Bertrand's paradox: "... centre point lies in one of the two centre quarters of the diameter perpendicular to this direction ..."? I am currently studying Photogrammetric Computer Vision – Statistics, Geometry, Orientation and Reconstruction by Förstner and Wrobel. Chapter 2 Probability Theory and Random Variables gives the following example:

In the case of alternatives which are not countable, e.g., when the event is to be represented by a real number, we have difficulties in defining equally probable events. This is impressively demonstrated by Bertrand’s paradox (Fig. 2.1), which answers the question: What is the probability of an arbitrarily chosen secant in a circle longer than the side of an inscribing equilateral triangle? We have three alternatives for specifying the experiment:

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*Choose an arbitrary point in the circle. If it lies within the concentric circle with half the radius, then the secant having this point as centre point is longer than the sides of the inscribing triangle. The probability is then $1/4$.

*Choose an arbitrary point on the circle. The second point of the secant lies on one of the three segments inducing sectors of $60^\circ$. If the second point lies in the middle sector the secant through these points is longer than the side of the inscribing triangle. The probability is then $1/3$.

*Choose an arbitrary direction for the secant. If its centre point lies in one of the two centre quarters of the diameter perpendicular to this direction the secant is longer than the side of the inscribing triangle. The probability is then $1/2$.


I don't quite understand the description given by 3.:

If its centre point lies in one of the two centre quarters of the diameter perpendicular to this direction the secant is longer than the side of the inscribing triangle. The probability is then $1/2$.

What is meant by "two centre quarters of the diameter perpendicular to this direction", and how is the probability $1/2$? A clearer explanation would be greatly appreciated.

*

*In the hexagram picture on the right, the chosen direction of the secant is horizontal


*so the centre of the secant is on the vertical line


*and the secant will be too short if the centre is the top quarter of the vertical line or the bottom quarter (the thin parts)


*but the secant will long enough if the centre is the middle two quarters (the thick parts of the vertical line)
