Compare Strategies Using Expected Value
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A gamer chooses between two loot boxes to open repeatedly. Net payoff is the value of the item received minus the purchase price.
Strategy A: Costs $4. Outcomes: with probability $0.70$ you get an item worth $5 (net $1); with probability $0.30$ you get an item worth $0 (net $-4).
Strategy B: Costs $4. Outcomes: with probability $0.10$ you get an item worth $30 (net $26); with probability $0.90$ you get an item worth $0 (net $-4).
Which strategy would result in a higher average net payoff over many repetitions?
Strategy A, because it is more likely to give a positive net payoff (0.70).
Strategy B, because it has the higher maximum net payoff ($26).
Strategy B, because it treats the two outcomes as equally likely and averages $26 and $-4.
Strategy A, because its expected net payoff is higher over many repetitions.
Explanation
This question involves comparing strategies using expected value in statistics and probability. Expected value represents the long-run average outcome if the strategy is repeated many times. For each strategy, compute the expected value by multiplying each possible net payoff—item value minus cost—by its probability and summing them up. In this case, Strategy A has a higher expected net payoff of $-0.50$ compared to Strategy B's $-1$. Over many trials, Strategy A would result in a higher average net payoff, favoring it for long-term gains. A tempting distractor is choosing Strategy B for its higher maximum net payoff of $26, but this ignores the low probability. To compare strategies, always multiply outcomes by their probabilities and sum to find and compare the totals, rather than looking at individual outcomes.
A company is choosing between two quality-control tests to run on many batches. Payoffs are net dollars to the company per batch.
Strategy A: Test costs $3,000. Outcomes: with probability 0.95 the batch passes and there is no further cost (net payoff $-3,000); with probability 0.05 the test catches a defect and prevents a $100,000 recall (net payoff $100,000-3,000=\$97,000).
Strategy B: Test costs $6,000. Outcomes: with probability 0.98 the batch passes (net payoff $-6,000); with probability 0.02 the test catches a defect and prevents the $100,000 recall (net payoff $94,000).
Which strategy would result in a higher average net payoff over many repetitions?
Strategy A, because it subtracts the test cost twice when computing expected value.
Strategy B, because it is more likely to avoid a recall (0.98 pass rate).
Strategy A, because it has the higher expected net payoff over many repetitions.
Strategy B, because it has the larger payoff in the defect-caught case ($94,000).
Explanation
This question involves comparing strategies using expected value in statistics and probability. Expected value represents the long-run average outcome if the strategy is repeated many times. For each strategy, compute the expected value by multiplying each possible net payoff—savings minus test cost—by its probability and summing them up. In this case, Strategy A has a higher expected net payoff of $2,000 compared to Strategy B's -$4,000. Over many trials, Strategy A would yield a higher average net payoff, making it preferable in the long run. A tempting distractor is choosing Strategy B for its higher pass rate of 0.98, but this overlooks the costs and payoffs. To compare strategies, always multiply outcomes by their probabilities and sum to find and compare the totals, rather than looking at individual outcomes.
A customer is considering two extended warranty plans for a laptop over many similar purchases. Payoffs are measured as net dollars to the customer (positive means money saved compared to paying repairs out-of-pocket; negative means money spent).
Strategy A (Cheaper plan): Pay $60 now. Outcomes: with probability 0.85 no repair is needed (net payoff $-60$); with probability 0.15 a repair is needed and the plan covers a $400 repair (net payoff $400 - 60 = 340$).
Strategy B (Pricier plan): Pay $90 now. Outcomes: with probability 0.85 no repair is needed (net payoff $-90$); with probability 0.15 a repair is needed and the plan covers the $400 repair (net payoff $400 - 90 = 310$).
Which strategy has the greater expected value (average net payoff) over many repetitions?
Strategy B, because the repair outcome still gives a positive payoff ($310$).
Strategy A, because it ignores the no-repair outcome and compares only $340$ vs $310$.
Strategy B, because paying more upfront must mean better coverage.
Strategy A, because it has the greater expected value over many repetitions.
Explanation
This question involves comparing strategies using expected value in statistics and probability. Expected value represents the long-run average outcome if the strategy is repeated many times. For each strategy, compute the expected value by multiplying each possible net payoff—savings minus warranty cost—by its probability and summing them up. In this case, Strategy A has a higher expected net payoff of $0$ compared to Strategy B's $-30$. Over many trials, Strategy A would yield a higher average net payoff, making it preferable in the long run. A tempting distractor is assuming the pricier plan is better due to higher cost, but this overlooks the actual expected values. To compare strategies, always multiply outcomes by their probabilities and sum to find and compare the totals, rather than looking at individual outcomes.
A player repeatedly chooses between two strategies in a video game to earn points. The player wants the higher average points per attempt (expected value).
