Suppose my system has $N$ particles, and I want to find a distribution for $n_i$, the number of particles in the $\epsilon_i$ energy state.
What I do know is the boltzmann probability, which tells me the probability of a single particle being in a certain energy state, and is denoted using $p_i$. I also know the expected number of particles in a system, given by $p_iN=\mu$.
According to some people that I asked, the distribution of particles would follow a binomial distribution, where the number of particles in the state would be a random variable. Hence, it would be given by :
$$p(n_i)={^{N}C_{n_i}}\space p_i^{n_i}(1-p_i)^{N-n_i}$$
Using this I can find the probability of each possible occupation number, through a binomial distribution.
Physically, this is equivalent to picking $N$ particles out of the system and counting the number of particles in $n_i$, and then repeating this experiment many many times and checking the frequency of each value of $n_i$. Because of fluctuations, we can be sure that $n_i$ keeps on changing every time I do this trial. It is basically like the distribution of heads in a $100$ coin tosses.
However, there is a second distribution that came to my mind. We know that the system can have $n_i$ particles, where $n_i$ ranges from $0$ to $N$. So, we consider $N+1$ different systems, labelled from $0$ to $N$, where the label denotes the number of particles in $\epsilon_i$ state of this system at a particular timestamp. Now what we do is, we check the probability of each of these systems of being the original system. Since these systems have a fixed number of $n_i$ each, we are indirectly getting the distribution of the number of particles $n_i$ of the original system.
So, what we check is, the probability of each of these systems, to give us a $p_i$ probability of picking up a particle at random, and finding it is in the $i$ state. In a sense, this is probability of probabilities.
How do we do that ?
Well, you know the expected number of particles in the energy state is given by $\mu$. So, we take each of these systems, and take out $N$ particles with replacement, and count the number of particles in that state, and repeat this experiment many many times. Whenever $\mu$ out of $N$ particles are in the desired state, we consider it a success. We check the frequency of success for each of these $N+1$ systems. This would give us a probability distribution of which system is how likely to give us a $p_i$ chance of picking up a random particle and finding it in a particular state. This distribution would be given by :
$$p(n_i)={^NC_\mu} \space\space(\frac{n_i}{N})^\mu \space\space (1-\frac{n_i}{N})^{N-\mu}$$
This distribution is not normalized, and looks a lot like the Binomial distribution. However, this takes into account, that for each system, the probability of getting a certain particle to be in a certain state, would be different. Moreover, the physical analogy also makes more sense in this case.
As discussed in my previous question, the boltzmann distribution is more like an estimator of the true probability of finding a particle in a certain state, as written by @Roger Vadim in this answer. So, instead of the distribution of the particles being a perfect boltzmann distribution, should it not be the distribution of different systems based on how likely the probability of getting a single particle in $i$ state in that system matches the boltzmann probability of the same.
In other words, shouldn't the second distribution be more accurate ?
I've checked that for large numbers, the two distributions produce more or less the same result, but I wanted to know which one is more accurate.
Best Answer
Counting how many times the system's actual $n_i$ value comes out to its expected value, as you are doing in the second part, is finding the distribution for the estimator of another distribution's parameter. This is where all the business about T, F and $\chi^2$ distributions comes from in statistics that is supposedly focused on Gaussian random variables.For the distribution of $n_i$ which you are actually looking for, binomial is correct. I will take $i = 1$ without loss of generality and write it as \begin{equation} P(n_1) = \binom{N}{n_1} \left ( \frac{e^{-\beta E_1}}{Z_1} \right )^{n_1} \left ( 1 - \frac{e^{-\beta E_1}}{Z_1} \right )^{N - n_1}. \end{equation} To check this, we can go back to the brute force definition. What is the probability that there are $n_1$ particles in level $1$? It's the probability that the system is in a microstate with $n_1$ particles in level $1$, summed over all such microstates. This means we take the $N$ particle partition function \begin{equation} Z_N = Z_1^N = \sum_{n_1 + \dots + n_k = N} \binom{N}{n_1, \dots, n_k} e^{-\beta(n_1 E_1 + \dots + n_k E_k)} \end{equation} and perform a very similar sum where $n_1$ is held fixed. \begin{align} P(n_1) &= Z_N^{-1} \sum_{n_2 + \dots + n_k = N - n_1} \binom{N}{n_1, \dots, n_k} e^{-\beta(n_1 E_1 + \dots + n_k E_k)} \\ &= Z_N^{-1} e^{-\beta n_1 E_1} \binom{N}{n_1} \sum_{n_2 + \dots + n_k = N - n_1} \binom{N - n_1}{n_2, \dots, n_k} e^{-\beta(n_2 E_2 + \dots + n_k E_k)} \\ &= Z_N^{-1} e^{-\beta n_1 E_1} \binom{N}{n_1} \left ( Z_1 - e^{-\beta E_1} \right )^{N - n_1} \\ &= Z_1^{-N} e^{-\beta n_1 E_1} \binom{N}{n_1} \left ( Z_1 - e^{-\beta E_1} \right )^{N - n_1} \end{align} The claim now follows. To go from line 2 to 3, we had to remember that the level $E_1$ was no longer present, making the combinatorial sum equal to a power of a corrected $Z_1$.
Discussion of the alternative scenario
After sleeping on it, I think the experiment you describe is a little more puzzling than constructing an estimator. First, when I see
$${^NC_\mu} \space\space(\frac{n_i}{N})^\mu \space\space (1-\frac{n_i}{N})^{N-\mu}$$
that is clearly the probability for something to happen $\mu$ times. Yet you are interpreting it as the probability for a certain variable to be equal to $n_i$ by saying that it needs to be normalized by hand. This is a red flag. It is common for physics arguments to only tell us "what probabilities are proportional to" but ideas like the law of large numbers are mathematical, not physical.
So I will try to make an analogy with coins. It will actually be an analogy to the formula
$${^NC_\mu} \space\space(\frac{1}{N})^\mu \space\space (1-\frac{1}{N})^{N-\mu}$$
because I don't know how you get $n_i$ to appear in the way it did. One of the $N$ copies has $n_i$ particles in state $i$. Not $n_i$ of the $N$ copies.
Anyway, suppose you bought a coin where heads comes up with probability $p$ and tails with probability $1 - p$. After $N$ flips, $\mu = Np$ is the expected number of heads. Now suppose you go to the same store and buy the same coin again. With your two identical coins, you paint over the head of one and the tail of the other and put them in a hat.
You can reach into this hat and draw coins with replacement many times and build up a distribution. But there is one coin of each type so this will be the distribution of a fair coin. I.e. every draw from the hat is like a 50-50 flip, not a $p$ biased flip. So nothing you do with the hat will give information about the probability of heads coming up a certain number of times in the original problem.