All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

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.)
 Complete proof finished. Comment 10 of 10 |
(In reply to re(2): A further improvement by Jer)

With the result that the ratio is at least 2, we can show my weightings hits a lower bound and is thus optimal.

P(sum=2) = P(sum=12)= p
P(sum=7) = 2p

The difference is at least p so we seek to minimize p.

It would be higher if one of the other eight sums had a probability less than p or greater than 2p.  So we set them all at 2p.

The probability of each of the nine sums from 3 to 11 is 2p, for a grand total of 20p.

Being a probability distribution this must equal 1.
20p=1
p=1/20

So the minimum difference is 1/20.

 Posted by Jer on 2011-01-24 16:44:44

 Search: Search body:
Forums (0)