Question 1

A gambler who keeps placing $1 bets on roulette will, after a very large number of bets, find that his average winnings per bet are close to $0.947. (The house keeps the other $0.053 per bet.) The statistical term for the number $0.947 is
A) the probability of winning a bet.
B) the bias of a bet.
C) a random number.
D) the expected value of a bet.

Accepted Answer

The number $0.947 is the expected value of a bet, which represents the average winnings per bet over a large number of bets. The expected value takes into account both the probability of winning and the amount that is won, so it is a more informative measure than simply the probability of winning (A), which only considers the likelihood of winning without considering the amount won or lost. The bias of a bet (B) refers to any systematic deviation from randomness in the game, and is not relevant in this context. A random number (C) is simply a number generated by a random process, and is not an appropriate description of $0.947, which has a specific meaning as the expected value of a bet.

Question 2

An employee's office window overlooks a busy four-way stop, and the employee decides to track, per minute, the number of drivers who fail to stop. From data taken over many randomly selected times, the employee gets this probability model:

In Question 14, the employee found the expected number of drivers failing to stop, per minute. The law of large numbers says:
A) observe whether each of many drivers fails to stop. The proportion who fail to stop will be close to the number calculated in the previous question.
B) if five observations of failure to stop are made in a row, the next several observations are likely to have four or five drivers failing to stop because the average must stay close to the expected value.
C) the expected value is correct only in a randomized comparative experiment.
D) probabilities can be found by simulation.
E) many observations should be made and the number of drivers failing to stop should be recorded in each case. The average will be close to the number calculated in the previous question.

Accepted Answer

A

Question 3

A friend rolls a cheap six-sided die many times. He reports that the probabilities of the possible outcomes are about as follows:

Using the probabilities above, what is the expected outcome when rolling this die?
A) 3
B) 3.3
C) 3.5
D) 4
E) 4.5

Accepted Answer

B

Question 4

A standard deck of cards contains 52 cards, of which 4 are aces. You are offered the following wager: draw one card at random from the deck. You win $10 if the card drawn is an ace. Otherwise, you lose $1. If you make this wager very many times, what will be the mean outcome?
A) about -$1, because you will lose most of the time
B) about $9, because you win $10 but lose only $1
C) about -$0.15, that is, on the average you lose about 15 cents per play
D) about $0.77, that is, on the average you win about 77 cents per play
E) about $0, because the random draw gives you a fair bet

Accepted Answer

The mean outcome is calculated by multiplying the probability of each outcome by its value and summing these products. With 4 aces in a 52-card deck, the probability of drawing an ace is 4/52 (or 1/13), and the probability of not drawing an ace is 48/52 (or 12/13). The expected value is (1/13)*$10 + (12/13)*(-$1) = $10/13 - $12/13 = -$2/13, which is approximately -$0.15.

Question 5

A poker player is dealt poor hands for several hours. He decides to bet heavily on the last hand of the evening on the grounds that after many bad hands he is due for a winner.
A) He's right, because the winnings have to average out.
B) He's wrong, because successive deals are independent of each other.
C) He's right, because successive deals are independent of each other.
D) He's wrong, because his expected winnings are $0 and he's below that now.

Accepted Answer

The outcome of each hand in poker is determined by chance and is independent of the previous hands. Therefore, the fact that the player has had poor hands for several hours does not increase the likelihood of a good hand in the next deal. Each hand is an independent event, and the odds of getting a good hand are the same regardless of what happened in the previous hands. The idea that one is "due" for a good hand is a fallacy known as the gambler's fallacy.

Question 6

A standard deck of 52 cards contains 13 hearts. Here is another wager: draw one card at random from the deck. If the card drawn is a heart, you win $2. Otherwise, you lose $1. Compare this wager (call it Wager 2) with that of the previous question (call it Wager 1). Which one should you prefer?
A) Wager 1, because it has a higher expected value
B) Wager 2, because it has a higher expected value
C) Wager 1, because it has a higher probability of winning
D) Wager 2, because it has a higher probability of winning
E) Both wagers are equally favorable.

Accepted Answer

The answer of A standard deck of 52 cards contains...

Question 7

In government data, a family consists of two or more persons who live together and are related by blood or marriage. Choose an American family at random and count the number of people it contains. Here is the assignment of probabilities for your outcome:

Using the probabilities above, what is the expected size of the family you draw?
A) 2 people
B) 3 people
C) 3.14 people
D) 3.5 people
E) 4.5 people

Accepted Answer

The answer of In government data, a family consists of...

