I finished reading a book about Paul Erdös "The Man Who Loved Only Numbers" and I came to couple of questions which I need to check if I am understanding them well or not, one of them about what has become known as Happy Ending Problem.
What I understand that according to the formula
$$ 2^{n-2} + 1 $$
you can know the least numbers of vertices to be able to construct n sided-polygon. If that is right then e.g. let's take $$n = 5 \to 2^{5-2}+1 = 9$$ which means 9 vertices needed to guarantee a pentagon (5 sided-polygon). However, if you have even less vertices you are still able to construct a pentagon (it is not mentioned if it should be regular or irregular) so according to the illustration below we need just as much vertices as the needed-polygon requires.
Please correct me if I misunderstood the problem and feel free to post any link that may clarify the subject and add more details.
You should carefully read the Wikipedia entry you linked to.
Note that a pentagon here is a closed simple, i.e., not selfintersecting, circuit consisting of five segments and five vertices. It is neither assumed regular nor even convex.
It is a theorem that among $\geq9$ different points in the plane, assumed in general position, i.e., no three in a line, you can always select $5$ that form the vertices of a convex pentagon. On the other hand the devil can place $8$ points in the plane such that no convex pentagon can be selected from them. This is the case for the $8$ points in your figure. It follows that $9$ is the "Ramsey number" for this little geometric problem.
Erdös and Szekeres 1935 conjectured the following generalization: Given $\geq2^{n-2}+1$ points in the plane, assumed in general position, you can always select $n$ of them forming the vertices of a convex $n$-gon. In its full generality this conjecture is still open.