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Chapter 5: Problem 115

Scott offers you the following game: You will roll two fair dice. If the sum of the two numbers obtained is \(2,3,4,9,10,11\), or 12, Scott will pay you \(\$ 20\). However, if the sum of the two numbers is 5 , 6,7, or 8 , you will pay Scott \(\$ 20\). Scott points out that you have seven winning numbers and only four losing numbers. Is this game fair to you? Should you accept this offer? Support your conclusion with appropriate calculations.

Short Answer

Expert verified
No, the game is not fair. Although there are more winning numbers, the probability of getting a losing sum is higher. The probability of winning is \(4/9\) whereas the probability of losing is \(5/9\). Therefore, accepting Scott's offer wouldn't be advisable.

Step by step solution

01

Analyze Possible Outcomes of Dice Rolls

Analyzing the potential outcomes of rolling two dice, it can be noticed that there are a total of \(6 * 6 = 36\) possible outcomes. The reason for this is there are 6 sides on each die and two dice are being rolled.
02

Calculate Winning Probabilities

Next, the frequency of the sums that lead to winning the game need to be calculated. These sums are 2,3,4,9,10,11,12. The outcomes can be split into the following possibilities: \n Sum of 2 : 1 way (1,1) \n Sum of 3 : 2 ways (1,2), (2,1) \n Sum of 4 : 3 ways (1,3), (2,2), (3,1) \n Sum of 9 : 4 ways (3,6), (4,5), (5,4), (6,3) \n Sum of 10 : 3 ways (4,6), (5,5), (6,4) \n Sum of 11 : 2 ways (5,6), (6,5) \n Sum of 12 : 1 way (6,6) \n Adding these winning possibilities, we get a total of 16 winning ways out of 36 possible outcomes. The probability of winning is thus \(16/36\), which reduces to \(4/9\).
03

Calculate Losing Probabilities

We now calculate the frequency of the sums that lead to a loss in the game. This includes the sums: 5, 6, 7, and 8. We split these outcomes into the following possibilities: \n Sum of 5 : 4 ways (1,4), (2,3), (3,2), (4,1) \n Sum of 6 : 5 ways (1,5), (2,4), (3,3), (4,2), (5,1) \n Sum of 7 : 6 ways (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) \n Sum of 8 : 5 ways (2,6), (3,5), (4,4), (5,3), (6,2) \n Adding these losing possibilities, we get a total of 20 losing ways out of 36 possible outcomes. The probability of losing is hence \(20/36\), which reduces to \(5/9\).
04

Evaluate Fairness

Given that the probability of losing (\(5/9\)) is higher than the probability of winning (\(4/9\)), it can be concluded that the game is not fair. The odds are stacked against you, and hence, it would not be advisable to accept Scott's offer.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Probability
Probability is a fundamental concept in mathematics that measures the likelihood of a specific outcome occurring. It is quantified as a number between 0 and 1, where 0 indicates impossibility and 1 signifies certainty.

When dealing with dice games, such as the game Scott offers, it's important to calculate the probability of winning and losing outcomes based on the number of favorable results against all possible results.
  • In our exercise, rolling two dice results in 36 possible outcomes (6 sides on each die, 6 * 6 = 36).
  • To figure out the probability of winning, you count the number of outcomes that allow you to win. For Scott's game, this is when the dice sum to 2, 3, 4, 9, 10, 11, and 12, totaling 16 winning outcomes.
  • The probability of winning is then the number of winning outcomes divided by the total number of outcomes: \(\frac{16}{36}\), simplifying to \(\frac{4}{9}\).
Understanding these calculations helps determine whether a game is favorable or not by comparing the probabilities of winning versus losing outcomes.
Basics of Dice Games
Dice games are often used in probability exercises because of their simple yet varied outcomes. Each die has six faces, numbered from 1 to 6, making probability computations both straightforward and diverse.

