Probability when playing roulette A roulette from a casino contain 18 red cases, 18 black cases and 1 green cases. A player shows up with $10$dollars. He decides to bet $1$dollar on red 10 consecutives times. If it's red, he wins a dollar and if not he loses his dollar. Let $S$ be the the amount of money the player has after 10 bets. Find the value $S$ can have, then calculate $P(S<3|S\leq18)$.
So the value S can take is every natural numbers $0\leq S\leq20$
but after that I'm confused on how to proceed. I guess we could do it using a density function but i don't know how.
Any help to point me in the right direction would be greatly appreciated.
Thank you.
 A: The value of $S$ cannot take every natural number in that range; though the range itself is okay.
In ten games the count of wins will be somehow distributed.   $W$ is the count of wins in $10$ games with success probability $p$.   What is $p$, and what is the distribution of $W$.  
$$\mathsf P(W=k) = \underline{\qquad}\quad[k\in\{0,1,2,3,4,5,6,7,8,9,10\}]$$
Assume the player started with $\$10$, and in each game bets one dollar and gets back two on a win.   So the amount of money $S$ taken away will be $S= \underline{\quad}$ (somehow related to $W$), and thus have distribution:
$$\mathsf P(S=s) = \underline{\qquad}\quad[s\in\{\underline{\qquad}\}]$$
Then by definition of conditional probability:
$$\begin{align}\mathsf P(S< 3\mid S\leq 18) & = \dfrac{\mathsf P(S< 3 \cap S\leq 18)}{\mathsf P(S\leq 18)} \\[1ex] & = \dfrac{\mathsf P(S< 3)}{1-\mathsf P(S> 18)} \\[1ex] & = \underline{\dfrac{(\qquad)}{(\qquad)}}\end{align}$$
Fill in the blanks and complete.
A: Some hints: First we can use the converse probability: 
$P(S\leq 18)=1-P(S>18)=1-P(S=20)-P(S=19)$
In case of $S=20$ the player has to win 10 times. $P(S=19)$ is zero, beause it is not possible.

$P(S<3)=P(S=0)+P(S=1)+P(S=2)$
For $S=0$ the player has to loose 10 times. 
$S=1$ is not possible. 
For $S=2$ the player has to win once.
Bayes theorem:
$P(S<3|S\leq 18)=\frac{P(S<3 \cap S\leq 18)}{P(S\leq 18)}$
$P(S<3 \cap S\leq 18)=P(S<3)$
Therefore $P(S<3|S\leq 18)=\frac{P(S<3)}{P(S\leq 18)}$

A calculation example:
$P(S=16)=\binom{10}{8}\cdot \left(\frac{18}{37}\right)^8\cdot \left(\frac{19}{37}\right)^2$
