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Mean-variance scoring

Expected valueHard

A gambler scores any lottery as its mean minus a tenth of its variance. Lottery A pays the sum of $2$ fair $6$ -sided dice, and lottery B pays the larger of the same kind of $2$ dice. What is the score of A minus the score of B?

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Hint

You need the mean and variance of both the sum and the maximum.

Solution

For A the mean is $7$ and the variance is $\frac{35}{6}$ , so the score is $7 - \frac{7}{12} = \frac{77}{12}$ .
For B the mean is $\frac{161}{36}$ and $E [max^{2}] = \frac{791}{36}$ , so the variance is $\frac{2555}{1296}$ and the score is $\frac{11081}{2592}$ .
The difference is $\frac{77}{12} - \frac{11081}{2592} = \frac{16632}{2592} - \frac{11081}{2592}$ .
That equals $\frac{5551}{2592}$ .

Answer: $\frac{5551}{2592}$

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