Probability a Three-Legged Disk Table Stands
You have a perfectly round disk. You attach three legs to the disk at three points chosen independently and uniformly at random on the circumference. The legs are perpendicular to the disk and firmly attached.
You flip the disk over so it rests on the three legs. The table stands upright if and only if the center of the disk lies inside the triangle formed by the three leg positions. Otherwise, the table tips over.
What is the probability that the table stands?
Hints
- The table stands exactly when the center of the disk lies inside the triangle formed by the three legs. Reframe this as a geometric probability question about points on a circle.
- Three points on a circle enclose the center if and only if no arc between consecutive points exceeds a semicircle ($\pi$ radians). Think about when all three points are bunched on one side.
- Fix one point at angle $0$ and let the other two be uniform on $[0, 2\pi)$. Condition on their order, write out the three arc-length constraints, and integrate over the feasible region.
Worked Solution
How to Think About It: The disk's center of mass is at its geometric center. The table stays upright exactly when that center point lies inside the triangle formed by the three legs. So this is a pure geometry question: if you pick three points uniformly at random on a circle, what is the probability the center of the circle is enclosed by the resulting triangle? Before doing any math, think about what configurations fail -- the triangle misses the center whenever all three points are bunched on one side of some diameter. That happens more often than you might guess.
Quick Estimate: Think about it with a symmetry argument. Fix one point at the top of the circle. The other two points each land on the left or right semicircle with equal probability. If both land on the same semicircle, the triangle cannot contain the center (all three points lie in a half-plane that does not contain the center). That happens with probability $1/2$, so the table stands at most half the time. But even when the two points land on opposite semicircles, the triangle still might not contain the center if one of the arcs is too wide. So the answer should be noticeably less than $1/2$. A reasonable guess is somewhere around $1/4$.
Approach: We use the arc-length characterization: three points on a circle enclose the center if and only if no arc between consecutive points exceeds $\pi$ (a semicircle). We then compute the probability by integration.
Formal Solution:
Place three points independently and uniformly on a circle of circumference $2\pi$. Fix one point at angle $0$ (by rotational symmetry, this costs nothing). Let the other two points have angles $U$ and $V$, each uniform on $[0, 2\pi)$.
Sort the three points around the circle. The three arcs between consecutive points sum to $2\pi$. The center lies inside the triangle if and only if every arc is strictly less than $\pi$.
Condition on $U < V$ (which happens with probability $1/2$). The three arc lengths are:
$$U, \quad V - U, \quad 2\pi - V$$
We need all three less than $\pi$:
$$U < \pi, \quad V - U < \pi, \quad 2\pi - V < \pi$$
The third condition gives $V > \pi$. Combined with the second, $V < U + \pi$. So the feasible region is:
$$0 < U < \pi, \quad \pi < V < U + \pi$$
The joint density is $1/(2\pi)^2$. The area of this region is:
$$\int_0^{\pi} \int_{\pi}^{U + \pi} dV\, dU = \int_0^{\pi} U\, dU = \frac{\pi^2}{2}$$
So the probability in the $U < V$ case is:
$$\frac{\pi^2 / 2}{(2\pi)^2} = \frac{\pi^2}{2 \cdot 4\pi^2} = \frac{1}{8}$$
By symmetry, the $V < U$ case contributes another $1/8$.
Answer:
$$P(\text{table stands}) = \frac{1}{8} + \frac{1}{8} = \boxed{\frac{1}{4}}$$
A three-legged table with legs placed uniformly at random on the rim of a circular disk stands with probability $1/4$.
Intuition
The key geometric fact is clean: three points on a circle enclose the center if and only if no semicircle contains all three points, or equivalently, every arc between consecutive points is less than $\pi$. Once you know this characterization, the problem reduces to a short integration. The answer $1/4$ is surprisingly low -- three random points on a circle fail to surround the center three-quarters of the time. Intuitively, it is easy for all three points to drift onto one half of the circle, leaving the center exposed.
This result shows up more broadly in geometric probability and random polytope theory. The probability that $n$ random points on a circle contain the center is $1/2^{n-1}$, so each additional point only doubles your chance. In practical settings, the lesson is that "random" placement of supports is a terrible strategy for stability -- you want to spread legs as evenly as possible (e.g., $120^\circ$ apart for three legs) to guarantee the center is enclosed. This is a nice example of how symmetry and a clean geometric criterion turn a vague physical question into a crisp calculation.