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 $2(1 + 4 + 6) = 22$ of the $32$ sign patterns. Beware $5 \cdot (1/2)^4 = 5/16$: that is the answer for five points on a circle, where "everyone else lies in the semicircle clockwise from point $i$" gives disjoint events. On a sphere one point does not pin down a hemisphere, so that count misses most of the good configurations.
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 (small-$n$ anchors): Anchor with cases you can reason out without the general formula. Any three points fit in a hemisphere: take the plane through the origin parallel to the plane through the three points, and all three sit on one side of it, so $P(3)=1$. For four points, the classic Putnam result is that the random tetrahedron contains the center with probability $1/8$, and the four points miss every closed hemisphere exactly when it does, so $P(4)=7/8=0.875$. A fifth point can only make containment harder, so $P(5)<0.875$; the exact answer $11/16\approx 0.69$ below fits.
Approach: Use antipodal symmetry, count the sign regions that $5$ planes through the origin cut $\mathbb{R}^3$ into, and divide by the $2^5$ equally likely sign patterns.
Formal Solution:
Step 1 (antipodal pairs). Generate the points in two stages: first draw $5$ random diameters of the sphere (each an antipodal pair $\{Q_i,-Q_i\}$), then flip an independent fair coin for each $i$ to set $P_i=+Q_i$ or $P_i=-Q_i$. This produces the same uniform distribution, and given the diameters, the $2^5=32$ sign configurations $(\pm,\pm,\pm,\pm,\pm)$ are equally likely.
Step 2 (hemisphere $\leftrightarrow$ region). All five points lie in the closed hemisphere with pole $\vec v$ iff $\vec v\cdot P_i\ge 0$ for all $i$. Fix the diameters and consider the five planes $\{x:x\cdot Q_i=0\}$. The sign choice $(\epsilon_1,\ldots,\epsilon_5)$ puts all five points in a common closed hemisphere iff the region $\{x:\epsilon_i\,(x\cdot Q_i)>0\ \text{for all } i\}$ is nonempty (any $x$ in it serves as a pole $\vec v$). Each region of the plane arrangement corresponds to exactly one such sign pattern, so the number of sign choices that fit in some hemisphere equals the number of 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.