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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.)
re(2): A further improvement | Comment 9 of 10 |
(In reply to re: A further improvement by broll)

The ratio of 2 is a minimum:

P(die 1 rolls 1) = a
P(die 1 rolls 6) = b
P(die 2 rolls 1) = x
P(die 2 rolls 6) = y

P(sum=2) = ax
P(sum=12) = by
P(sum=7) ≥ ay+bx (With equality if there is no other way to get a 7)

The ratio sought: P(sum=7)/P(sum=2) ≥ (ay+bx)/(ax) = y/x + b/a

Case 1.  Assume ax = by
a/b = y/x
The ratio becomes a/b + b/a
Which is always greater than or equal to 2.
It equals 2 when a=b, hence x=y which is the case in our solutions.

Case 2.  ax > by
a/b > y/x
The ratio becomes even bigger when we substitute ax for by.

Case 3. ax < by
same as above.

  Posted by Jer on 2011-01-24 16:31:50

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