Dominating Dice

We need to count the number of ways that the biggest score on the A die beats the biggest score on the B die, and at the same time the lowest score on the A die beats the lowest score on the B die.

If one or more of the A dice show a 1, then this is impossible.

If the A dice show a 'double $n$ ', then there are $(n-1)^2$ ways.

We only now need to investigate $(2,3) (2,4) (3,4)$ which have 3, 5 and 8 ways respectively.

The matrix below shows these results. There are 46 successful outcomes out of the 256 possible outcomes on both pairs of dice. $46/256 = 23/128$ so $a+b=151$

$\left( \begin{array}{ccc} 0 & 0 & 0&0 \\ 0 & 1 & 3&5 \\ 0 & 3 & 4&8 \\ 0&5&8&9 \end{array} \right)$

Using a pair of 4-sided dice player A can have the results of the following table (column 1) with the corresponding probabilities (column 2). Note that we put $a_1\le a_2$ and $(1,1)$ has probability $1/16$ , while $(1,2)$ has probability $2/16$ , because it includes 2 cases: $(1,2)$ and $(2,1)$ .

Player B , in order that $a_1>b_1$ and $a_2>b_2$ must have one of the results of the following table (column 3) with the corresponding probabilities (column 4).

The probability that $a_1>b_1$ and $a_2>b_2$ , in the case that player A has the result of column (1), is the product of columns (2) and (4), which is shown in column (5). The sum of all cases is shown in the last row of the table and it is ${46\over256}={23\over128}$ . So the correct answer is $23+128=\boxed{151}$ .