# Conditions required to yield exactly one solution in Bertrand's Paradox

Given two concentric circles ($S_1$, $S_2$) with radii $R_1=r$ and $R_2=\frac{r}2$, what is the probability, upon choosing a chord $c$ of the circle $S_1$ at random, that $c\:\cap\: S_2 \neq \phi$ ?

choose a chord of the larger circle at random and find the probability that it will intersect the smaller circle.

The same problem can also be stated as:

Given an equilateral triangle inscribed in a circle, find the probability of randomly choosing a chord of the circle greater than the length of a side of the triangle.

Bertrand's Paradox provides answers to the stated problem as $\frac12,\:\frac13,$ and even $\frac14$ at the same time, and all three answers are valid.

## The Question

Having said this, I intend to ask: what are the additional condition(s) and/or premises that one must provide while stating this question so that any ordinary mathematician would reach exactly one specific answer to this question?

This is a classical paradox showing that the ordinary, non-mathematical sense of the phrase "at random" is ambiguous. In mathematics, when you say "at random", you must specify a probability distribution. For example, in everyday life you can say "pick a random number from $1$ to $10$". You probably mean that every number should be picked with the same probability $1/10$. But strictly speaking, that doesn't follow from the phrase itself, and if I pick the number $1$ with probability $1/2$, and each of the others with probability $1/18$, it is still a random choice of number. To be precise, in mathematics you must say "pick a random number from $1$ to $10$ with equal probability" or "pick a number from $1$ to $10$ with respect to the uniform distribution".
Where we are talking about picking random points on a curve or some figure on the plane, the things get even more complicated. For example, let us pick a random point on the circle uniformly (each point with equal probability). What is the probability that we choose some concrete point $A$? It must be zero, because otherwise the sum of all points' probabilities will be infinite. But if the probability of picking each concrete point is zero, how we can pick a point at all?
Measure theory can give an answer to this question, but it is not very intuitive. We say that we assign to each possible event of the type "we picked a point in the set $A$" some probability $P(A)$ in such a way that $P(S)=1$, where $S$ is the set of all given points (circle in our case), and that we if have events $A_1,A_2,A_3,\cdots$, none of pairs $A_j,A_i$ of which cannot happen simultaneously, then $$P(\bigcup_i A_i)=\sum_i(P(A_i))\ \ (*)$$ This can be done on the circle in such a way that the probability of the set $A$ will be proportional to the intuitive notion of its length, and when we are talking about uniform distribution on the circle, we are in fact talking about this assignment $P$ (Lebesgue measure). All points get the probability zero, but that's not an obstacle, because we can't add a countable set of points to obtain something of non-zero length, and we cannot add uncountable quantities of them in $(*)$.
Shortly, when we say "pick a point on the circle uniformly", this means that the probability of getting into the set $A$ is the length of this set, divided by the length of the whole circle. However, it appears that the notion of length is not so easy to define mathematically, but it was done, as well as the area and the volume. Having all this in mind, we can formulate the Bertrand Problem as such: let us choose two random points on the circle uniformly with respect to the length and independently, then connect them by a chord. What is the probability that the chord intersects the inner circle? This formulation is strict, and a mathematician can give absolutely correct and unique answer: $1/3$, using the definitions of uniformity, sketched above, and independence (which means that the probability of point $a$ go to the set $A$, and $b$ to the set $B$ is the product of the probabilities $P(A)P(B)$). However, you could also give a different distribution on all chords: pick a random point in the circle uniformly with respect to the area and draw a chord, orthogonal to the radius-vector of this point, through this point. What is the probability that the chord intersects the inner circle? This is a different distribution and a different problem, as in the example with numbers from $1$ to $10$, so it's not a surprise that the answer can now be different: $1/4$. There are plenty other distributions all which are "choosing a chord at random", but of different nature and thus giving different answers to the question.