<s>[INST] What is the expected value of the largest of the three dice rolls? 
You toss a fair die three times. What is the expected value of the largest of the three outcomes?

My approach is the following:
calculate the probability of outcome when $\max=6$, which is
$$P(\text{at least one $6$ of the three rolls}) = 1 - P(\text{no }6) = 1 - (5/6)^3$$
and then calculate the probability of outcome when $\max=5$, which is
$$P(\text{at least one $5$ of the three rolls & $5$ is max}) = 1 - P(\text{no $5$ & $5$ is max}) = 1 - (4/6)^3.$$
I wonder if this approach is right.
 [/INST] Picture the cube of possible outcomes. The cells that represent a maximum of $6$ lie in a greedy half of the outer layer, which has $6^3-5^3=216-125=91$ cells in it.
The next layer represents max $5$, and has $5^3-4^3=125-64=61$ cells in it.
We can proceed in a similar manner and arrive at the sum of the whole cube:
$$6\cdot(6^3-5^3)+5\cdot(5^3-4^3)+4\cdot(4^3-3^3)+3\cdot(3^3-2^3)+2\cdot(2^3-1^3)+1\cdot(1^3-0^3)$$
$$=6^3(6)-5^3(6-5)-4^3(5-4)-3^3(4-3)-2^3(3-2)-1^3(2-1)-0^3(1)$$
$$=6\cdot6^3-5^3-4^3-3^3-2^3-1^3-0^3$$
$$=1296-225$$
$$=1071$$
Divide by the number of cells in the cube $6^3=216$, and the answer is:
$$\frac{1071}{216}=\frac{119}{24}\approx4.96$$
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