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Chapter 6: Problem 30

Connecticut Lottery In the Cash Five Lottery in Connecticut, a player pays \(\$ 1\) for a single ticket with five numbers. Five balls numbered 1 through 35 are randomly chosen from a bin without replacement. If all five numbers on a player's ticket match the five chosen, the player wins \(\$ 100,000 .\) The probability of this occurring is \(\frac{1}{324,632} .\) If four numbers match, the player wins \(\$ 300 .\) This occurs with probability \(\frac{1}{2164} .\) If three numbers match, the player wins \(\$ 10 .\) This occurs with probability \(\frac{1}{75}\). Compute and interpret the expected value of the game from the player's point of view.

Short Answer

Expert verified
The expected value of the game is \(0.579, meaning on average a player will lose about \)0.421 per ticket.

Step by step solution

01

Understand the Scenario

In the Cash Five Lottery, a player pays \(1 for a ticket with five numbers. Various winnings are associated with different number matches: \)100,000 for all five numbers, \(300 for four numbers, and \)10 for three numbers. Each prize has a given probability of being won.
02

Define the Probabilities and Outcomes

List the winning amounts and their associated probabilities:- Matching all 5 numbers: wins \(100,000 with a probability of \(\frac{1}{324,632}\)- Matching 4 numbers: wins \)300 with a probability of \(\frac{1}{2164}\)- Matching 3 numbers: wins \(10 with a probability of \(\frac{1}{75}\)- No win: wins \)0.
03

Calculate the Expected Value

The expected value (EV) is calculated by multiplying each outcome by its probability and summing these values.\[ \text{EV} = \text{P(win 5)} \times \text{prize 5} + \text{P(win 4)} \times \text{prize 4} + \text{P(win 3)} \times \text{prize 3} + \text{P(no win)} \times \text{prize no win} \]Calculate each term:\( \text{P(win 5)} \times \text{prize 5} = \frac{1}{324,632} \times 100,000 = 0.308 \)\( \text{P(win 4)} \times \text{prize 4} = \frac{1}{2164} \times 300 = 0.138 \)\( \text{P(win 3)} \times \text{prize 3} = \frac{1}{75} \times 10 = 0.133 \)\( \text{P(no win)} \times \text{prize no win} = (1 - \frac{1}{324,632} - \frac{1}{2164} - \frac{1}{75}) \times 0 = 0 \)
04

Sum the Calculated Values

Sum the results from the calculations to find the total expected value:\[ \text{EV} = 0.308 + 0.138 + 0.133 = 0.579 \]
05

Interpret the Expected Value

The expected value of the game from the player's point of view is \(0.579. This means that on average, a player is expected to lose approximately \)0.421 per ticket purchased (\(1 purchase cost - \)0.579 expected return).

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Key Concepts

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

Probability
Probability is a measure of how likely an event is to occur. It ranges from 0 to 1, where 0 means the event will not happen, and 1 means it definitely will.
To calculate probability, you divide the number of desired outcomes by the number of possible outcomes. For example, in the Cash Five Lottery, the probability of matching all five numbers is \[ \frac{1}{324,632} \]. This tiny number shows that winning the jackpot is extremely unlikely.

Whenever you're dealing with probabilities, it's important to understand that each event is independent. This means that previous outcomes don鈥檛 affect future ones. For example, just because you didn't win last time doesn't mean your chances improve the next time.
Practicing with different probability scenarios can help you become comfortable with these calculations.
Lottery Probabilities
Lottery probabilities involve understanding the odds of winning based on different possible outcomes. In games like the Cash Five Lottery, you have different prize categories, each with its own probability of success.

For our specific lottery example, here are the key probabilities:
  • Matching all 5 numbers: \[ \frac{1}{324,632} \]
  • Matching 4 numbers: \[ \frac{1}{2164} \]
  • Matching 3 numbers: \[ \frac{1}{75} \]
Understanding these probabilities helps set realistic expectations about your chances of winning and how often you might win smaller prizes versus the jackpot.

