Getting Rich Off Roulette If you bet 1 dollar on number 13 at roulette then you win $35 if that number comes up, an event of probability 1/38, and you lose your dollar otherwise. You play roulette 70 times.
(a) What is the probability that you have lost $70?
(b) What is the probability that you won $2?
(c) Use the Poisson approximation to show that if you play 70 times, the probability you will have won more money than you have lost is larger than 1/2
I have calculated the expected value of a single roll as (1/38)(35)-(37/38)(1) = -$.05. I am stuck after this, and I do not see how part C is possible, given a negative expected value of each trial.
Thank you for your help.
 A: First note that for the you to be losing 70 dollars you need to lose every time. 
For you to win 2 dollars you have to win 2 times and lose 68 times (finding this values is just a matter of solving a system of two equations)
Now model your experiment as a binomial random variable (it is in fact the sum of 70 independent Bernoulli variables)
In your case $N=70$, $p= \frac 1{38}$
You want to find the probability of $X=0$ and $X=2$
Finally your last problem is just $P(X\ge 2) = 1 - P(X=0)-P(X=1)$ which you may calculate directly or with the poisson distribution for a good approximation. 
It should result in a probability of $0.5528$

This counter intuitive result may be explained if you think that you just need to win $2$ times out of $70$;
It is true that the mean value is $70/38= 1.84$ which is less than $2$, but for you to be losing you should instead win at most $1$ time, which is farther from $1.84$ than $2$ is.
Think at a simple head or tail game; you win 1 dollar if it is head, you lose 1 dollar if it is tail.
The expected value of the game is clearly $0$; but this does not mean that the scenario after 70 games of you having exactly $0$ dollars is the most likely; instead accounts for just $0.095$. So even if the expected value of the game is $0$, after some games you'll probably we winning or losing, and you probably won't be tied up 
In your problem though there is an asymmetry that makes one outcome more likely (that you will be winning)
Then you ask "Hey but this way I'm going to get rich playing roulette! Why aren't casinos closing down?"
Well because you just have a probability of winning, you're not certain about it.
Sometimes you win, sometimes you lose, but on average you lose more than you win. You win more frequently, but accounting for the time you lose you'll find an average win of $-70 \cdot 0.05 = -3.5$
