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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.

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    $\begingroup$ Use central limit theorem and approximate the total outcome after many game (appropriately scaled) by standard normal distribution $\endgroup$ – A.S. Oct 23 '15 at 16:25
  • $\begingroup$ what do you mean you tried to take $P(X_n\geq -100)$.? Chebyshev, Markov inequality? $\endgroup$ – Lost1 Oct 23 '15 at 16:38
  • $\begingroup$ Just need to show that total winnings are greater than -100. In other words the fact you haven't lost more that $100 $\endgroup$ – Jim Gross Oct 23 '15 at 16:42
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Hint:

Let $X_i\sim \begin{pmatrix} 0 & 1\\ 19/20 & 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.

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