Chapter 7, Problem 25PT2

(a)

To determine

To describe how you will carry out a simulation of this game using the random digit table.

Explanation of Solution

In the question it is given that five cards each with different symbol are shuffled and we choose one. If it is diamond we win. Thus, we will carry out a simulation of this game using the random digit table as labeling are assigned based on the order they appeared in the picture but it will be arbitrary. So, first we need numeric labels for each shape. As,

  $\text{Star=0, Hexagon=1, Circle=2, Triangle=3, Diamond=4}$

Since $5$ is the maximum number we only need one-digit. We will cycle through the random digits in chunks of one if the number is $0-4$ then the cards is selected otherwise we move to the next random number.

(b)

To determine

To perform ten repetitions of your simulation.

Explanation of Solution

In the question it is given that five cards each with different symbol are shuffled and we choose one. If it is diamond we win. Thus, the payout per repetition will be the number of random numbers it took the amount of draws subtracted from $5$ . For example, if it took $3$ draws, then $5-3=2$ . If it took ten draws, then $5-\text{1}0=-5$ . This is because you need to pay $5$ dollars every selection and you get $5$ dollars at the end of the game. So, each repetition will be drawing a card then moving to the next random number until a diamond is drawn. We will calculate the payouts per each repetition. And numbers $5-9$ are excluded as mentioned in part (a). Thus, we have,

  $\begin{array}{l}\text{Repetition 1: 1, 2, 1, 3, 2, 1, 3, 0, 4 }\left(\text{Stop at 4}\right)\text{. Length: 9}\text{. Profit = }5-9=-4\\ \text{Repetition 2: }\left(\text{continue where we left off}\right)\text{ 4 }\left(\text{stop}\right)\text{. Length = 1, Payout = }5-2=4\\ \text{Repetition 3: 1, 4 }\left(\text{stop}\right)\text{. Length = 2, payout = 5}-3=3\\ \text{Repetition 4: 4 }\left(\text{stop}\right)\text{. Length = 1, payout = 5}-1=4\\ \text{Repetition 5: 2, 3, 2, 1, 1, 4 }\left(\text{stop}\right)\text{. Length = 6}\text{. Payout = 5}-6=-1\\ \text{Repetition 6: 0, 0, 0, 3, 0, 0, 2, 4 }\left(\text{stop}\right)\text{, length = 8, payout = 5}-8=-3\\ \text{Repetition 7: 2, 3, 4 }\left(\text{stop}\right)\text{. Length = 3}\text{. Payout = 5}-3=2\\ \text{Repetition 8: 2, 3, 2, 2, 0, 1, 2, 4 }\left(\text{stop}\right)\text{. Length = 8}\text{. Payout = 5 }-8=-3\\ \text{Repetition 9: 2, 3, 2, 2, 0, 1, 2, 4 }\left(\text{stop}\right)\text{. Length = 8}\text{. Payout = 5 }-8=-3\\ \text{Repetition 10: 1, 1, 4 }\left(\text{stop}\right)\text{. Length = 3}\text{. Payout = 5}-3=2\end{array}$

(c)

To determine

To find out what is the average number of cards you would need to draw to obtain a diamond.

Answer to Problem 25PT2

The average length of the simulation was $4.9$ draws.

Explanation of Solution

In the question it is given that five cards each with different symbol are shuffled and we choose one. If it is diamond we win. Thus, the number of cards in each of our simulation was titled length. To find the average we would add up all the ten repetitions and divide by $10$ . And the sum of all lengths in the previous part is $49$ . So, we have,

  $\begin{array}{l}\text{Average length}=\frac{49}{10}\\ =4.9\end{array}$

(d)

To determine

To explain is this a fair game.

Answer to Problem 25PT2

Yes, this is a fair game.

Explanation of Solution

In the question it is given that five cards each with different symbol are shuffled and we choose one. If it is diamond we win. Thus, to answer this question, we will use the simulation results. In order to get a more accurate answer, we would want to increase the repetition sizes. While the numbers are sampled at random, the sample size is low enough that sampling error may be large. And the decision rule would be, if we draw on average of five cards per game, we would break even. Based on our simulation results, we drew about $4.9$ cards on average. Since we drew $4.9$ cards on average we would say this is fairly close to $5$ . If the numbers were truly random we had have a $1\text{ out of 5}$ probability of getting the diamond and the expected loss would be zero. This is supported by the simulation and would be a fair game.