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

Home > Probability
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.)
proof of no exact solution | Comment 6 of 10 |

I managed this with a hint from a friend at a key point.

Assume there's a solution.

Let a1-a6 = probability of rolling 1-6 using die 1 and b1-b6 = probability of rolling 1-6 using die 2.

P(2) = a1*b1 = 1/11.

P(12) = a6*b6 = 1/11.

(Hint from friend follows. Thanks, friend.)

P(7) = a1b6 + a6b1 + some positive terms = 1/11.

So a1b6 + a6b1 < 1/11

a1(1/11a6) + a6(1/11a1) < 1/11

a1/a6 + a6/a1 < 1.

This is not possible with positive values, so the assumption there is a solution is false.


  Posted by xdog on 2011-01-21 12:47:56
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information