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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?

 See The Solution Submitted by Brian Smith Rating: 3.0000 (1 votes)

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 possible solution | Comment 2 of 10 |
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

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