There are many problems that can give multiple contradictory answers, particularly in geometric probability and the application of probability to physical situations. In some famous problems often only one method is cited when in fact similar ambiguities could arise if various assumptions are made differently. In the problem of Buffon’s Needle, as well as question A6 on the 1992 paper of the notoriously difficult Putnam examination, we must be given extra information as to how the random variable is selected in order for there to be a unique solution. For instance, in the verbatim statement of the Putnam problem –‘Four points are chosen independently and at random on the surface of a sphere (using the uniform distribution). What is the probability that the center of the sphere lies inside the resulting tetrahedron?’- we are told where to apply the uniform distribution and are not given the opportunity to explore other methods for choosing random points which may affect the answer. As previously stated, there is a broad consensus to use the uniform distribution (U) in cases where a preference for any particular outcome does not exist – hence indicating randomness. The problem arises when there are different options for which random variable this is to be applied to. In Bertrand’s Paradox, method 1 assigns a uniform distribution to the angle between the chord and tangent, so we can say 𝜃 ~ U (0 ° , 180 ° ). Method two assigns the uniform distribution to the chord midpoints in a line which forms a radius i.e. L ~ U (0, r) and method three assigns the uniform distribution to midpoints chosen from an area. The fact that these three random variables and their associated sample spaces: angle, length and area, are of different types is the key to understanding the ambiguity. If one of these random variables is chosen to be uniformly distributed it is often the case that the others necessarily can’t be. This is made clearer in the case of another problem. We are asked to draw a random cube with side length (L) somewhere between 3 and 5 cm (so that the surface area is between 54 cm 2 and 150 cm 2 and the volume is between 27 cm 3 and 125 cm 3 ). Most people would assume that there is a uniform distribution over the side length itself, L ~ U (3, 5). For a uniformly distributed random variable X ~ U (a, b) the average or expected value is E(X) = + 2 , which in this case would give a side length of 4 cm. Based on this the average surface area would be 96 cm 2 and the average volume would be 64 cm 3 . However, if we instead assume a uniform distribution over the surface area such that A ~ U (54, 150) the expected value will be 102 cm 2 , but this corresponds to an average side length of 4.123 cm. The problem worsens if instead we assume a uniform distribution over the volume, V ~ U (27, 125), which yields an expected volume of 76 cm 3 giving an average side length of 4.236 cm. Even though length, surface area and volume are functions of each other and so as random variables are wholly dependant, this does not mean that they have the same distribution. In fact, if one of them is uniformly distributed, the other two necessarily cannot be. This should highlight how variations in solutions can arise due to subtle differences in the application of the uniform distribution and therefore in the interpretation of randomness. But how can we know which random variable the uniform distribution should be applied to? The Issue
How Do We Choose?
One proposed solution to this issue is to take the average of all the different numerical solutions obtained by the different methods. This is called a meta-average or universal average, which in the case of Bertrand’s
( 1 2
+ 1 3
+ 1 4
)
= 13 36 ≈ 0.361 . This seems simple enough but actually leads to even more
paradox would give
3
questions. Is it problematic that the meta-average is different to any of the individual solutions? What if there are infinitely many solutions so that we can’t find an average? Can we assume that our alternative methods are to contribute equally to the meta-average or should we use a weighted average? Even in this case where we have three unique values, should we be so hasty to give them equal weighting? Casually assigning uniform distributions is the oversight that led us into this quandary, and applying it here, at the level of alternative methods rather than alternative outcomes of random variables, may be particularly hard to justify. And we are no better off if specific solutions arise more frequently than others; should their weighting in the meta-average be increased? It would seem unnatural to count a solution equally if it is more
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