>m$ that the bijection holds.
Now, we adjoin an index $i_{m+1} = n-m - (i_1+...+i_m)$, so we get that the above set is in bijection with : $$ \{(i_1,i_2,...,i_m,i_{m+1}) \in \{0,1,...,m\}^{m+1} : i_1+i_2+...+i_m +i_{m+1} = n-m\} $$
by projection onto the first $m$ coordinates.
This last set is calculated using stars and bars. Indeed, imagine a set of $n$ blanks, each filled with a $|$ or a $\circ$ (ball), so that there are exactly $m$ of $|$ and $n-m$ of $\circ$. $$ \underbrace{|\circ||\circ\circ\circ|\circ||...|\circ| \circ \circ}_{\text{$n$ blanks}} $$
The bars separate the circles into $m+1$ parts, each containing $i_1,...,i_{m+1}$ number of circles, which total to $n-m$ circles, and each of which is non-negative. In the example above, $i_1 = 0,i_2 = 1,i_3 =0 , i_4 = 3$ and so on.
It is easy to see that the number of ways of placing these bars and circles is the same as the number of elements in the set mentioned earlier. But we are basically placing $m$ bars into $n$ slots : the answer to this is just $\binom{n}{m}$.
Thus, we basically have $K = \binom{n}{m}$, with $Pr(S \geq m+1) = \frac{\binom{n}{m}}{n^m}$.
The expectation of $S$ is equal to $\sum_{i=0}^n \binom ni n^{-i} = (1 + \frac 1n)^n \to e$ as $n \to \infty$ (i.e. as we approach the uniform distribution)