A factory produces cubes with side-length between 0 and 1 unit. What is the probability that a randomly chosen cube has side-length between 0 and 1/2 a unit? The tempting answer is 1/2, as we imagine a process of production that is uniformly distributed over side-length. But the question can be given an equivalent restatement. A factory produces cubes with face-area between 0 and 1 square-units. What is the probability that a randomly chosen cube has face-area between 0 and 1/4 square-units? Now the tempting answer is 1/4, as we imagine a process of production that is uniformly distributed over face-area. So we get two different answers depending on whether we regard the randomisation as applying to length or to area. If we regard volume as being randomised then we get a third answer.
How can we test this in the real world? We can use a program to produce a "random number" (apparently no such thing exists, but let's ignore this for now). This number is then used as the basis for drawing a cube. How is it used? The random number can specify the length, the area, or the volume. Actually, there are infinitely many other possibilities, eg the power of 5.13 of the length of one side. This seems arbitrary, but it is no more so than using length or area. The point is that there is no canonical way to do this, no unique and obvious way that invalidates all the others.
Since different ways of using the random number to specify a cube produce different probabilities, it follows that the method of randomisation, specifically what you are randomising (length, area or volume), determines what answer you get. In other words, how you ask the question determines what sort of answer you get. This is generally true in life!
So you cannot phrase the question in a general way. If you ask "What's the probability of producing a cube with side less than 1/2 a unit if they are produced randomly?" then there is no sensible answer. The way the last question is phrased suggests that you should use the linear dimension to calculate probability, but this is an unwarranted assumption - you should not allow the wording to induce you to assume that length is being randomised.
In general, to be able to ask what is the probability of some random event, you have to specify how the randomness it to be effected. You have to tell the computer program used to do the randomisation exactly how it is to be done. Only then will you receive a unique answer.
The cubes paradox illustrates how tricky is the concept of randomness. Interestingly, it does not depend on that old chestnut - counting infinite sets - but is purely a probability problem. You could specify that the side length of the cubes must be in multiples of 0.1 units, so that there would be only 10 possible cubes. Yet the paradox would still be essentially the same. However, I think that infinity is still there, only hidden. The probability of an event is the expectation we have of how often it would occur if the trial were executed many times, or more exactly, the expected proportion as the number of trials tends to infinity.
The Ejection Paradox
Why not reformulate the cubes paradox again (groan)? I came up with the Ejection Paradox (patents pending). Instead of the cubes being made in some unspecified way, why not eject the material that constitutes them from a hose? A nozzle ejects a certain amount of glass, which is then moulded into a cube. A computer generates a random number between 0 and 10 that determines how long the nozzle is open. The side length of the cubes varies from 0 to 10 units.
So how does that relate to the cubes paradox? If the nozzle diameter is kept constant and the random number determines how long it is open, then we seem to have the volume randomisation version of the cubes paradox. If the nozzle is kept open for 1, 2... or 10 seconds then the volume of ejaculate, sorry, ejected material varies in direct proportion. In other words volume is randomised.
What about if we vary the diameter of the nozzle but keep it open for a fixed time? In this case, the volume of material ejected grows in proportion to the square of the diameter (or radius) of the nozzle. What are the chances of the side-length of the resulting cube being 5 or less? The table below shows the values of the nozzle diameter and the corresponding volumes of the cubes created.
nozzle diameter 1 2 3 4 5 6 7 8 9 10
volume of cube 10 40 90 160 250 360 490 640 810 1000
The side-length will be less than 5 only in the first 3 cases, as the volume must be 125 (= 5 x 5 x 5) or less. So the answer is roughly 35%. This is only a guess, but I think that randomising the diameter of the nozzle is equivalent to randomising the square root of the volume.
How about if we vary the time and the diameter, both randomly and independently of each other? To work this out, think of a grid where the horizontal dimension comes from the diameter varying from 1 to 10 and the vertical part represents "time open" varying from 1 to 10.
So the first row would be volumes of 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, ie the squares of the numbers from 1 to 10. The row below would be twice these numbers, and so forth. The last row would be 10, 40, 90, 160, 250, 360, 490, 640, 810, 1000. This covers all the combinations of time and diameter values.
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| increasing diameter ->
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time = 1 | 1 4 9 16 25 36 49 64 81 100
time = 2 | 2 8 18 32 50 72 98 128 162 200
time = 3 | 3 12 27 48 75 108 147 192 243 300
time = 4 | 4 16 36 64 100 144 196 256 324 400
time = 5 | 5 20 45 80 125 180 245 320 405 500
time = 6 | 6 24 54 96 150 216 294 384 486 600
time = 7 | 7 28 63 112 175 252 343 448 567 700
time = 8 | 8 32 72 128 200 288 392 512 648 800
time = 9 | 9 36 81 144 225 324 441 576 729 900
time = 10 | 10 40 90 160 250 360 490 640 810 1000
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This table of 100 results represents all the possible outcomes (admittedly looking only at integer values), and they are equiprobable. What is the chance of side-length being 5 or less? This will happen for all volumes in the table that are 125 or less. Counting the volumes below 125 in the table we see it is exactly 50%, which means the one-dimensional version of the cubes puzzle.
All this merely emphasises that (1) the way in which we randomise something determines the probability, and (2) that there are myriad ways of doing so.
Back to Bertrand
Is the cubes paradox equivalent to Bertrand's? Bertrand's three methods randomise the chord using angle, area and length respectively. I believe that the two paradoxes are equivalent. In any event, the answer to Bertrand's question is that the probability depends on how you select the random chords.
PS I hoped to obtain a unique answer to Bertrand's paradox by removing infinity from the problem, ie by limiting the number of chords to 10, or 100, or 1000 and so on. It turned out that, just as in the cubes paradox, removing infinity did not change the problem at all. However, I did work out a fourth solution.
Since the length of a side of the inscribed triangle is the square root of 3 times r, ie about 1.7321r, it follows that all chords longer than this will satisfy the condition. All the chords of the circle vary in length from zero to 2r, so it follows that (2 - 1.7321) / 2 of these chords will be longer than 1.7321r. This gives the answer 13.40%, ie a little over 1/8.
On reflection, I realised that I have merely chosen a different way of randomising. I have randomised chord length in a way different from Bertrand's three solutions, and hence have generated an answer that is no more (and no less) valid than the three answers given by Bertrand. This merely confirms the conclusion given above.