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Reloading the Dice (Posted on 2011-01-20) Difficulty: 3 of 5
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)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts A further improvement | Comment 3 of 10 |
(In reply to possible approach by broll)

broll's idea is definitely better than mine.  P(7)=2/12 and all the others are 1/12 for a difference of 1/12.

We can improve upon this if we let the first die roll only one and six with equal probability, but refine the second die.  It rolls 1 and 6 with probability 1/10 and 2,3,4,5 each with probability 1/5.

P(1)=P(12)=1/20
all the others are 1/10 for a difference of 1/20.

Edited on January 24, 2011, 4:12 pm
  Posted by Jer on 2011-01-20 20:00:10

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