4
$\begingroup$

I am reading a book titled QBism, a description of the Bayesian interpretation of Quantum mechanics, The author describes the “... baffling cube factory paradox...”. A factory makes perfect cubes with edges randomly distributed between 0 and 1 centimeters. If you are given a bin of these and measure the edge length, what is the probability of getting a length less than 0.5 cm. Most people say 1/2 for this probability as 0.5 cm is in the middle of the range (they assume a uniform distribution of edge length). The “paradox” arises because the question might have been what is the probability that a face produced by the factory would have an area less than 0.5*0.5 cm^2 = 0.25 cm^2. The claim is that it would be 1/4 as the measures value is “1/4 of the total range or areas”.

As I see it, this is not even close to being a paradox. The problem is that nobody specified the distribution of the random process governing the cube manufacture. So no wonder we don’t get a well defined answer. The book kind of says this in the space of a couple of pages but I still don’t understand why this story is anything but trivial. Perhaps the full message is that you need to know the actual distribution before answering questions.

Here is the question: can anyone give an example where has this well-known paradox ever been used to illuminate any non-trivial issue? My research into this used google and I was not able to find anything, just descriptions similar to the above text, but my experience with probability theory is not extensive.

Improve this question
$\endgroup$
3
  • $\begingroup$ "The “paradox” arises because the question might have been what is the probability that a face produced by the factory would have an area 0.5*0.5 cm^2 = 0.25 cm^2." Did you mean "less than"? Because the probability of them being exactly equal to $0.25\operatorname{cm}^2$ is zero for any smoothly varying probability density. $\endgroup$
    – Sean E. Lake
    Commented Jul 2, 2018 at 3:26
  • $\begingroup$ Yes, less than...sorry. I will fix that. $\endgroup$
    – abby yorker
    Commented Jul 2, 2018 at 3:30
  • 4
    $\begingroup$ You are absolutely right that this is not remotely a paradox. If $X$ is a random variable, then the distributions of $X$ and $X^2$ are in general different. Who would ever have expected otherwise? $\endgroup$
    – WillO
    Commented Jul 2, 2018 at 4:25

2 Answers 2

Reset to default
1
$\begingroup$

First of all you misstated the "paradox." From B. C. Van Fraassen (1989). Laws and Symmetry. New York, NY: Oxford University Press. Pge 303.

A precision tool factory produces iron cubes with edge length ≤ 2 cm. What is the probability that a cube has length ≤ 1 cm given that it was produced by that factory?

You stated

A factory makes perfect cubes with edges randomly distributed between 0 and 1 centimeters.

The fact that you stated that the edge length was distributed randomly defines the problem. The "paradox" as you noted is that you need to know the distribution before you can answer the question.

Van Frassen's point was that you could "naturally" chose between a uniform distribution of the side length, the face area, or the volume of the cube. Each choice of course yields a different answer.

This all goes back to problems made famous by Joseph Bertrand in his work Calcul des probabilités (1889). The most famous of the problem's is Bertrand's Chord Paradox.

The gist here is Laplace showed how to handle "uniform probability" over a discrete set of values. But how should probabilities be handled for variables which can have infinite values? It wasn't until 1933, that Kolmogorov published his book, Foundations of the Theory of Probability, which laid down the modern axiomatic foundations of probability theory. So before then the problem was indeed a "paradox."

Improve this answer
$\endgroup$
6
  • 1
    $\begingroup$ It remains a problem (maybe not a paradox), how to approach probabilistic circumstances when "total ignorance" doesn't define a unique distribution - which one to use? Jayne's started working on this a bit, using symmetry groups, but I don't know what the current status is. For example, to get probabilities of trajectories after a particle collision is makes sense to use the uniform distribution on $\theta$ (rather than on $x,y$) but can this be grounded more firmly? What about a less obvious case? $\endgroup$
    – DPatt
    Commented Jul 2, 2018 at 17:39
  • $\begingroup$ Jaynes used entropy theory to assert that Bertand's paradox had a unique solution. Paper was Foundations of Physics, 3, (1973), pp. 477-493 "The Well Posed Problem." The gist is that entropy theory says that a 50/50 chance is more random than say a 75/25 chance. $\endgroup$
    – MaxW
    Commented Jul 2, 2018 at 17:46
  • $\begingroup$ @Maxw: The numbers that I put in the example were from the book I was referencing the book QBism. Your answer was what I was looking for - putting this story in a context where it made some sense. I’ll mark it as answered although I have not yet read the references that you supply. I still maintain that calling this a “baffling paradox” in present day, as the author does, is pointless hyperbole. That said, I do not have a firm opinion on the book yet as I am still reading. $\endgroup$
    – abby yorker
    Commented Jul 2, 2018 at 19:17
  • $\begingroup$ @MaxW : That was a good read, thanks. However, even though he manages to solve Bertrand's problem, he admits (and gives an example) of further problems that his methods still cannot solve (the transformation group actually overdetermines the distribution). This is getting off-topic, but I feel there remains a paradox at work which amounts to the tension between our philosophical inclinations and the real possibility that not every problem even admits a probability distribution, let alone a unique one. $\endgroup$
    – DPatt
    Commented Jul 2, 2018 at 19:27
  • $\begingroup$ @Dpatt - I agree that semantics and logic don't necessarily mix. $\endgroup$
    – MaxW
    Commented Jul 2, 2018 at 20:48
-1
$\begingroup$

Not sure but seems to ask anyone side being less than $0.5$cm. Since cube has $12$ sides and there is a $50\%$ chance shorter maybe just same as as flipping coin $12$ times and getting say heads on any one toss. Heads representing the $50\%$ and single heads representing less than $0.5$cm.

Easier to use binomial for $(1-P(\mathrm{no\, heads}))$ as equal to at least one heads.

$$1-(12!/(12-0)!*0!)*(0.5^{12})(0.5^0) = 1-0.5^{12} (very small \#)$$

I likely made some assumptions that may not be correct ?

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.