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Compare Strategies Using Expected Value - Statistics and Probability

  1. A player is choosing between two carnival games. Compute expected net winnings per play (prize value minus entry fee) to decide which is better over many plays. Strategy A (entry fee $5): - 0.75 probability of winning a $0 prize - 0.20 probability of winning a $10 prize - 0.05 probability of winning a $50 prize Strategy B (entry fee $3): - 0.60 probability of winning a $0 prize - 0.35 probability of winning a $6 prize - 0.05 probability of winning a $30 prize Which strategy has the greater expected value (average net winnings) over many repetitions?
    Answer: Strategy B, because its expected net winnings are $0.60 per play. This question involves the skill of comparing strategies using expected value to decide which carnival game offers higher average net winnings over many plays. Expected value is the long-run average outcome, here net as prize minus fee, averaged per play. Compute it by finding the weighted average prize for each, then subtracting the fee. Strategy B has higher expected net winnings of $0.60, while Strategy A has -$0.50. Over many trials, Strategy B will yield higher average net winnings, though single plays vary. A tempting distractor is chasing the highest prize, like $50 in A as in choice A, but expected value incorporates fees and probabilities. To transfer, calculate weighted prize totals minus costs and compare, not isolated highs.
  2. A student can choose one of two ways to earn points on a practice quiz each day. Strategy A: Guaranteed 7 points. Strategy B: With probability 0.50 you earn 20 points; with probability 0.50 you earn 0 points. Over many repetitions, which strategy has the greater expected value (average points per day)?
    Answer: Strategy B, because its expected value is 10 points versus 7 points for Strategy A This problem compares strategies using expected value to find the higher average points per day. For Strategy A, the expected value is simply 7 points since it's guaranteed every time. For Strategy B, we calculate the expected value by multiplying each outcome by its probability: (0.50 × 20) + (0.50 × 0) = 10 + 0 = 10 points. Since 10 points is greater than 7 points, Strategy B has the higher expected value over many repetitions. A tempting mistake is thinking Strategy A must be better because 7 points is guaranteed, but this ignores that Strategy B averages 10 points in the long run. Another error is thinking Strategy B's expected value is 5 points by incorrectly averaging 20 and 0, rather than properly weighting by probabilities. The transfer strategy is to multiply each outcome by its probability and sum them, not just compare single outcomes or guarantees.
  3. A freelancer must choose between two payment plans for the same type of job. Compare the expected payment per job over many jobs. Strategy A (Guaranteed + bonus chance): - With probability 0.70, get $120 - With probability 0.30, get $180 Strategy B (Higher upside, lower typical pay): - With probability 0.20, get $300 - With probability 0.80, get $110 Which strategy has the greater expected value (average payment) per job over many jobs?
    Answer: Strategy B, because its expected payment is higher than Strategy A's. This problem asks us to compare payment strategies using expected value, which tells us the average payment per job over many repetitions. Expected value is calculated by multiplying each outcome by its probability and summing the results. For Strategy A: EV = (0.70 × $120) + (0.30 × $180) = $84 + $54 = $138 expected payment. For Strategy B: EV = (0.20 × $300) + (0.80 × $110) = $60 + $88 = $148 expected payment. Since Strategy B has a higher expected value ($148 vs $138), it provides better average earnings over many jobs. A common mistake is choosing Strategy A because the most likely outcome ($120 with 70% probability) seems more reliable, but this ignores the full probability distribution. The key insight is that expected value weights all outcomes by their probabilities—Strategy B's occasional high payment of $300 more than compensates for its typically lower $110 payment.
  4. A gamer can choose between two loot-box strategies each time they play, and they care about average profit over many plays (profit can be negative). Strategy A (costs $4 to open): - 0.60 probability of getting an item worth $2 - 0.30 probability of getting an item worth $6 - 0.10 probability of getting an item worth $20 Strategy B (costs $4 to open): - 0.80 probability of getting an item worth $3 - 0.15 probability of getting an item worth $8 - 0.05 probability of getting an item worth $25 Which strategy has the greater expected value (average profit) over many repetitions?
