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Standing Table I

Akuna CapitalCitadelFive RingsJane Street

Three legs are attached to a round tabletop at points chosen independently and uniformly at random around its edge. What is the probability that the table stands?

Try it yourself first. Open a hint if you get stuck.

Hint 1

The table stands exactly when the centre of the tabletop lies inside the triangle formed by the three legs.

Hint 2

Fix the first leg. Measure the second leg by the angle $\theta$ between them along the shorter arc. Where can the third leg go?

Show solution

When does it stand? The table's weight acts at the centre $O$ . It stands exactly when $O$ is inside the triangle formed by the three legs. (Cases where $O$ lands exactly on an edge have probability $0$ .)

Fix the first two legs. Put leg $A$ anywhere. Measure leg $B$ by the angle $\theta$ from $A$ along the shorter arc, so $0 \le \theta \le \pi$ . Whether $B$ is clockwise or anticlockwise of $A$ doesn't matter by symmetry, so $\theta$ is uniform on $[0, \pi]$ .

Where must the third leg go? Let $A'$ and $B'$ be the points directly opposite $A$ and $B$ . The triangle contains $O$ exactly when leg $C$ lands on the arc between $A'$ and $B'$ , the one facing the short arc $AB$ . If $C$ lands anywhere else, all three legs fit in one half of the table and $O$ is outside the triangle. That arc has angle $\theta$ , so

P(\text{stands} \mid \theta) = \frac{\theta}{2\pi}.

Legs A and B are an angle θ apart. The arc between their opposite points A′ and B′ also has angle θ, and the table stands when C lands on it.

Average over $\theta$ . Since $\theta$ is uniform on $[0, \pi]$ ,

E[\theta] = \frac{1}{\pi}\int_0^{\pi} \theta \, d\theta = \frac{\pi}{2},

P(\text{stands}) = \frac{E[\theta]}{2\pi} = \frac{\pi/2}{2\pi} = \frac{1}{4}.

Answer: $0.25000$ .

Quick check: the table falls exactly when all three legs fit in some half of the table. For $n$ random points on a circle that happens with probability $n/2^{n-1}$ , which is $3/4$ for $n = 3$ . That leaves $1/4$ .

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