Probability of choosing any real number in the range $[0,1]$?

Question

Basically I want to give a counter argument (if possible) to the statement:

"It is unlikely to randomly pick any real number in the interval $[0,1]$"

The statement implies that there is a small probability. So if you could either prove that the probability is $0$, or if you can prove that you cant define the probability in the first place, then that should be enough to make a counter argument?

I've done a bunch of reading and there seems to be multiple ways to tackle this, like measure theory, logical possibility, conceptual possibility etc.... Some guy said that the probability of choosing one specific number on a uniform distribution is $0$, but here we are talking about infinitely many numbers in the range $[0,1]$.

I'm assuming there is someone smarter than me here who knows this stuff, any help is appreciated.

Answer 1

It's simple. A single real number is a single point. [0,1] has lenght 1. [0,1/2] has lenght 1/2 so the probability of picking a number out of that interval is 1/2/1=1/2. A point has no (measurable) lenght. So the probability of picking it is 0 /1=0. Also picking out any rational number is zero, regardless of the fact that there is an infinite number of them, chances of landing on a rational number on a real line are 0. Because each single rational has on its own no measurable lenght. Sets like all rationals in [0,1], have no defined lenght because they have discontinuities, lots of them, they are missing all the irrationals like pi, square of 2 etc. Therefore only reals are smooth and complete enough to have a meaningful definition of lenght defined on their intervals.

Answer 2

If i understood correctly, then you are asking wheter there exists a uniform probability distribution $P$ over $\mathbb{R}$ and if so, what is $P([0,1])$.

The answer is no. Suppose however towards a contradiction, that such a measure exists. Now by basic probability axioms we can write $$P(\mathbb{R}) = \sum_{k\in \mathbb{Z}}P([k,k+1))$$ since $\{[k,k+1) \: : \: k\in \mathbb{Z}\}$ is a (countable) partition of $\mathbb{R}$. If the distribution is uniform, then $P([k,k+1))=P([n,n+1))$ for all $k,n \in \mathbb{Z}$, so we have two cases to cover:

$$P([k,k+1))=\alpha > 0 \Rightarrow P(\mathbb{R})= \sum_{k\in \mathbb{Z}} \alpha = \infty$$ and $$P([k,k+1))= 0 \Rightarrow P(\mathbb{R})= \sum_{k\in \mathbb{Z}} 0 = 0$$ but both cases give a contradiction, since we assumed $P$ to be a probability measure.