Question 346689
There are {{{36}}} total outcomes({{{6^2}}}).
The rolls that sum to {{{7}}} are: 
({{{1}}},{{{6}}})
({{{2}}},{{{5}}})
({{{3}}},{{{4}}})
({{{4}}},{{{3}}})
({{{5}}},{{{2}}})
({{{6}}},{{{1}}})
That makes {{{6}}} possible outcomes.
P(sum=7)={{{6/36=1/6}}}.