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The envious dice

These says seem comparable in a normal way, but they hide a circular relationship: each one can be defeated by another. The advantage is not in choosing the “best” die, but in choosing later.

There are four special dice: - A: 4, 4, 4, 4, 0, 0

  • B: 3, 3, 3, 3, 3, 3
  • C: 6, 6, 2, 2, 2, 2
  • D: 5, 5, 5, 1, 1, 1 You choose a die first. The dealer then chooses one of the remaining three. You both roll your says once and whoever gets the highest number wins. Is there one die better than all the others? What should the second player choose?

Hints

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  1. Don't just compare the averages: compare the says in pairs.
  2. It may happen that one says wins against another with an advantage, but loses against a third party.
  3. Find a cycle: A beats B, B beats C, C beats D, and D beats A.

Solution

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Answer: there is no one die that is better than all. The second player can always choose a die that beats the one chosen by the first with probability $2/3$. The key comparisons are: Comparison Why it wins Probability
A beats B A wins if he rolls 4; B always rolls 3 $4/6=2/3$
B beats C B wins if C rolls 2 $4/6=2/3$
C beats D C wins if he rolls 6, or if he rolls 2 and D rolls 1 $2/6+(4/6)(3/6)=2/3$
D beats A D wins if he rolls 5, or if he rolls 1 and A rolls 0 $3/6+(3/6)(2/6)=2/3$ This is how a cycle is formed: $$

A>B>C>D>A.
$$ Therefore, the second player always responds with the die that beats the one chosen by the first: - against A, choose D;

  • against B, choose A;
  • against C, choose B;
  • against D, choose C. In all cases he wins with probability $2/3$.

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