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?

I assumed that the two dice have the same weights as each other.  I also assumed that 1 and 6 have the same probability and 2 and 5 have the same probability.  I dont know for sure that the minimum difference occurs under these conditions, but they seem reasonable to me.

The problem then has only two variables: 
x=the probability a single die rolls a 1 (or a 6)
y=the probability a single die rolls a 2 (or a 5)
and the probability of a 3 or 4 = .5-x-y

Now the probability of each total can be found:
P(2)=x^2
P(3)=2xy
P(4)=y^2+2x(.5-x-y)
P(5)=2y(.5-x-y)+2x(.5-x-y)
P(6)=(.5-x-y)^2+2y(.5-x-y)+2xy
P(7)=2(.5-x-y)^2+2y^2+2x^2
[P(8) through P(12) just mirror these]

A little playing around shows P(7) is generally the largest and P(2) or P(3) the smallest.

The optimum seems to be x=1/4, y=1/8, .5-x-y=1/8
P(2)=P(3)=1/16
P(4)= 5/64
P(5)= 3/32
P(6)= 7/64
P(7)= 3/16

P(7)-P(2)= 1/8

To summarize the weights of the sides: 1 and 6 come up 1/4 of the time; 2,3,4,5 come up 1/8 of the time.

Posted by Jer on 2011-01-20 18:11:52