Bertrand's Paradox confusing So I recently heard about Bertrand's Paradox
So I have 2 worries
1) I don't understand the reasoning behind the $\frac{1}{3}$ answer
I mean when a point is taken at one of the vertices of the triangle and the probability is calculated is that the probability that any chord will be longer than the side of the triangle?
Or is it instead the probability that any chord with one end at the vertex of the triangle will be longer than the side?
So when one point is taken at one of the vertices it doesn't include all the chords of the circle
I heard someone say by symmetry but I don't really understand
2) When the problem was posed to me just as an exercise I said the length of the chord follows a uniform distribution on $[0,D]$ where D is the Diameter of the circle
Then I calculated the probability $P(X>s)$ where X is the random variable describing the length of the chord and s is the side of the triangle 
Knowing that the Diameter of that circle is $\frac{2s}{\sqrt{3}}$ I got a probability of $1-\frac{\sqrt{3}}{2}$
So my second worry is for someone to explain where I went wrong
Thanks in advance 
have a nice day y'all 
 A: 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:


*

*Choose each endpoint uniformly on the circumference.

*Choose the midpoint uniformly along a radius whose direction is uniformly chosen over $[0,2\pi).$

*Choose the midpoint uniformly over the area inside the circle.


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.
