[Guest]

imagine you have 3 dice, with the following sides.

A: 1,4,4,4,4,4

B: 3,3,3,3,6,6

C: 2,2,2,5,5,5

A\C 2 2 2 2 5 5  B\C 2 2 2 5 5 5  A\B  3 3 3 3 6 6   

 1  C C C C C C   3  B B B C C C   1   B B B B B B

 4  A A A C C C   3  B B B C C C   4   A A A A B B

 4  A A A C C C   3  B B B C C C   4   A A A A B B 

 4  A A A C C C   3  B B B C C C   4   A A A A B B

 4  A A A C C C   6  B B B B B B   4   A A A A B B

 4  A A A C C C   6  B B B B B B   4   A A A A B B

if you made it into a two player game with player 1 picking first, and player 2 picking second, and they each roll 10 times, player two would win the majority of the games, because he can always pick a winning dice. such a set of dice is called non-transitive. can you find a set of dice for three players? that is, no matter what two dice the first two players pick, the third player always wins?

it will require 7 dice.

Edited September 21, 2010 by phillip1882

[bonanova]

You need each die to beat three others in the set, and to lose to three others in the set. That way, no matter which two your opponents choose, there is a third die that will beat them both. So there are at least seven dice in the set. Let's look for a set of seven dice. For simplicity, assume opposite faces have the same number. There are three distinct numbers on each die and each number is repeated. So between two dice, there are 9 match-ups with equal likelihood. We look for ways one die will have either a 4/9 or 5/9 chance to beat all the other dice.

One die will beat another with 5/9 probability four different ways.

1. Its smallest number loses to all three of the other numbers; its middle number loses to the largest of the other die.
2. Its smallest two numbers each lose to the largest two of the other die.
3. Its smallest number loses to the two largest of the other die; its largest two number lose only to the largest of the other die.
4. Its largest two numbers lose to the largest of the other die; the smallest loses to the two largest of the other die.

Let the smallest numbers be 1 2 3 4 5 6 7 the middle numbers be 8 9 10 11 12 13 14 the largest numbers be 15 16 17 18 19 20 21 in some order. Then every middle number beats every lowest number and every largest number beats every lowest and middle number.

Label and order the dice by descending largest number: 21 > 20 > 19 > 18 > 17 > 16 > 15.

To complete the cycle, 15 > 21 and that can happen only by case 4: a b 15 > c d 21 means c < a < d > b, and so forth. A little trial and error, [and a conversation with Oskar!] gives this:

3 9 21 > 1 12 20 [5/9]

1 12 20 > 6 8 19 [5/9]

6 8 19 > 4 11 18 [5/9]

4 11 18 > 2 14 17 [5/9]

2 14 17 > 7 10 16 [5/9]

7 10 16 > 5 13 15 [5/9]

5 13 15 > 3 9 21 [5/9]

That is only one of the three cycles.

Here are all three.

21 beats 20 19 17

20 beats 19 18 16

19 beats 18 17 15

18 beats 17 16 21

17 beats 16 15 20

16 beats 15 21 19

15 beats 21 20 18

[Guest]

well done bonanova!