I have an unfair die whose sides are known to have the following probabilities:{.02,.04,.08,.12,.24,.5}
Assign the numbers {1,2,3,4,5,6} to these sides so the expected roll is the same as for a fair die.

Can you solve this without brute force?

(In reply to The answer is ... by Bob Smith)

No.

I looked at this like 6 kids on a see-saw (Or forces on a lever). The farther from the center value (3.5) the bigger the force applied.  Thus the .5 should go near the center and the .24 near the center on the opposite side and brute force the rest of the numbers until it works.

 1       2     3      4     5      6

.04   .08   .50   .12   .24   .02

.04   .16   1.50  .48  1.20  .12   = 3.5

or mirrored

6      5      4       3       2      1

.04   .08   .50   .12   .24   .02

.24   .40   2.00  .36   .48   .02   = 3.5

Posted by Leming on 2006-04-28 13:39:07