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I'm having trouble understanding the solution to a problem in the text book I am using. The problem goes as follows:

On each bet, a gambler loses $1$ with probability $.7$, loses $2$ with probability $.2$, or wins $10$ with probability $.1$. Approximate the probability that the gambler will be losing after his first 100 bets.

I know that I have to use two pieces of information: For any sequence of independent and identically distributed random variables, each having finite mean $\mu$, then for any $\epsilon>0$ $P\{|\frac{X_1+X_2+...}{n}-\mu|\ge\epsilon\}\to0$ as $n\to\infty$ and the central limit theorem. I am just not sure as to how to use them.

This is what I have figured out so far: Since there are $100$ bets, $X=\sum_{i=1}^{100}$. Also, I know that to get the expected value for $\mu$ is $-.1$ since $E[X]=(1\times-.7)+(2\times-.2)+(10\times.1)=-.1.$ Now to get $Var(X)$ you do $E[X^2]-(E[X])^2$. At this point I begin to get confused because the textbook gets a $Var(x)=11.49$, but I get $Var(X)=8.49$. Moreover, the book arrives to the answer by

$P\{X_1+...+X_{100}\le-.5\}=P\{ \frac{X_1+...X_{100}+10}{\sqrt{1149}}\le\frac{-.5+10}{\sqrt{1149}}\} \approx P\{Z\le.2803\} \approx.6104$

How did they get to $Var(x)=11.49$? Why did the author choose $\epsilon=-.5$ Why did they add $10$ to $-.5$? How did they achieve the value $\sqrt{1149}$?Im terribly confused and any help is appreciated.

Oscar Flores
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The variance of $X$ is given by $$\eqalign{ E(X^2)&=1\times0.7+4\times0.2+100\times0.1=11.5\cr {\rm Var}(X)&=E(X^2)-E(X)^2=11.49\ .\cr}$$ If $X_1,\ldots,X_n$ are independent and all have the same distribution as $X$, and if write $S=X_1+\cdots+X_n$, then $$E(S)=nE(X)\quad\hbox{and}\quad {\rm Var}(S)=n{\rm Var}(X)\ .$$ Since $S$ is a discrete random variable we use the continuity correction $$P(S<0)=P(S\le-{\textstyle\frac{1}{2}})\ .$$ Finally, the central limit theorem says in effect that $S$ is approximately normal, so $$P(S\le-0.5)=P\Bigl(Z\le\frac{-0.5-\mu}{\sigma}\Bigr)$$ where $$\mu=E(S)=100E(X)=-10\quad\hbox{and}\quad \sigma^2={\rm Var}(S)=100{\rm Var}(X)=1149\ .$$ Hope this clears up all your issues.

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David
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