The probability of coming out ahead in a european roulette Suppose I've bring 500 dollar to a roulette game, and spend 5$ every spin placing on a corner(4/37 winning), what's my chance to coming out ahead after 89 spins? what about 91 spins?
 A: Technically, you never specified when you stopped playing, and if you keep playing until you cant (you run out of money) then you have a 100% chance of losing your money if you keep playing indefinitely. Probably not the answer you are looking for, but the mathematically correct one to your question.
A: On any bet, you lose $5$ dollars with probability $\frac{33}{37}$, and win $40$ dollars with probability $\frac{4}{37}$. Your $\$500$ is enough to make sure that you don't get wiped out. 
After $90$ plays, we would come out "even" if we had $10$ wins and $80$ losses. So with $89$ plays we come out ahead if we have $10$ or more wins. Let the probability of this be $p$. It is easier to calculate first the probability $1-p$ of the complementary event that we get $9$ or fewer wins. The probability of $k$ wins is 
$$\binom{89}{k}\left(\frac{4}{37}\right)^k \left(\frac{33}{37}\right)^{89-k}.$$
Thus 
$$1-p=\sum_{k=0}^9 \binom{89}{k}\left(\frac{4}{37}\right)^k\left(\frac{33}{37}\right)^{89-k}.$$
It is feasible but not pleasant to evaluate $1-p$ by hand. Wolfram Alpha will compute the above sum easily. But if we use Alpha, we might as well not think, and just calculate the sum from $10$ to $89$. 
