CareerCup

As stated equal probability for each sum.
Since 36 combinations are possible 36/12 = 3 each sum should occur 3 times for equal probability.
1 initially was occurring 0 times, so to occur 3 times there should be 3 zeros in second dice.
Now 12 should also occur 3 times, that can be only possible with 6,6 combinations
which means 6 should also be present 3 times in the other dice.

0 0 0 6 6 6

Ans. Label should be 0 0 0 6 6 6

I think this can be done by labeling 3 faces of the dice with 0 and remainning 3 with 6..

Without relabelling, what is the probability of getting numbers from 1 to 12?
I think it is 1/12 (1/6 X 1/6)
So, if relabelling is a must, then i would just shift it...
1 2 3 4 5 6 to 2 3 4 5 6 1
Does my logic look fit? :-S

Without relabelling, what is the probability of getting numbers from 1 to 12?
I think it is 1/12 (1/6 X 1/6)
So, if relabelling is a must, then i would just shift it...
1 2 3 4 5 6 to 2 3 4 5 6 1
Does my logic look fit? :-S

How stupid am I !!! :(
Did not read the question correctly... i 'oxygen' is right

Hmmmm. Well, as stated, the simplest solution is to just relabel the dice with all zeros. The problem states that the sum must be between 1 and 12. Well, from 1 to 6 is between 1 and 12. Since every number on the untampered dice is of equal probability, and the tampered dice is guaranteed to show 0, we have satisfied the equal probability part as well.

I realize that we are "playing" a bit with the problem definition. However, stating that the sum must be "between 1 and 12" IS ambiguous...