TITLE: Expected value of the max of $2$ dice
QUESTION [9 upvotes]: What is the expected value of the max of two dice?
I just wonder if there's a better way to get the answer to this question than listing out all possible outcomes and determining the expected value from there as this is actually an interview question?
Thanks

REPLY [5 votes]: Reasoning as for the time allowed to answer to an interview:
a) in a square $6 \times 6$, the number of cases with max no greater than $m$, will be the square $m \times m$.
b) so $m^2/36$ gives the CDF.
c) the median will be at $CDF = 1/2$ that is for a square of area $=18$, i.e. about $4.2$.
d)  to be a bit more precise, the pmf is $(m^2-(m-1)^2)=(2m-1)/36$, which is linear and the average is therefore
$$
\sum\limits_{1\, \le \,m\, \le \,6} {m\left( {2m - 1} \right)} /36 = 161/36 \approx 4.47
$$