Losing less than $100 in a game of chance. In a game you win \$10 with probability $\frac{1}{20}$ and lose \$1 with probability $\frac{19}{20}$. Approximate the probability that you lost less than \$100 after the first 200 games. How will this probability change after 300 games?

Since I need to show both winning and losing together, I have come up with $$X_n=10S_n+(n-S_n)$$ where $X_n$ is the amount of winnings, and $S_n$ is the number of times you win the game. I also have $$E(X_n)=10, Var(X_n)=9.5$$ But when I want take $P(X_n\ge-100)$ I can never get a realistic answer.
Hint:

Let $X_i\sim \begin{pmatrix} 
0 &amp; 1\\
 19/20 &amp; 1/20 \\ 
\end{pmatrix} $
and 
$S=X\cdot10-(n-X)\cdot1$, where $X=\sum_{i=1}^nX_i$

It's easy to show that $E(X_i)=\frac{1}{20}$ and $Var(X_i)=\frac{19}{400}$.

Calculate $E(S)$ and $Var(S)$ and 
then apply central limit theorem.
Hint:

Let $X_i\sim \begin{pmatrix} 
0 &amp; 1\\
 19/20 &amp; 1/20 \\ 
\end{pmatrix} $
and 
$S=X\cdot10-(n-X)\cdot1$, where $X=\sum_{i=1}^nX_i$

It's easy to show that $E(X_i)=\frac{1}{20}$ and $Var(X_i)=\frac{19}{400}$.

Calculate $E(S)$ and $Var(S)$ and 
then apply central limit theorem.