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.
Simply calculate the expected value of the sum of these dice:
$$\Bbb E[d_1]=\sum^{12}_{n=1}\frac1{12}n=\frac{13}2=6.5$$ $$\Bbb E[2d_2]=2\left(\sum^6_{n=1}\frac16n\right) = 2\cdot\frac72=7$$ $$\Bbb E[3d_3]=3\left(\sum^4_{n=1}\frac14n\right) = 3\cdot\frac52=7.5$$
Therefore, $3d_3$ is to be the best choice.
EDIT:
I might have misinterpreted this question. My answer is correct if the number referred to in "highest number" is the sum. Alternatively, with two players, this could mean the highest number rolled on any die. I will answer this now:
Expected value of the highest die rolled is relatively simple:
$$\Bbb E[d_1]=\sum^{12}_{n=1}\frac1{12}n=\frac{13}2=6.5$$ $$\Bbb E[2d_2]=\sum^6_{n=1}\frac{n^2-(n-1)^2}{6^2}n = 4.47\bar2$$ $$\Bbb E[3d_3]=\sum^4_{n=1}\frac{n^3-(n-1)^3}{4^3}n = 3.4375 $$
So the answer in the case of the maximum die winning, you should choose the $12$-sided die.