Expected Max of Two Dice

Compute $P(\max \le k)$ for each $k$ first, then use $E[\max ]={\sum }_{k}k\cdot P(\max =k)$.

$P(\max \le k)=(k/6{)}^2$; differences give $P(\max =k)$.

1. 1. CDF of the max

   $P(\max \le k)=(k/6{)}^2$ for $k=1,\dots ,6$, so $P(\max =k)=(2k-1)/36$.

2. 2. Sum the weighted outcomes

   Compute $E[\max ]={\sum }_{k=1}^{6}k\cdot \frac{2k-1}{36}$.

   $$E[\max ]=\frac{1}{36}\sum _{k=1}^{6}k(2k-1)=\frac{161}{36}$$

Answer: 161/36