Background
Here is first some vocabulary:
Diploid: phase in the life cycle where the individuals carry two chromosomes of each type, just like in humans (exception of the sexual chromosomes).
allele: gene variant
locus: position in the genome
autosome: chromosome that is not a sexual chromosome
Hardy-Weinberg rule
In the first introductory courses in evolutionary biology, students usually learn about the Hardy-Weinberg rule.
Imagine an infinite (infinite number of individuals) diploid population with overlapping generation and where mating is random (individuals don't chose their partner). Given $x$ the frequency of the allele $A$ and $1-x$ the frequency of the allele $B$ at a bi-allelic locus located on a autosome, the frequency of the genotypes $AA$, $AB$, $BA$ and $BB$ in the next generation is given by $x^2$, $x(1-x)$, $x(1-x)$, $(1-x)^2$.
Finite population
In a finite population of size $N$, the frequencies of the 4 possible genotypes are not necessarily equal to those given by Hardy-Weinberg equilibrium but may vary from it.
My question is.
What is the probability density function of the genotypes frequency given $N$?
It seems to me that this distribution should look similar to a binomial distribution but I fail to wrap my head around this question because we are looking at 4 different dependent variables.
For small sample sizes, each of the four possible genotypes has its own pdf based on a binomial distribution, given by
$${N\choose k} p^k(1-p)^{N-k}$$
where $N$ is the population size, $k$ is the frequency of the given genotype, and $p$ is its probability.
For example, in a population of size $N=4$, the pdf of the genotype $AA$, assuming a given probability $x^2=0.16$ (where $x=0.40$ is the frequency of the allele $A$) is given by
$${4\choose k} 0.16^k(0.84)^{4-k}$$
So the probability that the genotype $AA$ is found in $k=0,1,2,3,4$ individuals are:
$$0 \rightarrow 0.84^{4}\approx 0.59$$ $$1 \rightarrow 4 \cdot 0.16 \cdot 0.84^3 \approx 0.38$$ $$2 \rightarrow 6 \cdot 0.16^2 \cdot 0.84^2 \approx 0.11$$ $$3 \rightarrow 4 \cdot 0.16^3 \cdot 0.01$$ $$4 \rightarrow 0.16^4\approx 0$$
The same considerations can be applied to calculate the pdf of the other three genotypes.