## Expedia Interview Question

Software Engineer / DevelopersHmmmm. 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...

"Me" Next time u appear for a written and wen u see this question.. Dont answer it.. Instead write on d answer sheet that question is ambiguous..

As stated equal probability for each sum.

- PR October 07, 2010Since 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