Pyrknot Planet Visibility Problem

An alien on planet Pyrknot (a perfect sphere of radius $R$) observes six planets that appear nightly in uniformly random locations across the sky. The alien says:

"If at a given moment there exists somewhere on Pyrknot that all six planets are visible, then from my friend's position on the surface they have an $\alpha$ probability of also seeing all of the planets."

A tower is then considered. It aggregates visibility: the tower can see every planet that is visible from at least one surface point within distance $r$ of the tower's base.

Task: In the limit where $r \ll R$, the new observation probability is $\alpha + \beta \cdot (r/R)$. Find exact values of $\alpha$ and $\beta$.

Setup and Key Definitions#

Planet Visibility#

Pyrknot is a sphere of radius $R$. A planet at direction $\mathbf{d} \in S^2$ (unit vector from the center) is visible from surface point $\mathbf{p} \in S^2$ if it is above the local horizon, i.e.:

$\mathbf{p} \cdot \mathbf{d} > 0$

The set of surface points from which planet $i$ is visible is the open hemisphere:

$H(\mathbf{d}_i) = \{ \mathbf{p} \in S^2 : \mathbf{p} \cdot \mathbf{d}_i > 0 \}$

The All-Visible Region#

The region where all 6 planets are simultaneously visible is:

$V = \bigcap_{i=1}^{6} H(\mathbf{d}_i)$

The condition "there exists somewhere on Pyrknot that all six planets are visible" is exactly $V \neq \emptyset$.

Geometrically, $V \neq \emptyset$ iff the six direction vectors $\mathbf{d}_1, \ldots, \mathbf{d}_6$ all lie in some common open hemisphere — since a point $\mathbf{p}$ sees all 6 iff $\mathbf{p}$ is in the hemisphere "opposite" to all $\mathbf{d}_i$.

Part 1: Computing $\alpha$ (No Tower)

The problem asks for a conditional probability:

$\alpha = P(\text{friend sees all 6} \mid \exists\, \mathbf{p} \in S^2 : \text{all 6 visible from } \mathbf{p})$

By Bayes' theorem (since "friend sees all 6" $\Rightarrow$ "someone sees all 6"):

$\alpha = \frac{P(\text{friend sees all 6})}{P(V \neq \emptyset)}$

Numerator: $P(\text{friend sees all 6})$

The six planets are placed independently and uniformly on $S^2$. For a fixed friend position $\mathbf{q}$, each planet independently satisfies:

$P(\mathbf{q} \cdot \mathbf{d}_i > 0) = \frac{1}{2}$

by symmetry (half the sphere is above any horizon). By independence:

$P(\text{friend sees all 6}) = \left(\frac{1}{2}\right)^6 = \frac{1}{64}$

Denominator: $P(V \neq \emptyset)$

We need the probability that 6 uniform random points on $S^2$ all lie in some common open hemisphere. This is a classical result.

Wendel / Cover–Efron Formula: For $n$ i.i.d. uniform points on $S^{d-1}$ (a sphere in $\mathbb{R}^d$), the probability they all lie in some open hemisphere is:

$p(n, d) = \frac{1}{2^{n-1}} \sum_{k=0}^{d-1} \binom{n-1}{k}$

For our sphere $S^2$ (so $d = 3$) with $n = 6$ planets:

$p(6, 3) = \frac{1}{2^5} \left[ \binom{5}{0} + \binom{5}{1} + \binom{5}{2} \right] = \frac{1}{32}(1 + 5 + 10) = \frac{16}{32} = \frac{1}{2}$

So $P(V \neq \emptyset) = \dfrac{1}{2}$.

Result#

$\boxed{\alpha = \frac{1/64}{1/2} = \frac{1}{32}}$

Part 2: Computing $\beta$ (With Tower)

Tower Visibility Condition#

The tower at base position $\mathbf{q}$ aggregates visibility from all surface points within distance $r$ of the base. Letting $\varepsilon = r/R$ (the angular radius of the disk), the tower sees planet $i$ iff:

$\exists\, \mathbf{p} \in D(\mathbf{q}, \varepsilon) : \mathbf{p} \cdot \mathbf{d}_i > 0$

where $D(\mathbf{q}, \varepsilon) = \{ \mathbf{p} \in S^2 : \angle(\mathbf{q}, \mathbf{p}) < \varepsilon \}$ is the spherical cap of angular radius $\varepsilon$ centered at $\mathbf{q}$.

