There is no paradox as, as has been noted below, the probability is defined, in this case, on sets in the plane and a point (or a line) in the plane is a set of measure zero (unless the probability distribution is atomic). In this case, I think that you're confounding the sample space with the sigma algebra of events (for which the measure is defined). If the probability is uniformly distributed over the plane, entropy is indeed at a maximum (conversely, if the distribution is a Dirac delta function, it's at its minimum). Finally, I'll note that the typical measure of information, Shannon Information, is only valid for finite, discrete alphabets. It's continuous counterpart, differential entropy, is NOT and information measure as it is not invariant under a bijective transformation (i.e., violates the axiomatic definition of information) and so a Kullback-Liebler divergence type measure is more commonly considered in this case as it does not violate any information axioms (Note the connection between KL divergence and the Radon-Nikodym derivative).

My feeling is a probability of zero. The reason is that a mathematical point has no extension (size) so you get an infinite number of locations in your square and as a consequence no chance to hit the same point.

Zero is the correct answer : it's one over infinity.

This is a common occurrence - the probability of your existence is also zero, yet there you are. Oddly enough, a probability of zero does not mean impossibility.

I cant see the paradoxon. Your point doesnt bear maximal information. In a information theoretical point of view a probability of zero is correlated with information content of infinity. So the simple formula for the entropy of your system (product of probability p times information content -lg(p)) is wrong.

0

This is an apparent paradoxe because the problem is not well posed. Actually, the measure (http://en.wikipedia.org/wiki/Measure_%28mathematics%29) of a point on a plane is zero and so, the probability of select him is zero. The only way to avoid this difficulty is to define probabilities on subset of the plane which include your point (http://en.wikipedia.org/wiki/Borel_measure).

Olivier, that does not work because any subset of the plane also has an infinity of points - specifically an Aleph naught infinity, and 1 over Aleph naught is zero.

If you label the points you must label them with number IDs, which run to infinity regardless of whether you use N or R numbers.

Since you say the plane is a square and since a point is also a coordinate (x,y), x=y and so there are simply xy=x^2=y^2 points in that square.

Probability not exactly zero but near to zero. It is still a paradox.

@H Chris Ransford :

" that does not work because any subset of the plane also has an infinity of points "

> Of course it works, the problem is that you want to measure the probability of the subset by adding the probability of each point which are part of this subset. But this approach is wrong. If you are considering a set of points E. Then probability are not defined on each subset of E but only on a particular one which is called a sigma-algebra B (this is a discovery of Kolmogorov).

http://en.wikipedia.org/wiki/Sigma-algebra

As a consequence, you must not work on E but on (E,B,P) where P is a probability defined on B. You can find how construct B for continuous space in the ref:

http://en.wikipedia.org/wiki/Probability_space

Especially:

http://en.wikipedia.org/wiki/Probability_space#Non-atomic_examples

Sorry Olivier, Kolmogorov complexity has to do with the ease of computation - not with probabilities .

As many have noted, it indeed decomplexifies infinities : if we take ordinary real numbers as an example, a computer program generating all of the real numbers is much shorter (in fact quite trivial) than a computer program that would generate even a single, arbitrarily very long real number. This argument has been used in a number of areas, for instance by proponents of the Multiverse to argue that a multiverse is, in fact, simpler than a single universe.

But it does not make infinities finite.

Read up on Aleph numbers, the number of points on any segment or any square or for that matter any three dimensional cube * of any size * is aleph naught - an infinite transfinite number.

Cj, for once I cannot follow your reasoning at all.

I'm sorry but I had never talk about Kolmogorov complexity. but about Kolmogorov axioms (it is written in the ref that I had given and also in:

1. http://en.wikipedia.org/wiki/Probability_axioms

2. http://www.princeton.edu/~achaney/tmve/wiki100k/docs/Probability_axioms.html

3. Andrei Nikolajevich Kolmogorov (1950). Foundations of the Theory of Probability.

Is it even possible to define such a propability? - As far as I remember from my maths lectures, a propability space should be made up of a countable (maybe infinite) number of subsets and a plane in Euclidean space contains an uncountable infinite number of points (subsets)...

@Cj Nev: unfortunately that is not true, if you were correct, a square with a side length of 1 should only contain 1 point, but still it contains an infinite number...

