Duck in the round pond

There is an interesting question:

There are 4 ducks in a circular pond. The position of these ducks in the pond are random. We assume that the pond is large enough and the size of the ducks is small enough. What is the probability that all ducks appear in the same half of the pond?

The most important thing is that the semi-pond is not fixed. So how to draw a line to seperate the pond is the most important question.

We consider the ducks are the points follows the law of $U(B_1)$, where $B_1$ is the unit ball in $\mathbb R^2$.

Since $P((0, 0)) = 0$, we know that $U(B_1)$ is the same as $U(B_1\setminus\{0\})$. Without loss of generality, we consider the radial angle to the first point. We can see $\theta_0 = 0$, and $\theta_i$ follows the law of $U((-\pi, \pi])$, $i=1, \cdots, n-1$.

We reordered these random variables $\theta_1, \cdots, \theta_{n-1}$, such that,

$$\theta_{(1)}\leq \theta_{(2)}\leq\cdots\leq \theta_{(n-1)}$$

If the n ducks are in the same half circle, $\theta_{(n)}-\theta_{(1)}< 1/2$.

If $\theta_{(n)}-\theta_{(1)}< \pi$, $\max(0, \theta_{(n-1)}) - \min(0, \theta_{(1)}) < \pi$. It means all of the ducks are in the fan-shape which angle from $\min(0, \theta_{(1)})$ to $\max(0, \theta_{(n-1)})$ and they of course in a half circle.

Otherwise, if $\theta_{(n)}-\theta_{(1)} > \pi$, we cant find a half circle covers the n points.

And

$$\begin{aligned} &\mathbb P\{\text{n ducks in one semi-disk}\} \\ =& \mathbb P\{\theta_{(n-1)} - \theta_{(1)}<\pi\} \\ =& \int_{y-x<\pi} d[\mathbb P(\theta_{(1)}<x, \theta_{(n-1)}<y)]. \end{aligned}$$

We should calculate $\mathbb P(\theta_{(1)}<x, \theta_{(n-1)}<y)$ first. (formally)

$$\begin{aligned} &d \mathbb P(\theta_{(1)}<x, \theta_{(n-1)}<y)\\ =&\mathbb P(x<\theta_{(1)}<x+dx, y<\theta_{(n-1)}<y+dy)\\ =&\frac{(n-1)!}{(n-2)!1!1!} (\frac{y-x}{2\pi})^{n-3}\frac{dx}{2\pi}\frac{dy}{2\pi},\quad (y > x) \end{aligned}$$

This is because that there are $n-3$ of $n$ points in the area of $(x+dx, y)$, the pobability of one point in the $(x, y)$ is $\frac{1}{2\pi}(y-x)$, and the probability of a point in the inteval $(x, x+dx)$ is $\frac{dx}{2\pi}$.

Therefore

$$\begin{aligned} &\mathbb P\{\text{n ducks in one semi-disk}\} \\ =& \mathbb P\{\theta_{(n-1)} - \theta_{(1)}<\pi\} \\ =& \int_{y-x<\pi} d[\mathbb P(\theta_{(1)}<x, \theta_{(n-1)}<y)]. \\ =&\int_{x\leq y\leq x + \pi} \frac{(n-1)!}{(n-2)!1!1!} (\frac{y-x}{2\pi})^{n-2}\frac{dx}{2\pi}\frac{dy}{2\pi}\\ =&\int_0^{\pi}\frac{dx}{2\pi}\int_x^{x + \pi} + \int_{\pi}^{2\pi}\frac{dx}{2\pi}\int_x^{2\pi}(n-1)(n-2)(\frac{y-x}{2\pi})^{n-3}\frac{dy}{2\pi}\\ =&\int_0^{\pi}\frac{n-1}{2^{n-2}}\frac{dx}{2\pi} + \int_{\pi}^{2\pi}(n-1)(\frac{1-x}{2\pi})^{n-2}\frac{dx}{2\pi}\\ =&\frac{n-1}{2^{n-1}} + \frac{1}{2^{n-1}} = \frac{n}{2^{n-1}} \end{aligned}$$