Probability Question-May be challenging?

Question

"we have discovered that by rolling 2 six sided dices and taking the sum all possibilities are not equally probable. 7s come far too often and 2s and 12s are rare
.....Your job is to relabel the 2 dice so that each of the sum from 2-12 is equally probable
what number should be on each dice?

Answer 1

Interesting problem. I think that you should look at converting 4s and 3s to smaller and larger numbers. I'm really not sure that you can get all the sums from 2 to 12 to have equal probability.

Convert 3s to 1s and 4s to 6s and check.

Btw, do the two dice have to be the same once you're finished relabeling them?

Answer 2

I don't see how this is possible. There are eleven numbers 2 thru 12. That means the probability for each number coming up would be 1/11. There are 6² = 36 possible rolls of the two dice. For probability to be 1/11 we need:

x/36 = 1/11
x = 36/11

But x is not an integer. There is no way each number can have 36/11 different ways to be rolled. I suppose you could have three ways to roll a number outside of 2 thru 12 leaving 33/11 = 3 ways for each of the numbers to be rolled, but I don't think that the question contemplates that kind of answer.

Answer 3

Easy

Put the numbers 20 - 25 on each die.

The probability of the sum being 2-12 is zero for each possibility.

Equally probable

QED

Answer 4

I think its not possible. i justify my answer as follows
no of sums from 2 to 12 is 11
no of dice combo for two six-sided face is 36
36 is not divisible by 11
hence they cannot be equiprobable