Q:

How many reflex angles can a polygon have?

I think the first of the polygons that can have a reflex angle is the pentagon. For a hexagon, a maximum of 2 reflex angles is possible. 
   I tried to draw many concave polygons to find out a relation, but I got stuck. Is there a formula for this?   

A:

The maximum number of reflex angles a simply connected $n-$gon can have is $n-3$.  This comes from the fact that the sum of the internal angles of a simply connected $n-$gon is $(n-2)180^\circ$ so there can be at most $n-3$ angles greater than $180^\circ$.  To prove we can achieve this, take an equilateral triangle and bend one edge inward to make $n-3$ angles, all slightly greater than $180^\circ$.  Below is a heptagon with four reflex angles.  I think it is clear how to get as many reflex angles you want up to $n-3$

A:

In non-Euclidean space, a digon can have both of its interior angles be reflex angles. On a sphere, it could look like a pac-man. This can be extended to any polygon which means you can have all sorts of polygons that are entirely reflex angles in large quantity.
In Euclidean space with $n_s$ polygon sides, you're limited to $n_s-3$ reflex angles. Why? First, consider that a reflex angle is between 180 and 360 degrees. Then consider the simplest possible polygon: a triangle. In Euclidean space, it is impossible for any of a triangle's interior angles to be larger than $180^\circ$, given the sum interior angle of a polygon is: $(n_s – 2)180$
If we take the lower limit of the reflex angle as $180^\circ$ in order to fit the most reflex angles possible into our polygon, we can then express this as an inequality. Given a number of reflex angles $n_r$, 
$n_r 180 \lt (n_s-2) 180 \to n_r \lt n_s-2$
As we can only have integer numbers of angles, that means the maximum solution that doesn't violate the inequality is an increment:
$n_r \lt n_s-2$
∴ $n_r + 1 = n_s-2 \to n_r = n_s-3$

