What's the chance of an acute triangle?

[superbonzo]

if I tied (1), I’d get some ratio-like expressions that wouldn’t really tell me much more that what I had without them?

Right, I suppose you known how to construct the set of rational numbers from the set of natural numbers ? you can do something similar, but with some technical difficulties. Anyway, as you said, the resulting "ratios" would be useless in modeling the concept of probability.

>> If I tried (2), someone might argue that this is only one of possible results, since there's really no "correct" probability distribution function.

Yes that's totally right, but...

>> ...this notion that there is some natural probability related to each outcome (that a result of a die roll, or of a point selection has some inherent probability), but this was wrong.

the key is what do you mean by "natural". Actually, modern mathematics has a well defined and sophisticated meaning of that word ( that comes from category theory ) capable of modeling a very general concept of structure.

Anyway, intuitively, given an object and its structure you can sometimes specify a set of natural properties or naturally related objects. These might include a probability structure.

For example, every scientist will understand you if you say "consider a uniform probability density over the interval [0,1]" or "over a sphere" or "over a finite set" or "over a thorus" or over "a projective space", "a flag manifold", "a Moebius strip" or even over the set of "right continuous left bounded paths"... but every scientist will NOT understand you if you say "consider a uniform probability density over the set of reals", or over "the set of triangles", "the set of continuous paths", "the set of chords of a circle", "the set of natural numbers"...

what does the former geometrical objects have in common that the latter have not ?

it's not cardinality: excluding finite sets, common cardinalities appear in both situations ...
it's not an intuitive notion of "boundedness": like, bounded interval vs the whole real line; but the set of chords of a circle is bounded ( or better, compact would be the right term ) ...

Most geometrical objects in both lists ( and every object in the first ) specify (maybe implicitly, sometimes it depends on the context in which they appear) a group of symmetries ("compatible" with their structure, this might imply continuity, measurability, differentiability, or whatever): a finite set has its permutation group, a sphere its group of rotations, the real line its group of translations, a bounded interval its semigroup of local translations, a 2-dimensional projective space its group of Moebius transformation and so on...

Now, each symmetry carries a notion of "uniformity", in the sense that "uniform things" will be somehow compatible with and invariant with respect to that symmetry.

Finally, using nuzzle's words, this is the heart of the problem: Does every geometrical object admit a probability structure that is compatible with and invariant with respect to its symmetry group ?

The answer is NO.

And this is why we cannot get an unanimous/unique answer to the question "what's the probability that of a triangle being acute?" : because we consider rotated\scaled\translated triangles naturally equivalent AND because the set of triangles does not admit a probability measure invariant with respect to its group of "natural" transformations.

[nuzzle]

About nuzzle’s infinite sided die - yes, it would be a sphere, but why would it roll forever? Introduce friction, and it would eventually stop. The topmost point is the result of the roll. But, the surface of a sphere is a continuum, so what would be the right way to model such a die? Would the sample space be the set of, say, non-negative reals [0, +inf. ), in respect to its physical surface, or the set of natural numbers {1, 2, 3, ...}, in relation to the concept of "infinite-sided" object?

I think the round dice quite well illustrates the problem here. It has no outcome just like drawing uniformly distributed points at random in the infinite plane has no outcome.

To get an outcome you need to introduce different measures, such as friction, to get the dice to stop. It will be the stopping measures that determines the outcome because the dice is still round no matter what you do. You get different results depending on how you make the dice stop. It's the same with the triangles.

[TheGreatCthulhu]

I think the round dice quite well illustrates the problem here.

I'll have to respectfully disagree - because I feel that, if things are left as they are, your example can be rather misleading.

To get an outcome you need to introduce different measures, such as friction, to get the dice to stop. It will be the stopping measures that determine the outcome because the dice is still round no matter what you do. You get different results depending on how you make the dice stop. It's the same with the triangles.

