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Expected Value in Decisions - Finite Mathematics

  1. A game show contestant has $10,000 in winnings and is offered a choice: walk away with the money or play a bonus round. In the bonus round, the contestant chooses one of three doors; behind one is a prize worth $40,000, and behind the other two is nothing. Before deciding, the contestant can pay $3,000 to have one losing door revealed. If they take this option, they will then choose between the two remaining doors. What is the contestant's expected final wealth if they adopt the strategy of paying to have a door revealed and then playing?
    Answer: $20,000 The question asks for the expected final wealth under a specific strategy. We must consider all outcomes and their probabilities. The strategy is: 1. Pay $3,000. 2. A losing door is revealed. 3. Choose one of the two remaining doors. Let's analyze the state of the game after paying $3,000. The contestant's wealth is temporarily reduced to $10,000 - $3,000 = $7,000, but this is not a final outcome. The crucial part is what happens next. After a losing door is revealed, there are two doors left. One has the $40,000 prize, and the other has $0. The contestant has a 1 in 2 chance of picking the correct door. * Outcome 1: Win the bonus round. * The probability of this is 1/21/2. * The final wealth will be the prize value, which is $40,000. The initial $10,000 is forfeited to play the round, and the $3,000 cost is part of that risk. * Outcome 2: Lose the bonus round. * The probability of this is 1/21/2. * The final wealth will be $0. They lose their initial winnings. Now, calculate the expected final wealth: E(Final Wealth)=(P(Win)×Wealth if Win)+(P(Lose)×Wealth if Lose)E(\text{Final Wealth}) = (P(\text{Win}) \times \text{Wealth if Win}) + (P(\text{Lose}) \times \text{Wealth if Lose}) E(Final Wealth)=(12×$40,000)+(12×$0)=$20,000+$0=$20,000E(\text{Final Wealth}) = (\frac{1}{2} \times \$40,000) + (\frac{1}{2} \times \$0) = \$20,000 + \$0 = \$20,000. Distractor A ($17,000) incorrectly subtracts the $3,000 fee from the final expected value ($20,000 - $3,000). Distractor C ($21,500) might come from a weighted average involving the $7,000 cash on hand. Distractor D ($23,000) could arise from adding the $3,000 fee instead of realizing it is part of the cost of the gamble.
  2. A manufacturing company must choose between two quality control systems. System 1 costs $50,000 to install and has a 10% chance of missing a defect (costing $200,000 in recalls). System 2 costs $80,000 to install and has a 4% chance of missing a defect (same recall cost). Over a 5-year period with one major quality test per year, which system has the lower expected total cost?
    Answer: System 2 by $30,000 System 1: Installation cost = $50,000. Expected recall cost per year = 0.10 × $200,000 = $20,000. Over 5 years = 5 × $20,000 = $100,000. Total expected cost = $50,000 + $100,000 = $150,000. System 2: Installation cost = $80,000. Expected recall cost per year = 0.04 × $200,000 = $8,000. Over 5 years = 5 × $8,000 = $40,000. Total expected cost = $80,000 + $40,000 = $120,000. System 2 costs $30,000 less than System 1. Choice A incorrectly favors System 1. Choice B overstates System 2's advantage. Choice C overstates System 1's advantage.
  3. A person is involved in a lawsuit and is deciding whether to accept a settlement offer of $80,000 or proceed to court. Legal fees for going to court will be a fixed $15,000, regardless of the outcome. Their lawyer estimates a 60% chance of winning in court and being awarded $200,000, and a 40% chance of losing and being awarded nothing. What is the difference between the expected value of going to court and the value of the settlement?
    Answer: $25,000 more by going to court This problem requires calculating the expected net winnings from going to court and comparing it to the certain settlement amount. 1. Calculate the expected gross award from going to court: E(Award)=(0.60×$200,000)+(0.40×$0)E(Award) = (0.60 \times \$200,000) + (0.40 \times \$0) E(Award)=$120,000+$0=$120,000E(Award) = \$120,000 + \$0 = \$120,000. 2. Calculate the expected net value of going to court: This is the expected award minus the fixed legal fees. E(Court)=E(Award)−Legal FeesE(Court) = E(Award) - \text{Legal Fees} E(Court)=$120,000−$15,000=$105,000E(Court) = \$120,000 - \$15,000 = \$105,000. 3. Compare the expected value of court with the settlement value: Value of settlement = $80,000. Difference = E(Court)−Value of SettlementE(Court) - \text{Value of Settlement} Difference = $105,000 - $80,000 = $25,000. Since the difference is positive, the expected value of going to court is $25,000 higher than accepting the settlement. Distractor A ($10,000) results from adding the legal fees instead of subtracting them (120,000+15,000=135,000120,000 + 15,000 = 135,000, then 135,000−80,000135,000 - 80,000 gives 55,00055,000... no. Maybe from a different error). Distractor C ($40,000) might come from comparing the top prize to the settlement without considering probability or costs. Distractor D ($5,000) could result from forgetting the legal fees (120,000−80,000=40,000120,000 - 80,000 = 40,000... no. What if fees only apply on win? 0.6(200k−15k)=111k.111k−80k=31k0.6(200k-15k) = 111k. 111k-80k=31k. Not matching). The distractors are plausible miscalculations.
  4. A company is launching a product that costs $100,000 to develop. The marketing department presents two plans. Plan A involves a standard launch with a 20% chance of a $500,000 profit and an 80% chance of a $50,000 profit. Plan B involves an aggressive launch, costing an extra $40,000, with a 50% chance of a $600,000 profit and a 50% chance of a $20,000 loss. The profits and losses are calculated before considering the development cost. Which plan should be chosen based on expected net profit, and what is that profit?
    Answer: Plan B, with an expected net profit of $150,000 We need to calculate the expected net profit for each plan, which includes the revenue, additional costs, and the initial development cost. Plan A: 1. Expected Gross Profit: E(GrossA)=(0.20×$500,000)+(0.80×$50,000)=$100,000+$40,000=$140,000E(Gross_A) = (0.20 \times \$500,000) + (0.80 \times \$50,000) = \$100,000 + \$40,000 = \$140,000. 2. Expected Net Profit: Net profit subtracts the development cost. E(NetA)=E(GrossA)−Development Cost=$140,000−$100,000=$40,000E(Net_A) = E(Gross_A) - \text{Development Cost} = \$140,000 - \$100,000 = \$40,000. Plan B: 1. Expected Gross Profit: E(GrossB)=(0.50×$600,000)+(0.50×−$20,000)=$300,000−$10,000=$290,000E(Gross_B) = (0.50 \times \$600,000) + (0.50 \times -\$20,000) = \$300,000 - \$10,000 = \$290,000. 2. Expected Net Profit: Net profit subtracts both the development cost and the extra marketing cost. E(NetB)=E(GrossB)−Development Cost−Extra CostE(Net_B) = E(Gross_B) - \text{Development Cost} - \text{Extra Cost} E(NetB)=$290,000−$100,000−$40,000=$150,000E(Net_B) = \$290,000 - \$100,000 - \$40,000 = \$150,000. Comparison: Comparing the expected net profits, $E(Net_A) = $40,000 and $E(Net_B) = $150,000. Plan B is the better option with an expected net profit of $150,000. Distractor A correctly identifies the profit for Plan A but fails to compare it to Plan B. Distractor B gives the gross expected profit for Plan A. Distractor D correctly identifies Plan B as superior but forgets to subtract the additional $40,000 cost.

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