    Answer: Strategy A, because its expected profit is $1.00 per play. This question involves the skill of comparing strategies using expected value to determine which loot-box option offers higher average profit over many plays. Expected value is the long-run average outcome, representing the typical profit per play if repeated many times. To compute it for each strategy, subtract the cost from each possible item value to get profits, then multiply each profit by its probability and sum the results. Strategy A has a higher expected profit of $1.00, while Strategy B has $0.85. Over many trials, Strategy A will provide a higher average profit, though individual plays may vary. A tempting distractor is focusing on the maximum payoff, like the $20 item in choice A, but this ignores probabilities of lower outcomes. To apply this skill, always multiply outcomes by their probabilities and compare the totals, rather than judging by single best or worst cases.
  5. 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?
    Answer: Strategy A, because it has the greater expected value over many quizzes. 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.
  6. 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?
    Answer: Strategy B, because its expected net value is higher over many repetitions. 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.
  7. A player can choose one of two strategies each time they play a carnival game. Strategy A: Pay $4 to play. Outcomes: with probability 0.70 you win $6; with probability 0.30 you win $0. Strategy B: Pay $4 to play. Outcomes: with probability 0.20 you win $20; with probability 0.80 you win $0. Over many repetitions, which strategy has the greater expected value (net profit in dollars)?
    Answer: Strategy A, because its expected net profit is $0.20 versus $0.00 for Strategy B This problem asks us to compare strategies using expected value, which is the long-run average outcome when playing many times. For Strategy A, the expected value is calculated by multiplying each outcome by its probability: (0.70 × $6) + (0.30 × $0) - $4 = $4.20 - $4 = $0.20 net profit. For Strategy B, the expected value is (0.20 × $20) + (0.80 × $0) - $4 = $4 - $4 = $0 net profit. Since Strategy A has an expected value of $0.20 compared to $0 for Strategy B, Strategy A is favored over many trials. A common mistake is choosing Strategy B because it offers the larger maximum payout of $20, but this ignores that you only win 20% of the time. The key insight is to multiply outcomes by probabilities and compare totals, not just look at single outcomes.
  8. 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 0.700.70 you get an item worth $5 (net $1); with probability 0.300.30 you get an item worth $0 (net $-4). Strategy B: Costs $4. Outcomes: with probability 0.100.10 you get an item worth $30 (net $26); with probability 0.900.90 you get an item worth $0 (net $-4). Which strategy would result in a higher average net payoff over many repetitions?
    Answer: Strategy A, because its expected net payoff is higher over many repetitions. 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 −0.50-0.50 compared to Strategy B's −1-1. 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.
  9. A delivery company is choosing between two routing software packages. For each day, exactly one traffic condition occurs, producing the listed net profit (after all costs) for that day. Strategy A: - Net profit $900 with probability 0.50 - Net profit $300 with probability 0.30 - Net loss $200 with probability 0.20 Strategy B: - Net profit $700 with probability 0.70 - Net profit $100 with probability 0.20 - Net loss $500 with probability 0.10 Which strategy has the greater expected value (higher average net profit per day) over many days?
    Answer: Strategy A, because its expected value is higher over many days. The skill here is comparing strategies using expected value to pick the routing software with higher average daily profit. Expected value is the long-run average net profit over many days. Compute it for each strategy by multiplying each profit or loss by its probability and summing the products. In this case, Strategy A has a higher expected value than Strategy B. Over many days, Strategy A would lead to higher average net profits, though daily results fluctuate. A common distractor is choosing Strategy A just for its highest possible profit of $900, but the decision relies on the probability-weighted average. For other situations, multiply all outcomes by their probabilities and compare the totals instead of isolated highs or lows.
  10. A café is choosing between two coupon strategies to distribute. Consider the café's perspective: expected cost per coupon equals the expected discount given. Strategy A (Simple Coupon): - 1.00 probability the customer gets $1 off Strategy B (Spin Wheel): - 0.50 probability the customer gets $0 off - 0.40 probability the customer gets $2 off - 0.10 probability the customer gets $8 off Which strategy has the greater expected value of discount (higher average cost to the café) over many coupons?
