I have another problem, that is so:
The probability that a roulette wheel stops on a red number is 18/37. For each bet on “red” you are returned twice your bet (including your bet) if the wheel stops on a red number, and lose your money if it does not.
If you bet $1 on each of the 10 consecutive plays, what is your expected winnings?
I started the question by declaring my random variables. I also know that each spin of the roulette wheel is independent of each other.
let X be the number of wins in the 10 consecutive plays
X ~ Binomial(10, 18/37), for any x such that $1 \leq x\leq 10$
By definition, my expected value function will look:
$E(x)=\displaystyle\sum\limits_{x=1}^{10} 3xf(x)$
What I am confused about is, how I determine my upper bound for my wins (how many wins can I get in 10 rolls), so that I can split my summation into two summations.
Since I do not know my probability distribution function, is my assumption that X follows a Binomial distribution correct?
Best Answer
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 $10\cdot\frac{18}{37}.$