Assuming that the 120 faces have all numbers from 1 to 120 printed on them, we have a dice that can output all numbers between 1 and 120 with the same probability of 1/120.
So, the expected value is
[tex]\displaystyle \sum_{i=1}^{120}i\cdot p(i) = \sum_{i=1}^{120}i\cdot \dfrac{1}{120}=\dfrac{1}{120}\sum_{i=1}^{120}i=\dfrac{1}{120}\dfrac{120\cdot 121}{2}=\dfrac{121}{2}=60.5[/tex]