Strategy A: With probability 0.30 you earn 40 points; with probability 0.70 you lose 10 points.
Strategy B: With probability 0.80 you earn 8 points; with probability 0.20 you lose 5 points.
Which strategy has the greater expected value over many repetitions?
Strategy B, because it has the greater expected value over many attempts.
Strategy A, because its expected value is $0.30(40) + 0.70(10) = 19$ points.
Strategy B, because earning points is more likely (0.80).
Strategy A, because it can earn as many as 40 points.
Explanation
The skill here is comparing strategies using expected value, which helps determine the better choice for long-term average outcomes. Expected value is calculated as the long-run average points per attempt, found by multiplying each possible points outcome by its probability and summing them up. For Strategy A, the expected value is computed by weighting the outcomes of 40 and -10 by their probabilities of 0.30 and 0.70, resulting in 5. For Strategy B, weighting 8 and -5 by 0.80 and 0.20 gives 5.4. Since Strategy B has the higher expected value, it yields better average points over many attempts, though individual results may vary. A tempting distractor is favoring Strategy A for its maximum of 40 points, but expected value accounts for losses too. To apply this, always multiply outcomes by their probabilities and sum for each strategy, then compare the totals rather than isolated high or low values.
A charity is choosing between two fundraising phone scripts. The charity wants the higher expected net donation per call over many calls. (Net donation = donation received minus the cost of the call.)
Strategy A: Each call costs $1. With probability 0.15, the person donates $20; with probability 0.85, the person donates $0.
Strategy B: Each call costs $2. With probability 0.10, the person donates $35; with probability 0.90, the person donates $0.
Which strategy has the greater expected value of net donation over many repetitions?
Strategy B, because it has the larger possible donation ($35).
Strategy B, because its expected value is $0.10(35) - 2 = $5.5$.
Strategy A, because a donation is more likely (0.15 vs 0.10).
Strategy A, because it has the greater expected net donation over many calls.
Explanation
The skill here is comparing strategies using expected value, which helps determine the better choice for long-term average outcomes. Expected value is calculated as the long-run average net donation per call, found by multiplying each possible net donation by its probability and summing them up. For Strategy A, the expected value is computed by weighting the net donations of $20 - $1 and $0 - $1 by their probabilities of 0.15 and 0.85, resulting in $2. For Strategy B, weighting $35 - $2 and $0 - $2 by 0.10 and 0.90 gives $1.50. Since Strategy A has the higher expected value, it yields better average net donation over many calls, though individual results may vary. A tempting distractor is selecting Strategy B for its larger possible donation of $35, disregarding the costs and probabilities. To apply this, always multiply outcomes by their probabilities and sum for each strategy, then compare the totals rather than isolated high or low values.
A student repeatedly chooses between two quiz strategies. The student wants the higher expected score change per quiz (points).
Strategy A: With probability 0.50 you gain 12 points; with probability 0.50 you lose 6 points.
Strategy B: With probability 0.90 you gain 4 points; with probability 0.10 you lose 20 points.
Which strategy has the greater expected value over many repetitions?
Strategy B, because gaining points is more likely (0.90).
Strategy B, because it has the larger maximum gain (4 points) compared to losing 6.
Strategy A, because its expected value is $0.50(12) + 0.50(6) = 9$ points.
Strategy A, because it has the greater expected value over many quizzes.
Explanation
The skill here is comparing strategies using expected value, which helps determine the better choice for long-term average outcomes. Expected value is calculated as the long-run average score change per quiz, found by multiplying each possible change by its probability and summing them up. For Strategy A, the expected value is computed by weighting the changes of 12 and -6 by their probabilities of 0.50 and 0.50, resulting in 3. For Strategy B, weighting 4 and -20 by 0.90 and 0.10 gives 1.6. Since Strategy A has the higher expected value, it yields better average score change over many quizzes, though individual results may vary. A tempting distractor is choosing Strategy B for its high probability of gaining points, ignoring the severe loss impact. To apply this, always multiply outcomes by their probabilities and sum for each strategy, then compare the totals rather than isolated high or low values.
A company is deciding between two marketing emails. The company earns profit in dollars depending on the response. The company wants the higher expected profit per email sent over many repetitions.
Strategy A: With probability 0.05, profit is $200; with probability 0.95, profit is $0.
Strategy B: With probability 0.20, profit is $60; with probability 0.80, profit is $5.
Which strategy has the greater expected value (average profit) over many trials?
Strategy B, because it has the greater expected profit over many emails.
Strategy A, because its expected value is $0.05(200) = $10$, which is greater than Strategy B.
Strategy A, because the $0 outcome is more likely in Strategy A.
Strategy A, because $200 is the largest possible profit.
