The Distribution of Ducks on a Pond

Question

Ducks are uniformly distributed on a large pond. What is the probability that a duck is its nearest neighbor’s nearest neighbor?

Answer

Interpreting the Problem

The key is to construct the right probabilistic model.

Suppose the two ducks being tested for mutual nearest-neighbor status are A and B, with $AB=r$. In a pond where ducks are uniformly distributed, the probability that A and B are mutual nearest neighbors decreases as $r$ grows. If we can express this probability as a function $P(r)$, an integral will give the expected probability of mutual nearest-neighbor status.

The problem can therefore be restated as follows:

Ducks are uniformly distributed. Consider ducks A and B at distance $r$—itself a probabilistic event—find the probability that they are nearest neighbors under this condition, and then take the expectation.

Equivalently, use this procedure:

1. Choose a central duck A.
2. Find A’s nearest neighbor B and calculate the density $p(r)$ for their distance.
3. For this B, calculate the probability $q(r)$ that A is also B’s nearest neighbor.
4. Integrate the product $p(r)q(r)$.

The first three steps are equivalent to calculating $P(\text{mutual nearest neighbors}\mid r)$.

Modeling

How do we obtain this probability? The only information available is uniformity, so let the constant density be $\lambda$. What is the probability that a region of area $S$ contains no ducks?

Divide the region into infinitely many tiny cells. The probability of a duck occupying any one cell approaches zero, while the expected total remains the constant $\lambda$. This is exactly a Poisson distribution.

A Poisson distribution describes the number of times a random event occurs in a fixed interval of time or space. Its defining features are that events occur independently, randomly, and at a constant average rate.

It is the limiting form $\frac{\mu^k}{k!} e^{-\mu}$ of the binomial distribution $\binom{n}{k} p^k (1-p)^{n-k}$.

Here, $P(X=0) = \frac{(\lambda S)^0}{0!} e^{-\lambda S} = e^{-\lambda S}$ and $S=\pi r^2$. Therefore $P(R>r)=P(X=0)$ and $P(R\le r)=1-P(X=0)$.

Differentiating $P(R\le r)$ gives the density $f(r) = \frac{d(1-e^{-\lambda\pi r^2})}{dr} = 2\pi\lambda r \cdot e^{-\lambda \pi r^2}$. The quantity $f(r)dr$ is the probability that the distance lies in $[r,r+dr]$, which can be viewed as the infinitesimal analogue of $P(R=r)$.

We now have $p(r)$. The next step is $q(r)$, the probability that A is also B’s nearest neighbor.

In the diagram, the radius-$r$ neighborhood around A is already empty. For A to be B’s nearest neighbor, only the crescent-shaped region $S_2$ on the right must also contain no ducks. We need only calculate its area and substitute it into the probability formula; under a uniform distribution, the shape itself does not matter.

Finally, multiply $p$ and $q$ and integrate:

$$P = \int_0^\infty 2\lambda\pi r \, e^{-\lambda(\pi r^2 + S_2)} \, dr$$

Substituting all values gives $P = \frac{6\pi}{8\pi + 3\sqrt{3}}$.

Another observation: the exponent in the integrand corresponds to $S_1+S_2$, the total area. Geometrically this is the figure-eight union of two overlapping circles of radius $r$, and the expression gives the probability that this union is empty.