Statistics Quiz: Find Expected Value Of A Game

What this quiz covers

This quiz focuses on Find Expected Value Of A Game, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. \2.00$
2. \0.50$ (correct answer)
3. \1.50$
4. -\0.50$

Explanation: Expected value (EV) for this coin flip wager is the average net payoff over many flips, guiding long-run expectations. Net payoffs are $2 for heads (probability 1/2) and -$1 for tails (1/2), giving EV = (2)(1/2) + (-1)(1/2) = 0.50. Over repeated plays, you'd expect to gain $0.50 per flip on average, tying EV to prolonged results. This positive EV makes the game favorable for the player. A common misconception is equating EV to the median outcome, but here EV is positive while half the time you lose$1. This distinction clarifies EV's role in probability.

Question 2

1. \1$ (correct answer)
2. -\4$
3. -\1$
4. \4$

Explanation: The expected value (EV) is the long-term average net gain or loss from playing the game repeatedly. The expected payout is (5/10)×$8 + (5/10)×$0 = $4. Subtract the$3 cost: $4 -$3 = $1. Over many plays, the player averages$1 gain per game. In long-run terms, 100 plays might net about $100 profit. A common misconception is not accounting for the cost, leading to an EV of$4 instead of $1. Additionally, people often mistake EV for the most likely single outcome, but here both$8 and $0 are equally likely, not$1.

Question 3

1. \3.50$
2. -\3.50$
3. -\0.50$ (correct answer)
4. \0.50$

Explanation: Expected value (EV) measures the long-run average net gain or loss per ticket over many purchases. Expected payout: (1/6)×$9 + (1/3)×$3 + (1/2)×$0 =$2.50. Net EV: $2.50 -$3 = -$0.50. This implies an average loss of$0.50 per ticket in the long term. Over 200 tickets, you'd expect to lose about $100. A misconception is omitting the cost, leading to$2.50 as EV. Another is confusing EV with the most probable outcome, which is $0, not -$0.50.

Question 4

1. -\\tfrac{1}{3}$ (correct answer)
2. -\\tfrac{5}{3}$
3. \\tfrac{5}{3}$
4. \\tfrac{1}{3}$

Explanation: Expected value (EV) indicates the average net gain or loss in the long run over many plays. Expected payout: (1/6)×$10 + (5/6)×$0 = $10/6 ≈$1.67. Net EV: $1.67 -$2 = -$1/3 ≈ -$0.33. This means a long-term average loss of about $0.33 per roll. For example, over 300 rolls, you'd expect to lose around$100. One misconception is ignoring the cost, yielding an EV of $1.67 instead. Another is believing EV means you'll lose$0.33 every time, but actual outcomes are $10 or$0, with EV as the average.

Question 5

1. \4.00$
2. -\1.60$
3. -\0.20$ (correct answer)
4. \1.60$

Explanation: The expected value (EV) for this marble draw measures the average net payoff over many repeated plays. Net payoffs are $4 for red (probability 4/10) and -$3 for blue (probability 6/10), resulting in EV = (4)(4/10) + (-3)(6/10) = -0.20. In the long run, you'd anticipate losing $0.20 per draw on average, linking EV to overall trends rather than single results. This negative EV indicates the game benefits the house over time. A common misconception is averaging payoffs without probabilities, such as simply averaging$7 and $0, ignoring costs and chances. Grasping EV correctly reveals the game's inherent bias.

Question 6

1. $1.40
2. $-0.60
3. $-1.40
4. $0.60 (correct answer)

Explanation: Expected value in this wager is the average net gain or loss per play over the long run. Net gains are $7 (probability 1/5) and -$1 (probability 4/5) after subtracting $2 cost, giving (1/5)×7 + (4/5)×(-1) = 0.60. This positive value means an expected gain of 60 cents per play in the long term, beneficial for the player. It connects to repeated plays by averaging outcomes to show overall trend. A common misconception is not subtracting the play cost, resulting in expected receipt of$2.60, which exaggerates the positivity. This concept is key for assessing game fairness.

Question 7

1. $0.00
2. $-0.40 (correct answer)
3. $-1.60
4. $1.60

Explanation: The expected value in a game like this lottery represents the long-run average net gain or loss per ticket if you were to buy many tickets over time. To compute it, calculate the net gain for each outcome by subtracting the $2 cost from the prize, then multiply by the respective probabilities and sum them up. Here, the net gains are$8 (probability 0.10), $1 (probability 0.20), and -$2 (probability 0.70), yielding an expected value of 0.10×8 + 0.20×1 + 0.70×(-2) = -0.40. This negative value indicates that, on average, you lose 40 cents per ticket in the long run, making the game unfavorable for the player. A common misconception is forgetting to subtract the ticket cost from each prize, which would incorrectly give an expected prize of $1.60 instead of the net expected value. Recognizing this distinction helps players understand the true financial implications of participating in such games.

