I love this die

Let $k$ be the proportionality constant.
$\begin{aligned}P(r)&=\dfrac{k}{r} ;r=1,2,3,4,5,6 \\\sum P(r)&=1 \quad \quad \quad \quad (\text{Exhaustive Events})\\\Rightarrow k&=\dfrac{60}{147}\end{aligned}$
Probability of getting 4 or 5 on the die in two throws is.
$\begin{aligned}P(E)&=P(4)\times P(5) \times 2!\\&= \dfrac{k}{4}\times\dfrac{k}{5}\times 2 \\&=\dfrac{{\left(\dfrac{60}{147}\right)}^2}{4\times 5}\times 2 \\&\huge\color{#3D99F6}{=\boxed{\dfrac{40}{2401}}}\end{aligned}$

Not a pretty solution but... the chance of rolling a 4 on one roll is (1/4)/(1+1/2+1/3+1/4+1/5+1/6), which we will define as X. The chance of rolling a 5 on one roll is (1/5)/(1+1/2+1/3+1/4+1/5+1/6), which we will define as Y. To roll a 4 and then a 5 the odds are XY. But we can also roll a 5 and then a 4 so the overall odds must be 2XY, which equals 0.01666.