What's the expectation in this casino game? Elementary probability 
You start with S coins in a casino game. If you win a round (P = 3/4)
  you double your money. if you lose a round (P = 1/4) you decrease it by
  a factor of 10. How much money approximately will you have after n rounds?

I've tried using the random variable on $\Omega = \{win, lose\}$ $X(win) = 2, X(lose) = 1/10$ and to manipulate it somehow to get the "average" factor by which S increases or decresess, but wasn't able to find the right way. 
I'll appreciate some hints!
 A: After $n$ rounds you have $$ W = S\prod_{i=1}^n X_i,$$ where $X_i$ are i.i.d. random variables that have value $2$ with probability $3/4$ and $1/10$ with probability $1/4.$ We have $E(X) = 61/40$ your average amount of money is $$ SE(X)^n = S (61/40)^n$$ which balloons exponentially as $n$ increases.
However, this is not "approximately how much money you will have". We have $$ \log(W) = \log(S) + \sum_{i=1}^n \log(X_i)$$ and we have $E(\log(X)) = \frac{1}{4}\log(8/10) < 0,$ so this is a random walk with negative drift and $\log(W)$ goes to negative infinity almost surely, so after many rounds you will be broke with probability one.
What is happening is that due to the exponential growth from doubling up a bunch, the rare paths that win an extraordinary amount contribute an outsized amount to the expected value. So while there are a few versions of you in parallel universes that are a gazillionaires and this drags up the expected value, you have vanishing probability of being them.
A: Every $4$ throws, your money decreases by $\frac{2^3}{10}=0.8$, so after $4k$ rounds, you should expect to have $0.8^k\cdot S$ coins left.