Question 8

1. $0.60 (correct answer)
2. $-0.60
3. $-1.60
4. $1.60

Explanation: The expected value for this wager reflects the average net outcome per play in the long run. Find net gains by subtracting the $1 cost from receipts:$3 (probability 0.40) and -$1 (probability 0.60), then compute 0.40×3 + 0.60×(-1) = 0.60. This positive EV suggests a long-term gain of 60 cents per play, favoring the player. It connects to long-run outcomes by averaging results over many wagers, showing overall profitability. A common misconception is overlooking the cost deduction, giving an expected receipt of$1.60, which overstates the benefit. Appreciating this helps evaluate if a game offers a genuine advantage.

Question 9

1. $-0.40
2. $-1.00
3. $0.00 (correct answer)
4. $0.40

Explanation: For this prize wheel, expected value is the long-run average net gain or loss per spin. Calculate net gains by subtracting $1 cost:$6 (1/10), $0 (3/10), -$1 (6/10), resulting in (1/10)×6 + (3/10)×0 + (6/10)×(-1) = 0.00. Zero EV indicates a fair game with no long-term advantage or disadvantage. It relates to long-run outcomes as the average settles near zero over many spins. A common misconception is ignoring cost in nets, leading to expected prize of $1.00, mistakenly implying profit. Understanding this prevents overoptimism in game evaluations.

Question 10

1. $0.75
2. $1.50
3. $-1.50
4. $-0.75 (correct answer)

Explanation: The expected value represents the average net gain or loss per spin if you played many times. Since the player pays $3 to spin, we must subtract this cost from any winnings to find net gains. With 8 equal sections, the probability of landing on the$9 section is $1/8$ (net gain of $6), the probability of landing on a$5 section is $2/8$ (net gain of $2), and the probability of landing on a$0 section is $5/8$ (net loss of $3). The expected value is$(\frac{1}{8} \times 6) + (\frac{2}{8} \times 2) + (\frac{5}{8} \times (-3)) = 0.75 + 0.50 - 1.875 = -0.75$. This negative expected value indicates that players will lose an average of 75 cents per spin in the long run. Students often mistakenly calculate the expected winnings without accounting for the cost to play.

Question 11

1. $0.00
2. $-0.90 (correct answer)
3. $-1.10
4. $1.10

Explanation: To find the expected value from the player's perspective, we calculate the average net gain/loss by considering all possible outcomes and their probabilities. The player pays $2 to play and has a$1/20$chance of winning$10 (net gain of $8), a$3/20$chance of winning$4 (net gain of $2), and a$16/20$chance of winning$0 (net loss of $2). The expected value is $\frac{1}{20} \times 8 + \frac{3}{20} \times 2 + \frac{16}{20} \times (-2) = 0.40 + 0.30 - 1.60 = -0.90$. This negative expected value means that over many plays, a player would lose an average of 90 cents per ticket. A common misconception is forgetting to subtract the cost of playing when calculating net gains, which would incorrectly give a positive expected value.

Question 12

1. $-3.00 (correct answer)
2. $-5.00
3. $3.00
4. $15.00

Explanation: The expected value tells us the average net gain or loss per play over many repetitions. Since the player pays $5 to play, we must subtract this from any winnings to find the net result. With probability 0.10, the player wins$20 (net gain of $15), and with probability 0.90, the player wins$0 (net loss of $5). The expected value is: 0.10×($20-$5) + 0.90×($0-$5) = 0.10×$15 + 0.90×(-$5) =$1.50 - $4.50 = -$3.00. This negative expected value means that on average, players lose $3 per play in the long run. A common misconception is calculating only the expected winnings ($2) without accounting for the cost to play, which would miss the fact that this is a losing game.

