Bertrand’s paradox in probability theory

bertrand paradox prob theoryBertrand Paradox. 

We are given a circle C of radius r and we wish to determine the probability p that the length l of a “randomly selected” chord AB is greater than the length r \sqrt {3} of the inscribed equilateral triangle.

We shall show that this problem can be given at least three reasonable solutions

I. If the center M of the chord AB lies inside the circle C_{1}  of radius r/2 shown in figure a (attached), then l > r \sqrt{3}. It is reasonable, therefore, to consider as favourable outcomes all points inside the circle C_{1} and as possible outcomes all points inside the circle C. Using as measure of their numbers the corresponding areas \pi r^{2}/4 and \pi r^{2}, we conclude that

p= \frac {\pi r^{2}/4}{\pi r^{2}}=1/4

II. We now assume that the end A of the chord AB is fixed. This reduces the number of possibilities, but it has no effect on the value of p because the number of favourable locations of B is reduced proportionately. If B is on the 120 degree arc of DBE of figure b (attached), then l > r \sqrt{3}. The favourable outcomes are now the points on this arc and the total outcomes all points on the circumference of the circle C. Using as their measurements, the corresponding lengths 2\pi r/3 and 2\pi r, we obtain

p= \frac {2\pi r/3}{2 \pi r}=1/3.

III. We assume finally that the direction of AB is perpendicular to the line FK of figure c (attached). As in II, the restriction has no effect on the value of p. If the center M of AB is between G and H, then l >r \sqrt{3}. Favourable outcomes are now the points on GH and possible outcomes all points on FK. Using as their measures the respective lengths r and 2r, we obtain

p= r/2r=1/2.

We have thus found not one but three different solutions for the same problem! One might remark that these solutions correspond to three different experiments. This is true but not obvious and, in any case, it demonstrates the ambiguities associated with the classical definition, and the need for a clear specification of the outcomes of an experiment and the meaning of the terms “possible” and “favourable.”

More later,

Nalin Pithwa

why is an empty set a subset of itself

Note first of all the definition of a subset: We say A \subset B only when: if x \in A then x \in B.

The empty set is the set with no elements. It is denoted by the symbol \phi. A source of confusion to beginners is that although the empty set consists of nothing, it itself is something (namely, some particular set, the one characterized by the fact that nothing is in it). The set \{\phi \} is a set containing exactly one element, namely the empty set. (In a similar way, when dealing with numbers, say with ordinary integers, we must be careful not to regard the number zero as nothing; zero is something, a particular number, which represents the numbers in “nothing”. Thus, zero and \phi are quite different, but there is a connection between them in that the set has zero elements.) Note that for any set X, we have

\phi \subset X and X \subset X.

A special case of both of these statements is the statement  which occasions difficulty if, as is often improperly done, one reads “is contained in” for both of the symbols \phi and \in The statement

\phi \subset \phi

is true because the statement “for each x \in \phi, we have x \in \phi” is obviously true, and also because it is “vacuously true”, that is there is no for which the statement must be verified, just as the statement “all pigs with wings speak Chinese” is vacuously true.


Do send your questions, suggestions, comments, more such seemingly paradoxical statements to me.

More later, happy new year 🙂

Nalin Pithwa