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Phil
Phil
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I've heard this brainteaser before, and usually its phrased that 2-12 must come up equally likely (no comment about other sums). With this formulation (or interpretation) it becomes possible. Namely, {0,0,0,36,36,36} and {1,2,3,4,5,6}. In this case, you can also generate the sum 1, but 2-12 are equally likely (1-12 are equally likely). Without allowing for other sums I do suspect its impossible (and looks like proofs have been given). I arrived at this answer by noting we are asking for equal probability for 11 events that come from 36 (6*6) possible outcomes, which immediately seems unlikely. However equal probability for 12 events from 36 outcomes is far more manageable :)

Phil

I've heard this brainteaser before, and usually its phrased that 2-12 must come up equally likely (no comment about other sums). With this formulation (or interpretation) it becomes possible. Namely, {0,0,0,3,3,3} and {1,2,3,4,5,6}. In this case, you can also generate the sum 1, but 2-12 are equally likely (1-12 are equally likely). Without allowing for other sums I do suspect its impossible (and looks like proofs have been given). I arrived at this answer by noting we are asking for equal probability for 11 events that come from 36 (6*6) possible outcomes, which immediately seems unlikely. However equal probability for 12 events from 36 outcomes is far more manageable :)

Phil

I've heard this brainteaser before, and usually its phrased that 2-12 must come up equally likely (no comment about other sums). With this formulation (or interpretation) it becomes possible. Namely, {0,0,0,6,6,6} and {1,2,3,4,5,6}. In this case, you can also generate the sum 1, but 2-12 are equally likely (1-12 are equally likely). Without allowing for other sums I do suspect its impossible (and looks like proofs have been given). I arrived at this answer by noting we are asking for equal probability for 11 events that come from 36 (6*6) possible outcomes, which immediately seems unlikely. However equal probability for 12 events from 36 outcomes is far more manageable :)

Phil

Source Link
Phil
Phil
  • 31
  • 2

I've heard this brainteaser before, and usually its phrased that 2-12 must come up equally likely (no comment about other sums). With this formulation (or interpretation) it becomes possible. Namely, {0,0,0,3,3,3} and {1,2,3,4,5,6}. In this case, you can also generate the sum 1, but 2-12 are equally likely (1-12 are equally likely). Without allowing for other sums I do suspect its impossible (and looks like proofs have been given). I arrived at this answer by noting we are asking for equal probability for 11 events that come from 36 (6*6) possible outcomes, which immediately seems unlikely. However equal probability for 12 events from 36 outcomes is far more manageable :)

Phil