Probability of selecting an even natural number from the set $\Bbb N$.

I confirmed on this thread that there are as many as even natural numbers as there are natural numbers.

Question : Suppose I have selected a number $n \in \mathbb N$; what is the probability that $n$ is even?

My Thought :

$\text{Probability} = \dfrac{\text{n(E)}}{\text{n(S)}}$

Here $\text{n(S)}$ is the set of all natural numbers i.e. $\mathbb N$, and $\text{n(E)}$ is set of all even natural numbers.

Since it is proved that number of elements is the set $\mathbb N$ is exactly the same as the number of elements in the set of natural numbers

(it’s very easy to put the set of natural numbers, $\Bbb N=\{0,1,2,3,\dots\}$, into one-to-one correspondence with the set $\text{E}=\{0,2,4,6,\dots\}$ of even natural numbers; the map $\Bbb N\to \text{E}:n\mapsto 2n$ is clearly a bijection.) ;

Thus, Probability $= \boxed 1$

I know this is definitely wrong.Probability must be $0.5$. But where am I wrong?

Can anyone explain ?

Thanks!

When you write Probability = $\dfrac{\text{n(E)}}{\text{n(S)}}$, you're assuming that you're drawing a number uniformly at random, which means that every number has the same probability to be drawn. This formula is valid if $\text{E}$ is a finite set, but not if $\text{E}$ is infinite. In fact, we can show that there is no way to draw uniformly at random over $\mathbb{N}$ or $\mathbb{Z}$, as said in the comments. You can't do that and at the same time satisfy the properties that are expected from probabilities. See Yikai's answer for a proof of this fact, if you have some knowledge of measure theory.

So if you want to compute some probability of getting an even number among natural numbers, you first have to specify what is the distribution on $\mathbb{N}$, but this cannot be the uniform distribution.

You first have to define a probability measure over the sample space $\Omega = \mathbb{N}$. In your question, the sigma algebra contains all singleton sets, i.e., $\{i\}$ for $i \geq 0$.

Suppose there is a probability measure on $\Omega$ such that $\epsilon =\Pr(\{0\})= \Pr(\{1\}) = \Pr(\{2\}) = \cdots$

By definition of probability, we have $$ 1 = \Pr(\mathbb{N}) = \sum_{i=0}^\infty \Pr(\{i\}) = \sum_{i=0}^\infty \epsilon \tag{$1$} $$ This is a contradiction since if $\epsilon > 0$, then $\sum_{i=0}^\infty \epsilon$ can not be $1$ and if $\epsilon = 0$, then $\sum_{i=0}^\infty \epsilon = 0$ violating $(1)$.

What you're missing here is that when you say $$\text{Probability}=\frac{n(E)}{n(S)}$$ You're forgetting that $N(E)$ and $N(S)$ are both infinite, so you're claiming: $$\text{Probability}=\frac\infty\infty$$ You can't make the assumption that this equals one.