Three sheep

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?

Here we have to be a bit less handwavy, but let's assume that the sheep are located at points $A=(a_1,a_2),$ $B=(b_1,b_2)$ and $C=(c_1,c_2)$, where the measure of the angle $m\angle CAB = \alpha,$ $m\angle ABC = \beta,$ and $m\angle BCA = \gamma.$ In this case the sheep at $A$ can only see $B$ if it is looking up to an angle of $\frac{\pi}{2}$ to the left or right of the line between $A$ and $B$. Similarly, it can only see $C$ if it is looking up to an angle of $\frac{\pi}{2}$ to the left or right of the line between $A$ and $C.$ Let's assume that the line between $A$ and $B$ forms an angle of $\theta$ with respect to the positive $x$-axis. Then since $m\angle CAB = \alpha,$ we see that the line between $A$ and $C$ must either form an angle of $\theta-\alpha$ or $\theta+\alpha.$ That means that either the sheep must be staring in the direction of $\phi \in \left[ \theta - \frac{\pi}{2}, \theta + \frac{\pi}{2} \right] \cap \left[ \theta - \alpha - \frac{\pi}{2}, \theta - \alpha + \frac{\pi}{2} \right] = \left[ \theta - \frac{\pi}{2}, \theta - \alpha + \frac{\pi}{2} \right]$ or $\phi \in \left[ \theta - \frac{\pi}{2}, \theta + \frac{\pi}{2} \right] \cap \left[ \theta + \alpha - \frac{\pi}{2}, \theta + \alpha + \frac{\pi}{2} \right] = \left[ \theta + \alpha - \frac{\pi}{2}, \theta + \frac{\pi}{2} \right],$ so in either case we have that the probability of the sheep at $A$ looking at both the sheep at $B$ and the sheep at $C$ is $$\frac{ \pi - \alpha }{ 2\pi} = \frac{1}{2} - \frac{\alpha}{2\pi}.$$ Since there was nothing special about $A$ and everything is independent, we see that the probability of all of the sheep looking at each other given the angles $\alpha,$ $\beta$ and $\gamma$ are $$p(\alpha, \beta, \gamma) = \left( \frac{1}{2} - \frac{\alpha}{2\pi}\right) \left( \frac{1}{2} - \frac{\beta}{2\pi} \right) \left( \frac{1}{2} - \frac{\gamma}{2\pi}\right).$$ We see that the maximum probability for any particular shape would be for an equilateral triangle $(\alpha=\beta=\gamma=\frac{\pi}{3})$, where the probability is $\frac{1}{27} = 3.\overline{703}\%,$ whereas the minimum probability is for three collinear points where say $\alpha=\pi$ and $\beta=\gamma=0,$ where the probability is $0.$

So if we know where $A$, $B$ and $C$ are all then we can get the side lengths of the triangle $a=\|B-C\|,$ $b=\|A-C\|,$ and $c=\|A-B\|.$ From this we can use the law of cosines to get \begin{align*}\alpha &= \cos^{-1} \left( \frac{b^2 + c^2 - a^2}{2bc} \right) = \cos^{-1} \left( \frac{ \|A-C\|^2 + \|A-B\|^2 - \|B-C\|^2}{ 2 \|A-B\| \|A-C\| } \right)\\ \beta &= \cos^{-1} \left( \frac{a^2 + c^2 - b^2}{2ac} \right) = \cos^{-1} \left( \frac{ \|B-C\|^2 + \|A-B\|^2 - \|A-C\|^2}{ 2 \|A-B\| \|B-C\| } \right)\\ \gamma &= \cos^{-1} \left( \frac{a^2 + b^2 - c^2}{2ab} \right) = \cos^{-1} \left( \frac{ \|B-C\|^2 + \|A-C\|^2 - \|A-B\|^2}{ 2 \|A-C\| \|B-C\| } \right),\end{align*} so we get $$p(A,B,C) = \left( \frac{1}{2} - \frac{\alpha(A,B,C)}{2\pi} \right) \left( \frac{1}{2} - \frac{\beta(A,B,C)}{2\pi} \right) \left( \frac{1}{2} - \frac{\gamma(A,B,C)}{2\pi} \right)$$ Putting this all together we get the probability that all three sheep see each other as $$P = \iint_{A \in [0,1]^2} \iint_{B \in [0,1]^2} \iint_{C \in [0,1]^2} p(A,B,C) \,dA\,dB\,dC,$$ which we will just use some Monte Carlo to estimate this probability.

After 10,000 Monte Carlo simulations for $A$, $B$, and $C,$ we obtain an estimated probability of all three sheep looking at one another $P \approx 2.72\%$. Knowing the way that these problems work out I wouldn't be surprised if somehow the integral works out to roughly $\frac{e}{100},$ but looking at the gnarliness of the integral, I would also be relatively surprised if it did.