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Chapter 14: Problem 15

A Game of Chance \(A\) box contains 100 envelopes. Ten envelopes contain \(\$ 10\) each, ten contain \(\$ 5\) each, two are "unlucky," and the rest are empty. A player draws an envelope from the box and keeps whatever is in it. If a person draws an unlucky envelope, however, he or she must pay \(\$ 100 .\) What is the expectation of a person playing this game?

Short Answer

Expert verified
The expectation of the game is -$0.50, meaning an average loss of $0.50 per play.

Step by step solution

01

Identify Possible Outcomes and Probabilities

The possible outcomes of drawing an envelope are: getting \(10, \)5, getting unlucky (losing \(100), or getting nothing. The probabilities are:- Probability of drawing a \)10 envelope = \( \frac{10}{100} = 0.1 \)- Probability of drawing a $5 envelope = \( \frac{10}{100} = 0.1 \)- Probability of drawing an unlucky envelope = \( \frac{2}{100} = 0.02 \)- Probability of drawing an empty envelope = \( \frac{78}{100} = 0.78 \).
02

Calculate Expected Value for Each Outcome

To find the expected value, multiply the value of each outcome by its probability:- Expected value for \(10 envelopes: \( 10 \times 0.1 = 1 \)- Expected value for \)5 envelopes: \( 5 \times 0.1 = 0.5 \)- Expected value for unlucky envelopes: \( -100 \times 0.02 = -2 \)- Expected value for empty envelopes: \( 0 \times 0.78 = 0 \).
03

Sum the Expected Values

Add the expected values from all possible outcomes to find the overall expected value of playing the game:\[1 + 0.5 + (-2) + 0 = -0.5 \].

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Key Concepts

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

Probability
Probability is a mathematical concept that measures the likelihood of an event occurring. It is represented as a number between 0 and 1. A probability of 0 means the event cannot happen, while a probability of 1 indicates certainty.
In the context of our game of chance exercise, probability is used to assess the chance of drawing different types of envelopes from the box. For example, the probability of drawing a $10 envelope is calculated by dividing the number of $10 envelopes (10) by the total number of envelopes (100). This gives us a probability of 0.1.
Similarly, other probabilities in the game are determined by dividing the count of each type of envelope by 100.
  • The $5 envelopes also have a probability of 0.1.
  • The chance of drawing an unlucky envelope is 0.02, as there are only 2 such envelopes.
  • With 78 empty envelopes, the probability of selecting one is 0.78.
Understanding how probability works allows players to make informed decisions when engaging in games of chance.
Game of Chance
A game of chance is one in which the outcome is highly influenced by random factors or luck. It contrasts with games of skill, where the player's abilities significantly affect the result.
In the provided game scenario, players draw envelopes, making it a classic example of a game of chance. The random selection means that players must depend on probability rather than skill to determine their winnings or losses.
Games of chance often offer a mix of potential rewards and risks:
  • There is a potential to gain monetary rewards, such as the $10 or $5 envelopes.
  • A significant downside exists, like the possibility of selecting an unlucky envelope and incurring a $100 loss.
  • Drawing an empty envelope results in neither gain nor loss.
While luck is the primary factor in determining success in this game, understanding the probabilities can help players better weigh their risks and rewards.
Mathematical Expectation
Mathematical expectation, also known as expected value, is a key concept in probability and statistics that quantifies the average outcome of a random event over many trials.
In our game, the expected value is computed by summing up the products of each possible outcome's value and its probability of occurring. This value gives players an idea of what they can expect to gain or lose on average per play.
For this game:
  • The expected value from drawing a \(10 envelope is \)1.
  • For a \(5 envelope, it's \)0.5.
  • The unlucky envelope contributes an expected value of -\(2, reflecting a potential loss.
  • Drawing an empty envelope has an expected value of \)0.
The overall expected value is then calculated as the sum of these individual expected values: \[ 1 + 0.5 - 2 + 0 = -0.5 \]This means, on average, a player would lose $0.50 per play. Understanding the expected value helps players make more informed decisions when deciding whether or not to engage in games of chance.

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Why Is \(\left(x^{n}\right)\) the Same as \(C(n, r) ?\) This exercise explains why the binomial coefficients \(\left(\begin{array}{c}{n} \\\ {r}\end{array}\right)\) that appear in the expansion of \((x+y)^{n}\) are the same as \(C(n, r),\) the number of ways of choosing \(r\) objects from \(n\) objects. First, note that expanding a binomial using only the Distributive Property gives $$ \begin{aligned}(x+y)^{2} &=(x+y)(x+y) \\ &=(x+y) x+(x+y) y \\ &=x x+x y+y x+y y \\\\(x+y)^{3} &=(x+y)(x x+x y+y x+y y) \\ &=x x x+x x y+x y x+x y y+y x x \\ &+y x y+y y x+y y y \end{aligned} $$ (a) Expand \((x+y)^{5}\) using only the Distributive Property. (b) Write all the terms that represent \(x^{2} y^{3}\) together. These are all the terms that contain two \(x^{\prime}\) s and three \(y^{\prime} s .\) (c) Note that the two \(x\) 's appear in all possible positions. Conclude that the number of terms that represent \(x^{2} y^{3}\) is \(C(5,2) .\) (d) In general, explain why \((r)\) in the Binomial Theorem is the same as \(C(n, r) .\)
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