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Chapter 14: Problem 11

Pick a card, any card You draw a card from a deck. If you get a red card, you win nothing. If you get a spade, you win $$\$ 5$$. For any club, you win \(\$ 10\) plus an extra \(\$ 20\) for the ace of clubs. a. Create a probability model for the amount you win. b. Find the expected amount you'll win. c. What would you be willing to pay to play this game?

Short Answer

Expert verified
a) The probability model is: P(Winnings=$0) = 0.5, P(Winnings=$5) = 0.25, P(Winnings=$10) = \(\frac{11}{52}\), P(Winnings=$30) = \(\frac{1}{52}\). b) The expected winnings is approximately $2.88. c) A player should be willing to pay less than $2.88 to play this game.

Step by step solution

01

Create a Probability Model

There are total 52 cards in a deck - 26 red cards (which doesn't win anything), 13 spades cards ($5 winning), 12 clubs cards ($10 winning) and 1 specific club - 'Ace of Clubs' ($30 winning). Each card has a probability to be picked of \(\frac{1}{52}\). For red cards, the winnings are $0, for spade cards, the winnings are $5, for club cards (except the 'Ace of Clubs'), the winnings are $10, and for 'Ace of Clubs', the winnings are $30.
02

Calculate Expected Winnings

To calculate the expected winnings, multiply each potential outcome by its probability and sum these values. The expected winnings is E[X] = \(\sum x \cdot P(x)\), where \(x\) is the winnings and \(P(x)\) is the probability of \(x\). Therefore, E[X] = 0 * \(\frac{26}{52}\) + 5 * \(\frac{13}{52}\) + 10 * \(\frac{11}{52}\) + 30 * \(\frac{1}{52}\) = $\approx \$2.88.
03

Determine the Amount to Pay

Based on the expected returns of approximately $2.88, a player should be willing to pay less than this amount to play this game, since the price of playing should be less than or equal to the expected return to ensure non-negative profit expected.

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Key Concepts

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

Expected Value
The concept of expected value is fundamental in probability theory because it gives a single number to summarize the value expected from a random event. When playing card games where the outcomes can lead to monetary gains or losses, understanding expected values can help in making informed decisions.

In this card picking game, we find the expected value by considering all possible outcomes and their probabilities. For each possible draw from the deck, we multiply the value won by the probability of drawing that card. The sum of these products gives us the expected value, which represents the average winnings per single play over a long period.

This calculation helps to determine whether a game 鈥渇avors鈥� the player or not. If the expected gain is positive, it's likely beneficial for the player to play the game frequently.
Probability Distribution
A probability distribution in card games showcases the probabilities of different outcomes in a structured way. For this particular exercise, the probability distribution involves:
  • a 50% chance (probability of \( \frac{26}{52} \)) of drawing a red card and winning \(0,
  • a 25% chance (\( \frac{13}{52} \)) of drawing a spade card and winning \)5,
  • a 21.15% chance (\( \frac{11}{52} \)) of drawing a club card (excluding Ace of Clubs) and winning \(10,
  • and a 1.92% chance (\( \frac{1}{52} \)) of drawing the Ace of Clubs and winning \)30.
The probabilities add up to 1, covering all possible outcomes. When decisions are based on probability distributions, it's clearer to see how likely certain results are compared to others, which supports strategic decision-making.
Card Games
Playing card games is not just about luck but involves understanding probabilities and strategies to maximize winnings. Each deck contains 52 cards divided into four suits: hearts, diamonds, clubs, and spades. Each suit has 13 cards each.

Red cards (hearts and diamonds) typically represent one kind of outcome in this exercise, meaning winning nothing. Conversely, black cards, spades, and clubs present opportunities for monetary wins, enhancing the game's dynamics by assigning monetary values and thus impacting strategies.

By assigning these values, you can create a game model that combines entertainment with chances of gain, blending fun with elements of gambling and analysis.
Probability Theory
Probability theory is the mathematical framework that deals with the likelihood of different outcomes. It provides tools and methods to predict future events based on past occurrences and statistical principles.

In card games, probability theory helps in calculating winning chances which guides bet decisions. By knowing probabilities, like those in this exercise, players can decide how much to stake based on potential results. The exercise illustrates practical application of these concepts in determining the expected winnings and whether or not to participate in the game.

Overall, an understanding of probability theory paves the way for strategic advantage and better decision-making in games of chance and investment.

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Most popular questions from this chapter

The bicycle shop in Exercise 50 will be offering 2 specially priced children's models at a sidewalk sale. The basic model will sell for $$\$ 120$$ and the deluxe model for $$\$ 150 .$$ Past experience indicates that sales of the basic model will have a mean of 5.4 bikes with a standard deviation of 1.2 , and sales of the deluxe model will have a mean of 3.2 bikes with a standard deviation of 0.8 bikes. The cost of setting up for the sidewalk sale is $$\$ 200$$. a. Define random variables and use them to express the bicycle shop's net income. b. What's the mean of the net income? c. What's the standard deviation of the net income? d. Do you need to make any assumptions in calculating the mean? How about the standard deviation?

A company selling vegetable seeds in packets of 20 estimates that the mean number of seeds that will actually grow is \(18,\) with a standard deviation of 1.2 seeds. You buy 5 different seed packets. a. How many bad (non-growing) seeds do you expect to get? b. What's the standard deviation? c. What assumptions did you make about the seeds? Do you think that assumption is warranted? Explain.

Day trading An option to buy a stock is priced at $$\$ 200$$. If the stock closes above 30 on May \(15,\) the option will be worth $$\$ 1000$$. If it closes below \(20,\) the option will be worth nothing, and if it closes between 20 and 30 (inclusively), the option will be worth $$\$ 200$$. A trader thinks there is a \(50 \%\) chance that the stock will close in the \(20-30\) range, a \(20 \%\) chance that it will close above 30 , and a \(30 \%\) chance that it will fall below 20 on May \(15 .\) a. Should she buy the stock option? b. How much does she expect to gain? c. What is the standard deviation of her gain?

Software A small software company bids on two contracts and knows it can only get one of them. It anticipates a profit of $$\$ 50,000$$ if it gets the larger contract and a profit of $$\$ 20,000$$ on the smaller contract. The company estimates there's a \(30 \%\) chance it will get the larger contract and a \(60 \%\) chance it will get the smaller contract. Assuming the contracts will be awarded independently, what's the expected profit?

The amount of cereal that can be poured into a small bowl varies with a mean of 1.5 ounces and a standard deviation of 0.3 ounces. A large bowl holds a mean of 2.5 ounces with a standard deviation of 0.4 ounces. You open a new box of cereal and pour one large and one small bowl. a. How much more cereal do you expect to be in the large bowl? b. What's the standard deviation of this difference? c. If the difference follows a Normal model, what's the probability the small bowl contains more cereal than the large one? d. What are the mean and standard deviation of the total amount of cereal in the two bowls? e. If the total follows a Normal model, what's the probability you poured out more than 4.5 ounces of cereal in the two bowls together? f. The amount of cereal the manufacturer puts in the boxes is a random variable with a mean of 16.3 ounces and a standard deviation of 0.2 ounces. Find the expected amount of cereal left in the box and the standard deviation.
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