Yeah doing a brute force count of the possibilities the (6,6) option is double counted by the 1/6 + 1/6 method as two distinct possibilities. Or that's what it

Yeah doing a brute force count of the possibilities the (6,6) option is double counted by the 1/6 + 1/6 method as two distinct possibilities. Or that's what it looks like is happening. Been too long since my college combinatorics class.