Contains the Center? 3

A variation on my solutions to this problem and this problem .

The center of the square does not lie in the triangle if one can draw a "diameter", i.e. a line segment from one side of the square to the opposite side through the center, that does not intersect with the triangle. The end points of this "diameter" divide the square into two sections of equal length $L/2$ . The three vertices of the triangle must lie entirely in one segment.

Suppose this is the case. Then one of the chosen points $P_a$ is such that the other two points lie within a distance $L/2$ measured in clockwise direction along the perimeter. The probability that these two points happen to be there is $(\tfrac12)^2 = \tfrac14$ . The role of $P_a$ may be played by any of the three points, so the odds that the center of the square does not lie in the triangle is $3\times \tfrac14 = \tfrac34$ .

The probability that the center of the square lies inside the triangle is the complement, $1 - \tfrac34 = \tfrac14 = \boxed{0.25}$ .