Probability Theory – Problem 62: find the probability that all ducklings fall within the same semicircle

(7+8 points) This problem discusses the probability that $n \geqslant 3$ ducklings swimming independently in a circular pond all fall within the same semicircle. Suppose their positions (treated as point masses) are $\{P_i\}_{i = 0}^{n-1}$. Fix a point $O^*$ on the boundary circle. For each duckling's position $P_i$, draw the ray from the center $O$ of the pond through $P_i$, intersecting the circle at $P_i^*$. Let $\Theta_i$ denote the central angle measured counterclockwise from $O^*$ to $P_i^*$. We may assume $\Theta_i \overset{\text{i.i.d.}}{\sim} U(0,2\pi)$. Define $\displaystyle X_i = \frac{\Theta_i}{2\pi}$, called the ``absolute coordinate'' of each duckling. Fix one duckling $P_0$, called the ``lead duckling'', and set $\hat{\Theta}_0 := 0$. For each other duckling $P_i$, let $\hat{\Theta}_i$ be the central angle measured counterclockwise from $P_0^*$ to $P_i^*$. Define the ``relative coordinate'' $\displaystyle \hat{X}_i = \frac{\hat{\Theta}_i}{2\pi}$.

Write the transformation formula from ``absolute coordinates'' to ``relative coordinates''. Then find the probability that all $n$ ducklings fall within the same semicircle.

The absolute coordinate $X_i$ and the central angle $\Theta_i$ satisfy $X_i = \dfrac{\Theta_i}{2\pi}$, where $X_i \in [0, 1)$. The relative coordinate $\hat{X}_i$ is determined by the angle $\hat{\Theta}_i$ measured counterclockwise from $P_0^*$ to $P_i^*$, i.e., $\hat{X}_i = \dfrac{\hat{\Theta}_i}{2\pi}$. When $X_i \geqslant X_0$, the counterclockwise arc corresponds to the proportion $X_i - X_0$; when $X_i < X_0$, the counterclockwise path crosses the angle $0$ (i.e., $2\pi$), corresponding to the proportion $1 + X_i - X_0$. Therefore the transformation formula from absolute to relative coordinates is: $$\hat{X}_i = \begin{cases} X_i - X_0, & X_i \geqslant X_0 \\ X_i - X_0 + 1, & X_i < X_0 \end{cases}$$ This formula can also be written uniformly as $\hat{X}_i = (X_i - X_0) \pmod 1$.

Regarding the probability that all $n$ ducklings fall within the same semicircle, this is equivalent to the existence of an arc gap of length greater than $\dfrac{1}{2}$ among the $n$ points uniformly distributed on a circle of circumference $1$. Let the $n$ points divide the circle into $n$ arcs of lengths $L_1, L_2, \dots, L_n$, satisfying $\sum_{k=1}^n L_k = 1$. The event "all points lie in the same semicircle" is equivalent to "there exists some $k$ such that $L_k > \dfrac{1}{2}$". Since the total length is $1$, it is impossible for two arcs to simultaneously have length greater than $\dfrac{1}{2}$, so the events $\{L_k > \dfrac{1}{2}\}$ for different $k$ are mutually exclusive.

Consider the arc length $L_i$ starting from point $P_i$ in the counterclockwise direction. The event $L_i > \dfrac{1}{2}$ means that the counterclockwise arc of length $\dfrac{1}{2}$ starting from $P_i$ contains no other points, or equivalently, all other $n-1$ points lie in the clockwise semicircle from $P_i$. For a fixed $P_i$, the probability that each of the remaining $n-1$ independently distributed points falls in the specified semicircle is $(\dfrac{1}{2})^{n-1}$. Since there are $n$ points and the corresponding mutually exclusive events number $n$, by the addition rule for mutually exclusive events, the probability that all points lie in the same semicircle is $P = \sum_{i=1}^n (\dfrac{1}{2})^{n-1} = n(\dfrac{1}{2})^{n-1} = \dfrac{n}{2^{n-1}}$.

The transformation formula is $\hat{X}_i = (X_i - X_0) \pmod 1$; the probability is $\dfrac{n}{2^{n-1}}$.