a= number or dice 1 b= number on dice 2 (a,b)=(1,3),(3,1),(2,2),(2,3),(3,2),(1,4),(4,1) Required probability $\begin{aligned}
& =\frac{2}{6} \times \frac{2}{6}+\frac{1}{6} \times \frac{1}{6}+\frac{2}{6} \times \frac{2}{6}+\frac{2}{6} \times \frac{2}{6}+\frac{1}{6} \times \frac{2}{6}+\frac{2}{6} \times \frac{1}{6}+\frac{1}{6} \times \frac{2}{6} \
& =\frac{18}{36}=\frac{1}{2}
\end{aligned}$