Suppose we have a roulette wheel with $38$ total slots ($18$ black/red, $2$ neither). The ball lands in one slot selected uniformly at random, independent of all previous spins. An $\$x$ bet on "black" pays $\$2x$ if the ball lands on black, and $\$0$ otherwise.
If you bet $1 on black for 100 consecutive spins, how much money will you end up with in expectation?
You walk up to the Roulette table with $\$15$ and intend to walk away with $\$16$ using the following betting strategy. You will first bet $\$1$ on black. If you win you have $\$16$ and walk away, otherwise you have $\$14$ and bet $\$2$ on black on the next spin. If you win that bet you walk away, otherwise you bet $\$4$ on the next spin. If you win that bet you walk away, otherwise you bet $\$8$ on the next spin and walk away, either with $\$16$ if you win or $\$0$ if you lose.
What's the probability you will walk away with $\$16$?
How much money are you expected to walk away with?
My biggest problem with these kinds of questions is figuring out how to choose the random variable to calculate the expected values. I understand the formula for calculating the expected value, but translating a problem into those terms has been giving me a hard time.
Best Answer
Let's first look at the first question:
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:
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.]