Dicey puzzle: Solution

The full puzzle statement may be found here.

In short "What are the odds that n=2 v/s n=3 dice are rolled, given that the sum is 7?"

Solution:

For n=2, there are 6 ways of rolling a 7 (1+6, 2+5, 3+4, 4+3, 5+2, and 6+1), out of a total of 6^2=36 total outcomes.

Therefore p(sum = 7 | n = 2) = 6/36 = 36/216.

For n = 3, there are 15 ways of rolling a 7 (1+1+5, 1+2+4, 1+3+3, 1+4+2,1+5+5, 2+1+4, 2+2+3, 2+3+2, 2+4+1, 3+1+3, 3+2+2, 3+3+1, 4+1+2, 4+2+1, 5+1+1), out of a total of 6^3 = 216 total outcomes.

Therefore p(sum = 7 | n = 3) = 15/216.

Thus, the odds of n = 2 v/s n = 3 are 36/15. That is it is about 2.5 times more likely that n = 2.

What happened? There were more ways of getting a 7 with n = 3?