Sheep That See Each Other

Problem

Two sheep are at two random points inside a square pen. They are munching grass and staring in two random directions. Each sheep has a field of view that’s 180 degrees. What is the probability that they both see each other?

Extra credit. Now, three sheep are at three random points inside a square pen. They are munching grass and staring in three random directions. As before, each sheep has a field of view that’s 180 degrees. What is the probability that all three sheep see each other?

Credit: The Fiddler on the Proof.

What it means to see

Each sheep stands at a point chosen uniformly in the unit square and faces a direction chosen uniformly on the circle, independently of everything else. A field of view of $180^\circ$ means the sheep sees the entire half-plane in front of it: the half whose boundary runs through the sheep at right angles to its facing direction. In vectors, a sheep at $P$ facing the unit vector $\mathbf{u}$ sees a point $Q$ exactly when $Q$ lies forward of that boundary, $(Q - P)\cdot \mathbf{u} \ \ge\ 0.$

Two sheep

Fix the two positions $A$ and $B$ for a moment and look only at the directions. Sheep $A$ sees sheep $B$ when its facing vector points into the half-plane containing $B$, that is, when the facing direction lies within $90^\circ$ of the direction from $A$ to $B$. Those facing directions fill exactly half of the circle, and the facing direction is uniform, so $\mathbb{P}(A \text{ sees } B) = \frac{180^\circ}{360^\circ} = \frac12.$ Crucially this does not depend on where $A$ and $B$ are: wherever $B$ sits, half of $A$‘s possible facings point towards it. By the same argument $\mathbb{P}(B \text{ sees } A) = \tfrac12$. The two facings are independent, so $\mathbb{P}(\text{both see each other}) = \frac12 \cdot \frac12 = \frac14.$ The positions never entered the calculation, so the square is a red herring: the answer is $\tfrac14$ for a pen of any shape.

Three sheep: each must see two

With three sheep the directions still do all the work, but now each sheep has to see two others at once. Consider sheep $A$, and let $\alpha$ be the interior angle at $A$ in the triangle $ABC$, the angle between the direction to $B$ and the direction to $C$. Sheep $A$ sees $B$ when its facing lies in the half-circle of directions centred on $A \to B$, and sees $C$ when its facing lies in the half-circle centred on $A \to C$. To see both, the facing must lie in the overlap of these two half-circles.

Two half-circles whose centres are $\alpha$ apart overlap in an arc of $180^\circ - \alpha$. (Slide one half-circle round by $\alpha$ and it loses exactly $\alpha$ of its overlap with the other.) Since the facing is uniform, $\mathbb{P}(A \text{ sees both } B \text{ and } C) = \frac{180^\circ - \alpha}{360^\circ} = \frac{\pi - \alpha}{2\pi},$ writing $\alpha$ in radians. The same holds at $B$ and $C$ with their interior angles $\beta$ and $\gamma$.

Averaging over the triangle

Each sheep’s success depends only on its own facing, and the three facings are independent. So, with the positions fixed, $\mathbb{P}(\text{all three see each other} \mid \text{triangle}) = \frac{\pi-\alpha}{2\pi}\cdot\frac{\pi-\beta}{2\pi}\cdot\frac{\pi-\gamma}{2\pi} = \frac{(\pi-\alpha)(\pi-\beta)(\pi-\gamma)}{8\pi^3}.$ Because $\alpha + \beta + \gamma = \pi$, each factor is the sum of the other two angles, $\pi - \alpha = \beta + \gamma$, and the same identity expands the product neatly: $(\pi-\alpha)(\pi-\beta)(\pi-\gamma) = \pi(\alpha\beta + \beta\gamma + \gamma\alpha) - \alpha\beta\gamma.$ What remains is to average this over the triangle made by three uniform points in the square, $\mathbb{P}(\text{all three see each other}) = \frac{1}{8\pi^3}\,\mathbb{E}\bigl[(\pi-\alpha)(\pi-\beta)(\pi-\gamma)\bigr].$ The distribution of the angles of a random triangle in a square has no tidy closed form, so this last expectation is evaluated numerically. To high precision, $\mathbb{P}(\text{all three see each other}) \approx 0.02721,$ that is, about $2.72\%$. The factor $(\pi-\alpha)(\pi-\beta)(\pi-\gamma)$ vanishes whenever one angle approaches $\pi$, so the long, thin triangles that three random points so often make contribute almost nothing, which is why the answer sits well below the value $1/27 \approx 0.037$ that a perfectly equilateral pen would give.