In the case of games involving multiple dice, like Scott's game, analyzing the chances of different sums results in a deeper understanding:
  • For two dice, the smallest sum is 2 (with a combination of 1+1), and the largest is 12 (with a combination of 6+6).
  • Different sums will have different frequencies, i.e., some outcomes will be more likely than others. For instance, a sum of 7 can be rolled in 6 different ways, making it more common than a 2 or 12, which each have only one combination.
The variety in potential outcomes is what makes dice games educational tools for learning about probability and expected value.
Explaining Expected Value
Expected value is a key concept in probability that provides a prediction of the average outcome of a random event, based on all possible outcomes and their probabilities. In games like the one Scott offers, knowing the expected value can help decide whether the game is beneficial or not.

To compute the expected value of participating in Scott鈥檚 game, consider:
  • For each winning sum (2,3,4,9,10,11,12), you earn \(20. The probability of each event is \(\frac{4}{9}\).
  • In contrast, for each losing sum (5,6,7,8), you owe \)20. The probability of these events is \(\frac{5}{9}\).
The expected value (EV) is calculated as:\[EV = 20 \times \frac{4}{9} - 20 \times \frac{5}{9}= \frac{80}{9} - \frac{100}{9} = -\frac{20}{9}\]A negative expected value signifies that, on average, you're expected to lose money in the long run, indicating it's not a favorable game to play in this case.
Understanding Gambling Mathematics
Gambling mathematics involves applying mathematical principles to understand and predict gambling outcomes. These principles help analyze if games offer fair chances or if they are biased against the player.

In gambling, and games like Scott's dice game, mathematics can reveal the truth behind seemingly attractive offers. Even though Scott proposes more winning sums than losing sums, mathematical analysis uncovers an imbalance:
  • Though there are seven favorable outcomes compared to four unfavorable ones, the probabilities and expected values reveal a different tale.
  • A higher risk (losing more frequently and losing money overall) demonstrates how a basic understanding of probability isn't always sufficient without further mathematical inquiry.
Applying gambling mathematics ensures that players make informed decisions based on thorough probability and expected value assessments rather than misleading odds or enticing offers.

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Most popular questions from this chapter

Let \(x\) be the number of magazines a person reads every week. Based on a sample survey of adults, the following probability distribution table was prepared. $$ \begin{array}{l|cccccc} \hline x & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline P(x) & .36 & .24 & .18 & .10 & .07 & .05 \\ \hline \end{array} $$ Find the mean and standard deviation of \(x\).

According to a Harris Interactive poll, \(52 \%\) of American college graduates have Facebook accounts (http://www.harrisinteractive.com/harris_poll/pubs/Harris_Poll \(200904_{-16}\).pdf). Suppose that this result. is true for the current population of American college graduates. a. Let \(x\) be a binomial random variable that denotes the number of American college graduates in a random sample of 15 who have Facebook accounts. What are the possible values that \(x\) can assume? b. Find the probability that exactly 9 American college graduates in a sample of 15 have Facebook accounts.

In a 2009 poll of adults 18 years and older, (BBMG Conscious Consumer Report) about half of them said that despite tough economic times, they are willing to pay more for products that have social and environmental benefits. Suppose that \(50 \%\) of all such adults currently hold this view. Suppose that a random sample of 20 such adults is selected. Use the binomial probabilities table (Table I of Appendix \(\mathrm{C}\) ) or technology to find the probability that the number of adults in this sample who hold this opinion is a. at most 7 b. at least 13 c. 12 to 15

The following table gives the probability distribution of a discrete random variable \(x\) $$ \begin{array}{l|lllllll} \hline x & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline P(x) & .11 & .19 & .28 & .15 & .12 & .09 & .06 \\ \hline \end{array} $$ Find the following probabilities. a. \(P(x=3)\) b. \(P(x \leq 2)\) c. \(P(x \geq 4)\) d. \(P(1 \leq x \leq 4)\) e. Probability that \(x\) assumes a value less than 4 f. Probability that \(x\) assumes a value greater than 2 g. Probability that \(x\) assumes a value in the interval 2 to 5

Classify each of the following random variables as discrete or continuous. a. The time left on a parking meter b. The number of bats broken by a major league baseball team in a season c. The number of cars in a parking lot d. The total pounds of fish caught on a fishing trip e. The number of cars crossing a bridge on a given day f. The time spent by a physician examining a patient
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