So, Simon Wadsley’s comment clearly answers this question but a hypothetical future user will not have his/her eye drawn to that answer. That’s why I’m posting this, which is all Simon’s idea, in the hope that the OP will come back on at some point and accept this answer (thus preventing this problem from being bumped up to the front-page by the Mathoverflowbot). If the OP is reading this, please click the check mark next to this box so it'll count as being answered. I'm giving a CW answer so I don't get rep (in line with the recommended procedure on [meta][1])

The homological and cohomological dimensions of a group do NOT have to agree. As you point out, if they did agree then the projective and injective dimensions of $\mathbb{Z}$ as a $\mathbb{Z}[G]$ module would agree (this is basically just the definition of Ext). An example that this could fail, take $G$ to be the trivial group. Then the projective dimension of $\mathbb{Z}$ as a $\mathbb{Z}$-module is zero because any ring is projective over itself. But the injective dimension is not zero because $\mathbb{Z}$ is not a divisible abelian group. Indeed, the injective dimension is 1, as can be seen from the fact that $\mathbb{Z}$ is a PID and hence has global dimension 1. Or you can just write down an injective resolution. Or you can read Dummit-Foote for their treatment.

  [1]: http://mathoverflow.tqft.net/discussion/328/should-we-do-anything-if-a-question-is-answered-well-in-the-comments/#Item_0