Quantity	Value
$\mathbb{P}(\text{two sheep see each other})$	$1/4 = 0.25$
$\mathbb{P}(\text{three sheep all see each other})$	$\approx 0.02721$

Computational verification

We confirm both answers by simulation: place the sheep uniformly, draw their facings uniformly, and test the half-plane condition $(Q - P)\cdot\mathbf{u} \ge 0$ for every ordered pair. This uses nothing but the raw rule of the problem, no angles and no formula, so it checks the whole argument independently. As a faster cross-check, which assumes the $(\pi-\alpha)/2\pi$ step rather than testing it, we also average the per-triangle formula $(\pi-\alpha)(\pi-\beta)(\pi-\gamma)/8\pi^3$, with the directions already integrated out.

import numpy as np
rng = np.random.default_rng(20260529)
N = 20_000_000

def dirs(n):                       # uniform facing directions
    t = rng.uniform(0, 2*np.pi, n)
    return np.stack([np.cos(t), np.sin(t)], -1)

def sees(P, u, Q):                 # viewer at P facing u sees Q?
    return ((Q - P) * u).sum(-1) >= 0

# two sheep
A, B = rng.uniform(0,1,(N,2)), rng.uniform(0,1,(N,2))
both = sees(A, dirs(N), B) & sees(B, dirs(N), A)
print(both.mean())                 # -> 0.2500

# three sheep
A, B, C = (rng.uniform(0,1,(N,2)) for _ in range(3))
uA, uB, uC = dirs(N), dirs(N), dirs(N)
allsee = (sees(A,uA,B) & sees(A,uA,C) &
          sees(B,uB,A) & sees(B,uB,C) &
          sees(C,uC,A) & sees(C,uC,B))
print(allsee.mean())               # -> 0.0272

Quantity	Exact / formula	Monte Carlo
two sheep	$0.25000$	$0.25005$
three sheep	$0.02721$	$0.02721$

A pen where the answer is exact: the circle

The square resists a closed form because the angles of a random triangle inside it have no tractable distribution. One pen escapes this: a circular fence with the sheep on the rail. Three points on a circle make a triangle inscribed in it, and inscribed triangles pin the angles down exactly.

The angles are half the opposite arcs

Fix the circle’s centre $O$. The inscribed-angle theorem says the interior angle at a vertex is half the central angle standing on the same chord. At $A$ this is $\angle BAC = \tfrac12 \angle BOC$, the central angle of the arc $BC$ opposite $A$.

Write $\hat a$ for the central angle of the arc opposite $A$, and likewise $\hat b, \hat c$. Then $\alpha = \tfrac12\hat a, \qquad \beta = \tfrac12\hat b, \qquad \gamma = \tfrac12\hat c.$ The three opposite arcs tile the circle, $\hat a + \hat b + \hat c = 2\pi$, which simply re-derives $\alpha + \beta + \gamma = \pi$.

The arcs are uniform on the simplex

The three points are independent and uniform on the circle. By rotational symmetry we may hold one of them fixed; the other two are then uniform on the cut circle, an interval of circumference $2\pi$. The three arcs are exactly the gaps left by two uniform points on that interval, and the gaps of $k$ uniform points on an interval are uniformly distributed over the simplex of fractions that sum to one. With $k = 2$, the three normalised arcs, and hence the normalised angles $(a, b, c) = \Bigl(\tfrac{\alpha}{\pi}, \tfrac{\beta}{\pi}, \tfrac{\gamma}{\pi}\Bigr), \qquad a + b + c = 1,$ are uniform over the triangle $\{a, b, c \ge 0,\ a + b + c = 1\}$.

Two integrals

On that triangle the density is constant. The triangle has area $\tfrac12$, so the density is $2$. Writing $c = 1 - a - b$ and integrating over $a \ge 0,\ b \ge 0,\ a + b \le 1$, $\mathbb{E}[ab] = 2\int_0^1\!\!\int_0^{1-a} ab \, db \, da = \int_0^1 a(1-a)^2 \, da = \frac{1}{12},$ $\mathbb{E}[abc] = 2\int_0^1\!\!\int_0^{1-a} ab(1 - a - b) \, db \, da = \frac13\int_0^1 a(1-a)^3 \, da = \frac13\cdot\frac{1}{20} = \frac{1}{60},$ the inner integrals being $\int_0^{1-a} b\,db = (1-a)^2/2$ and $\int_0^{1-a} b(1-a-b)\,db = (1-a)^3/6$. By symmetry $\mathbb{E}[ab + bc + ca] = 3\,\mathbb{E}[ab] = \tfrac14$.

Assembling the answer

For any pen, the per-triangle probability is $(\pi-\alpha)(\pi-\beta)(\pi-\gamma)/8\pi^3$. In normalised angles $\pi - \alpha = \pi(1 - a)$, so this is $\frac{\pi^3 (1-a)(1-b)(1-c)}{8\pi^3} = \frac{(1-a)(1-b)(1-c)}{8} = \frac{(ab + bc + ca) - abc}{8},$ using $(1-a)(1-b)(1-c) = 1 - (a+b+c) + (ab+bc+ca) - abc$ with $a + b + c = 1$. Taking the expectation over the uniform-simplex angles, $\mathbb{P}(\text{all three see each other}) = \frac18\Bigl(\mathbb{E}[ab + bc + ca] - \mathbb{E}[abc]\Bigr) = \frac18\Bigl(\frac14 - \frac{1}{60}\Bigr) = \frac18\cdot\frac{14}{60} = \frac{7}{240} \approx 0.02917.$ The independent direction-level simulation, with the sheep placed on the circle and their facings drawn at random, gives $0.02916$, in agreement.

A round pen is therefore a little kinder to mutual sightlines than a square one, $7/240 \approx 2.92\%$ against the square’s $\approx 2.72\%$. Points on a rim never fall almost in a line, so the triangles stay fatter, the binding overlaps $\pi - \alpha_i$ stay larger, and the product survives more often. Two sheep, as always, see each other with probability exactly $\tfrac14$, on a pen of any shape at all.