Probability Paradox (1 Viewer)

[Dumsum]

An interesting problem highlighting our inherent lack of intuition for randomness and probability (and the concept of infinity).

Let x be a randomly chosen element in R.
i) what is the probability that x is rational?
ii) what is the probability that it is irrational?

 

[Trebla]

Isn't ii) just 1 because x is an element of the real numbers?

 

[Dumsum]

Well...technically all rational numbers are real, but I mean real and not rational (edit: by which I clearly mean irrational. My brain is pretty much fried...kill me now).

 

[umop 3pisdn]

Well Q is countably infinite and R isn't... I don't know how that helps though

 

[TesseracT22]

just had a thought.... is i) zero probability?

 

[Dumsum]

I suspect that the answer is i) zero; ii) one, but you'd probably have to ask someone who knows something about measure theory about that.

Alternately, you can read what Norman Wildberger has to say about infinite set theory.

http://web.maths.unsw.edu.au/~norman/views2.htm

(NB: Norman has many interesting ideas, but he is very much in the minority.)

Wildberger is fantastic, I'm hoping to take his honours geometry subject next year.

Your suspicions are correct and it is due to the fact that though the cardinalities of Q and R are both infinite, the cardinality of R is infinitely greater than the cardinality of Q. This is an odd concept in itself, since if I'm not mistaken, between any two rational numbers there is an irrational number, and between any two irrational numbers there is a rational number.

Despite this, however, is it not conceivable that a rational number could in fact be chosen in the earlier scenario? It would lead to the conclusion that a probability of 0 does not necessarily mean an event is impossible, or that our belief merely highlights a massively misunderstood conception of true randomness.

 

[TesseracT22]

I was sort of thinking that it has to do alot with the "random" part. Also i thought you could have infinite possible irrational numbers between each rational number, does that make any sense or just wrong? So comparing the number of rational numbers to irrational numbers is like comparing 1 to infinity and/or 0 to 1.

So if you randomly chose a number it would have to be irrational as comparitively there are infinite irrational to each rational or 1 irrational for 0 rationals so even though a sample of rational numbers(infinite amount) may exist to chose from, it's the same as having 0 rational and 1 irrational number to choose from, i think?

Just wondering if my understanding of this works

 

[D47]

how about this:

There is an infinite number of integers, and also an infinite number of irrational numbers. Therefor there is an equal chance that a random number will be one of the two. 50:50.

 

[D47]

i realize that, but each set of infinity spans on... to infinity... its infinity. one infinity cant be bigger than another...

 

[pyrrha]

i realize that, but each set of infinity spans on... to infinity... its infinity. one infinity cant be bigger than another...

Some infinities are bigger than others. It's a great topic in maths and philosophy, and there's a myth that trying to grapple with infinity sent Cantor mad. I agree with the 0/1 argument.

 

[Dumsum]

Does no one else have an issue with the idea that events with a probability of 0 can still occur?

Another example: The probability of me being precisely as tall as I am is 0. And yet I am that tall.

 

[TesseracT22]

I don't see any problems with that becuase of the fact that the selection is random. I mean if you had a bag of the numbers and randomly picked one out it would never be a rational one even though you had rational numbers in the bag.
Due to the random nature of selection the chance of getting a rational number is impossible. I think the example you have given with the height is different becuase it is not random and so the logic of having 0 probability there doesn't work. Though if your height was selected randomly then then you could not possibly be the height you are now as the prob would be 0 in that case where the height is random.

 

[pyrrha]

Does no one else have an issue with the idea that events with a probability of 0 can still occur?

Another example: The probability of me being precisely as tall as I am is 0. And yet I am that tall.

I have no problem with the probability of a point in a continuous random variable being 0.

 

[darkliight]

Hehe, I think you just gave the discussion a big push in exactly that direction Is anything ever truely random? Let alone the choice of a number from set.

To the OP, I'm no authority (I haven't done a course on measure theory) so this will hardly be precise, but ... the rational numbers are countable subset of R and so have zero measure. As far as the real numbers go, the compliment of the set of rational numbers (keeping in mind it has zero measure) is the set of irrational numbers, and we can use this as a definition for 'almost all'. That is, a property holds for almost all elements of a set, if the elements that do not possess the property have zero measure. Moving in the direction of probability theory, the probability of choosing an element with a property, such that almost all elements have that property, is 1. Likewise, the probability of choosing an element with a property, such that the set of all elements that share this property have zero measure, is 0. That is, the probability of choosing a rational number from the set of reals is 0.

A final note, an uncountable subset does not correspond to a subset whose members are considered to be almost all members of the whole set. An example of this is the Cantor set - its elements are uncountable, is a subset of [0,1], but has zero measure. This isn't what we'd expect I guess.

Maybe that makes sense, but take it with a grain of salt (not an authority! )

 

[SeDaTeD]

I don't believe you can have such a probability measure defined on the entire real number line which is uniform (which I suppose would reflect our intuitive understanding of 'at random'.)

