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L'lanmal
Daedalian Member

 Posted: Sun Dec 02, 2012 1:00 pm    Post subject: 1 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
Amb the Hitched.

 Posted: Sun Dec 02, 2012 6:04 pm    Post subject: 2 0 - 6, or, 1-6?
Trojan Horse
Daedalian Member

 Posted: Sun Dec 02, 2012 8:28 pm    Post subject: 3 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
Daedalian Member

 Posted: Sun Dec 02, 2012 9:28 pm    Post subject: 4 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
Daedalian Member

 Posted: Fri Dec 07, 2012 3:52 pm    Post subject: 5 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
Daedalian Member

Posted: Sun Dec 09, 2012 1:41 pm    Post subject: 6

 lostdummy wrote: 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

This is my solution for #1 too.

I'm not saying you are wrong, but the "no duplicated numbers" portion allows for the numbers 0-12 rather than just 0-6, and also permits (in fact, requires) the occasional tie. So it doesn't follow immediately.

The implied question is whether Buffet selected those requirements in order to allow for a solution with no repeated sides on the faces of the dice. Or just to obfuscate the simple solution.

I think (spoiler)you can avoid repeats on several, but not all the dice. But I wanted to see if someone else could manage it.
lostdummy
Daedalian Member

 Posted: Tue Dec 11, 2012 4:22 pm    Post subject: 7 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
Daedalian Member

 Posted: Wed Dec 12, 2012 11:39 am    Post subject: 8 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.
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