*A probability puzzle, from TAoCP.*

Look at these two dice.

The first one is a normal dice, the second one is a cheat's dice. Clearly, the cheat's dice is better. In a contest, chances are $5/6$ that the cheat wins, and $1/6$ that there is a draw.

Let's say that dice $A$ is *better* than dice $B$ when $A$ is expected to win over $B$ in more than half the contests.

**Question.** Is it possible to find dice $A$, $B$, and $C$ such that $A$ is better than $B$, which is better than $C$, which is better than $A$?

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