    Answer: Strategy B, because its expected discount is $1.60. This question involves the skill of comparing strategies using expected value to determine which coupon approach costs the café more on average over many distributions. Expected value is the long-run average outcome, here the typical discount per coupon. Compute it for each by multiplying each discount amount by its probability and summing. Strategy B has a higher expected discount of $1.60, compared to Strategy A's $1.00. Over many trials, Strategy B will cost the café more on average, though individual discounts vary. A tempting distractor is preferring the certain low cost, like $1 in A as in choice A, but expected value includes potential higher discounts. To apply, calculate probability-weighted totals and compare, avoiding bias toward guarantees or maximums.
  11. A student is deciding between two study apps that give random scholarship credits each month. The student cares about the expected credits earned per month over many months. Strategy A (App A): - With probability 0.50, earn 8 credits - With probability 0.30, earn 4 credits - With probability 0.20, earn 0 credits Strategy B (App B): - With probability 0.10, earn 30 credits - With probability 0.40, earn 3 credits - With probability 0.50, earn 0 credits Which strategy has the greater expected value in credits per month over many months?
    Answer: Strategy A, because its expected credits per month are higher than Strategy B's. This problem requires comparing strategies using expected value to determine average scholarship credits earned per month. Expected value represents the long-run average outcome when a random process repeats many times. For Strategy A (App A): EV = (0.50 × 8) + (0.30 × 4) + (0.20 × 0) = 4 + 1.2 + 0 = 5.2 credits. For Strategy B (App B): EV = (0.10 × 30) + (0.40 × 3) + (0.50 × 0) = 3 + 1.2 + 0 = 4.2 credits. Since Strategy A has a higher expected value (5.2 vs 4.2 credits), it's the better choice over many months. A tempting mistake is choosing Strategy B because it offers a chance at 30 credits, focusing on the maximum possible outcome rather than the average. The key strategy is to multiply each credit amount by its probability and sum the results, comparing total expected values rather than being swayed by individual large payouts with low probabilities.
  12. 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?
    Answer: Strategy A, because it has the greater expected net donation over many calls. 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.
  13. 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?
    Answer: Strategy B, because it has the greater expected value over many spins. 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.
  14. An investor is comparing two one-day trading strategies. Payoff is net profit in dollars. Strategy A: With probability 0.60 you gain $50; with probability 0.40 you lose $30. Strategy B: With probability 0.30 you gain $120; with probability 0.70 you lose $10. Over many repetitions, which strategy has the greater expected value (higher average net profit)?
    Answer: Strategy B, because its expected value is $29 versus $18 for Strategy A This problem asks us to compare trading strategies using expected value to find the higher average net profit. For Strategy A, the expected value is: (0.60 × $50) + (0.40 × -$30) = $30 + -$12 = $18. For Strategy B, the calculation is: (0.30 × $120) + (0.70 × -$10) = $36 + -$7 = $29. Since $29 is greater than $18, Strategy B has the higher expected value over many trading days. A common mistake is choosing Strategy A because it has a higher probability of gain (0.60 vs 0.30), but this ignores the magnitude of gains and losses. Another error is focusing on Strategy B's larger maximum gain of $120 without considering probabilities. The transfer strategy is to multiply each outcome by its probability and sum them all, considering both positive and negative values to find the true long-run average.
  15. A delivery driver is choosing between two routes each day. Measure payoff as net earnings in dollars (tips minus extra fuel cost). Strategy A (Highway): Extra fuel cost is $2. Outcomes: with probability 0.80 you get $10 in tips; with probability 0.20 you get $0 in tips. Strategy B (Downtown): Extra fuel cost is $0. Outcomes: with probability 0.50 you get $12 in tips; with probability 0.50 you get $2 in tips. Over many repetitions, which strategy has the greater expected value (average net earnings)?
