8 > 2x4 ?

The probability is given by

$P(8 > 2 \times 4) = \sum_{2}^{8}P_4(n)P_8(n)$

Where,

$P_4(n) =$ Probability of rolling a total of $n$ with two four sided die.

$P_8(n) =$ Probability of rolling something higher than $n$ with an eight sided die.

So

$P(8 > 2 \times 4) = \frac{1}{16}\cdot\frac{6}{8} + \frac{2}{16}\cdot\frac{5}{8} + \frac{3}{16}\cdot\frac{4}{8} + \frac{4}{16}\cdot\frac{3}{8} + \frac{3}{16}\cdot\frac{2}{8} + \frac{2}{16}\cdot\frac{1}{8} + \frac{1}{16} \cdot 0 = \dfrac{3}{8}$

$3+8 = \boxed{11}$

The expected value for Danny is $\frac{8+1}{2}$ = 4.5

The expected value for Ted is $\frac{4+1}{2}$ x 2 = 5

The probability that Danny's roll exceeds Ted's expected value is when Danny rolls a 6, 7, or 8. The probability is = $\frac{3}{8}$

(I am not entirely sure if this method will always work.)