I knew the setup immediately but fumbled counting the cases.
Break down the problem by calculating the probabilities of each outcome (all three match, exactly two match, all different) using combinatorics, then compute the expected value as the sum of each outcome's probability times its payoff. Verify that the probabilities sum to 1 and present the final expected value clearly.
Pro tip: After computing the expected value, briefly discuss how the result would change if the dice were biased or if the payouts were different, showing you can generalize and think about edge cases.
Identify the three possible outcomes and calculate their probabilities using counting principles. For three fair 6-sided dice, the total number of outcomes is 6^3 = 216.
There are 6 outcomes where all three dice show the same number (e.g., 1-1-1, 2-2-2, etc.), so probability = 6/216 = 1/36.
Choose the matching number (6 ways), choose which two dice match (3 ways), and choose the different number (5 ways), giving 6*3*5 = 90 outcomes, so probability = 90/216 = 5/12.
The remaining outcomes are all different: 6*5*4 = 120 outcomes, so probability = 120/216 = 5/9. Check that 1/36 + 5/12 + 5/9 = 1.
Multiply each probability by its payoff and sum: EV = (1/36)*20 + (5/12)*10 + (5/9)*(-2) = 20/36 + 50/12 - 10/9 = 5/9 + 25/6 - 10/9 = 25/6 - 5/9 = (75-10)/18 = 65/18 ≈ $3.61.
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