Round table on three randomly placed legs

A table top is a circular disc, and three legs are attached at points chosen independently and uniformly on its rim. The table stands only if its centre lies inside the triangle formed by the three legs. What is the probability that it stands?

$1/4$. The centre is inside the triangle iff the three points are not all within one semicircle. The probability that $n$ random points lie in some semicircle is $\tfrac{n}{2^{n-1}}$, which is $\tfrac34$ for $n=3$, so the table stands with probability $\tfrac14$.