    Answer: Strategy B, because its expected net earnings are $7.00 versus $6.00 for Strategy A This problem compares delivery route strategies using expected value, measuring net earnings (tips minus fuel cost). For Strategy A, we subtract the $2 fuel cost from expected tips: (0.80 × $10) + (0.20 × $0) - $2 = $8 + $0 - $2 = $6.00. For Strategy B, with no extra fuel cost, the expected value is simply: (0.50 × $12) + (0.50 × $2) = $6 + $1 = $7.00. Since $7.00 is greater than $6.00, Strategy B has the higher expected value over many deliveries. A tempting error is choosing Strategy A because it has a higher probability (0.80) of getting $10 in tips, but this ignores both the fuel cost and the probability weighting of all outcomes. The key is to calculate the complete expected net earnings by multiplying each outcome by its probability and subtracting all costs, not just focus on individual probabilities or maximum values.
  16. 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 −60-60); with probability 0.15 a repair is needed and the plan covers a $400 repair (net payoff 400−60=340400 - 60 = 340). Strategy B (Pricier plan): Pay $90 now. Outcomes: with probability 0.85 no repair is needed (net payoff −90-90); with probability 0.15 a repair is needed and the plan covers the $400 repair (net payoff 400−90=310400 - 90 = 310). Which strategy has the greater expected value (average net payoff) over many repetitions?
    Answer: Strategy A, because it has the greater expected value over many repetitions. 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 00 compared to Strategy B's −30-30. 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.
  17. 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?
    Answer: Strategy A, because it has the higher expected net payoff over many repetitions. 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.
  18. A phone buyer is deciding whether to purchase a protection plan. Consider the buyer's expected total cost over many similar buyers. Strategy A (Buy plan): Pay $30 now. - 0.90 probability of no damage (no additional cost) - 0.10 probability of damage; repair cost would be $200 but the plan covers it (no additional cost) Strategy B (No plan): Pay $0 now. - 0.90 probability of no damage (no cost) - 0.10 probability of damage and paying the $200 repair Which option would result in a lower average total cost for the buyer over many repetitions?
    Answer: Strategy B, because its expected cost is $20, which is lower than $30. This question involves the skill of comparing strategies using expected value to find which protection plan choice leads to lower average total cost over many buyers. Expected value is the long-run average outcome, representing typical total cost if situations repeat. For each, compute it by multiplying each possible total cost by its probability and summing. Strategy B has a lower expected cost of $20, compared to Strategy A's $30. Over many trials, Strategy B will result in lower average costs, despite risk of high repair bills. A tempting distractor is avoiding the worst-case $200, like in choice A, but expected value weighs the low damage probability. In practice, multiply all outcomes by probabilities to compare totals, not just guarantees or likeliest scenarios.
  19. 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?
    Answer: Strategy B, because it has the greater expected value over many attempts. 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.
  20. A student can choose between two grading schemes for a course project. Payoff is final project score in points. Strategy A: You take a single presentation. Outcomes: with probability 0.75 you score 80 points; with probability 0.25 you score 40 points. Strategy B: You submit a written report. Outcomes: with probability 0.50 you score 90 points; with probability 0.50 you score 60 points. Over many repetitions, which option would result in a higher average outcome (expected project score)?
    Answer: Strategy B, because its expected score is 75 points versus 70 points for Strategy A This problem compares grading schemes using expected value to find the higher average project score. For Strategy A (presentation), the expected score is: (0.75 × 80) + (0.25 × 40) = 60 + 10 = 70 points. For Strategy B (written report), the calculation is: (0.50 × 90) + (0.50 × 60) = 45 + 30 = 75 points. Since 75 points is greater than 70 points, Strategy B yields the higher expected score over many repetitions. A tempting mistake is thinking Strategy A is better because the most likely outcome (80 points with 0.75 probability) is higher than 60 points, but this ignores the complete probability distribution. Another error is calculating Strategy A's expected value using only the 0.75 outcome, which incorrectly ignores the 0.25 probability of scoring 40. The key is to weight all outcomes by their probabilities to find the true average.

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