Median of three uniforms on (0,3) in the middle third

Three independent random variables are each uniform on $(0,3)$. What is the probability that their median lies between $1$ and $2$?

$13/27$. Each variable falls in $(0,1)$, $(1,2)$ or $(2,3)$ with probability $1/3$. The median lands in the middle third unless at least two fall in the same outer third: $1-2\left[3\cdot\tfrac{2}{27}+\tfrac{1}{27}\right]=1-\tfrac{14}{27}=\tfrac{13}{27}$.