There is no paradox as, as has been noted below, the probability is defined, in this case, on sets in the plane and a point (or a line) in the plane is a set of measure zero (unless the probability distribution is atomic). In this case, I think that you're confounding the sample space with the sigma algebra of events (for which the measure is defined). If the probability is uniformly distributed over the plane, entropy is indeed at a maximum (conversely, if the distribution is a Dirac delta function, it's at its minimum). Finally, I'll note that the typical measure of information, Shannon Information, is only valid for finite, discrete alphabets. It's continuous counterpart, differential entropy, is NOT and information measure as it is not invariant under a bijective transformation (i.e., violates the axiomatic definition of information) and so a Kullback-Liebler divergence type measure is more commonly considered in this case as it does not violate any information axioms (Note the connection between KL divergence and the Radon-Nikodym derivative).

Johannes, I wouldn't agree that a square with side length 1 necessarily contains 1 point. Per Derek's question, the square is in a Cartesian plane. A Cartesian plane has points, as I mentioned above, identified as x-y coordinates, allowing then any point on the plane and so any point or points on the square in the Cartesian plane also to be easily identified and ordinary probabilities to be calculated for some same point selected in each square as Derek poses. Hope this is clearer.

@Cj! Ever heard the terms integer number, rational number, real number etc.? Or are you just joking?

Just a brief note Olivier,

you write Quote "you want to measure the probability of the subset by adding the probability of each point ...." Unquote - Of course, probabilities never add but multiply, I trust you mistyped what you meant to say.

Johannes, you make an excellent point - the probability of "finding" a point is not trivially definable, because a point is dimensionless, and 'finding' something dimensionless is at the very least a misnomer

Cj, sorry but your point seems to be becoming ever less clear ?

Hanno, please note I am providing ONE answer to the question that in my answer counters any seeming 'paradox.' Of course, there are infinite answers (0, 1, or infinitesimal and infinity, etc., as all of those numbers exist on the question's CARTESIAN plane-two REAL orthonormal number lines) and even questions that can uphold any 'paradox' for the sake of arguments, but doubtful for any knowledge or learning.

Hi Cj,

to be honest I really can´t follow your argumentation. If you are calculating areas of rectangles or squares by muliplating the side lengths you don´t get any information about the number of points and the related probabilities. Please help me to understand your informations and ideas!

Sure, Hanno -- Yes I am calculating (specifically to the question posed above only) the area of the square in the CARTESIAN plane, where ALL POINTS BEING EQUAL gives you all the information, in fact, infinite information. I have picked a point, or x-y coordinate in that square on a CARTESIAN plane. The question asks for the probability of your selecting the same point from essentially the same square in the same plane, "an exact copy," Derek stipulates. As that same point I selected does not move or change, the probability is AT LEAST one (never zero in THIS stipulated scenario) in however many SELF-SAME (DIMENSIONLESS) points are in the square, determined by the square's area (x*y=x^2=y^2) on the stipulated CARTESIAN plane.

Making the chances zero would require ignoring the square or drawing the square infinitely in size as the Cartesian plane, accomplished only by drawing a Cartesian plane, rendering the square posed in the question superfluous and the question trivial.

I'll add here, this is precisely why "with probability zero" does not mean impossible, merely that the corresponding event (a point usually) is a set of measure zero. Any measurement (assuming infinite precision) of a (non-atomically distributed) RV has probability zero yet clearly, the RV takes some value. Similarly, with probability one does not mean absolute certainty. Rather, the contradictory events occupy a set of measure zero.

@Cj: Dereks requirement is that different persons meet the same point by chance, not you without moving your finger or brain away from the position or coordinates. So your conclusion is definitely wrong except if you exchange the necessary information with the second person!

The simple and correct answer for the posed questions is: zero. If someone wants this probability to be positive, then in place of mathematical points the "physical" dots, having finite diameter, should be used.

@Cj: would you tell us exactly the same arguments if instead of square we would be considering a circle (2 dim sphere)? This is also a part of CARTESIAN plane, by the way.

Okay, thank you.

Zero. Unless the diameter of the "dot" is positive (non-zero).

Zero is the probability of finding a point on a plane.

Zero

Zero, which is the 2-dimensional measure of a point.

One only need to read up some book in Real Analysis to be convinced also that:

-The probability of finding a line in a plane is zero

-The probability of finding a plane in 3D space is zero

-The probability of find a point with rational coordiates in a (line/plane/space) is zero

-The probability of finding an invertible real/complex matrix is one.

....

Dear Prof. Carriegos,

Your last statement, the one concerning invertibility of matrices, appeared surprising to me. Looks nice, but I am unable to proof it. And here is why:

Suppose I have a matrix M, having no inverse. It is easy to see that in its epsilon surrounding there exists uncountable infinity of invertible matrices. Why? The determinant of original matrix is necessarily equal to zero, so changing slightly just one arbitrary component of M, say M_11, will make it invertible, as the determinant is n-linear and antisymmetric function of all matrix elements. This is working for any epsilon, so we are done. But are we?

