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!
 A: You can define a probability law on the set of natural numbers such that $P(Even)=1$ by many ways, but it is completely unrelated to the existence of one-to-one mapping between even natural numbers and all natural numbers. You seem to be assuming that probability is always defined as a fraction like $\dfrac{n(E)}{n(S)}$ - it is not. You can define probabilities however you like, given the probability axioms are not violated. By defining probabilities as fractions you seem to suggest the uniform probability law, but as others already said the uniform probability law on an infinitely countable set is impossible.
Returning to your question "where I am wrong?" - you are wrong in assuming that one-to-one mapping between even and all natural numbers is somehow related to defining a probability law on the set of natural numbers.
A: The OP implicitly required equiprobability for every natural. But let's try something else. We ask the following (starting from $1$, not $0$)

What is the probability of choosing an even natural number if the probability of each number $n$ being chosen is $p_n
= 2^{-n}?$

The good thing here is that we have just provided ourselves with a proper probability measure over $\mathbb N$, because
$$\sum_{n=1}^{\infty}p_n = \sum_{n=1}^{\infty}\frac {1}{2^n} = 1$$
The sum of probabilities related to even numbers under this scheme is
$$P[n \;\;\text{is even}] = \frac 1{2^2} + \frac 1{2^4} + \frac 1{2^6} +... =\frac 1{(2^2)} + \frac 1{(2^2)^2} + \frac 1{(2^2)^3}+... $$
$$=\frac 1{4} + \frac 1{4^2} + \frac 1{4^3}+... = \frac 13$$
...only. This is in reality a special case of the Geometric distribution with parameter $p=1/2$.
A: 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.
A: 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 $(1)$.
A: Define the natural density of any set $S$ of positive integers as follows:
Given natural number $n,$ let $s(n) =$ number of elements of $S$ that are $\le n.$
(Note that $s(n) =$ cardinality of the intersection of $S$ and $\{1, 2, \dots, n\}.$)
Now define the natural density of $S: d(S) =$ limit of $s(n)/n$ as $n\to\infty $
(assuming this limit exists for the moment).
We now define the probability that a natural number drawn is from $S$ by $d(S).$
Example: $S = \{\text{evens}\}.$ It is easy to prove that $d(S) = 1/2$ here.
Example: $S = \{\text{multiples of pos. int.} \;m\}.$ Now $d(S) = 1/m.$
Example: $S$ is finite, so $d(S) = 0.$
I believe this is the sense in which we say that the probability of picking an even number is $1/2,$ and it would conform to our intuition.
Dr. Michael W. Ecker/
Associate Professor of Mathematics
(retired, effective July 1, 2016)/
Pennsylvania State University
Wilkes-Barre Campus/
Lehman, PA 18627 
A: 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.
A: The Laplace formula is only valid for finite probabilistic spaces in which the postulate of Indifference is fulfilled.
It would be necessary to define a probability (or measure) in N that fulfills the axioms of Kolmogorov.
A: I'm going to approach this question the other way around. What is the probability that you will pick one number among natural numbers? If we go by your formula:
$$\text{Probability}=\frac1\infty=0$$
Picking one number has the probability of zero, but what if we tried picking $n$ numbers and expanded $n$ to infinity? So let $i$ be a number and the probability of picking it $Pr(i)$:
$$\lim_{n \rightarrow \infty}(\text{Pr(i)}\times n) = \lim_{n \rightarrow \infty}(0\times n) = 0 $$
This isn't really an answer and we really shouldn't be going at it this way around, but I wanted to show you that, if you tried to answer it, this question would have more than one answer, therefore it can't have an answer.
Also, I've already upvoted Yikai's answer and Mitchell Faas' answer, which are the two right ways of approaching this matter.
A: Obviously, if you limit yourself to a drawing from any bounded and contiguous subset of the integers, i.e. a range $[m,n]$, the number of odd and even numbers will differ by at most one and the probabilities can be exactly or very close to $\frac12$.
This must be why our intuition tells us that the drawing will be balanced, because we actually reason in terms of finite sets, and much less in terms of infinite ones.
It is also true that in the limit for $n\to\infty$, the probabilities are one half.
