Weight, That’s Not Fair!

Let the probability of rolling a number $n$ (1 through 6) with the fair die be $p_n$ and that with the weighted die be $q_n$ . Then the probability of rolling a sum of 4 with both dice is given by:

$\begin{aligned} p_1q_3 + p_2q_2 + p_3q_1 & = \frac 1{12} \\ \frac 16 \left(q_1+q_2+q_3 \right) & = \frac 1{12} \\ \implies q_1+q_2+q_3 & = \frac 12 \end{aligned}$

It is given that $q_4 = \frac 1{10}$ . Then

$\begin{aligned} {\color{#3D99F6}q_1+q_2+q_3} + {\color{#D61F06}q_4} +q_5 + q_6 & = 1 \\ {\color{#3D99F6}\frac 12} + {\color{#D61F06} \frac 1{10}} +q_5 + q_6 & = 1 \\ \implies q_5 + q_6 & = 1 - \frac 12 - \frac 1{10} = \frac 25 \end{aligned}$

Now the probability of rolling 11 with both dice is given by:

$\begin{aligned} p_5q_6 + p_6q_5 & = \frac 16 \left(q_5+q_6 \right) = \frac 16 \times \frac 25 = \frac 1{15} \end{aligned}$

Therefore, $a+b = 1+15 = \boxed{16}$ .

Great Title!