A pair of dice, when rolled, produces sums of 2 to 12, with varying probabilities. Can the dice be reweighted (each face assigned a probability other than 1/6) so that all 11 sums occur with the same frequency?
If so how, if not how close can the difference between the least and most likely sum be made?

Just some ideas to get things going:-

Reweighting.

The answer seems to be no. Consider this simplification:    
Two coins are black on one side, white on the other.    
The probability of two black is pq.    
the probability of two white is (1-p)(1-q)    
(we can just ignore the mixtures).   
Now we need pq=(1-p)(1-q)=1/3    
But if so then:  p = 1/6 (3-3^(1/2)i), q = 1/6 (3+3^(1/2)i) which suggests that there is no real solution to the problem.

Least difference

Since we are only interested in the sums, not the 'realism' of the dice, assume that one die is weighted so that it always rolls a one or a six, with equal probability, while the other is unweighted. Then 7 will occur just twice as often as any other number, all of which occur with the same frequency.

Posted by broll on 2011-01-20 17:33:23