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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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Physical intervention thoughts. | Comment 8 of 10 |
Many years ago I had occasion to play a simulation game.  I wasn't really serious when I quipped that the die was loaded.  In actual fact it was but not in the conventionally accepted way; one of two surfaces had been very slightly curved to advantage certain events.

We notice that the sums of 2 and 6 occur least while 7 has the greatest frequency.

Might we not turn these around somewhat by actually physically weighting each face in favour of the lower frequencies?

Arbitrarily I'm assigning a weight of 3 to the faces of 1 and 6, one of 2 to faces 2 and 5 and a 3 weight to faces 3 and 4.

I'm not going to try to analyse this further (the weighting system may to need to be adjusted). 

If dice throws could be programmed as in "Snake-Eyes Joe" one could manipulate the probability.  I don't have a text directly available, but I recall seeing where one call skew events by placing conditions upon the seed value returned by a random number generator.

Sure, neither of these approaches would yield uniformity "across the board" in terms of integer values, but would be possible, if the weighting systems suggested were not in a simple sequential order, and not even integers?

  Posted by brianjn on 2011-01-21 19:37:20
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