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?


1 Answer 1


This is just an example of

nontransitive dice.


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.

  • $\begingroup$ Yeah, that´s it. Thanks for the reference, I was not aware that this riddle even has a name. $\endgroup$
    – ThomasL
    Feb 13, 2020 at 21:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.