Suppose it did:

Note that a requirement of being a probability measure in this scenario would be for the probability measure of the entire real number line to equal 1.

What we'd hopefully expect is that everything 'looks the same' everywhere. So all points should have the same probability measure, all finite intervals should have probability measure proportional to their length.

I guess it's clear that any singleton set (ie. just one point) should have probability zero, otherwise we could take the union of sufficiently many of these which would result in a set which has probability greater than 1. This would also imply the prob measure of a countable set equals zero.

Now look at the interval [0,1). Suppose it had positive probability measure, c>0 say. Then take n > 1/c and look at the interval [0,n). This is a union of the intervals [0,1), [1,2), ... , [n-1,n) which are disjoint. By countable additivity (measure of a countable disjoint union is equal to the sum of the measures of each set), we see that

the prob of [0,n) = n times prob of [0,1) (using the uniform assumption)
= n/c >1 , which is not possible.

So we must conclude that the prob of [0,1) is zero, and all finite intervals must follow suit (we can add or remove the endpoints without worry since the prob of a singleton is zero).

I guess it's clear that the entire number line can be written as a countable union of disjoint intervals of the form [a,b). (eg. all the [n-1,n) )
Again by coutable additivity, this would imply that the prob measure of the entire number line would be equal to zero, contradicting the assumption that it is a probability measure.

Well... that was a bit of a pain to write out. Perhaps some simpler, less technical arguements could be used:

Look at the prob. density function: Since it is uniform, it must be constant everywhere. We require the 'area' under the curve to be equal to 1. This can't possibly be possible. (Check - I guess it uses pretty much the same arguement without being too technical.)

Edit: I can't believe I wrote "This can't possibly be possible."

 

[Templar]

I can't say much about random, but I can elaborate on what exactly pseudorandom is (at least for the purpose of random number generators).

 

[milton]

I think, Dumsum, that it is probably even worse than that: your chance of choosing an algebraic number (not just a rational number) is zero, since the algebraic numbers are also denumerable, and the chance of choosing a transcendental number is one.

It is odd, I will agree, but the probability of picking any specific number is zero, however, some number will be picked. I think it probably demonstrates a lack of clarity concerning the concept of infinity, rather than problems with the idea of randomness.

First of all, I have to point out that theres a difference between impossibility and probability 0
impossibility implies probability 0, but not necessarily the converse
flipping a coin and getting heads forever is conceivably possible, but has limiting probability 0
something similar applies for a continous random variable- each value has probability 0, but some value WILL be chosen, hence each value is still "possible", just has infinitesmal probabiltiy

and the original question doesnt really make sense, unless you can integrate the probability density function for the random variable.

and it even gets even more stranger than probabiltiy of choosing a algebraic number being 0,
the probability of choosing a COMPUTABLE number is actually 0 (since number of computable numbers = number of integers)
so how would we actually "choose" such a random number from the reals? naturally, we would get a computer to pick such a number, yet by definition, any number the computer generates is computable. so even if we could choose such a number, chances are, it would be impossible to know what it actually its (even approximately)

and finally, personally, i find it hard to swallow the philosophical concept of randomness. how can anything be inherently random?
after all, nearly all fields of human intellectual endearvour is about linking cause and effect, to understand the relationship between the 2 and making predictions. if we assume that there are some situations where cause and effect are totally decoupled, then isnt the whole world totally fucked up?

if you think hard about it, it seems that pretty much all of the time, we adopt the critical axiom that cause and effect are totally connected, that the same cause must, will and can have one and only one effect. that if we repeated experiements in exactly the same place, time and situation, it would wind up the same way. how is it possible that the exact same inputs, then processed the exact same way under the same circumstances can produce different outputs? to me, to deny this axiom is totally unfathomable. there would be no point learning then, when the same question would have different answers and theres no way to know when which one is right.

to me, randomness is a mystery that i dont thinmk i'l ever understand

 

[§eraphim]

The measure of the rationals Q is zero.

First, note that Q is countable (as it can be enumerated {q_1,q_2,q_3,...) and denote the Lebesgue measure lambda(.) (Over [0,1], this corresponds to a uniformy distributed random variable). Recall that the "measure" is a function mapping a set to the real line which measures the size of a set (or in other words, the probability that a random variables takes values in a certain set). By intuition (or standard proof)

lambda(union of sets_i) <= sum over i of lambda(sets)

(ie, the measure of a union of sets is less than or equal to the sum of the measures of the individual sets)

Fix eps>0 and consider the intervals of length eps/2^i centred at a rational no. q_i:

S_i = (q_i - eps/2^i,q_i + eps/2^i).

the lambda(Q) <= lambda (union of S_i) <= sum over i of lambda(S_i)

Lambda(S_i)= 2*eps/2^i

so the last sum is a geometric sum with answer something like eps or eps/2 (depending on the indexing of the rationals)

thus, for all eps>0, lambda(Q)<= eps/2 say. Let eps->0 to get the answer.