Suppose we have triangle $ABC$ with area $k$ with a point $P$ chosen inside $ABC$. What is the expected area of triangle $PBC$?
I'm pretty sure if we let $P$ be the centroid we get $k/3$. Also, how would you solve this question for an $n$-sided polygon?
Triangle
Let the vertices of the triangle be ${\bf a}$, ${\bf b}$, ${\bf c}$. Then the map$${\bf g}:\quad (u,v)\mapsto\left\{\eqalign{x(u,v)&=a_1+u(b_1-a_1)+v(c_1-a_1) \cr y(u,v)&=a_2+u(b_2-a_2)+v(c_2-a_2)\ ,\cr}\right.\tag{1}$$ maps the arbitrary triangle from $(x,y)=(a_1,a_2),(b_1,b_2),(c_1,c_2)$ bijectively onto a right triangle with $(u,v)=(0,0),(1,0),(0,1)$ (make a sketch for visualization). Therefore the transformation formula for multiple integrals can be used, and we obtain $$\int_T f(x,y)\>{\rm d}(x,y)=\int_{T'}f\bigl(x(u,v),y(u,v)\bigr)\>\bigl|J_{\bf g}(u,v)\bigr|\ {\rm d}(u,v)\ .$$ The Jacobian determinant is a constant: From $(1)$ we obtain $$J_{\bf g}(u,v)=\det\left[\matrix{x_u&x_v\cr y_u&y_v\cr}\right]=(b_1-a_1)(c_2-a_2)-(c_1-a_1)(b_2-a_2)\ .$$ Therefore we can write $$\int_T f(x,y)\>{\rm d}(x,y)=\bigl|J_{\bf g}\bigr|\>\int_0^1\>\int_0^{1-u}f\bigl(x(u,v),y(u,v)\bigr) dv\ du\ .$$
The area of an arbitrary triangle ${\bf a},{\bf b}, {\bf p}$ is $f=1/2\cdot\text{base}\cdot \text{height}$. In $u,v$ coordinates the base is 1 and the height is $v$. For normalization the integral has to be divided by the total area that is $1/2$. The expected area of a random triangle ${\bf a},{\bf b}, {\bf p}$ is therefore
$$\mathbb{E}[A_{{\bf a},{\bf b}, {\bf p}}]=\frac{\frac{1}{2}}{\frac{1}{2}}\bigl|J_{\bf g}| \int_0^1 \int_0^{1-u} v dv\ du\ =\frac{1}{6}\bigl|(b_1-a_1)(c_2-a_2)-(c_1-a_1)(b_2-a_2)\bigr| \tag{2}$$
As the area of the triangle ${\bf a},{\bf b}, {\bf c}$ is
$$A_{{\bf a},{\bf b}, {\bf c}}=\frac{1}{2}\bigl|(b_1-a_1)(c_2-a_2)-(c_1-a_1)(b_2-a_2)\bigr|$$ the expected area of the random triangle is 1/3 of the area of the original triangle, independently which side is used as a base.
Polygon
The question for the polygon is not clearly defined. In any case it must be a convex polygon. If convexity is not present then some triangles lie not completely in the polygon. Also the OP has not defined if for any random triangle a random polygon side is select or always the same edge is selected. For the last case one has to split the n-sided polygon in n-2 triangles and then use (2) to sum up over every triangle.