Standing Table I

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

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

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

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

so

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 $1/4$.