Microcanonical ensemble confusion I am a bit confused about the difference between macrostate and microstate in the microcanonical ensemble. So I have read that for the microcanonical ensemble, the probabilities of each microstate are equal $$ p = 1/\Omega $$ where $\Omega$ is the number of microstates. For a given number of particles $N$ the number of microstates of particles distributed over discreet energy levels is given by 
$$\Omega = \frac{N!}{\prod_i n_i} $$
where $n_i$ is the number of particles in the $i$th energy level. Maximising $\Omega$ given the constraint that the particle number is constant $\sum_i n_i = N $ and the energy is constant $\sum_i \varepsilon_i n_i = E$ gives
$$ p_i  \propto e^{-\varepsilon_i/kT}$$
This gives a probability for a particle to be in the $i$th energy level. I thought if we were in the microcanonical ensemble all probabilities are equal?
Thanks
 A: Here is how to derive the microcanonical and the canonical distributions. In all cases
$$
   \Omega(\{n_i\}) = \frac{N!}{\prod_i n_i!}
$$
Microcanonical 
All $n_i$ in all distributions of the ensemble have the same energy: $E_i=E$ for all $i$. The problem is to find the distribution that maximizes $\Omega$ under the sole constraint
$$
   \sum_i n_i = N
$$
The solution is
$$
   \frac{n_i^*}{N} = \frac{1}{\Omega} \doteq p_i
$$
is the multiplicity of distirbution $\{n_i\}$.
Canonical 
We don't know the energy if microstate $i$ but we know that the average energy per microstate is the same in all distributions of the ensemble: $E_\text{tot}/N = \bar E$ for all distributions $\{n_i\}$. The problem now is to find $\{n_i\}$ that maximizes $\Omega$ under the following two conditions:
$$
   \sum_i n_i = N,\quad
   \sum_i n_i E_i = N \bar E
$$
The solution now is
$$
  \frac{n_i^*}{N} = \frac{e^{-\beta E_i}}{Q} \doteq p_i, 
$$
The key difference is that all microstates in the microcanonical ensemble have the same energy while in the canonical ensemble the have the same energy only on average. 
