perplexus.info :: Probability : Reloading the Dice

Reloading the Dice (Posted on 2011-01-20)

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?

Submitted by Brian Smith

Rating: 3.0000 (1 votes)

Solution: (Hide)

Die 1: 1=1/2, 2=0, 3=0, 4=0, 5=0, 6=1/2
Die 2: 1=1/10, 2=1/5, 3=1/5, 4=1/5, 5=1/5, 6=1/10
2 and 12 occur with 1/20 probability and 3 through 11 occur with 1/10 probability.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
Complete proof finished.	Jer	2011-01-24 16:44:44
re(2): A further improvement	Jer	2011-01-24 16:31:50
Physical intervention thoughts.	brianjn	2011-01-21 19:37:20
Nice proof!	Steve Herman	2011-01-21 16:21:36
proof of no exact solution	xdog	2011-01-21 12:47:56
re: A further improvement	broll	2011-01-21 03:07:04
degrees of freedom	Steve Herman	2011-01-21 01:36:23
A further improvement	Jer	2011-01-20 20:00:10
possible solution	Jer	2011-01-20 18:11:52
possible approach	broll	2011-01-20 17:33:23

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