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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)

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re: A further improvement | Comment 5 of 10 |
(In reply to A further improvement by Jer)

Jer,

Yes indeed, but the ratio between the most likely sum and the least likely sum remains 2 (twice 1/12 is 1/6 and twice 1/20 is 1/10). Is there any way to further reduce that ratio?

One thought that occurs to me is to use a coin for the first die (which is what it effectively is anyway) with heads=1 tails=6. Then by convention if tails are thrown and the die shows 1 the result is a rethrow until some other result is obtained. This arrangement would produce all outcomes with equal probability of 1/11, and does not require any fiddling with weights.


  Posted by broll on 2011-01-21 03:07:04
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