Chapter 17.6, Problem 14RE

a.

To determine

Fill in the blanks.

Answer to Problem 14RE

The house will make money on 52.9412 of the pairs of bets, give or take 4.9993 or so.

Explanation of Solution

Expected value for the sum of the draws:

The expected value for the sum of the draws made at random with replacement from the box equals $\left(\text{number of draws}\right)\times \left(\text{average of box}\right)$.

There are 38 possible outcomes for a roulette wheel. The numbers are 0,00 and numbers from 1 to 36. The numbers 0 and 00 are green and 18 of the 36 numbers are red and the remaining are black.

Let 1 represents a red number and 0 represents a green or black number.

The number of wins is equivalent to drawing 100 tickets at random with replacement from the box containing 18 tickets labeled 1 and 20 tickets labeled 0.

The pair of bets, $1 on red and $1 on black, is made 100 times. There are two numbers. The average is obtained as follows:

$\begin{array}{c}\text{Average }=\frac{1}{38}\left(\stackrel{18\text{ times}}{\overbrace{1+1+\mathrm{...}+1}}+\stackrel{20}{\overbrace{0+0+\mathrm{...}+0}}\right)\\ =\frac{18}{38}\\ =\frac{9}{17}\end{array}$

Thus, the expected value for the sum of the draws is as follows:

$\begin{array}{c}\text{Expected value =}\left(\text{number of draws}\right)\times \left(\text{average of box}\right)\\ =100\times \frac{9}{17}\\ =52.9412\end{array}$

That is, the expected value for the sum of the draws is 52.9412.

Standard error for the sum of the draws:

The standard error for the sum of the draws made at random with replacement from the box equals $\sqrt{\text{number of draws}}\times \left(\text{SD of box}\right)$.

The standard deviation is obtained as follows:

$\begin{array}{c}\left(\begin{array}{l}\text{Standard deviation }\\ \text{of the box}\end{array}\right)=\left(\text{big number}-\text{small number}\right)\sqrt{\left(\begin{array}{l}\text{fraction with big number}\times \\ \text{fraction with small number}\end{array}\right)}\\ =\left(1-0\right)\sqrt{\frac{18}{38}\times \frac{20}{38}}\\ =0.4993\end{array}$

Thus, the standard error for the sum of the draws is as follows:

$\begin{array}{c}\text{Stanadard error =}\sqrt{\text{number of draws}}\times \left(\text{SD of box}\right)\\ =\sqrt{100}\times 0.4993\\ =4.993\end{array}$

The house will make money on 52.9412 of the pairs of bets, give or take 4.9993 or so.

b.

To determine

Fill in the blanks.

Answer to Problem 14RE

The net gain for the house from the 100 pairs of bets will be around $5, give or take $9.986 or so.

Explanation of Solution

There are 38 possible outcomes for a roulette wheel. The numbers are 0,00 and numbers from 1 to 36. The numbers 0 and 00 are green and 18 of the 36 numbers are red and the remaining are black.

The house wins –$1 when a red number is obtained and wins $1 when a green or black is obtained.

The number of wins is equivalent to drawing 100 tickets at random with replacement from the box containing 18 tickets labeled -$1 and 20 tickets labeled $1.

The average is obtained as follows:

$\begin{array}{c}\text{Average }=\frac{1}{38}\left(\stackrel{18\text{ times}}{\overbrace{\left(-\$1\right)+\left(-\$1\right)+\mathrm{...}+\left(-\$1\right)}}+\stackrel{20}{\overbrace{\left(\$1\right)+\left(\$1\right)+\mathrm{...}+\left(\$1\right)}}\right)\\ =\frac{-\$18+\$20}{38}\\ =\frac{\$2}{38}\\ =\$0.05\end{array}$

Thus, the expected value for the sum of the draws is as follows:

$\begin{array}{c}\text{Expected value =}\left(\text{number of draws}\right)\times \left(\text{average of box}\right)\\ =100\times \$0.05\\ =\$5\end{array}$

That is, the expected value for the sum of the draws is $5.

Standard error for the sum of the draws:

The standard error for the sum of the draws made at random with replacement from the box equals $\sqrt{\text{number of draws}}\times \left(\text{SD of box}\right)$.

The standard deviation is obtained as follows:

$\begin{array}{c}\left(\begin{array}{l}\text{Stanadard deviation }\\ \text{of the box}\end{array}\right)=\left(\text{big number}-\text{small number}\right)\sqrt{\left(\begin{array}{l}\text{fraction with big number}\times \\ \text{fraction with small number}\end{array}\right)}\\ =\left(1-\left(-1\right)\right)\sqrt{\frac{18}{38}\times \frac{20}{38}}\\ =0.9986\end{array}$

Thus, the standard error for the sum of the draws is as follows:

$\begin{array}{c}\text{Stanadard error =}\sqrt{\text{number of draws}}\times \left(\text{SD of box}\right)\\ =\sqrt{100}\times 0.9986\\ =9.986\end{array}$

The net gain for the house from the 100 pairs of bets will be around $5, give or take $9.986 or so.