Question 8

Kevin thinks he can use ESP to predict the outcome of rolling a fair die. You agree to pay him $3 if he can correctly predict the results of the next roll. Kevin has to pay you $1 if he is wrong. If Kevin doesn't have any psychic powers, which of the following is closest to the expected value of your net winnings on this bet?
A) -$0.67
B) -$0.33
C) $0
D) $0.33
E) $0.67

Accepted Answer

The expected value of your net winnings can be calculated by considering the probability of each outcome and the corresponding payoff. Since the die is fair, Kevin has a 1/6 chance of predicting the roll correctly and a 5/6 chance of being wrong. If he's right, you pay him $3, and if he's wrong, he pays you $1. The expected value (EV) is calculated as follows: EV = (1/6)*(-$3) + (5/6)*($1) = -$0.50 + $0.83 = $0.33.

Question 9

Suppose you know the percentage of foul shots a basketball player makes during the season. You want to estimate the expected number of shots made in 10 shots. You simulate 10 shots 25 times and get the following numbers of shots made:

Your estimate is:
A) 5.5
B) 6
C) 6.4
D) 6.7
E) 7

Accepted Answer

The answer of Suppose you know the percentage of foul...

Question 10

A home game for a certain National League Central Division team is chosen at random. Here is the probability distribution for the number of hits the leadoff batter has had:

What is the expected value of the number of hits the leadoff batter will have in a randomly chosen game?
A) 1.00
B) 1.02
C) 1.42
D) 1.97
E) 2.5

Accepted Answer

The answer of A home game for a certain National...

Question 11

You play a game with two possible outcomes. Outcome A has probability 0.4 and outcome B has probability 0.6. When B occurs you win $2.00; otherwise, you lose $1.00. What is your expected value for this game?
A) $2.00
B) -$0.10
C) $0.20
D) -$0.80
E) $0.80

Accepted Answer

The answer of You play a game with two possible...

Question 12

A multiple choice exam offers four choices for each question. Paul just guesses the answers, so he has probability 1/4 of getting any one answer right.
Paul's guess on any one question gives no information about his guess on any other question. The statistical term for this is
A) absence of bias.
B) absence of confounding.
C) law of large numbers.
D) independence.
E) expected value.

Accepted Answer

The scenario describes independence, which means that the outcome of one event has no impact on the outcome of another event.

Question 13

A multiple choice exam offers four choices for each question. Paul just guesses the answers, so he has probability 1/4 of getting any one answer right.
What is the expected number of right answers Paul will get if the test has 20 questions?
A) 5
B) 4
C) 2
D) 1/4
E) Can't tell from the information given.

Accepted Answer

The expected number of right answers is calculated by multiplying the probability of getting an answer right (1/4) by the total number of questions (20). This gives 20 * (1/4) = 5.

Question 14

Which of the following statements about a table of random digits is true?
A) Statement I: If each line contains 40 digits, there will be exactly 4 zeros in every line.
B) Statement II: The probability that there are exactly 4 zeros in a line of 40 digits is exactly 0.5.
C) Statement III: The expected number of zeros in a line of 40 digits is 4.
D) Statement IV: There can never be 4 zeros in a row because that pattern isn't random.
E) Both statements III and IV are true.

Accepted Answer

Statement I is false because a table of random digits is just that - random. There is no guarantee that there will be exactly four zeros in every line. Statement II is also false because the probability of getting exactly four zeros in a line of 40 digits is not 0.5, but can be calculated using binomial probability. Statement III is true, however, because the expected number of zeros in a line of 40 digits can be calculated as 40 * 0.1 = 4. Statement IV is false because there can be 4 zeros in a row in a table of random digits, just as there can be any other pattern that may seem non-random.

Question 15

A grocery chain runs a prize game by giving each customer a ticket that may win a prize when a box is scratched. Printed on the ticket are the following probabilities for a customer who shops once a week:

If several thousand customers play the grocery store game, you expect that the mean amount they win will be close to
A) the probability of winning something.
B) the median amount they win.
C) the expected value of a customer's winnings.
D) $100.

Accepted Answer

The expected value is calculated by multiplying each prize amount by its probability, and then adding up all the values. Since the mean amount is an average of the winnings, it is expected to be close to the expected value of the winnings. Therefore, C is the best choice. A is incorrect because the probability of winning something does not take into account the prize amounts. B is incorrect because the median amount does not consider the probability of winning each amount. D is incorrect because there is no information to support a mean amount of $100.

Question 16

A game involving a pair of dice pays you $4 with probability 16/36, costs you $2 with probability 14/36, and costs you $6 with probability 6/36.
What is your expected net result, in dollars, per play?
A) -$8
B) -$1.78
C) $0
D) $1.78
E) $8

Accepted Answer

The expected net result is calculated by multiplying each outcome by its probability and summing these products: (4 * 16/36) + (-2 * 14/36) + (-6 * 6/36) = 0.

