jmc

Since 9 out of the 12 possible results are one digit numbers, each die will show a one digit number with probability of $\frac{3}{4}$ and a two digit number with probability of $\frac{1}{4}$. The probability that two particular dice will show 2 two digit numbers and 2 one digit numbers is thus ${\left(\frac{1}{4}\right)}^2{\left(\frac{3}{4}\right)}^2$. There are $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ ways to select which two dice will show one digit numbers, so we multiply to get the probability that we want: $6\cdot {\left(\frac{1}{4}\right)}^2{\left(\frac{3}{4}\right)}^2=\frac{54}{256}=\boxed{\frac{27}{128}}$.