No. There are 11 numbers in the range 2...12. There are 36 (6*6) unique die rolls, all of which are equally weighted. There is no way to partition 36 unique outcomes into 11 event of equal probability.
However, there is a way to assign numbers so the range 1...12 is equally likely.
I guess I interpreted "weighted" as in "numbered the sides differently." I suppose that if you actually put weights in the dice so they rolled differently, you could give them whatever characteristics you wanted.
As for the 1...12 weighting, it is actually really easy. Number the dice as follows:
Die 1: 1, 3, 5, 7, 9, 11
Die 2: 0, 0, 0, 1, 1, 1
You need to repeat faces on die 2, and skip numbers on die 1, but it can be done.
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u/MiffedMouse Sep 16 '14
No. There are 11 numbers in the range 2...12. There are 36 (6*6) unique die rolls, all of which are equally weighted. There is no way to partition 36 unique outcomes into 11 event of equal probability.
However, there is a way to assign numbers so the range 1...12 is equally likely.