Keep in mind that calculating these odds involves combinatorics, which helps determine how likely certain combinations of numbers are.
Expected Value
Expected value (EV) is a way to determine the average outcome of a game of chance over the long run. You multiply each possible outcome by its probability and then sum these values. This gives you a single number that represents the average amount you can expect to win or lose per game.

In the Cash Five Lottery example, calculating the expected value involves: \[ \text{EV} = \frac{1}{324,632} \times 100,000 + \frac{1}{2164} \times 300 + \frac{1}{75} \times 10 \] After performing these calculations, we get:
  • \[ \frac{1}{324,632} \times 100,000 = 0.308 \]
  • \[ \frac{1}{2164} \times 300 = 0.138 \]
  • \[ \frac{1}{75} \times 10 = 0.133 \]
  • \[ \text{P(no win)} \times \text{prize no win} = (1- \frac{1}{324,632} - \frac{1}{2164} - \frac{1}{75}) \times 0 = 0 \]
Summing these values, the EV is approximately \[ 0.579 \]. This means on average, you would win \(0.579 for every \)1 ticket purchased. Since the tickets cost \(1, you would lose about \)0.421 each time.

This way, expected value helps you understand the fairness of the game and whether it's worth playing.

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Most popular questions from this chapter

A binomial probability experiment is conducted with the given parameters. Compute the probability of \(x\) successes in the \(n\) independent trials of the experiment. $$ n=40, p=0.99, x=38 $$

Historically, the probability that a passenger will miss a flight is 0.0995. Source: Passenger-Based Predictive Modeling of Airline No-show Rates by Richard D. Lawrence, Se June Hong, and Jacques Cherrier. Airlines do not like flights with empty seats, but it is also not desirable to have overbooked flights because passengers must be "bumped" from the flight. The Lockheed L49 Constellation has a seating capacity of 54 passengers. (a) If 56 tickets are sold, what is the probability 55 or 56 passengers show up for the flight resulting in an overbooked flight? (b) Suppose 60 tickets are sold, what is the probability a passenger will have to be "bumped"? (c) For a plane with seating capacity of 250 passengers, how many tickets may be sold to keep the probability of a passenger being "bumped" below \(1 \% ?\)

In a recent poll, the Gallup Organization found that \(45 \%\) of adult Americans believe that the overall state of moral values in the United States is poor. Suppose a survey of a random sample of 25 adult Americans is conducted in which they are asked to disclose their feelings on the overall state of moral values in the United States. (a) Find and interpret the probability that exactly 15 of those surveyed feel the state of morals is poor. (b) Find and interpret the probability that no more than 10 of those surveyed feel the state of morals is poor. (c) Find and interpret the probability that more than 16 of those surveyed feel the state of morals is poor. (d) Find and interpret the probability that 13 or 14 believe the state of morals is poor. (e) Would it be unusual to find 20 or more adult Americans who believe the overall state of moral values is poor in the United States? Why?

A probability distribution for the random variable \(X,\) the number of trials until a success is observed, is called the geometric probability distribution. It has the same criteria as the binomial distribution (see page 328 ), except that the number of trials is not fixed. Its probability distribution function (pdf) is $$ P(x)=p(1-p)^{x-1}, \quad x=1,2,3, \ldots $$ where \(p\) is the probability of success. (a) What is the probability that Shaquille O'Neal misses his first two free throws and makes the third? Over his career, he made \(52.4 \%\) of his free throws. That is, find \(P(3)\) (b) Construct a probability distribution for the random variable \(X,\) the number of free-throw attempts of Shaquille O'Neal until he makes a free throw. Construct the distribution for \(x=1,2,3, \ldots, 10 .\) The probabilities are small for \(x>10\) (c) Compute the mean of the distribution, using the formula presented in Section 6.1 (d) Compare the mean obtained in part (c) with the value \(\frac{1}{p}\). Conclude that the mean of a geometric probability distribution is \(\mu_{X}=\frac{1}{p} .\) How many free throws do we expect Shaq to take before we observe a made free throw?

In Problems \(17-28,\) a binomial probability experiment is conducted with the given parameters. Compute the probability of \(x\) successes in the \(n\) independent trials of the experiment. $$ n=10, p=0.4, x=3 $$
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