The classifying space of a group $G$ is given by taking a contractible space $E$ equipped with a free $G$-action, and looking at the quotient, which we dub $BG$. The homotopy type of this space (and thus its cohomology) depend only on $G$, and this gives us one definition of group cohomology.

Now, we can also look at the orbifold $[pt/G]$, and compute its orbifold cohomology (as its regular cohomology is rather uninteresting). For finite groups $G$, we have the isomorphism (see Adem-Leida-Ruan) $$ H^*_{orb}\big([pt/G],\mathbb{C}\big) \cong Z(\mathbb{C}G). $$

We can easily see that for finite groups the cohomology obtained is not the same as we would get from the usual group cohomology.

Why? I understand that the actual constructions are very different, but this seems very unsatisfying to me. Morally this space is a contractible one modulo a free group action, and it is a model of $BG$ if we look at it in the appropriate way, so why shouldn't their cohomologies be the same?