Five Points on a Hemisphere
Five points are placed independently and uniformly at random on the surface of a sphere. What is the probability that there exists some closed hemisphere such that all five points lie within it?
A closed hemisphere is half of the sphere, including its bounding great circle.
Hints
- Searching over all hemispheres is hopeless. Reframe it: the five points lie in the closed hemisphere with pole $\vec v$ exactly when $\vec v \cdot P_i \ge 0$ for every $i$ -- so the question is which sign patterns $(\pm, \pm, \pm, \pm, \pm)$ are achievable by some pole.
- By antipodal symmetry each point is equally likely to be $P_i$ or $-P_i$, giving $2^5 = 32$ equally likely sign configurations. The achievable ones correspond exactly to the regions that the five planes $\{x : x \cdot P_i = 0\}$ cut $\mathbb{R}^3$ into -- so count regions.
- $n$ planes through the origin in general position in $\mathbb{R}^d$ create $2\sum_{k=0}^{d-1}\binom{n-1}{k}$ regions; for $n = 5$, $d = 3$ that is $2(1 + 4 + 6) = 22$ out of $32$. Beware the tempting union-bound answer $5 \cdot (1/2)^4 = 5/16$: those events are NOT disjoint, and it badly undercounts.
Worked Solution
How to Think About It: Brute-force integration over all hemispheres is intractable. The elegant route is antipodal symmetry + region counting (Wendel's theorem): treat each point as defining a plane through the origin, and a hemisphere as a choice of which side of each plane you're on. The clean fact is that the answer depends only on how many *sign regions* $n$ planes cut space into. Name the heuristic: reduce a continuous geometric probability to a discrete count of arrangements.
Quick Estimate (extreme-case + small-$n$ anchors): Anchor the formula with cases you can reason out. For $n=1$ point on a sphere it trivially fits some hemisphere: $P=1$. For $n=2$, any two points always share a hemisphere: $P=1$. For $n=3$ on a sphere, three points *always* fit in a hemisphere too, $P=1$. The first time it can fail is $n=4$ (the tetrahedron-like spread), where Wendel gives $P=\tfrac{1}{2^{3}}\sum_{k=0}^{2}\binom{3}{k}=\tfrac{1+3+3}{8}=\tfrac78=0.875$. Adding a fifth point makes containment harder, so $P(5)<0.875$; Wendel's $\tfrac{1}{2^{4}}(1+4+6)=\tfrac{11}{16}=0.6875$ sits just below — consistent. So the answer should be a bit under $0.875$, and $11/16\approx 0.69$ lands right there.
Approach: Antipodal symmetry → count sign regions of $5$ planes in $\mathbb{R}^3$ → divide by $2^{n-1}$.
Formal Solution:
Step 1 (antipodal pairs). Each point $P_i$ and its antipode $-P_i$ are equally likely, giving $2^5=32$ equally likely sign configurations $(\pm,\pm,\pm,\pm,\pm)$ — one sign per point recording which side of the plane $\{x:x\cdot P_i=0\}$ a candidate pole lies on.
Step 2 (hemisphere $\leftrightarrow$ sign pattern). All five points lie in the closed hemisphere with pole $\vec v$ iff $\vec v\cdot P_i\ge 0$ for all $i$. So a set of points fits in *some* hemisphere iff the sign pattern of the poles realizes an all-$+$ (up to relabeling) region carved out by the five planes. Counting achievable sign patterns = counting the regions the planes create.
Step 3 (Schläfli / Wendel region count). Five planes through the origin in general position in $\mathbb{R}^3$ cut space into $$R = 2\sum_{k=0}^{d-1}\binom{n-1}{k} = 2\Big(\binom{4}{0}+\binom{4}{1}+\binom{4}{2}\Big)=2(1+4+6)=22$$ regions ($n=5,\ d=3$).
Step 4 (probability). Of the $32$ equally likely antipodal configurations, exactly $22$ correspond to a hemisphere-containable arrangement, so $$P=\frac{22}{32}=\boxed{\tfrac{11}{16}}=0.6875.$$
General (Wendel, 1962): for $n$ points on the sphere in $\mathbb{R}^d$, $P(n,d)=\dfrac{1}{2^{n-1}}\sum_{k=0}^{d-1}\binom{n-1}{k}$; here $d=3,n=5$ gives $11/16$.
Answer: $\dfrac{11}{16}$.
Intuition
This problem illustrates one of the most powerful techniques in geometric probability: replacing integration over geometric objects with a discrete counting argument via symmetry.
The key insight is the antipodal trick. Each random point $P_i$ defines an antipodal pair $\{P_i, -P_i\}$, and by symmetry, either member of the pair is equally likely. The question "do all five points fit in some hemisphere?" becomes "how many of the $2^5 = 32$ sign configurations correspond to a valid hemisphere?"
The answer comes from Schläfli's formula, which counts how many regions $n$ hyperplanes through the origin create in $\mathbb{R}^d$. Each region corresponds to one sign pattern that a single hemisphere can capture. For 5 planes in 3D, this gives 22 regions out of 32 total configurations, yielding the probability $11/16$.
The common mistake is the "anchor point" argument: fix one point as the hemisphere's boundary, note the other 4 each land on the correct side with probability $1/2$, giving $5 \times (1/2)^4 = 5/16$. This is wrong because the events are NOT mutually exclusive — multiple points can simultaneously serve as boundary points of a valid hemisphere. The correct answer $11/16$ is more than double $5/16$, which makes sense geometrically: five points on a sphere fit in a hemisphere more often than not.