Probability of coming out ahead in Roulette I am very much lost with this problem- If you bet $5 on red every time, what is the probability you'll be ahead (have more than what you started with) after 20 spins of betting on red?
American Roulette with an infinite amount to bet
 A: Let $W$ be a binomial random variable such that $W=k$ if and only if $k$ is the number of wins among $n=20$ spins. The probability of winning on any particular spin is $p = \frac{18}{38} = \frac{9}{19} \approx 0.4737$. The probability mass function $P$, which conveys the probability of winning $k$ of $20$ spins, is given by
$$ P(W=k) = {n \choose k} p^k (1-p)^{n-k} = {20 \choose k} 0.4737^k (1-0.4737)^{20-k} = {20 \choose k} 0.4737^k 0.5263^{20-k}$$
with cumulative distribution function $F$
$$ F(W=k) = P(W \le k) = \sum _{i=0} ^k {n \choose i} p^i (1-p)^{n-i} = \sum _{i=0} ^k {20 \choose i} 0.4737^i 0.5263^{20-i} $$
Now, consider that the amount you bet is irrelevant since it remains constant and you always gain or lose increments of it. Thus, I will continue this answer in terms of "betting units" that are gained or lost... For every win, you get your betting unit back plus $1$ more. For every loss, you lose $1$ betting unit. Hence, after $20$ spins, the total number of betting units you have gained (over and above the amount you started with) is given by $x = W - (20-W) = 2W - 20$. We wish to find the minimum number of wins such that $x > 0$. So,
$$ x > 0 $$
$$ \Leftrightarrow 2W - 20 > 0 $$
$$ \Leftrightarrow 2W > 20 $$
$$ \Leftrightarrow W > \frac{20}{2} = 10 $$
So, you will make money as long as you win more than $10$ times among $20$ spins. Hence, we must find the probability that $W>10$. In other words, we want to find $P(W>10) = 1 - P(W \le 10) $ which is given by
$$ 1 - \sum _{i=0} ^{10} {20 \choose i} 0.4737^i 0.5263^{20-i} = 1- 0.67761 = 0.32239$$
