Unconditional probability of for a system microstate in canonical ensemble

Question

In Kardar's 'Statistical Physics of Particles', it is stated that the unconditional probability for a microstate $\mu_S$ of system $S$ (in a canonical ensemble made using a system $S$ and reservoir $R$) is obtained by $p(\mu_S) = \sum_{\{\mu_R\}}p(\mu_S\otimes\mu_R)$.

Then it is claimed that this probability is proportional to the number of reservoir states having energy $E_{tot} - \mathcal{H}(\mu_S)$. So, the unconditional probability of a given system microstate $\mu_S$ then becomes

$p(\mu_S) = \frac{\Omega_R(E_{tot} - \mathcal{H}(\mu_S))}{\Omega_{S\oplus R}(E_{tot})}$

My question is, wouldn't this last expression overestimate the unconditional probability of a given microstate, because in general many microstates may correspond to the same internal energy $E_S = \mathcal{H}(\mu_S)$?

(Sorry if the formatting is off. This is my first post)

Answer 1

Short answer, No.

Because in Cannonical ensemble, the microstates are labelled by their energy. So all the microstates with same energy are degenerate(meaning the system cannot distinguish between them). Thus technically speaking, from the viewpoint of the system all the microstate with same e=internal energy is just one state. And the probability you're calculating is the probability that the system is in one of these states.