Question

The business of casino gaming. Casino gaming yields over \(35 billion in revenue each year in the United States. Chance (Spring 2005) discussed the business of casino gaming and its reliance on the laws of probability. Casino games of pure chance (e.g., craps, roulette, baccarat, and keno) always yield a “house advantage.” For example, in the game of double-zero roulette, the expected casino win percentage is 5.26% on bets made on whether outcomewill be either black or red. (This implies that for every \)5 bet on black or red, the casino will earn a net of about 25¢.) It can be shown that in 100 roulette plays on black/red, the average casino win percentage is normally distributed with mean 5.26% and standard deviation 10%. Let x represent the average casino win percentage after 100 bets on black/red in double-zero roulette.

a. Find $\mathbf{P}\left(x>0\right)$. (This is the probability that the casino wins money.)

b. Find $\mathbf{P}\left(5<x<15\right)$.

c. Find $\mathbf{P}\left(x<1\right)$.

d. If you observed an average casino win percentage of -25% after 100 roulette bets on black/red, what would youconclude?

Short answer

a. The value of$p\left(x>0\right)$ is 0.7673

b. The value of$p\left(5<x<15\right)$ is 0.3704

c. The value of$p\left(<1\right)$ is 0.2643

d. After observing an average casino win percentage of -25% after 100 roulette bets on black/red, it is 0.0006 hence we can conclude that it is a little bit unusual to observe such a winning percentage.

Step-by-step solution

Given information

100 rouletteplays on black/red the average casino win percentage is normally distributed with a mean 5.26% and a standard deviation 10%

X isthe average casino winpercentage after 100 bets on black/ redin double-zero roulette.

Calculating the value of P(x>0)

a.

Let X follows a normal distribution with$\mu =7.3$ and$\sigma =10$

We know that$Z=\frac{x-\mu }{\sigma }$

Therefore,

$$\begin{gathered} \mathrm{P}\left(\mathrm{x}>0\right)=\mathrm{P}\left(\mathrm{z}>\frac{\mathrm{x}-\mathrm{\mu }}{\mathrm{\sigma }}\right) \\ =\mathrm{P}\left(\mathrm{z}>\frac{5-7.3}{10}\right) \\ =\mathrm{P}\left(\mathrm{z}>-0.73\right) \\ \mathrm{P}\left(\mathrm{x}>0\right)=1-\mathrm{P}\left(\mathrm{z}<-0.73\right) \\ =0.7673 \end{gathered}$$

 Calculating the value of P(5<x<15)

b.

$$\begin{gathered} \mathrm{P}\left(5<\mathrm{x}<15\right)=\mathrm{P}\left(\frac{\mathrm{x}-\mathrm{\mu }}{\mathrm{\sigma }}<\mathrm{Z}<\frac{\mathrm{x}-\mathrm{\mu }}{\mathrm{\sigma }}\right) \\ =\mathrm{P}\left(\frac{5-7.3}{10}<\mathrm{Z}<\frac{15-7.3}{10}\right) \\ \mathrm{P}\left(5<\mathrm{x}<15\right)=\mathrm{P}\left(-0.23<\mathrm{Z}<0.77\right) \\ =0.3704 \end{gathered}$$

 Calculating the value of P(x<1)

c.

$$\begin{gathered} \mathrm{P}\left(\mathrm{x}<1\right)=\mathrm{P}\left(\mathrm{Z}-\frac{\mathrm{x}-\mathrm{\mu }}{\mathrm{\sigma }}\right) \\ =\mathrm{P}\left(\mathrm{Z}<\frac{1-7.3}{10}\right) \\ \mathrm{P}\left(\mathrm{x}<1\right)=\mathrm{P}\left(\mathrm{Z}<-0.63\right) \\ =0.2643 \end{gathered}$$

Calculating the value of P(x<-25)

d.

$$\begin{gathered} \mathrm{P}\left(\mathrm{x}<-25\right)=\mathrm{P}\left(\mathrm{Z}<\frac{\mathrm{x}-\mathrm{\mu }}{\mathrm{\sigma }}\right) \\ =\mathrm{P}\left(\mathrm{Z}<\frac{-25-7.3}{10}\right) \\ \mathrm{P}\left(\mathrm{x}<-25\right)=\mathrm{P}\left(\mathrm{Z}<-3.23\right) \\ =0.0006 \end{gathered}$$

After observing an average casino win percentage of 25% after 100 roulette bet on black/red that is 0.0006 we can conclude that it is a little unusual to observe such winning percentage.