Contains the center? 1

I will assume that the cyclic quadrilateral is convex, i.e. the sides do not intersect each other.

Call the given points $P_1, \dots, P_4$ .

In order for the center of the circle not to lie in the quadrilateral, there must be an arc of more than $180^\circ$ without points. Let $P_a$ be the point immediately after this long arc in clockwise direction. The other three points then lie within $180^\circ$ , or half a circle, clockwise from $P_a$ . The probability that they do so is $\tfrac12$ for each point, or $(\tfrac12)^3 = \tfrac18$ combined.

That first point $P_a$ may be any of the four chosen points. Therefore the total probability that the center of the circle does not lie in the quadrilateral is $4 \times \tfrac18 = \tfrac12$ .

The probability that the center of the circle does lie inside the quadrilateral is the complement, $1 - \tfrac12 = \boxed{\tfrac12}$ .