You mean - if the friction is stronger, the roll will end sooner? You think that with cube-shaped dice is different, that no "external" factors influence the result? Actually, there's no real difference. What if you threw a standard 6-sided die underwater? This would certainly shorten the time it takes for the roll to end. And if you could roll it in space, in vacuum? It would spin forever, at least in an idealized case. Just like your sphere if there are no "stopping measures". So, a die stops and gives a specific result not because it's cube-shaped, but because there's gravity, and because it has a certain effect on that specific shape. This is clearly a "stopping measure", as you called it.

It has no outcome just like drawing uniformly distributed points at random in the infinite plane has no outcome.

If you agree with what I said above, then you'll also have to agree that if a sphere can stop, there is an outcome.

The problem here has little to do with the plane being infinite. It has to do with the fact that we need to transform the set of triangles into something that we can work with (like your half American football). The transforms used are translation/rotation/scale; this is because these specific transformations preserve something that we deemed important - and that is the shape of the triangle - they give images that are similar to the original triangles.

But this is where our problem comes into play. There is no way to define such a transformation, apply it, and still get the same probability of acute triangles. This transformation changes our probability measure, and each specific transformation method changes it in a different way (so, in the end, we get different results).

I'll try to create an analogy here. (But, I'll admit on the start that it could be better.)

Imagine if you made a cube out of paper, something like those that kids might make at school. After you've written some dots/numbers, you can use it as a 6-sided die. Now, a chance of getting an even number is 1/2. So, you have a set of outcomes O = {1, 2, 3, 4, 5, 6}, and an event that you're interested in, where an even number is the result E_even = {2, 4, 6}. This is a subset of O. With our triangles, you have a set of outcomes (sample space) where each represents a formation of a triangle after 3 points have been picked, and you have an event "an acute triangle is formed", which is a subset of sample space containing all the elements that form such a triangle. An event has occurred if any of its outcomes is picked.

Now we apply the transformation. Imagine you take your paper cube and you unwrap or open it a bit. Imagine also that this is some strange or super-quality paper that keeps it shape. (See the attached image for reference.) Although this transformation doesn't have a real purpose, it's still a valid transformation - some translation, and some rotation. It's analogous to the one we apply in the other case; only, with our set of triangles, we apply the transformation for a reason - in order to solve the problem, and in such a way that the shape of the triangle doesn't change. With our paper cube, we keep the shape of the sides (square). However, what is the chance of getting an even number now? We altered the shape of our paper die, and it's not a cube anymore! We changed the probability measure. With triangles, we shape the set of all triangles (which doesn't really have a shape - the set, I mean) into a half American football, or into a circular area, or a rectangular area, or something else.

Again, this may not be the best analogy, but I guess it illustrates the point.

Also, there's another thing I'd like to point out. Consider the half American football approach. Here, the set of all triangles is mapped onto the half American football shape, and then one point is picked from there. What does this has to do with the act of picking points in a plane? There might be some correlation between the two, but you'll note that the plane doesn't figure anywhere in this approach.

Finally, I'll try to compile everything that superbonzo said in a few lines of quotes.

You'll see that he tries to tell, from the start, that the resulting probability is not invariant to the transformation that is applied.
You'll also see that he argues there's no "correct" probability assignment over the sample space, although there might be some "natural" (or natural) probability distribution arising from the symmetries of the structure in question (thus being more appealing to the human mind).

"BTW, note that R2 does not admit a translation invariant [...] probability measure" (src >)

"But again "picking randomly" from any unbounded set of an euclidean space in a translation invariant way is pure-non-sense." (src >)

"The problem of bad posedness has nothing to do with the use of infinite sets or infinite dimensional geometries of any cardinality [...]" (src >)

"Here we are not speaking of introducing a probability in the set of all triangles. Of course, this is possible. The problem comes when you try to define a natural notion of "probability uniformity" with infinite sets.

