http://statweb.stanford.edu/~susan/courses/s60/split/node65.html
Okay, so basically I was playing a board game and got into a debate on if a 3-11 is more likely than a 2-12. From my perspective I saw that a 2 was able to be a achieved by a 1 and a 1 being rolled on each die, but either of the two die may result in a 1. (Same applies for two 6's) A three can only be obtained by a 2 and a 1 (11 6 + 5), but either of the two die may be a 6. So when I looked up this chart it confused me further that the person I was debating claimed to be correct, but since the sum of the numbers is based on a combination and not a permutation I believe that the rolling of a 2,3,11, and 12 are all equally likely. I feel like I'm wrong in this, but can't come up with a logical reason as to why.
Color one die red and the other green.
There's one way to get a $2$ $(1R, 1G)$ and one way to get a $12$ $(6R, 6G)$.
There are two ways to get $3$ $\{(1R, 2G), (1G, 2R)\}$ and similarly for $11$.
Therefore, getting a $3$ or an $11$ is twice as likely as getting a $2$ or $12$.