<s>[INST] sample all pairs without repeats Assuming I have a very large number of $K$ colored balls and we know the fraction of each color. If we randomly sample all pairs so that all pairs have two balls of different colors then what is the fraction of pairs with a given color combinations?
For example if there are 3 colors with fractions, $f_{blue}=0.5$, $f_{red}=0.25$ and $f_{green}=0.25$ then if you sample all pairs of balls such that each pairs consists of two different colors. Then I would expect that $50\%$ of pairs will be $(blue/red)$, $50\%$ will be $(blue/green)$ and $0\%$ $(red/green)$ (if 50% is blue then there must be a blue in each pair). This scenario is easy since there is only one way to sample all unordered pairs. 
If more than $50\%$ of balls have the same color there will be no solution and if $50\%$ of balls are a certain color there is only one way to sample all balls (unordered) as above. 
If the fraction of 3 colors are then same $f_{blue}=1/3$, $f_{red}=1/3$ and $f_{green}=1/3$ then by symmetry I would expect the fraction of pairs to be  $1/3$ $(blue/red)$, $1/3$ $(blue/green)$ and $1/3$ $(red/green)$ if they where randomly sampled. 
Is there a general way to calculate the expected fraction of colored pairs given you know the fraction of each of the $K$ colors?
Edit/update
 Ertxiem gave the solution in the case of K=3 where you do not use the assumption of randomly drawing all pairs (pairs of different colors without replacement). 
Here is what I have tried so far. 
Let $f=(f_1,f_2,\ldots,f_K)$ be the fraction of each colored ball assuming K colors. 
For the case of K=3 then we can calculate the fraction of pairs of ball by solving the following 
$Ax=f$ where $A=    \left( \begin{array}{ccc}
0.5 & 0.5 & 0 \\
0.5 & 0 & 0.5 \\
0 & 0.5 &0.5
  \end{array}
\right)
$, $f=  \left( \begin{array}{c}
f_1\\
f_2 \\
f_3 
  \end{array}
\right)$, $x= \left( \begin{array}{c}
\pi_{12}\\
\pi_{13} \\
\pi_{23} 
  \end{array}\right)$ where $\pi_{ij}$is the probability of a pair of color $i$ and $j$. 
This gives the solution for $K=3$. 
For $K>3$ we cannot use the same approach because there will be multiple solutions for example for $K=4$. Solving $Ax=f$ where $A=    \left( \begin{array}{cccccc}
0.5 & 0.5 & 0.5 & 0 & 0 & 0  \\
0.5 & 0 & 0  &0.5 &0.5 &0 \\
0 & 0.5 &0  & 0.5 & 0 & 0.5 \\
0 & 0 &0.5  & 0& 0.5& 0.5
  \end{array}
\right)
$, $f=  \left( \begin{array}{c}
f_1\\
f_2 \\
f_3 \\
f_4 
  \end{array}
\right)$ and  $x= \left( \begin{array}{c}
\pi_{12}\\
\pi_{13} \\
\pi_{14} \\ 
\pi_{23} \\
\pi_{24} \\
\pi_{34}\end{array}\right)$ gives multiple multiple solutions. Is there a way to solve it by assuming the color combinations are independent given that they are different?  
update with example and simulations
For the $K=4$ case then I have tried to solve $Ax=f$ for $x$ using Moore-Penrose generalized inverse (pseudoinverse using least squared solution) however, this does not give the same results as simulations (rejection sampling using $5e7$ balls).  For the case of $f=(3/8,1/8,2/8,2/8)$ I get the following results
$$\begin{array}{c|c|c}
\hat{\pi} & pseudo inverse & simulations\\
\pi_{12} & 4/24 & 31/236\\
\pi_{13} & 7/24 &  73/236\\
\pi_{14} & 7/24 &  73/236\\ 
\pi_{23} & 1/24 &  14/236\\
\pi_{24} & 1/24 &  14/236\\
\pi_{34} & 4/24 &  31/236
  \end{array}$$. 
So I am still not able to find a analytical solution (for K>3). 
 [/INST] Suppose you have $K$ colours of balls with respective numbers $n_1,...,n_K$, with a total of $n = \sum n_i$ balls.  Let $\mathscr{S}$ denote the set of all pairs of distinct balls and let $\mathscr{C}$ denote the set of all pairs of distinct balls of the same colour.  Since $\mathscr{C} \subset \mathscr{S}$ the number of ways you can sample two balls of different colours is:
$$\begin{equation} \begin{aligned}
|\mathscr{S} - \mathscr{C}| = |\mathscr{S}| - |\mathscr{C}| 
&= {n \choose 2} - \sum_{k=1}^K {n_k \choose 2} \\[6pt]
&= \frac{n(n-1)}{2} - \sum_{k=1}^K \frac{n_k (n_k-1)}{2} \\[6pt]
&= \frac{1}{2} \Big[ n(n-1) - \sum_{k=1}^K n_k (n_k-1) \Big] \\[6pt]
&= \frac{1}{2} \Big[ (n-1) \sum_{k=1}^K n_k - \sum_{k=1}^K n_k (n_k-1) \Big] \\[6pt]
&= \frac{1}{2} \sum_{k=1}^K n (n-n_k). \\[6pt]
\end{aligned} \end{equation}$$
Let $\mathscr{M}_{a,b}$ denote the set of all pairs of distinct balls with colours $a \neq b$.  The number of ways you can sample two balls with a given (different) colour combination is:
$$|\mathscr{M}_{a,b}| = \frac{n_a n_b}{2}$$
Hence, the fraction of sample-pairs of different colours that are of the specified colour pair $a \neq b$ is:
$$P_n(a,b) = \frac{|\mathscr{M}_{a,b}|}{|\mathscr{S} - \mathscr{C}|} = \frac{n_a n_b}{\sum_{k=1}^K n_k (n-n_k)}.$$
Taking $n \rightarrow \infty$ and letting $p_1,...,p_K$ be the respective limiting sample proportions of the balls of each colour, you have:
$$P_\infty(a,b) = \lim_{n \rightarrow \infty} \frac{|\mathscr{M}_{a,b}|}{|\mathscr{S} - \mathscr{C}|} = \frac{p_a p_b}{\sum_{k=1}^K p_k (1-p_k)}.$$

Application to your problem: In your example you have $K=3$ colours with proportions $\mathbf{p} = (\tfrac{1}{2}, \tfrac{1}{4}, \tfrac{1}{4})$ for the respective colours $\text{Blue}, \text{Red}, \text{Green}$.  This gives:
$$P_\infty(a,b) 
= \frac{p_a p_b}{\tfrac{1}{2} \cdot \tfrac{1}{2} + \tfrac{1}{4} \cdot \tfrac{3}{4} + \tfrac{1}{4} \cdot \tfrac{3}{4}}
= \frac{p_a p_b}{5/8}
= \tfrac{8}{5} \cdot p_a p_b.$$
So you have:
$$\begin{equation} \begin{aligned}
P_\infty(\text{Blue}, \text{Red}) 
&= \tfrac{8}{5} \cdot \tfrac{1}{2} \cdot \tfrac{1}{4} = \tfrac{1}{5}, \\[6pt]
P_\infty(\text{Blue}, \text{Green}) 
&= \tfrac{8}{5} \cdot \tfrac{1}{2} \cdot \tfrac{1}{4} = \tfrac{1}{5}, \\[6pt]
P_\infty(\text{Red}, \text{Green}) 
&= \tfrac{8}{5} \cdot \tfrac{1}{4} \cdot \tfrac{1}{4} = \tfrac{1}{10}. \\[6pt]
\end{aligned} \end{equation}$$
</s>