Urn balls and probabilities Think of the following balls as individuals of populations. 

Say I have $U$ urns, and some balls. Both numbers are really large. So large, that authors like Blanchard and Diamond have approximated the binomial operations that follow with Poisson probabilities.

The balls are either red ($R$) or green ($G$). At the beginning of the period, every ball is (randomly, iid) tossed into an urn. There is no miss chance (i.e. every ball is in some urn). Some urns will have more than one ball, some might have none.

There are two exercises I want to do in this setup, and I'm not sure to what extent they're overlapping (i.e. helping me understand one would help me understand the other one as well), so I will post both.

My issue is kind of that I have a binomial thinking going on, and you can see that from the structure that I have imposed onto solving the following probabilities. Should I switch to Poisson probabilities instead? What is the neatest way to solve the following setups?

Red Ball Solo

I would like to compute the ex-ante probability of a red ball (from the perspective of a red ball) of being tossed into an urn where there is no other red ball.

So far, I was thinking about doing

$$ 
Prob(\text{sole red ball}) = \sum_{x = 1}^{R + G} Prob(\text{Urn has $x$ balls}) \cdot \sum_{y=0}^{x-1}Prob(\text{No other red ball }  | x \text{ balls})
$$ 

Red Ball Super Ball

Out of each urn, one red ball becomes a super ball. This probability is uniform. I would like to compute the probability of a red ball becoming a super ball.

My abstract idea was again similar:

$$ 
Prob(\text{red ball becomes super ball}) = \sum_{x = 1}^{R + G} Prob(\text{Urn has $x$ balls}) \cdot \sum_{y=0}^{x-1}Prob(\text{$y$ other red balls | $x$ total balls }) \frac{1}{1+y}
$$
(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).