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 managed this with a hint from a friend at a key point.
Assume there's a solution.
Let a1-a6 = probability of rolling 1-6 using die 1 and b1-b6 = probability of rolling 1-6 using die 2.
P(2) = a1*b1 = 1/11.
P(12) = a6*b6 = 1/11.
(Hint from friend follows. Thanks, friend.)
P(7) = a1b6 + a6b1 + some positive terms = 1/11.
So a1b6 + a6b1 < 1/11
a1(1/11a6) + a6(1/11a1) < 1/11
a1/a6 + a6/a1 < 1.
This is not possible with positive values, so the assumption there is a solution is false.
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Posted by xdog
on 2011-01-21 12:47:56 |