Question

A Las Vegas roulette wheel has 38 numbered slots. Roulette wheels are calibrated such that each outcome is

equally likely. To play roulette, you choose a number; you win$35 if the ball ends up in that slot and lose $1 otherwise.

d). Estimate the chance that you will win more than $12 in 100 plays.

Answer 1

The probability of winning is

p = P(win) = 1/38

Let X is a random variable shows the winning amount. Here X can take value 35 and -1. So,

P(X=35) = P(win) = 1/38

P(X=-1) = 1-P(win) = 37/ 38

The expected value of X is

E(X) = (1/38) * 35 + (37/38) * (-1)= -2 / 38 =-0.0526

Now,

$E(X^{2})=\frac{1}{38}\cdot 35^{2}+\frac{37}{38}\cdot (-1)^{2}=\frac{1188}{38}$

The standard deviation is

$\sigma=\sqrt{E(X^{2})-[E(X)]^{2}}=\sqrt{\frac{1188}{38}-\frac{4}{38^{2}}}=5.5911$

d)

Let T shows the total amount you win in 100 plays. Using CLT, T will have normal distribution with mean and SD as follow,

$\mu_{T}=100\cdot E(X)=100\cdot (-0.0526)=-5.26$

$\sigma_{T}=\sqrt{100}\cdot \sigma=10\cdot (5.5911)=55.911$

The z-score for T = 12 is

$T=\frac{12+5.26}{55.911}=0.31$

The  chance that you will win more than $12 in 100 plays is

$P(T>12)=P(z>0.31)=1-P(z\leq 0.31)=1-0.6217=0.3783$