Explanation
The skill here is comparing strategies using expected value, which helps determine the better choice for long-term average outcomes. Expected value is calculated as the long-run average profit per email, found by multiplying each possible profit by its probability and summing them up. For Strategy A, the expected value is computed by weighting the profits of $200 and $0 by their probabilities of 0.05 and 0.95, resulting in $10. For Strategy B, weighting $60 and $5 by 0.20 and 0.80 gives $16. Since Strategy B has the higher expected value, it yields better average profit over many emails, though individual results may vary. A tempting distractor is choosing Strategy A for its maximum profit of $200, overlooking the low probability. To apply this, always multiply outcomes by their probabilities and sum for each strategy, then compare the totals rather than isolated high or low values.
A player repeatedly chooses between two prize wheels. The player wants the higher expected winnings per spin (in dollars).
Strategy A: With probability 0.70, win $3; with probability 0.20, win $8; with probability 0.10, win $0.
Strategy B: With probability 0.50, win $4; with probability 0.25, win $10; with probability 0.25, win $1.
Which strategy has the greater expected value over many repetitions?
Strategy B, because it has the greater expected value over many spins.
Strategy A, because winning $3 is the most likely outcome (0.70).
Strategy A, because its expected value is $0.70(3)+0.20(8)= $3.7$, which is greater than Strategy B.
Strategy B, because $10 is the largest single payoff.
Explanation
The skill here is comparing strategies using expected value, which helps determine the better choice for long-term average outcomes. Expected value is calculated as the long-run average winnings per spin, found by multiplying each possible winnings by its probability and summing them up. For Strategy A, the expected value is computed by weighting the winnings of $3, $8, and $0 by their probabilities of 0.70, 0.20, and 0.10, resulting in $3.70. For Strategy B, weighting $4, $10, and $1 by 0.50, 0.25, and 0.25 gives $4.75. Since Strategy B has the higher expected value, it yields higher average winnings over many spins, though individual results may vary. A tempting distractor is favoring Strategy A for its most likely outcome of $3, without summing all weighted values. To apply this, always multiply outcomes by their probabilities and sum for each strategy, then compare the totals rather than isolated high or low values.
A student plays a carnival game many times and wants the higher average profit (expected value) per play.
Strategy A: Pay $4 to play. Outcomes: with probability 0.20 you win $20; with probability 0.80 you win $0.
Strategy B: Pay $4 to play. Outcomes: with probability 0.60 you win $8; with probability 0.40 you win $0.
Which strategy has the greater expected value (average profit) over many repetitions?
Strategy A, because it has the larger maximum payout ($20).
Strategy A, because its expected value is $0.20(20) - 4 = $1.
Strategy A, because $0 is more likely in Strategy B than in Strategy A.
Strategy B, because it has the higher expected value over many plays.
Explanation
The skill here is comparing strategies using expected value, which helps determine the better choice for long-term average outcomes. Expected value is calculated as the long-run average profit per play, found by multiplying each possible profit by its probability and summing them up. For Strategy A, the expected value is computed by weighting the profits of $16 and -$4 by their probabilities of 0.20 and 0.80, resulting in $0. For Strategy B, weighting the profits of $4 and -$4 by 0.60 and 0.40 gives $0.80. Since Strategy B has the higher expected value, it yields a better average profit over many plays, though individual results may vary. A tempting distractor is focusing on the maximum payout of $20 in Strategy A, but expected value considers all outcomes weighted by probability. To apply this, always multiply outcomes by their probabilities and sum for each strategy, then compare the totals rather than isolated high or low values.
A student is choosing between two raffle ticket bundles to buy repeatedly over many school fundraisers. Each bundle costs money up front and then has one of the listed outcomes.
Strategy A (cost $4):
- Win $0 with probability 0.70
- Win $10 with probability 0.25
- Win $50 with probability 0.05
Strategy B (cost $6):
- Win $0 with probability 0.55
- Win $15 with probability 0.40
- Win $30 with probability 0.05
Considering net winnings (prize minus cost), which strategy has the greater expected value?
Strategy B, because it has a higher probability of winning a positive prize (0.45 vs 0.30).
Strategy B, because its expected net value is higher over many repetitions.
Strategy A, because its expected net value is higher over many repetitions.
Strategy A, because its maximum prize ($50) is larger.
Explanation
The skill here is comparing strategies using expected value to determine which raffle bundle is better in the long run. Expected value represents the long-run average net winnings if you buy the bundle many times. To compute the expected value for each strategy, multiply each possible net outcome (prize minus cost) by its probability and sum the results. In this case, Strategy B has a higher expected net value than Strategy A. Over many fundraisers, choosing Strategy B would lead to higher average net winnings, though results vary in any single purchase. A common distractor is selecting Strategy A due to its larger maximum prize of $50, but this ignores the low probability of winning it. To apply this elsewhere, always multiply outcomes by their probabilities and compare the totals rather than focusing on the best or worst single outcome.