Some days ago I read this problem, and I haven't arrived yet at a solution: Imagine you have 3 kinds of regular number die:
d1 has 12 faces (from 1 to 12), d2 has 6 faces (from 1 to 6), d3 has 4 faces (from 1 to 4).
Now, you can choose between using one d1, two d2's, or three d3's. Which configuration of dice would you choose if the one who gets the higher number wins?
I would choose the first one, because the probability of winning (getting the highest number) is 1/12. Instead, throwing two d2's the probability is (1/6)^2, and for three d3's it's (1/4)^3. But I'm not sure.