Actually, the same issue happens with finite sets; but finite sets come equipped with a natural measure (the counting measure) that allows you speaking about "chance"." (src >)

"Of course, you can arbitrarely introduce symmetries and/or probability measure on such [infinite] sets. But there are inifinitely possible inequivalent choices." (src >)

"a quick calculation (non tested, but I suppose is correct) with wolfram mathematica gives me 0.360+-0.001 for your transformation and 0.248+-0.001 for the transformation above. Both angle preseving, different "chances"." (src >)

"BTW, you could think that there exist a "physically" intuitive notion of "uniformly random"; you'd be wrong" (src >)

"if f and f' preserve acuteness and F and F' are bounded then L{ p(f(j(A))) } / L{ F } = L{ p'(f'(j(A))) } / L{ F' }

this statement is false ( I provided a counter example in post #10, but it should be evident from the very definition of a measure )." (src >)

"What you cannot do is defining a translation invariant probability measure over that set [the set of triangles], because such a thing does not exist ( eg. there's no such probability space in the class of probability spaces )." (src >)

"[qouting me] >> ...this notion that there is some natural probability related to each outcome (that a result of a die roll, or of a point selection has some inherent probability), but this was wrong.

the key is what do you mean by "natural". Actually, modern mathematics has a well defined and sophisticated meaning of that word ( that comes from category theory ) capable of modeling a very general concept of structure.

Anyway, intuitively, given an object and its structure you can sometimes specify a set of natural properties or naturally related objects. These might include a probability structure.

[...]

each symmetry carries a notion of "uniformity", in the sense that "uniform things" will be somehow compatible with and invariant with respect to that symmetry.

Finally, using nuzzle's words, this is the heart of the problem: Does every geometrical object admit a probability structure that is compatible with and invariant with respect to its symmetry group ?

The answer is NO.

And this is why we cannot get an unanimous/unique answer to the question "what's the probability that of a triangle being acute?" : because we consider rotated\scaled\translated triangles naturally equivalent AND because the set of triangles does not admit a probability measure invariant with respect to its group of "natural" transformations." (src >)

[nuzzle]

The problem here has little to do with the plane being infinite. It has to do with the fact that we need to transform the set of triangles into something that we can work with

Well, in my opinion the fundamental problem is the infinite plane. The infinite plane simply has no notion of a uniform random distribution of points so picking a triangle at random using this distribution just doesn't have an outcome, just like a round dice doesn't.

There is nothing wrong with the transformations. The problem is that we apply them to an acute angle fraction that doesn't exist. We're using the transformations to "fix" a fraction in a region of the plane from a nonexisting fraction in the infinite plane. We're using the transformations as ways of making a round dice stop by giving it sides. It will be the transformations that determine the acute angle fraction.

The mathematically correct way to deal with this is to not assuming uniformly randomly distributed points in the plane. Instead one must switch to a probability distribution that is defined for the infinite plane such as the Gaussian.

[TheGreatCthulhu]

I don't want to be to insistent, and I'm not claiming I'm absolutely right.
But, I said "little to do with", not "nothing to do with". So I agree, but IMO, it's only a part of the story.

The way I see it, a similar problem might arise even with finite sets, if you had to apply a transformation for some reason (the word "transformation" having a rather broad sense here), and if the probability measure was not invariant to this particular transformation.

So, yes, there's nothing wrong with the transformation - there's something wrong with the probability measure - it's not invariant to it.
Anyway, whatever probability distribution you used, can you avoid translation/rotation/scaling in order to solve the problem?

[nuzzle]

Anyway, whatever probability distribution you used, can you avoid translation/rotation/scaling in order to solve the problem?

I think one has to accept that the problem lacks a solution. The reason is that the uniform probability distribution doesn't exist in the infinite plane. Picking triangles from this distribution simply has no outcome (again compare with the round dice). The transformations are just different ways to obscure this fact (in the round dice analogy they would artificially add sides to enforce an outcome).

The only way this problem has a solution is to use a distribution which does exist in the infinite plane, such as the Gaussian. Then there's no need for transformations. I've seen in the litterature that for such so called Gaussian triangles the acute triangle fraction is 1/4.