Here I will show that the diagram is correct

The chance that 1 will be thrown is the following

$$
\frac{1}{2}\cdot\left(\sum_{k=1}^\infty \frac{1}{4}^k\right)=
\frac{1}{8}\cdot\left(\sum_{k=0}^\infty \frac{1}{4}^k\right)=
\frac{1}{8}\cdot\frac{1}{1-1/4} = \frac{1}{6}
$$

It is quite obvious that it should be the same for all the other sides

$$
\frac{1}{8}\cdot \sum_{k=0}^\infty \left(\frac{1}{4})\right^k=
\frac{1}{8}\cdot \sum_{k=0}^\infty \left(\frac{1}{4})\right^k=
$$