Chapter 10.4, Problem 8E

Solution

To calculate: The expected value of the payoff and find for the play of this game need to pay $5 or not.

The expected value of the payoff is -$0.77 and for play this game no need to pay any amount.

Given information:

For $5, a carnival huckster will let you draw a card at random from a fair deck of 52.

He will pay you a dollar times the value of your card (ace = 1), but zero for a face card.

Formula used:

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

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

Calculation:

A standard deck of cards contains 52 cards, out of which 26 are red and 26 are black, 13 are of each suit (hearts, diamonds, spades, clubs) and 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 drawing a face card, then the payout is $0 while it costs $5 to draw a card and thus, make a loss of $5.

3·4=12 of the 52 cards in the standard deck of cards are face cards.

Substitute 12 for $E$ and 52 for $S$ in equation (1)

  $P\left(-\$5\right)=\frac{12}{52}=\frac{3}{13}$

When drawing a non-face card with value x, then the payout is $x while it costs $5 to draw a card and thus, make a profit of $x-$5.

4 of the 52 cards are non-face cards with value x.

Substitute 4 for $E$ and 52 for $S$ in equation (1)

  $P\left(-\$4\right)=\frac{4}{52}=\frac{1}{13}$

  $P\left(-\$3\right)=\frac{4}{52}=\frac{1}{13}$

  $P\left(-\$2\right)=\frac{4}{52}=\frac{1}{13}$

  $P\left(-\$1\right)=\frac{4}{52}=\frac{1}{13}$

  $P\left(\$0\right)=\frac{4}{52}=\frac{1}{13}$

  $P\left(\$1\right)=\frac{4}{52}=\frac{1}{13}$

  $P\left(\$2\right)=\frac{4}{52}=\frac{1}{13}$

  $P\left(\$3\right)=\frac{4}{52}=\frac{1}{13}$

  $P\left(\$4\right)=\frac{4}{52}=\frac{1}{13}$

  $P\left(\$5\right)=\frac{4}{52}=\frac{1}{13}$

Now,

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

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

Substitute $x\text{and}P\left(x\right)$ for different cases in equation (2) respectively,

  $\begin{array}{c}\mu =\left(-\$5\right)\times \frac{3}{13}+\left(-\$4\right)\times \frac{1}{13}+\left(-\$3\right)\times \frac{1}{13}+\left(-\$2\right)\times \frac{1}{13}\\ +\left(-\$1\right)\times \frac{1}{13}+\left(\$0\right)\times \frac{1}{13}+\left(\$1\right)\times \frac{1}{13}+\left(\$2\right)\times \frac{1}{13}\\ +\left(\$3\right)\times \frac{1}{13}+\left(\$4\right)\times \frac{1}{13}+\left(\$5\right)\times \frac{1}{13}\\ =\frac{-\$10}{13}\\ =-\$0.77\end{array}$

Hence, the expected value of the payoff is -$0.77 and for play this game no need to pay any amount.