Use linearity of expectation by defining an indicator for each possible winning value and summing the expected contributions. Alternatively, compute the expected maximum of two distinct rolls by enumerating the 30 ordered pairs where the rolls differ. Clearly state the final expected payout and verify with a quick symmetry or simulation check.
Pro tip: Mention that the expected value can also be derived by considering the expected maximum of two rolls minus the expected value when they are equal, but since equal rolls yield zero, it's simpler to condition on distinct rolls. This shows you understand multiple solution paths and can choose the most efficient one.
Let X be the payout. X = max(R1, R2) if R1 ≠ R2, and X = 0 if R1 = R2.
Express X as the sum over k=1 to 6 of k * I(X = k). Then E[X] = sum_{k=1}^6 k * P(X = k).
For a given k, X = k occurs when the maximum of the two distinct rolls is k. Count ordered pairs (a,b) with a ≠ b, max(a,b) = k. There are 2*(k-1) such pairs (one roll is k, the other is less than k, and order matters). Total outcomes = 36, so P(X = k) = 2(k-1)/36.
E[X] = sum_{k=1}^6 k * [2(k-1)/36] = (1/18) * sum_{k=1}^6 k(k-1) = (1/18) * (sum k^2 - sum k) = (1/18)*(91 - 21) = 70/18 = 35/9 ≈ 3.888... dollars.
Check by alternative method: condition on distinct rolls. Given distinct, the expected maximum is 14/3? Actually, compute E[max | distinct] = (sum over distinct pairs max)/30 = (sum_{k=1}^6 2k(k-1))/30 = (2*70)/30 = 140/30 = 14/3 ≈ 4.666..., then multiply by P(distinct)=30/36=5/6 to get (5/6)*(14/3)=70/18=35/9. Confirm answer.
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