Question

8.8. 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.

Short answer

The probability is $.6103$.

Step-by-step solution

Given information

A gambler loses $1$with probability $.7$, loses $2$ with probability $.2$, or wins $10$ with probability $.1$.

Explanation

Let $X_i$denotes the gambler's winnings on $i$th bet. Then, discrete random variable, takes the values, $-1,-2$and $10.$

Probabilities:

$P\left\{X_i=-1\right\}=.7,P\left\{X_i=-2\right\}=.2$,

And, $P\left\{X_i=10\right\}=.1$
Hence, the value of$X_i$ is:
$$\begin{gathered} \mu =E\left[X\right]=(-1)(.7)+(-2)(.2)+10(.1)=-.1 \\ E\left[X^2\right]=(-1{)}^2(.7)+(-2{)}^2(.2)+{10}^2(.1)=11.5 \end{gathered}$$
The variance of$X_i$ is
${\sigma }^2=Var\left(X_i\right)=11.5-(-.1{)}^2=11.49$

Explanation

Let, gambler's first$100$ bets:
$X=X_1+X_2+\cdots +X_{100}$
The probability that the gambler will be losing after his first$100$ bets is:
$P\{X<0\}$
Using the central limit theorem:
$$\begin{gathered} P\{X<0\}=P\left\{\frac{X-100\mu }{10\sigma }<\frac{(0-.5)-100\mu }{10\sigma }\right\} \\ \approx \Phi \left(\frac{-.5-100\mu }{10\sigma }\right)=\Phi \left(\frac{-.5-100(-.1)}{10\sqrt{11.49}}\right) \\ =\Phi (.28)=.6103 \end{gathered}$$

Hence, the probability is $.6103$