Standing Table II

Question

Five points are chosen independently and uniformly at random on the circumference of a circular table. Legs are attached at those five points. What is the probability that the table stands?

QuantSoc · Accepted Answer

The table will stand exactly when the pentagon formed by joining the five attachment points contains the centre of the circle. This occurs precisely when the five points are not all contained within any semicircular arc.

From the result of “same arc I”, the probability that five independently chosen points on a circle all lie within some semicircle is:

$$
\frac{5}{16}
$$

Therefore, the probability that they do not all lie within a semicircle is:

$$
1 - \frac{5}{16} = \frac{11}{16}
$$

Hence, the required probability is:

$$
\boxed{\frac{11}{16}}
$$