Question 13

1. $-0.50
2. $-2.50
3. $2.50
4. $0.50 (correct answer)

Explanation: The expected value tells us the average net gain or loss per play if this game is repeated many times. The player pays $2 to play and has a 1/4 probability of winning$10 (net gain of $8) and a 3/4 probability of winning$0 (net loss of $2). The expected value is: (1/4)×($10-$2) + (3/4)×($0-$2) = (1/4)×$8 + (3/4)×(-$2) =$2.00 - $1.50 =$0.50. This positive expected value means that on average, a player gains 50 cents per play in the long run. Students often mistakenly calculate the expected winnings ($2.50) instead of the expected net gain, forgetting to subtract the cost of playing from all outcomes.

Question 14

1. \4.00$
2. -\0.20$ (correct answer)
3. \1.60$
4. -\1.60$

Explanation: The expected value (EV) for this marble draw measures the average net payoff over many repeated plays. Net payoffs are $4 for red (probability 4/10) and -$3 for blue (probability 6/10), resulting in EV = (4)(4/10) + (-3)(6/10) = -0.20. In the long run, you'd anticipate losing $0.20 per draw on average, linking EV to overall trends rather than single results. This negative EV indicates the game benefits the house over time. A common misconception is averaging payoffs without probabilities, such as simply averaging$7 and $0, ignoring costs and chances. Grasping EV correctly reveals the game's inherent bias.

Question 15

1. $1.83
2. $0.83 (correct answer)
3. $-1.83
4. $-0.83

Explanation: Expected value represents the average net gain or loss per spin if you played this wheel many times. The player pays $1 to spin and has a 1/6 chance of winning$7 (net gain of $6), a 2/6 chance of winning$2 (net gain of $1), and a 3/6 chance of winning$0 (net loss of $1). The expected value calculation is: $(1/6) \times 6 + (2/6) \times 1 + (3/6) \times (-1) = 1.00 + 0.33 - 0.50 = 0.83$. This positive expected value of 83 cents means that over many spins, a player would gain an average of 83 cents per spin. A common error students make is forgetting to account for the cost of playing when determining net gains and losses from each outcome.

Question 16

1. -\1.40$
2. \1.40$
3. \0.60$
4. -\0.60$ (correct answer)

Explanation: Expected value (EV) is the long-term average net gain or loss per draw over repeated plays. Expected payout: (3/10) \times \4 + (2/10) \times $1 + (5/10) \times $0 = $1.40$. Net EV:$\1.40 - $2 = -$0.60$. This indicates an average loss of$0.60 per draw in the long run. Over 100 draws, you'd expect to lose about $60. One misconception is not subtracting the cost, resulting in$1.40 as EV. Another is mistaking EV for the median outcome, which here is $0, not -$0.60.

Question 17

1. $0.40
2. $8.00
3. $-4.00
4. $-0.40 (correct answer)

Explanation: Expected value tells us the average net gain or loss per play over many repetitions of the game. The player pays $4 to play and has a 0.30 probability of winning$12 (net gain of $8) and a 0.70 probability of winning$0 (net loss of $4). The expected value calculation is: EV = 0.30×($12-$4) + 0.70×($0-$4) = 0.30×$8 + 0.70×(-$4) =$2.40 - $2.80 = -$0.40. This negative expected value means that on average, a player loses 40 cents per play in the long run. A common error is calculating expected winnings ($3.60) instead of expected net gain/loss, which requires subtracting the cost to play from all outcomes.

Question 18

1. $-0.75
2. $-3.00
3. $0.00 (correct answer)
4. $3.00

Explanation: The expected value of this wager is the long-run average net gain or loss per play over numerous trials. Compute it by subtracting the $3 cost from each possible receipt to get net outcomes, then multiplying by probabilities and adding. Net gains are$9 (probability 0.25) and -$3 (probability 0.75), giving 0.25×9 + 0.75×(-3) = 0. This zero EV means, on average, you neither gain nor lose money in the long run, making it a fair game. A common misconception is calculating only the expected receipt without subtracting the cost, yielding$3.00, which doesn't represent the true net perspective. This understanding of EV helps distinguish fair games from those tilted against the player.

Question 19

1. -\1.50$
2. -\0.50$
3. \1.50$
4. \0.50$ (correct answer)

Explanation: The expected value (EV) represents the average net gain or loss if the coin flip is repeated many times. Expected payout: (1/2) \times \3 + (1/2) \times $0 = $1.50 $. Net EV:$ $1.50 - $1 = $0.50 $. In the long run, this means gaining$0.50 per flip on average. For 1,000 flips, expect about $500 profit. A common misconception is thinking EV is zero since outcomes are symmetric, ignoring the unequal payouts. Another is believing EV predicts a single flip's result, but each flip yields either$3 or $0, with$0.50 as the average.