phil1882 Posted September 11, 2012 Report Share Posted September 11, 2012 it's possible to construct a set of 3 dice, such that no matter what dice your opponent picks first, the dice you pick will beat it. 1v2 1 4 4 4 4 4 2 2 1 1 1 1 1 2 2 1 1 1 1 1 2 2 1 1 1 1 1 5 2 2 2 2 2 2 5 2 2 2 2 2 2 5 2 2 2 2 2 2 2v3 3 3 3 3 3 6 1 3 3 3 3 3 3 4 2 2 2 2 2 3 4 2 2 2 2 2 3 4 2 2 2 2 2 3 4 2 2 2 2 2 3 4 2 2 2 2 2 3 3v1 2 2 2 5 5 5 3 3 3 3 1 1 1 3 3 3 3 1 1 1 3 3 3 3 1 1 1 3 3 3 3 1 1 1 3 3 3 3 1 1 1 6 3 3 3 3 3 3 is it possible to construct a set of 5 dice such that the total of any two dice can be beaten by the total of two other dice? Quote Link to comment Share on other sites More sharing options...
0 Rainman Posted September 12, 2012 Report Share Posted September 12, 2012 Yes! Doing the math was hard work though. Die 1: 1 1 4 5 5 5 Die 2: 1 4 4 4 4 4 Die 3: 2 2 2 5 5 5 Die 4: 3 3 3 3 3 6 Die 5: 3 3 3 4 4 4 1,2 is beaten by 4,5. 1,3 is beaten by 2,4. 1,4 is beaten by 2,3. 1,5 is beaten by 2,3. 2,3 is beaten by 4,5. 2,4 is beaten by 1,5. 2,5 is beaten by 3,4. 3,4 is beaten by 1,5. 3,5 is beaten by 1,2. 4,5 is beaten by 1,3. Quote Link to comment Share on other sites More sharing options...
0 phil1882 Posted September 12, 2012 Author Report Share Posted September 12, 2012 nicely done rain man - A+ Quote Link to comment Share on other sites More sharing options...
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phil1882
it's possible to construct a set of 3 dice, such that no matter what dice your opponent picks first, the dice you pick will beat it.
is it possible to construct a set of 5 dice such that the total of any two dice can be beaten by the total of two other dice?
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