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1 vote
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

User Gerd Klima
by
7.1k points

1 Answer

4 votes

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 (1)/(120)=(1)/(120)\sum_(i=1)^(120)i=(1)/(120)(120\cdot 121)/(2)=(121)/(2)=60.5

User Rburny
by
7.5k points
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