Bertrand's paradox (probability)
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- This article is about Bertrand's paradox in probability theory. For Bertrand paradox related to economics, see Bertrand paradox (economics).
Bertrand's paradox is a problem within the classical interpretation of probability theory. Consider an equilateral triangle inscribed in a circle. Suppose a chord of the circle is chosen at random. What is the probability that the chord is longer than a side of the triangle?
This problem was originally posed by Joseph Bertrand in his work, Calcul des probabilités (1888). Bertrand gave three arguments, all apparently valid, yet yielding inconsistent results.
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The "random endpoints" method: Choose a point on the circumference and rotate the triangle so that the point is at one vertex. Choose another point on the circle and draw the chord joining it to the first point. For points on the arc between the endpoints of the side opposite the first point, the chord is longer than a side of the triangle. The length of the arc is one third of the circumference of the circle, therefore the probability a random chord is longer than a side of the inscribed triangle is one third.
Random chords, selection method 1; red = longer than triangle side, blue = shorter -
The "random radius" method: Choose a radius of the circle and rotate the triangle so a side is perpendicular to the radius. Choose a point on the radius and construct the chord whose midpoint is the chosen point. The chord is longer than a side of the triangle if the chosen point is nearer the center of the circle than the point where the side of the triangle intersects the radius. Since the side of the triangle bisects the radius, it is equally probable that the chosen point is nearer or farther. Therefore the probability a random chord is longer than a side of the inscribed triangle is one half.
Random chords, selection method 2 -
The "random midpoint" method: Choose a point anywhere within the circle and construct a chord with the chosen point as its midpoint. The chord is longer than a side of the inscribed triangle if the chosen point falls within a concentric circle of radius 1/2. The area of the smaller circle is one fourth the area of the larger circle, therefore the probability a random chord is longer than a side of the inscribed triangle is one fourth.
Random chords, selection method 3
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[edit] Classical Solution
The problem's solution thus hinges on the means by which a chord is chosen "at random". It turns out that once the method of random selection is specified, the problem has a well-defined solution. There is no unique selection method, so there cannot be a unique solution. The three solutions presented by Bertrand correspond to different selection methods, and in the absence of further information there is no reason to prefer one over another.
The selection methods can be visualized as follows. Other than a diameter, a chord is uniquely identified by its midpoint. Each of the three selection methods presented above yields a different distribution of midpoints. Methods 1 and 2 yield two different nonuniform distributions, while method 3 yields a uniform distribution. Other distributions can easily be imagined, many of which will yield a different proportion of chords which are longer than a side of the inscribed triangle.
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[edit] Common-Sense Solution using Transformation Groups
In his 1973 paper The Well-Posed Problem [1], E. T. Jaynes asked the question whether we really cannot tell what "at random" means. He writes that we know more about this "at random" than we think at first.
If we do a random experiment theoretically, we can get insight into it by thinking about how we would actually carry out that experiment. In this case, we would try to throw straws onto a coin that lies far enough away from us that it is sufficiently random how the straws fall onto that coin. It is then common sense that the solution is the same no matter whether we take a smaller coin or a larger coin, and no matter whether the coin is placed a bit more to the left or a bit more to the right.
If we permit this additional common-sense information, then there is a unique solution to the problem: method 2, the "random radius" method.
More precisely, the argument goes as follows: if we have chosen random chords in a circle of radius r, and then we inscribe another circle of radius r / 2 with the same centre point into that first circle, then some of the chords of the bigger circle will also cut the smaller circle. If the choice of chords is common-sense random, then we would expect that the set of randomly chosen chords would give the same probability in both the large and the small circle. The solution should be scale invariant.
Similarly, if the centre point of the small circle is moved away from the centre point of the large circle, this should also still give the same probability. The solution should also be translational invariant.
Being very precise, Jaynes also points out that the solution should be rotational invariant.
Jaynes proved that there is only one distribution of chord midpoints that is translational invariant and scale invariant. It can be derived directly from the integral equation for translational invariance. This distribution is the one described as "method 2" above.
So while we still cannot say that we have found the mathematically correct solution, it is clear that "method 2" is the only distribution that fulfils the common-sense idea of what "choosing a random chord" means in practice.
[edit] References
- Michael Clarke. Paradoxes from A to Z. London: Routledge, 2002.
- E. T. Jaynes. The Well-Posed Problem. Foundations of Physics, vol. 3, 1973, pp. 477-493. [2]
