Danny decides to roll an 8 sided die (with sides numbered 1-8)

Ted decides to roll two 4 sided dice (each with sides numbered 1-4)

What is the probability that Danny's roll will be more than the sum of Ted's rolls?

If the answer is a fraction, $\frac{a}{b}$ where $a$ and $b$ are coprime positive integers give the answer as $a+b$ .

The answer is 11.

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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

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

$3+8 = \boxed{11}$