The two best of three dice

Idea. Instead of finding the two best dice, subtract the worst one from the total. The expected minimum has a neat formula through its tail probabilities.

Step 1: the total. Three dice add up to $3\times 3.5=10.5$ on average.

Step 2: the minimum. The minimum is at least $k$ when all three dice are, with probability ${\left(\frac{7-k}{6}\right)}^3$. So $$\mathbb{E}[\min ]=\frac{6^3+5^3+4^3+3^3+2^3+1^3}{216},$$ which is $\frac{441}{216}$.

Step 3. $$\begin{array}{rl}10.5-\frac{441}{216}& =\frac{2268-441}{216}=\frac{1827}{216}\\ & =\frac{203}{24}\approx \mathrm{8.46.}\end{array}$$

As a market. A fair price is about 8.46; a market maker would quote around it, for example 8.3 at 8.6.