Expected Min of Two Dice

Use $P(\min \ge k)=((7-k)/6{)}^2$ and the tail-sum formula.

Alternatively, relate $E[\min ]$ to $E[\max ]$ using $\min +\max =X+Y$.

1. 1. Tail-sum approach

   $E[\min ]={\sum }_{k=1}^{6}P(\min \ge k)={\sum }_{k=1}^6{\left(\frac{7-k}{6}\right)}^2$.

2. 2. Evaluate

   Summing ${\left(\frac{6}{6}\right)}^2+\cdots +{\left(\frac{1}{6}\right)}^2$ gives $91/36$.

   $$E[\min ]=\frac{91}{36}$$

Answer: 91/36