/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 Consider the following card game... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the following card game with a well-shuffled deck of cards. If you draw 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 at this game. Also, find the expected winnings for a single game and the standard deviation of the winnings. (b) What is the maximum amount you would be willing to pay to play this game? Explain your reasoning.

Short Answer

Expert verified
Expected win: $4.13, Risk: \(\sigma \approx 4.73\). Max pay: $4.13 (to avoid loss).

Step by step solution

01

Define the Outcomes and Winnings

There are 52 cards: 26 red cards (hearts and diamonds, win nothing) and 26 black cards (spades and clubs). The winnings are structured as follows: Red cards - $0, Spades (13 cards) - $5 each, Clubs (12 cards) - $10 each (excluding ace), Ace of Clubs (1 card) - $30 ($10 + $20 extra).
02

Calculate Probabilities for Each Outcome

The probability of drawing a red card is \( \frac{26}{52} = \frac{1}{2} \). The probability of drawing a spade is \( \frac{13}{52} = \frac{1}{4} \). The probability of drawing a club (excluding the ace) is \( \frac{12}{52} = \frac{3}{13} \) and the probability of drawing the ace of clubs is \( \frac{1}{52} \).
03

Build the Probability Model

Assign winnings to outcomes in the probability model:- Red card: win \(0 with probability \( \frac{1}{2} \).- Spade: win \)5 with probability \( \frac{1}{4} \).- Non-ace club: win \(10 with probability \( \frac{3}{13} \).- Ace of Clubs: win \)30 with probability \( \frac{1}{52} \).
04

Calculate the Expected Winnings (Mean)

The expected winnings \( E(X) \) is calculated by summing the products of each outcome's winnings and its probability:\[ E(X) = 0 \times \frac{1}{2} + 5 \times \frac{1}{4} + 10 \times \frac{3}{13} + 30 \times \frac{1}{52} \]= 0 + 1.25 + 2.3077 + 0.5769 = 4.1346 (approximately $4.13).
05

Calculate the Standard Deviation of Winnings

First, calculate the variance \( \sigma^2 \):\[ \sigma^2 = (0 - 4.1346)^2 \times \frac{1}{2} + (5 - 4.1346)^2 \times \frac{1}{4} + (10 - 4.1346)^2 \times \frac{3}{13} + (30 - 4.1346)^2 \times \frac{1}{52} \]Evaluate each term, and sum them to find \( \sigma^2 = 22.35 \) (rounded to 2 decimal places).The standard deviation \( \sigma = \sqrt{22.35} \approx 4.73 \).
06

Determine Maximum Pay to Play

The maximum amount you should be willing to pay to play the game is the expected winnings, which is about $4.13. Paying more would result in a loss over time.

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Key Concepts

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

Expected Value
Expected value is a key concept in probability and statistics that represents the average outcome if an experiment or game were repeated a very large number of times. It provides a way to quantify the center of a probability distribution, which can help us understand what to expect over the long run.
In the card game exercise, the expected value was computed as the sum of all possible winnings, each weighted by their probability. We had four different outcomes based on which type of card was drawn, and each had an associated probability and amount won (from $0 to $30). By multiplying each amount by its probability and adding all these products together, we arrived at an expected value of approximately $4.13.
This means that if you played the card game many times, you would, on average, win about $4.13 per game. This is useful for determining how much you should be willing to pay to play the game, aiming to match or stay below this expected value in the long term.
Standard Deviation
Standard deviation is a measure of how dispersed or spread out a set of values is around the expected value. It gives us an idea of the variability or risk involved with a particular result.
In our card game, after determining the expected value of $4.13, we calculated how each winning scenario deviated from this average. We did this by finding the squared differences between each actual winnings and the expected value, then weighting these by their corresponding probabilities. The sum of these terms gives the variance, and by taking the square root of the variance, we find the standard deviation.
For this game, the standard deviation was found to be approximately $4.73. This tells us that the winnings per game tend to deviate by around $4.73 from the expected value. A higher standard deviation indicates more variability, meaning the outcomes can fluctuate noticeably from the average.
Statistical Reasoning
Statistical reasoning involves applying statistical tools and concepts to make informed decisions based on data. It helps to analyze situations quantitatively, providing insights that might not be obvious from raw numbers.
In the context of this card game, we used statistical reasoning to determine how much a player should be willing to pay to participate. By analyzing the expected value of $4.13 and linking it with the concept of long-term averages, statistical reasoning guided us to suggest that paying more than this amount might not be wise. This approach helps in understanding the financial implications and risks associated with decisions involving chance or uncertainty.
Additionally, it underscores the value of the standard deviation, as it outlines potential deviations from the average, helping in assessing risk more comprehensively.
Card Game Probability
Card game probability deals with understanding and calculating the likelihood of different outcomes occurring in a game involving cards. Each card drawn in a typical deck instructs a random event with specific probabilities that can be determined through the composition of the deck.
In our card game, the probability of drawing a red card, spade, club, or ace of clubs was calculated based on their respective counts in the deck. For instance, there was a 50% chance of drawing a red card since there are 26 red cards out of 52. Similarly, other probabilities were calculated based on the count of each type of card relative to the total cards.
By creating a probability model from this information, we were able to determine the expected value and standard deviation, thereby providing a deeper understanding and strategic insight into the mechanics and strategy of playing such a game.

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