The maximum of $\mathbf{p} \cdot \mathbf{d}_i$ over $\mathbf{p} \in D(\mathbf{q}, \varepsilon)$ is achieved at the point in the cap closest to $\mathbf{d}_i$. If $\theta = \angle(\mathbf{q}, \mathbf{d}_i)$:

$\max_{\mathbf{p} \in D(\mathbf{q}, \varepsilon)} \mathbf{p} \cdot \mathbf{d}_i = \cos\!\bigl(\max(0, \theta - \varepsilon)\bigr)$

This is positive iff $\max(0, \theta - \varepsilon) < \pi/2$, i.e., $\theta < \pi/2 + \varepsilon$, i.e.:

$\mathbf{q} \cdot \mathbf{d}_i > -\sin(\varepsilon)$

So: planet $i$ is visible from the tower iff $\angle(\mathbf{q}, \mathbf{d}_i) < \pi/2 + \varepsilon$.

Per-Planet Tower Probability (3D Geometry)#

Since $\mathbf{d}_i$ is uniform on $S^2$, the probability that a uniform point lies within angle $\pi/2 + \varepsilon$ of the fixed point $\mathbf{q}$ is:

$P\!\left(\angle(\mathbf{q}, \mathbf{d}) < \frac{\pi}{2} + \varepsilon\right) = \frac{1 - \cos(\pi/2 + \varepsilon)}{2} = \frac{1 + \sin(\varepsilon)}{2} \approx \frac{1}{2} + \frac{\varepsilon}{2}$

Note on 3D vs 2D: In a 2D cross-sectional model (circular sky), the analogous formula gives $\frac{1}{2} + \frac{\varepsilon}{\pi}$. This is incorrect for our 3D sphere. The correct coefficient is $\frac{1}{2}$, not $\frac{1}{\pi}$, because we integrate over the spherical measure $\sin\theta\, d\theta$, not the circular measure $d\theta$.

All-6 Tower Probability#

Since the 6 planets are independent:

$P(\text{tower sees all 6}) = \prod_{i=1}^{6} P\!\left(\mathbf{q} \cdot \mathbf{d}_i > -\sin\varepsilon\right) = \left(\frac{1 + \sin\varepsilon}{2}\right)^6$

Expanding to first order in $\varepsilon = r/R$:

$\left(\frac{1}{2} + \frac{\varepsilon}{2}\right)^6 = \frac{1}{64}(1 + \varepsilon)^6 \approx \frac{1}{64}\left(1 + 6\varepsilon\right) = \frac{1}{64} + \frac{6\varepsilon}{64} = \frac{1}{64} + \frac{3}{32}\cdot\frac{r}{R}$

Validity of the First-Order Approximation#

One must verify that the event $\lnot A$ (no common viewpoint exists) does not contribute at first order. When one planet is in the thin band $\mathbf{q} \cdot \mathbf{d}_1 \in (-\varepsilon, 0)$ and the remaining five are strictly above the equator of $\mathbf{q}$, a small tilt of the hemisphere toward $\mathbf{d}_1$ generically includes all six. The probability of the tilt failing (some $\mathbf{d}_j$ dropping below the horizon) requires two planets near the equator simultaneously, contributing $O(\varepsilon^2)$. Therefore, the correction from $\lnot A$ does not affect the first-order term.

Conditional Probability with Tower#

$P(\text{tower sees all 6} \mid V \neq \emptyset) \approx \frac{\dfrac{1}{64} + \dfrac{3}{32}\cdot\dfrac{r}{R}}{\dfrac{1}{2}} = \frac{1}{32} + \frac{3}{16}\cdot\frac{r}{R}$

$\boxed{\beta = \frac{3}{16}}$

Final Answers#

$\alpha = \frac{1}{32}, \qquad \beta = \frac{3}{16}$

Summary of Errors in the 2D Approach#

The denominator $P(V \neq \emptyset) = 1/2$ comes from the Wendel formula, a non-trivial combinatorial geometry result. Without it, the conditional probability structure of the problem is missed entirely.