Degenerate states, Boltzmann factor and statistical mechanics

The probability of finding a particle with energy $$E$$ according to Maxwell-Boltzmann distribution is: $$P(E) =\frac{1}{Z}g(E)e^{\frac{-E}{k_BT}} \qquad eq(1)$$ where g(E) is the degeneracy of the energy level $$E$$.

However, the deduction of this formula is the following:

-Using the following definition of entropy: $$S=k_Bln(\Omega(s))$$, where $$\Omega$$ is the multiplicity of a macrostate $$s$$.

-Considering also a system in contact with a large reservoir, and two possible states: $$s_1$$ and $$s_2$$. Then, the ratio of their probabilities is the ratio of the multiplicities of the reservoir that make the system be in that state.

$$\frac{P(s_1)}{P(s_2)} = \frac{\Omega_R(s_1)}{\Omega_R(s_2)}=\frac{e^{S_R(s_1)}}{e^{S_R(s_2)}}$$

For example, if $$\Omega_R(s_1)=10$$ and $$\Omega_R(s_2)=5$$, then its twice as likely to find the system in $$s_1$$ than $$s_2$$.

Then by using thermodynamic:

$$dS_R = \frac{1}{T}dU_R$$ $$dU_R = -dE$$

where E is the energy of the system and U the reservoir. Using these equations one can easily get to equation 1.

But, my question is the following:

Why in equation (1) there is the degeneracy term $$g(E)$$? Isn't that being accounted in the deduction when they count the multiplicity?