It really isn't a paradox. It is more of a warning that statistics is a quirky area of mathematics.
It hinges on how a "random chord" is chosen. With no process defined, then the solver of the problem is allowed to use whatever process she wants. Choosing different processes gives different probabilities.
But let's turn the problem on its head.
Let's define a random chord as the chord between two points chosen randomly from the circumference of a "measuring circle." We'll let the measuring circle be concentric with the circle of interest. Now let's see what happens as we increase the diameter of the measuring circle.
Diameter of measuring circle probability
< 0.25 1
>0.25 to <1 drops from 1 to 1/3
1 1/3
100,000 almost 1/2
Bertrand's 1/4 solution is "wrong" for a circle. It requires that the midpoint of the chord is evenly distributed over the area of the circle. He just hand waves away the center of the circle since it is just one point and one point has no area. But by definition the center of the circle has an infinite number of diameters passing through it, not 1. So the center of the circle must be an asymptote. Hence "circleness" prevents this solution.
There is also a symmetry wrinkle pointed out by Poincare. Consider the flip problem. We throw a circle of fixed diameter onto a line. Consider a hoop being tossed as a line on the floor. Given that the hoop intersects the line, then the only probability is 1/2.
Considering Bertrand's "paradox":
I've made a hobby of collecting references to this problem and I have about 500!
Note part of the conundrum in Bertrand's paradox is figuring out "chord" and "secant".
Choosing a midpoint over the area is just wrong. Bertrand didn't say "circular area" he explicitly said circle. We all learned in grade school that a circle has an infinite number of diameters. So the center isn't just any point in the area, it is an asymptote. Thus at the center of the circle the density of points must go to infinity. aka Gabriel's horn, or Gabriel's wedding cake. (oddly I think just one of my references points out this fact.)
Now consider choosing two points randomly over the circumference of the circle. However let's add a twist. Let's use a measuring circle (radius r) concentric with the chosen circle R (radius R). The midpoint of any diameter longer than the side of the inscribed equilateral triangle triangle must be within the circle inscribed in the triangle. So for two randomly chosen points on our measuring circle then there are three different intervals to consider for $r$:
- [0, 1/2R] the probability of a longer chord is 1
- (1/2R, R] the probability of a chord decreases from 1 to a 1/3
- (R, $\infty$] the probability rises from just over a 1/3 to 1/2
Note that splicing the various curves together does not result in a continuous function. In particular at r=R there is a discontinuity.
So there are two "magic" radii to consider. First when r=R, and also when r=$\infty$. Of course when counting chords for r>R one must also stipulate that the chord cuts circle R.
Obviously Bertrand's paradox relies on the symmetry of the circle. Now generalize the problem. Given an equilateral triangle of side $\sqrt{3}$ what is the probability that a random chord is longer than the radius of the inscribed circle? Now it isn't so easy to apply geometric probability when choosing two points on the circumference. It is just as easy for a Monte Carlo computer program.
Let's think deeper about this problem however. If the measuring circle has a radius of R, what happens at a circle R' 1 kilometer away? All of the random lines generated from the circle R when r=R will be essentially parallel at circle R'. So using two random points on the circumference of circle R doesn't generate a whole plane of random lines. However if we make our measuring circle 1 astronomical unit in diameter, the the lines are essentially isotropic and of the same density at both our circle R and at our circle R'.
Now let's also think of the problem another way. I have two sheets of rigid plastic. On one is a circle and on the other much larger one is a line. Is throwing the line on the circle different that throwing the circle on the line? For throwing a circle on a line the answer is unequivocally 1/2 withe condition that only throws where the circle intersects the line are counted. From Buffon's needle problem we can see that a large area covered with parallel lines space R distance apart will always yield one random chord.
The use of a measuring circle of infinite radius immediately follows from the work of Crofton who found that for for convex body the number of random lines of the plane cutting the convex body is equal to the perimeter of the body. Interestingly for a line segment the perimeter is two times the length - think about an infinitely thin rectangle.
Best Answer
As you know given a circle of diameter $D,$ the side of an inscribed equilateral triangle is $s = \frac{\sqrt3}{2} D.$
Answer to Part 1)
The $\frac13$ answer comes from a randomization procedure in which we choose a point $P$ with uniform probability on the circumference of the circle, choose another point $Q$ with uniform probability on the circumference of the circle independently of $P,$ and then construct the chord $PQ.$ How do you compute the probability that $\lvert PQ\rvert > s$ under these particular conditions?
You don't actually have to inscribe any triangle with a vertex at $P$ if you don't want to. You just have to be able to deduce, in some way or another, that if the shorter arc length between $P$ and $Q$ is less than or equal to $\frac13 \pi D,$ then $\lvert PQ\rvert \leq s$. Otherwise $\lvert PQ\rvert > s$.
