A Deal with Dice

[L'lanmal]

The book Warren Buffet Speaks: Wit and Wisdom from the World's Greatest Investor describes a wager that Warren Buffet offered Bill Gates.

Warren Buffet showed Gates four "unusual dice", with the faces labeled with the numbers 0-12. Each person would choose one die, and then the player who rolled the higher number (all ties rerolled) would win. Die A had an 11/17 chance of beating die B, die B had an 11/17 chance of beating C, die C had an 11/17 chance of beating D, and die D had an 11/17 chance of beating A. Gates examined the dice, then asked Buffet to pick first.

We will presume the dice were cubic, as their shape wasn't mentioned.
1) How do you label four dice to meet the slightly stricter requirements that only the numbers 0-6 are used, and the odds of winning are 2/3?
2) Modify your answer to 1 to create dice meeting the original restrictions.
3) Is it possible to meet the original restrictions and also not have the same number appear more than once on the same die?

[Amb]

0 - 6, or, 1-6?

[Trojan Horse]

Assuming L'lanmal has in mind the same solution that I've seen before, it's 0-6. There are indeed some 0s in the solution.

[L'lanmal]

Yes, I used 0-6. If you can find a way using only 6 distinct numbers, more power to you.

The real puzzle in my mind is why Buffet used 0-12.

[lostdummy]

Well, this is only solution for #1 that I found:
(3,3,3,3,3,3) -> (2,2,2,2,6,6)-> (1,1,1,5,5,5)-> (0,0,4,4,4,4)-> (3,3,3,3,3,3)
so I guess there was no solution without zeros or duplicated numbers ;p

[L'lanmal]

[lostdummy]

yes, my answer about "no duplicates" was referring to 0-6 case, which was only one I tried ;p

in meantime I tried 0-12 too, and I can confirm that there isn't any "no duplicates" solution. There are "no zeros" solutions, like :
(1,1,8,8,9,10)-> (5,5,6,6,7,8)-> (3,4,4,5,11,12)-> (2,2,3,10,10,11)

But no solution without duplicates. Depending how you define "duplicates", you could search for "minimal duplicates":

1) duplicate = count any side that repeats. Minimum I could find is 5 duplicated sides:
(0,1,7,7,8,9)-> (5,5,6,6,6,7)-> (3,4,4,5,11,12)-> (1,2,3,9,10,11)

2) duplicate = count any dice that has duplicated non-zero numbers. I didn't search for minimal here, but some example with only 2 dice with duplicate non-zero sides is:
(0,0,7,8,9,10)-> (5,5,6,6,7,7)-> (2,3,4,5,11,12)-> (1,1,2,10,10,11)

3) duplicate= count any dice that has duplicated any number (including zero). I didn't search for these at all, but both examples above are with 1 non-duplicate dice. Probably there are solutions with 2 or more ...

[lostdummy]

I found solution that have two dice without repeated side and zeros, and also only one dice with zeros. I don't think solution with three 'non-repeating' dice exists.