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On a table there are four standard six-sided dice with faces numbered 1, 2, 3, 4, 5, 6 each. Alice starts and chooses a die. Then Bob chooses a die and both throw the dice. The winner is the person with the greater number. It is a draw, if both show the same number.

Now the numbers on the faces of each of the four dice are changed according to the following rules:
1) number 0 is allowed on a face in addition to the numbers from 1-6.
2) a number can appear more than once on different faces of the same die.
For example you can have a die with faces 0,0,0,4,4,6.
After all four dice have been designed, Alice starts again, chooses a die. (same concept as before). Now Bob chooses a die (he knows which die was chosen by Alice) and both throw the dice.

How must the four dice be designed such that Bob wins with a probability twice as big as that of Alice?

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This is just an example of

nontransitive dice.

Particularly,

Efron's dice are exactly what this question is asking for. The dice must have the faces [444400], [333333], [662222], [555111]; each die beats the next in the list with probability 2/3.

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  • $\begingroup$ Yeah, that´s it. Thanks for the reference, I was not aware that this riddle even has a name. $\endgroup$ – ThomasL Feb 13 at 21:20

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