Let our initial matrix is not invertible beacuse two its rows are identical. Now I perturb it by adding arbitrary epsilon/2 to two elements in first column of both those identical rows. Evidently no effect - the matrix remains non-invertible. Even worse: the same effect (persistent non-invertibility) may be obtained by permuting rows of the starting matrix.

Conclusion: in epsilon surrounding of an un-invertible matrix we have uncountably infinite set of invertible matrices but also another set of non-ivertible matrices, also uncountably infinite number of elements. How do you know which one of them is more numerous?

@Marek. Low's explanation is sharp. Nothing to remark.

But let me quote that probability is not about cardinals in the infinite case: Think in the probability of finding an even integer into the set of natural numbers: It is 1/2 but the set of natural numbers and the set of even numbers have the same cardinal (map n --> 2n is a bijection).

Once again we can see how tricky the infinities can be. Extremal care is a must. 'Take home' mesage is: don't be afraid to jump into higher dimension (thank you Robert!). There's nothing to loose.

@prof. Carriegos: I like your second example, the one with even integers. And I know how to prove it, regardless of your highly confusing comment about the n --> 2n map!

Dear Professor Derek Abbott, A very very typical question. There are fundamental

contradictions inherent in assuming things about the infinite plane.

Dear Dr. Robert Low, Yes, its lack of uniform measure on the entire plane.

Richard Kolacinski, describes nicely the matters. I would like to add here a nonstandard answer. In a nonstandard framework, we may discretetize the plane using hypefinite partitions, so that if the point is x then it would belong to a infinitesimal square, and its probability would be an infinitesimal. In this way the paradox of having measure 1 for [0,1] whereas each point has measure zero, does not exist.

Robert Low: OK, in order to treat the problem more formally we would need Loeb Measures, which are build on external sets as well! So in this case transfer and pure internality does not apply. Any way, let as take any standard point x=(x_1,x_2). Having at our disposal an infinitesimal we ma form a=(x_1-dx_1, x_1+dx_1) x(x_2-dx_2,x_2+dx_2). a now contains x and has infinitesimal area. In one dimension, say [a,b], we may form an infinitesimal partition {a, a+dx,....,b}. By selecting appropriately the infinitesimal dx, I can manage to include into partition all rational points, and irrational point appears in the partition. This construction makes your assertion at least partly false. A complete answer with proofs etc. it is not appropriate for this forum.

Zero.

Robert Low: Dear Robert, I will try to as simple as possible. Let H be an hyper-natural such that H=h!, h another hyper-nutural. Let dx:=1/H and let us form the partition:

{0, dx, 2dx, ....,1}. This partition contains every standard rational in [0,1] and no irrational appears in this partition, i.e. we have constructed an S-dense set. Now every x\in [0,1] belongs to one and only one cell of the partition, and thus the probability of hitting it is dx. As for your second question, you can see that the hyperfinite sum Σ[kdx, (k+1)dx]=1. Even if you take now standard parts you will add up with 1!

Dear Robert: This is my last post to convince you!

Let us choose H:= K! with K\in (*N-N), and let m/n be a rational in [0, 1]. Then

m/n = (mK!)/(nK!)=(m(1.2.3...(not n),...K)/(1.2...n...K)=Mdx

where M:=m(1.2...(not n)....K)

and the above expression is an internal one.

Thus every rational in [0,1] can be expressed as the product of an infinitesimal time a hypernatural. Of course every time that we are in the monad of say zero, every cell of the form (nΔx, (n+1)Δx], n\in N, does not contain any standard number! But us you pas to hyperintegers, then you can scan all the monads around rational numbers. The said partition constitutes an S-dense set, i.e. for every standard a\in [0,1], there exists an x\in {0,dx,2dx,…ndx,…Mdx,… !} which is infinitely close to a.

Many interesting replies to this question. I'll just add: most of the terms you use can be given operational definitions (probability, plane, square). But I have not seen a generally accepted operational definition of "surprising". So one needs to decide what that term actually means. This leads (for example) to re-expression in terms of classical statistical hypothesis testing, or to Bayesian hypothesis testing. Travel this road, and I think you will find the apparent paradox begins to fade into the autumn mists.

I do not see a paradox:

Specifying an arbitrary point on a plane with infinite precision of course in most cases requires an infinite number of bits, and therefore carries infinite information - but only with respect to any integrable continuous distribution.