Question 17

A grocery chain runs a prize game by giving each customer a ticket that may win a prize when a box is scratched. Printed on the ticket are the following probabilities for a customer who shops once a week:

What is the expected value of a customer's winnings in this game?
A) $22.00
B) $10.00
C) $370.00
D) $0.31
E) $100

Accepted Answer

The answer of A grocery chain runs a prize game...

Question 18

Choose an American household at random and ask how many computers that household owns. Here are the probabilities as of 2009:

What is the expected number of computers owned by a randomly chosen household?
A) 1.412
B) 1.307
C) 2.214
D) 2
E) 2.5

Accepted Answer

The answer of Choose an American household at random and...

Question 19

A game involving a pair of dice pays you $4 with probability 16/36, costs you $2 with probability 14/36, and costs you $6 with probability 6/36.
If you play this game many times, in the long run how will your actual average gain per play compare with your answer to the previous question?
A) smaller
B) bigger
C) about the same
D) The question can't be answered.

Accepted Answer

By calculating the expected value of this game, we get:

E(X) = (16/36)($4) + (14/36)($-2) + (6/36)($-6) = $0.44

This means that on average, you can expect to gain $0.44 per play in the long run. Since this value is higher than $0 (meaning that on average, you'll make some money per play), your actual average gain per play will likely be lower due to fluctuations in individual outcomes. However, the overall average should still be close to $0.44, so the answer is about the same.

Question 20

An employee's office window overlooks a busy four-way stop, and the employee decides to track, per minute, the number of drivers who fail to stop. From data taken over many randomly selected times, the employee gets this probability model:

The expected number of drivers failing to stop, per minute, is:
A) 0.5.
B) 2.2.
C)2)
D)3)
E) varies from minute to minute.

Accepted Answer

To find the expected number of drivers failing to stop per minute, we need to find the weighted average of the number of drivers failing to stop and their probabilities.

Expected number of drivers failing to stop = (0 x 0.6) + (1 x 0.3) + (2 x 0.1) = 0 + 0.3 + 0.2 = 0.5 + 0.2 = 2.2

Therefore, the expected number of drivers failing to stop per minute is 2.2, and the best choice is B.

Quiz 20: The House Edge: Expected Values

A gambler who keeps placing $1 bets on roulette will, after a very large number of bets, find that his average winnings per bet are close to $0.947. (The house keeps the other $0.053 per bet.) The statistical term for the number $0.947 is

A friend rolls a cheap six-sided die many times. He reports that the probabilities of the possible outcomes are about as follows: Using the probabilities above, what is the expected outcome when rolling this die?

A standard deck of cards contains 52 cards, of which 4 are aces. You are offered the following wager: draw one card at random from the deck. You win $10 if the card drawn is an ace. Otherwise, you lose $1. If you make this wager very many times, what will be the mean outcome?

A poker player is dealt poor hands for several hours. He decides to bet heavily on the last hand of the evening on the grounds that after many bad hands he is due for a winner.

A standard deck of 52 cards contains 13 hearts. Here is another wager: draw one card at random from the deck. If the card drawn is a heart, you win $2. Otherwise, you lose $1. Compare this wager (call it Wager 2) with that of the previous question (call it Wager 1) . Which one should you prefer?

Suppose you know the percentage of foul shots a basketball player makes during the season. You want to estimate the expected number of shots made in 10 shots. You simulate 10 shots 25 times and get the following numbers of shots made: Your estimate is:

A home game for a certain National League Central Division team is chosen at random. Here is the probability distribution for the number of hits the leadoff batter has had: What is the expected value of the number of hits the leadoff batter will have in a randomly chosen game?

You play a game with two possible outcomes. Outcome A has probability 0.4 and outcome B has probability 0.6. When B occurs you win $2.00; otherwise, you lose $1.00. What is your expected value for this game?

A multiple choice exam offers four choices for each question. Paul just guesses the answers, so he has probability 1/4 of getting any one answer right. Paul's guess on any one question gives no information about his guess on any other question. The statistical term for this is

A multiple choice exam offers four choices for each question. Paul just guesses the answers, so he has probability 1/4 of getting any one answer right. What is the expected number of right answers Paul will get if the test has 20 questions?

Which of the following statements about a table of random digits is true?

A game involving a pair of dice pays you $4 with probability 16/36, costs you $2 with probability 14/36, and costs you $6 with probability 6/36. What is your expected net result, in dollars, per play?

A grocery chain runs a prize game by giving each customer a ticket that may win a prize when a box is scratched. Printed on the ticket are the following probabilities for a customer who shops once a week: What is the expected value of a customer's winnings in this game?

Choose an American household at random and ask how many computers that household owns. Here are the probabilities as of 2009: What is the expected number of computers owned by a randomly chosen household?

A game involving a pair of dice pays you $4 with probability 16/36, costs you $2 with probability 14/36, and costs you $6 with probability 6/36. If you play this game many times, in the long run how will your actual average gain per play compare with your answer to the previous question?