Question 1021315
There are {{{6^3}}} or {{{216}}} possible outcomes.
Six of them are three of a kind
111, 222, 333, 444, 555, 666
Half of them (108) have a sum that is odd, half of them have a sum that is even.
Three of them are three of a kind and have an odd sum (111,333,555).
So,
{{{P=(6+108-3)/216}}}
{{{P=111/216}}}
{{{P=37/72}}}