Question

The largest possible fair dice is a disdyakis triacontahedron, which has 120 sides. What is the expected value, E(X), of rolling a disdyakis triacontahedron dice?
a. 3.5
b. 60.5
c. 70
d. 1/120

Answer 1

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

$\displaystyle \sum_{i=1}^{120}i\cdot p(i) = \sum_{i=1}^{120}i\cdot \dfrac{1}{120}=\dfrac{1}{120}\sum_{i=1}^{120}i=\dfrac{1}{120}\dfrac{120\cdot 121}{2}=\dfrac{121}{2}=60.5$