Logic and Language

Copyright   James R Meyer    2012 - 2023 https://www.jamesrmeyer.com

The Bertrand paradox (Footnote: Bertrand, Joseph (1889), “Calcul des probabilités”, Gauthier-Villars, p. 5-6.) (Footnote: Not to be confused with the Bertrand economics paradox.) is the problem of giving a definitive answer to the question:

Given a circle and an equilateral triangle inscribed within it, where its vertices are points on the circle, if a chord of the circle is selected at random, what is the probability that the chord is longer than the side of the triangle?

This is considered to be paradoxical because different answers are obtained, depending on how the chord is “randomly selected ”. Before going further, we note that a side of the triangle divides the circumference of the circle into two arcs, one of which has twice the length of the other. In the following, the random chord will be called R and a chord of the length of the side of triangle is called T. Given that, then different answers that Bertrand gives are as follows (in the images below, red indicates that R is longer than T, and blue indicates that it is shorter: (Footnote: Images from Wikipedia by GNU Free Documentation License.)

(1)

By a random selection of endpoints on the circumference:

Here the argument is that two points are selected randomly on the circumference of the circle and the chord joins them. For any point P that is one endpoint of the chord R, if the chord R is less than the chord T, then there are two segments (to either side of P), whose arc is 1/3 of the circumference, where the other endpoint of the chord R can be. So the argument is that the probability that the chord R is less than a chord T is 2/3, and hence the probability that the chord R is longer than a chord T is 1/3.

(2)

By a random selection of a point within the circle:

Here the argument is that a point P is selected at random anywhere in the circle, and a radius is drawn through the point P. The chord R intersects the point P and is perpendicular to the radius. A perpendicular chord T intersects that radius at precisely half its length. The argument is that since half of the selected chords will be through a point P that gives a chord longer than a chord T, then the probability is 1/2.

(3)

By a random selection of a point within the circle:

Here the argument is that a point P is selected at random anywhere in the circle, and the chord is given by making that point P the midpoint of the chord. Now, every chord T has its midpoint on a circle C2 whose radius is exactly half of the original circle C1. Hence any chord R which is longer than a chord T will have its midpoint inside this circle C2. The argument is that since the area of the original circle C1 is 4 times the area of the circle C2, then the probability is 1/4. (Footnote: Note that the selection is the same in case (2) and (3). Since the point selected in (2) is necessarily the midpoint of the resultant chord, as is the case in (3), then the same chord is selected in both cases. That would seem to indicate that at least one of the methods of calculation of a probability in (2) or (3) cannot be correct.)

## Assumptions

All three methods require the notion of “selecting a real number at random ”, but at the same time, they have to make assumptions as to what this means.

In (1) given the first endpoint, the calculation of the probability of the other endpoint producing a chord longer than T is given by considering the selection of a point further around the circumference than 1/3 of that circumference, which is calculated by dividing the length of 1/3 of the circumference, by the circumference. This means that the assumption is that the probability of “selecting a real number at random ” from a part of a line segment is given by the length of that part divided by the total length of the line segment.

In (2) the calculation of the probability is by a consideration that if a point is randomly selected on an arbitrary radius, then the calculation is made based on whether that point is on a given side of the midpoint of the radius. That calculation is given by dividing the length of half the radius by the entire length of the radius. As for the above case, the assumption is that the probability of “selecting a real number at random ” from a part of a line segment is given by the length of that part divided by the total length of the line segment.

In (3), the calculation of the probability of “selecting a real number at random ” in a part of a circle is given by the area of the part divided by the total area of the circle.

But the use of such assumptions mean that one is not assuming a perfectly unbiased hypothetical selection method, since the method is tantamount to assuming that there are “more” numbers in a line length than there are in a part of the length, because the length in one case is greater than the other (or for case (3) that there are “more” numbers in a total area than there are in a part of that area). But the quantity of real numbers over the entire segment is infinite, and the quantity of real numbers over the partial length is infinite, and the real numbers in the partial length can be put in a one-to-one correspondence with the real numbers in the entire segment.

The “paradox” is simply the result of attempting to assign a valid meaning to the notion of “random selection ” when the sets in question are limitlessly large. Asking what the probability is like asking if there are “more” chords longer than a triangle chord T. And that is like asking are there “more” real numbers between 3/2 and 1 than there are between 0 and 3/2. For more on this see the page Random selection of numbers.

## Answers in terms of real situations

It is evident that many of the attempts to calculate an answer include an assumption that relates to some sort of real world situation. The reason why there are several different answers to the question is simply because there is no singular physical situation that is a definitive analogue of the hypothetical description - different people imagine different physical situations, each of which can be valid in its own way.

### Approximation of a chord

My preferred situation is where one might suppose that one has a circular tin whose base is covered with very thin sticks all aligned in the same direction. (Footnote: The calculated result is approximately valid when the sticks are much thinner than the diameter of the tin. The result becomes more accurate as the sticks get thinner.) The distribution of the sticks is uniform across the tin from one side to the other, so the distribution curve is a straight line given by:

y = 1/2 over   −1 < x < 1.

The total area under the distribution curve over  −1 < x < 1 is 1. The triangle chord length is 3/2 , which as y in our circle corresponds to an x value of 0.5. We want to find the area under the distribution curve between the x points of −0.5 to +0.5, which is given by the integral of the distribution curve between those points, and so this area is 1/2.

The probability value is this area divided by the total area, giving a probability value of 1/2 that a stick picked at random will be longer than 3/2. This happens to be the same answer as in (2) above.

### Approximation of a point

On the other hand, one might suppose that the base of the circular tin is covered with very small cubes, and one can calculate the probability that a cube picked at random will be within the area occupied by the sticks of the case above over a certain length. This can be said to be an approximation to the notion of “selecting a real number at random ” within the circle; the smaller the cube, the closer it is to the hypothetical situation. In this case, the distribution of the cubes is defined by a circle, and from this one can formulate an answer using the frequency of occurrence of the cubes across the width of the tin. From that we can deduce the probability of selecting a cube that is in a part of the tin where the sticks are greater than a given length.

The equation of a unit circle centered on the origin is

x2 + y2 = 1

and one can say that there is a distribution of the cubes given by:

y = 2 √(1 - x2) over -1 < x < 1.

The total area under the distribution curve is π. As above the triangle chord length is 3/2 , which corresponds to an x = 0.5. We want to find the area under the distribution curve between the x points of −0.5 to +0.5, which is given by the integral of the distribution curve between those points, and which is given by the evaluation of:

arcsin(x) + 1/2 Sin[2 arcsin(x)]

over the x values −0.5 to +0.5. This evaluates as π/3 + 3/2 (approx. 1.913). The probability value is this area divided by the total area (π), giving a probability value of:

1/3 + 3/ ≅ 0.61

that a cube picked at random will be from that part of the tin where the sticks are longer than  3/2. This result is quite different to the result given in (2) above, although this method might be considered to be a good approximation to the definition in (2) above. But this is not surprising, since it is a result that is specific for the physical situation it refers to. Here the selection of a small cube does not force that cube to be a midpoint of a stick, whereas (2) above forces the choice to be the midpoint of a chord. Other different and equally valid results can be calculated by assuming different physical situations. (Footnote: See, for example, the Wikipedia entry Bertrand paradox (probability).)

In summary, there isn’t really a paradox, it’s simply a question that is only meaningful in terms of a physical situation, but where the precise details of the physical situation are not defined.

Footnotes:

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Copyright   James R Meyer   2012 - 2023
https://www.jamesrmeyer.com