Chapter 5.1, Problem 15E

To determine

(a)

The Expected value of Mike’s bet.

Answer to Problem 15E

Solution:

The expected value of Mike’s bet is $-4.517$.

Explanation of Solution

Given Information:

Mike’s older brother, Jeff, bets him that he can’t roll two dice and get doubles three times in a row. If Mike does it, Jeff will give him $100.00. Otherwise, Mike has to give Jeff $5.00.

Formula Used:

The expected value for a discrete random variable X is equal to the mean of the probability distribution of X and is given by,

$E(X)={\displaystyle \sum \left[x_i.P\left(X=x_i\right)\right]}$

Where $x_i$ is the $i^{th}$ value of the random variable X.

Calculation:

There are 6 faces in a die. The total number of outcomes of rolling two dice is $36$. There are six different rolls that are doubles (double 1, double 2, double 3, double 4, double 5, double 6) out of a total of 36 possible outcomes when rolling two dice

The probability of rolling a double is

$\begin{array}{rllll}P(X& =& x)& =& \frac{6}{36}\\ & =& \frac{1}{6}\end{array}$

Each roll is independent of the others. So the probability of rolling two dice and getting doubles three times in a row is

$\begin{array}{rll}P(X& =& x)=\frac{1}{6}\times \frac{1}{6}\times \frac{1}{6}\\ & =& \frac{1}{216}\\ & =& 0.0046\end{array}$

Each roll is independent of the others. So the probability of rolling two dice and not getting doubles three times in a row is

$\begin{array}{rll}P(X& =& x)=1-\frac{1}{6}\times \frac{1}{6}\times \frac{1}{6}\\ & =& 1-\frac{1}{216}\\ & =& 1-0.0046\\ & =& 0.9954\end{array}$

X	$100	$\$-5$
$P(X = x)$	0.0046	0.9954

The expected value is given by

$E(X)={\displaystyle \sum \left[x_i.P\left(X=x_i\right)\right]}$

The expected value

$\begin{array}{rll}E(X)& =& \left(100\times 0.0046+(-5)\times 0.9954\right)\\ & =& -4.517\end{array}$

Conclusion:

Thus the expected value of Mike’s bet is $-4.517$.

To determine

(b)

The expected value of Jeff’s bet.

Answer to Problem 15E

Solution:

The expected value of Jeff’s bet is 4.517

Explanation of Solution

Given Information:

Mike’s older brother, Jeff, bets him that he can’t roll two dice and get doubles three times in a row. If Mike does it, Jeff will give him $100.00. Otherwise, Mike has to give Jeff $5.00.

Formula Used:

The expected value for a discrete random variable X is equal to the mean of the probability distribution of X and is given by

$E(X)={\displaystyle \sum \left[x_i.P\left(X=x_i\right)\right]}$

Where $x_i$ is the $i^{th}$ value of the random variable X.

Calculation:

There are 6 faces in a die. The total number of outcomes of rolling two dice is $36$. There are six different rolls that are doubles (double 1, double 2, double 3, double 4, double 5, double 6) out of a total of 36 possible outcomes when rolling two dice

The probability of rolling a double is

$\begin{array}{rllll}P(X& =& x)& =& \frac{6}{36}\\ & =& \frac{1}{6}\end{array}$

Each roll is independent of the others. So the probability of rolling two dice and getting doubles three times in a row is

$\begin{array}{rll}P(X& =& x)=\frac{1}{6}\times \frac{1}{6}\times \frac{1}{6}\\ & =& \frac{1}{216}\\ & =& 0.0046\end{array}$

Each roll is independent of the others. So the probability of rolling two dice and not getting doubles three times in a row is

$\begin{array}{rll}P(X& =& x)=1-\frac{1}{6}\times \frac{1}{6}\times \frac{1}{6}\\ & =& 1-\frac{1}{216}\\ & =& 1-0.0046\\ & =& 0.9954\end{array}$

X	$\$-100$	$5
$P(X = x)$	0.0046	0.9954

The expected value is given by

$E(X)={\displaystyle \sum \left[x_i.P\left(X=x_i\right)\right]}$

The expected value

$\begin{array}{rll}E(X)& =& \left((-100)\times 0.0046+5\times 0.9954\right)\\ & =& 4.517\end{array}$

Conclusion:

Thus the expected value of Jeff’s bet is 4.517.

To determine

(c)

The expected value of Mike’s bet.

Answer to Problem 15E

Solution:

If Mike and Jeff make the same bet 30 times, Mike is expected to lose $135.51

Explanation of Solution

Given:

Mike’s older brother, Jeff, bets him that he can’t roll two dice and get doubles three times in a row. If Mike does it, Jeff will give him $100.00. Otherwise, Mike has to give Jeff $5.00.

Formula Used:

The expected value for a discrete random variable X is equal to the mean of the probability distribution of X and is given by

$E(X)={\displaystyle \sum \left[x_i.P\left(X=x_i\right)\right]}$

Where $x_i$ is the $i^{th}$ value of the random variable X.

Calculation:

There are 6 faces in a die. The total number of outcomes of rolling two dice is $36$. There are six different rolls that are doubles (double 1, double 2, double 3, double 4, double 5, double 6) out of a total of 36 possible outcomes when rolling two dice

The probability of rolling a double is

$\begin{array}{rllll}P(X& =& x)& =& \frac{6}{36}\\ & =& \frac{1}{6}\end{array}$

Each roll is independent of the others. So the probability of rolling two dice and getting doubles three times in a row is

$\begin{array}{rll}P(X& =& x)=\frac{1}{6}\times \frac{1}{6}\times \frac{1}{6}\\ & =& \frac{1}{216}\\ & =& 0.0046\end{array}$

Each roll is independent of the others. So the probability of rolling two dice and not getting doubles three times in a row is

$\begin{array}{rll}P(X& =& x)=1-\frac{1}{6}\times \frac{1}{6}\times \frac{1}{6}\\ & =& 1-\frac{1}{216}\\ & =& 1-0.0046\\ & =& 0.9954\end{array}$

X	$100	$\$-5$
$P(X = x)$	0.0046	0.9954

The expected value is given by

$E(X)={\displaystyle \sum \left[x_i.P\left(X=x_i\right)\right]}$

The expected value

$\begin{array}{rll}E(X)& =& \left(100\times 0.0046+(-5)\times 0.9954\right)\\ & =& -4.517\end{array}$

If Mike and Jeff make the same bet 30 times, then the expected value of Mike is

$\begin{array}{rll}E(x)& =& 30\times (-4.517)\\ & =& -135.51\end{array}$

Conclusion:

Thus, if Mike and Jeff make the same bet 30 times, Mike is expected to lose $135.51.