Urn balls and probabilities
(If urns are vacancies and balls are unemployed, what distinction between unemployed workers does the Red/Green dichotomy reflects?)  

Each ball has in front of it an identical box, each with the exact same lottery tickets, its ticket has a number on it, and each number corresponds to an urn. 

We say "Go!" and each ball draws "randomly" (i.e. with equal probability) a ticket from the box in front of it. Note that the requirement that the probabilities are Uniform, impose the condition that, if we want to have proper distributions, the number of urns must be finite (not even countably infinite -see this post in statistics.se) Then:

RED BALL SOLO
From the point of view of a single Red ball, what is the probability that it will draw a number that no other Red ball draws?

Since each ball draws independently from the others, it is clear that we don't care at all about the existence of Green balls, to the degree that no upper limit of "number of balls in an urn" exists (if it existed the draws would not be independent). So Green balls (their number and what they draw from their boxes) can be kept out of the picture.

Since Red balls are indistinguishable, let's take $R_1$ as our hero. If it draws the number $1$ (i.e. conditional on) we want the probability

$$\Pr(R_j \neq 1 \mid R_1 =1)\;\;  j=2,...N_R$$
But draws are independent so we want simply

$$\Pr(R_j \neq 1) = \prod_{j=2}^{N_R}[1-\Pr(R_j = 1)] =  (1-1/N_U)^{N_R-1}$$

But $R_1$ can draw not just $"1"$, but any number, and each with probability $1/N_U$. So the probability of the event that we want, denote it $\Pr(\text{RBS})$ is (Law of Total Probability)

$$ \Pr(\text{RBS}) = \sum_{i=1}^{N_U}\left(\frac {1}{N_U}\right)\cdot \Pr(R_j \neq i \mid R_1 =i) = \left(\frac {1}{N_U}\right)\sum_{i=1}^{N_U}(1-1/N_U)^{N_R-1}$$

$$\implies \Pr(\text{RBS}) = \left(\frac {1}{N_U}\right) \cdot N_U \cdot (1-1/N_U)^{N_R-1} = (1-1/N_U)^{N_R-1}$$

...again.

This is intuitive: the greater the number of urns, the higher the probability. The greater the number of Red balls, the lower the probability.

Now, define job market tightness as usual by 
$$\theta = N_U/N_R \implies N_R = \frac 1 {\theta} N_U$$ 

(here "U" denotes "urns" so vacancies). Then

$$\Pr(\text{RBS}) = \left(1-\frac 1{N_U}\right)^{\frac 1 {\theta}  N_U-1}$$

If $N_U$ grows "large" then the above is well approximated by 

$$\Pr(\text{RBS}) \approx e^{-\frac 1 {\theta} }$$

The higher job market tightness, the higher the probability that an unemployed will be matched solo to a vacancy (again, ignoring what the Greens represent).