Expected value of the max of $2$ dice 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 
 A: Denote outcomes $(X_1,X_2)$ and $Z=\max\{X_1,X_2\}$. Then just use that the outcomes are independent and  break the event "$Z=a$" into "$(X_1=a \text{ and }X_2\leq a)\text{ or }(X_2=a\text{ and }X_1<a)$":
$$E(Z)=\sum_{a=1}^6 a\cdot P(Z=a)=\sum_{a=1}^6 a\cdot (P(X_1=a \text{ and }X_2\leq a )+P(X_2=a \text{ and }X_1<a))$$
$$=\sum_{a=1}^6 a\cdot \left(\frac{1}{6}\cdot\frac{a}{6}+\frac{1}{6} \cdot \frac{a-1}{6}\right)=\frac{161}{36}$$
A: 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
$$
\sum\limits_{1\, \le \,m\, \le \,6} {m\left( {2m - 1} \right)} /36 = 161/36 \approx 4.47
$$
A: Suppose the dices have $n$ faces. Let $X=\max(a,b)$
Note that:
$X=1$ only in the case (1,1), i.e. $p_X(1)=1/n^2$
$X=2$ in the cases (1,2), (2,2), (2,1), i.e. $p_X(2)=3/n^2$
$\vdots$
$X=k$ in $k+k-1=2k-1$ cases, i.e. $p_X(k)=(2k-1)/n^2$(this is easy to see when you draw all the cases).
$$E(X)=\frac{1}{n^2} \sum_{x=1}^{n}x(2x-1)=\frac{1}{n^2} \sum_{x=1}^{n}(2x^2-1)=\\ \frac{1}{n^2}\left(\frac{n(n+1)(2n+1)}{3}-\frac{n(n+1)}{2}\right)=\frac{4n^3+3n^2-n}{6n^2} = \dfrac{4n^2 + 3n - 1}{6n}$$
