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:

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

  • 1
    $\begingroup$ How exactly is this related to Economics? $\endgroup$
    – Giskard
    Aug 30, 2015 at 10:58
  • 2
    $\begingroup$ @denesp It's methods. If you think about urns as vacancies, and balls as unemployed, you could recover something similar to Blanchard-Diamond (1990). I've made better experiences with asking math-related questions on Econ.SE than on Math.SE, simply because people here are more likely to have a similar math background to me, and hence more likely to understand where I'm going with this. $\endgroup$
    – FooBar
    Aug 30, 2015 at 11:00
  • 2
    $\begingroup$ If you disagree on this being ontopic, you'd also have to disagree on questions on how to derive the OLS estimator as being off topic, as that's strictly speaking also mathematics only. In any case, the correct place for discussing that is meta. $\endgroup$
    – FooBar
    Aug 30, 2015 at 11:08
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    $\begingroup$ @suriv To the degree that Stochastic Economics is on-topic here, I don't see why any question related to probability theory but with specific economic applications in mind is not also on-topic in economics.se. What I would suggest is that the OP moves this economic application from the comments to the main body of the question. $\endgroup$ Aug 30, 2015 at 18:02

3 Answers 3


(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:

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)

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

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


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).

  • $\begingroup$ 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. What do you mean by proper distributions, discrete? I have to admit I didn't perfectly follow that Stats post. $\endgroup$
    – FooBar
    Aug 31, 2015 at 14:52
  • $\begingroup$ @FooBar Discrete, or continuous or a mix, proper distributions are those with a cdf having the "usual" properties. If a random variable can take an infinite number of discrete values (say, the natural numbers), it cannot have a Uniform distribution of probabilities over them, because necessarily each probability will be zero. If we nevertheless assume that it does follow a Uniform, then it usually is used as an "improper prior" (distribution) in the context of Bayesian statistics. $\endgroup$ Aug 31, 2015 at 19:22
  • $\begingroup$ I dont follow. Isn't that exactly the continuous uniform distribution? $\endgroup$
    – FooBar
    Aug 31, 2015 at 19:26
  • $\begingroup$ @FooBar. No. The Uniform distribution has bounded support, taking values for $a$ to $b$, whit both numbers finite. The "improper" version (in the continuous case), emerges when we let $a$ and $b$ go to infinity. $\endgroup$ Aug 31, 2015 at 21:21
  • $\begingroup$ Take a look at my fresh answer for that matter. $\endgroup$
    – FooBar
    Sep 11, 2015 at 8:35

Red ball solo

For any red ball in some urn, the probability that another red ball misses that urn is $1-\frac{1}{U}$. The probability that all other red balls miss that urn is the product of the chance they all miss that urn, or $\left(1-\frac{1}{U}\right)^{R-1}$.

Red ball super ball

The balls are symmetric, so $\Pr[\text{a red ball is a superball}] = \frac{\mathbb{E}[\text{# superballs}]}{R}$. The expected number of superballs is the expected number of urns with a nonzero number of red balls. The probability that an urn has $\geq 1$ red ball is $1 - \left(1-\frac{1}{U}\right)^R$, because the chance that every red ball misses the urn is $\left(1-\frac{1}{U}\right)^R$. The expected number of urns with a nonzero number of red balls is therefore $U$ times this probability. So $$ \Pr[\text{a red ball is a superball}] = \frac{U}{R}\left(1 - \left(1-\frac{1}{U}\right)^R\right) . $$


This is a complement/comment to Alecos' answer, who said that

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

Denote the total size of the world as $N$. Denote urns as $U_N$, balls as $B_N$, and say that we randomly toss every ball into an urn. Urns and balls are "scaled", i.e. $U_N = uN$, $B_N = bN$, for two constants $u$, $b$. Then the distribution of balls for any urn is Binomial

$$ Prob(k \text{ balls arrive}) = Binomial(k; n=B_N, p=1/U_N)$$

In the next step, I want to let the number of urns and balls go to infinity. Note that if $np$ converges to a finite number as we let $N\to \infty$, the distribution of ball arrivals for any particular urn is approximated (pretty accurately) by

$$ Prob(k \text{ balls arrive for N ``large''}) = Poisson(k; \lambda=np)$$

And indeed, due to the setup, $np = B_N/U_N = b/u$ converges as $N\to \infty$.

  • $\begingroup$ That's fine, because that's a limiting argument that starts from a finite set up. If you have tried to assume from the beginning, that $N=1,.....$ (which is translated "urns and balls are countably infinite"), you wouldn't be able to define the binomial in the first place, because then, $p$ would be zero from the start, while in what you write $p$ tends to zero as $N$ tends to infinity. $\endgroup$ Sep 11, 2015 at 8:53

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