Chapter 10.4, Problem 15E

Solution

To determine: The prize for drawing the ace of diamonds.

The prize for drawing the ace of diamonds is $75.

Given information:

The game is considered as a fair game when the expected value is 0.

Pay the $5 for draw a card from a deck.

If it’s black, you lose.

If it’s a heart, you win your $5 back.

If it’s any diamond except the ace, you win $10.

Formula used:

  $\mathrm{P}\left(X\right)=\frac{E}{S}$   ...... (1)

Here, E is the favourable outcomes and S is the total possible outcomes.

Calculation:

A standard deck of cards contains 52 cards, of which 26 are red and 26 are black, 13 are of each suit (hearts, diamonds, spade, clubs) and out of which 4 are of each denomination (A, 2 to 10, J, Q, K).

The face cards are the jacks J, queens Q and kings K.

When a black card is drawn, then lose the $5 that draw a card 26 of the 52 cards in the deck of cards are black.

Substitute 26 for $E$ , 52 for $S$ and - $\$5$ for $X$ in equation (1),

  $\mathrm{P}(-\$5)=\frac{26}{52}$

When a heart is drawn, then win our $5 back and thus a payoff of $5-$5 = $0.

13 of the 52 cards in the deck of cards are hearts.

Substitute 13 for $E$ , 52 for $S$ and $\$0$ for $X$ in equation (1),

  $\mathrm{P}(\$0)=\frac{13}{52}$

When a diamond is drawn that is not the ace of diamonds, then win $10 while already paid $5 to draw a card and thus, a payoff of $10-$5 = $5.

12 of the 52 cards in the deck of cards are diamonds but not the ace of diamonds.

Substitute 12 for $E$ , 52 for $S$ and $\$5$ for $X$ in equation (1),

  $\mathrm{P}(\$5)=\frac{12}{52}$

Let the prize of drawing the ace of diamonds be $\$x$ .

It costs $5 to draw a card, we would then have a payoff of $\left(\$x-\$5\right)=\$\left(x-5\right)$ .

1 of the 52 cards in the deck of cards is the ace of diamonds.

Substitute 1 for $E$ , 52 for $S$ and $\$\left(x-5\right)$ for $X$ in equation (1),

  $\mathrm{P}(\$(x-5))=\frac{1}{52}$

Now,

The expected value (or mean) is the sum of the product of each possibility x with its probability P(x):

  $\mu ={\displaystyle \sum xP(x)}$   ...... (2)

Substitute $-\$5\text{and}\frac{26}{52}$ for $x$ and P( $x$ ), $\$0\text{and}\frac{13}{52}$ for $x$ and P( $x$ ), $\$5\text{and}\frac{12}{52}$ for $x$ and P( $x$ ) and $\$\left(x-5\right)\text{and}\frac{1}{52}$ for $x$ and P( $x$ ) in equation (2) respectively,

  $\begin{array}{c}\mu =(-5)\times \frac{26}{52}+0\times \frac{13}{52}+5\times \frac{12}{52}+(x-5)\times \frac{1}{52}\\ =\frac{x-75}{52}\end{array}$

For the fair game, the expected value should be 0.

  $\begin{array}{c}\frac{x-75}{52}=0\\ x-75=0\\ x=75\end{array}$

Thus, the prize for drawing the ace of diamonds is $75.