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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

An employer pays a mean salary for a 5-day workweek of \(\$ 1250\) with a standard deviation of \(\$ 129 .\) On the weekends, his salary expenses have a mean of \(\$ 450\) with a standard deviation of \(\$ 57 .\) What is the mean and standard deviation of his total weekly salaries?

At a certain coffee shop, all the customers buy a cup of coffee; some also buy a doughnut. The shop owner believes that the number of cups he sells each day is Normally distributed with a mean of 320 cups and a standard deviation of 20 cups. He also believes that the number of doughnuts he sells each day is independent of the coffee sales and is Normally distributed with a mean of 150 doughnuts and a standard deviation of 12 . a. The shop is open every day but Sunday. Assuming day-to-day sales are independent, what's the probability he'll sell more than 2000 cups of coffee in a week? b. If he makes a profit of 50 cents on each cup of coffee and 40 cents on each doughnut, can he reasonably expect to have a day's profit of over $$\$ 300 ?$$ Explain. c. What's the probability that on any given day he'll sell a doughnut to more than half of his coffee customers?

1 is 0.8 . If you get contract #1, the probability you also get contract #2 will be 0.鈥� # Your company bids for two contracts. You believe the probability you get contract #1 is 0.8 . If you get contract #1, the probability you also get contract #2 will be 0.2, and if you do not get #1, the probability you get #2 will be 0.3 a. Are the two contracts independent? Explain. b. Find the probability you get both contracts. c. Find the probability you get no contract. d. Let \(X\) be the number of contracts you get. Find the probability model for \(X\) e. Find the expected value and standard deviation.

A delivery company's trucks occasionally get parking tickets, and based on past experience, the company plans that each truck will average 1.3 tickets a month, with a standard deviation of 0.7 tickets. a. If they have 18 trucks, what are the mean and standard deviation of the total number of parking tickets the company will have to pay this month? b. What assumption did you make in answering?

An insurance company estimates that it should make an annual profit of $$\$ 150$$ on each homeowner's policy written, with a standard deviation of $$\$ 6000$$. a. Why is the standard deviation so large? b. If it writes only two of these policies, what are the mean and standard deviation of the annual profit? c. If it writes 10,000 of these policies, what are the mean and standard deviation of the annual profit? d. Is the company likely to be profitable? Explain. e. What assumptions underlie your analysis? Can you think of circumstances under which those assumptions might be violated? Explain.
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