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Average of the larger die

ProbabilityMedium

Roll $2$ fair $6$ -sided dice independently. What is the expected value of the larger of the $2$ results? Give an exact fraction.

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Hint

First find the probability that the maximum is at most $k$ , then the probability that it equals $k$ .

Solution

The maximum is at most $k$ with probability $\frac{k ^{2}}{36}$ .
So $P (max = k) = \frac{k ^{2} - ( k - 1 ) ^{2}}{36} = \frac{2 k - 1}{36}$ .
The expected value is $\sum \frac{k ( 2 k - 1 )}{36}$ over $k = 1..6$ .
We have $\sum k (2 k - 1) = 2 \times 91 - 21 = 161$ , since $\sum k^{2} = 91$ and $\sum k = 21$ .
Hence the expected value is $\frac{161}{36}$ .

Answer: $\frac{161}{36}$