One way to deduce this fact involves inscribing an equilateral triangle (not necessarily "the" triangle of side $s$) with one vertex at $P.$ A more laborious way is to construct the triangle $\triangle OPQ$ where $O$ is the center of the circle, and use trigonometry to determine that if $\angle OPQ \leq \frac23 \pi$ then $\lvert PQ\rvert \leq s$ and if $\angle OPQ > \frac23 \pi$ then $\lvert PQ\rvert > s$. This second method does not involve any equilateral triangles.
Now to determine the probability that $\angle OPQ < \frac23 \pi$, one way (a laborious way) is to partition the circumference of the circle into many arcs of equal length before choosing either point $P$ or point $Q.$ Now choose $P$ and $Q$ so that each falls on one of the arcs (possibly the same arc both times). Then sometimes $P$ and $Q$ will fall on arcs such that if you choose any points $A$ and $B$ on those respective arcs, then $\angle AOB \leq \frac23 \pi,$ and therefore you know that $\angle OPQ \leq \frac23 \pi$. Sometimes $P$ and $Q$ will fall on arcs such that if you choose any points $A$ and $B$ on those respective arcs, then $\angle AOB > \frac23 \pi,$ and therefore you know that $\angle OPQ > \frac23 \pi$. And sometimes $P$ and $Q$ will fall on arcs such that the angle formed by points on the respective arcs might be either less than or greater than $\frac23 \pi,$ and you don't know whether $\angle OPQ > \frac23 \pi$. But you have a lower bound and upper bound on $P(\angle OPQ > \frac23 \pi),$ namely, the probability that $P$ and $Q$ land on a pair of arcs of the second kind, and the probability that $P$ and $Q$ do not land on a pair of arcs of the first kind, keeping in mind that $P$ is equally likely to fall on any of the small arcs and so is $Q$ (because each point is uniformly distributed over the circumference).
Now partition the circle into sets of smaller and smaller arcs to get a better approximation of the probability, and see what happens to the bounds on the probability as the size of the small arcs goes to zero. If you do all the arithmetic accurately you'll get the same result that the equilateral-triangle people got instantly: $P(\angle OPQ > \frac23 \pi) = \frac13.$ But you can do this without constructing any figures (other than the chord $PQ$) after the points $P$ and $Q$ are randomly chosen.
Answer to Part 2)
You can indeed select a random chord so that the probability distribution of its length is as you describe. Here's how:
First choose $X$ randomly with uniform distribution on the interval $[0,D].$ Then, to satisfy the intuition that the chord should be equally likely to sit in any orientation on the circle, choose $Y$ randomly with uniform distribution on the interval $[0,2\pi).$ Choose a radius of the circle at angle $Y.$ Choose a point $M$ on that radius at a distance $\frac12\sqrt{D^2 - X^2}$ from the center of the circle. Construct a chord of length $X$ through $M$ perpendicular to the chosen radius.
I believe that's a perfectly good probability distribution of chords on a circle, it is capable of producing any chord that exists, and it has exactly the distribution of chord length that you said. And you are correct that according to this way of selecting a random chord, $P(\lvert PQ\rvert > s) = 1 - \frac{\sqrt3}{2}.$
What this signifies is that you have found a fourth interpretation of the "random chord" in Betrand's paradox, which you can put alongside the three other interpretations that most people list:
And you have thereby reinforced the lesson of the "paradox," that when you say something is random you need to have a concept of how that randomness is defined.
Is your idea a reasonable one for randomizing a chord? I think so. You can implement it by cutting a magnetized needle to a random length and tossing it randomly inside an iron ring on a level table. Because the needle is magnetized, it is attracted to the ring and sticks to it, forming a chord of the circle inside the ring.
Summary
If someone indeed gave you Bertrand's paradox as an exercise without specifying any particular randomization procedure, I think it would be unfair of them to expect you to come up with the answer $\frac13.$ After all, there are two other well-known interpretations of a "random chord" that give different answers.
If no specific probability distribution was described, my opinion is that there is no mistake in your answer. It's just as valid as any of the well-known answers.
Whether your answer is complete depends on exactly how the problem was put to you. If your task was to find more than one way to randomize the chord and produce a second answer with a different probability (thereby illustrating the "paradox"), then of course a response with only one probability is incomplete. If the person setting the problem was careful to specify the random distribution in an unambiguous way, then a complete response should at least find the probability according to the specified distribution, not just according to another distribution you devised. But if you were just told "a random chord" and expected to come up with one probability, I would argue that your answer is one of several possible correct answers.