What is the probability distribution of a single genome base pair in the genome we have 4 nucleotides (A,T,C,G). Now given a nucleotide sequence like 
AGT CG TA CG ATCT CG ,
we can count the number of "CG" pairs. That's 3 in this case. (we count all the pairs so, ACT has pairs AC and CT)
Now I would like to test the significance of my results, or how likely is it that I would get 3 CG pairs if that sequence was random. I could test that with a permutation test, but that's not completely accurate and might also take time.
Now the question: What is the probability distribution of such CG pair, given the length of the sequence and the count of each element (A,C,T,G), so that I could calculate the exact probability that my result could come from a random sequence. 
 A: I'll assume that the intended question is this: Given the length of a sequence and the counts of all four nucleotides in this sequence (as opposed to their frequency in sequences in general), what is the probability that a sequence drawn randomly uniformly from all sequences fulfilling that description would contain exactly a certain number $k$ of CG pairs?
Denote the counts of the nucleotides by $\def\n#1{n_{\text #1}}\n A$, $\n C$, $\n G$ and $\n T$ and their sum, the length of the sequence, by $n$. Then we can form $k$ CG pairs and distribute these $k$ pairs and the remaining $n-2k$ individual nucleotides in
$$
\binom{n-k}{\n A,\n T,\n C-k,\n G-k,k}
$$
different ways (see multinomial coefficients). But this overcounts, since we're allowing the remaining C and G nucleotides to form pairs. Every combination with $m$ pairs is being counted $\binom mk$ times, where it shouldn't be counted at all. Making use of
$$
\sum_{j=k}^m\binom mj\binom jk(-1)^{j-k}=\delta_{km}\;,
$$
we can correct for the overcounting and calculate the desired count of sequences fulfilling the description as
$$
\begin{align}
&\sum_{j=k}^\infty\binom{n-j}{\n A,\n T,\n C-j,\n G-j,j}\binom jk(-1)^{j-k}\\=&\sum_{j=k}^\infty\binom{n-j}{\n A,\n T,\n C-j,\n G-j,j-k,k}(-1)^{j-k}\;,
\end{align}
$$
where the sum actually only runs up to $\min(\n C,\n G)$ and the remaining terms are zero. This count has to be divided by the total number of sequences fulfilling the description, which is
$$
\binom n{\n A,\n T,\n C,\n G}\;.
$$
In your example, with $\n A=3$, $\n C=\n G=\n T=4$, $n=15$ and $k=3$, the result would be
$$
\binom{15}{3,4,4,4}^{-1}\left(\binom{12}{3,4,1,1,3}-\binom{11}{3,4,1,3}\right)=\frac{44}{1365}\approx3\%\;.
$$
[Edit in response to comment:]
If you want to count the sequences with at least $k$ pairs, we still need to correct for overcounting, since each of the sequences with more than $k$ pairs is counted more than once, but the correction is slightly different. The required binomial coefficient identity is
$$
\sum_{j=k}^m\binom mj\binom{j-1}{k-1}(-1)^{j-k}=1\;,
$$
and the resulting sum is
$$
\sum_{j=k}^\infty\binom{n-j}{\n A,\n T,\n C-j,\n G-j,j}\binom{j-1}{k-1}(-1)^{j-k}\;.
$$
In your example, with $\n A=3$, $\n C=\n G=\n T=4$, $n=15$ and $k=3$, the result would be
$$
\binom{15}{3,4,4,4}^{-1}\left(\binom{12}{3,4,1,1,3}\binom22-\binom{11}{3,4,4}\binom32\right)=\frac{3}{91}\approx3\%\;.
$$
The change relative to the result for exactly $3$ pairs is less than one tenth of a percent. The difference in the counts,
$$
\left(\binom{12}{3,4,1,1,3}\binom22-\binom{11}{3,4,4}\binom32\right)
-
\left(\binom{12}{3,4,1,1,3}-\binom{11}{3,4,1,3}\right)
=
\binom{11}{3,4,4}
\;,
$$
is precisely the number of sequences with $4$ CG pairs.
