Number of "3-edge" triangles in a planar triangulation
I think only those vertices corresponding to triangles sharing an edge with boundary or holes of the polygon will be of degree less than 3. Now the number of triangles in the triangulation is fixed. So you can always get the exact ratio of the number of degree 3 vertices for a polygon.

If the polygon has $k$ vertices and $l$ holes vertices and the number of triangles in the total triangulation is $N$ 
then the upper bound should be $(N-K-l)/N$.