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 any one explain ?


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    $\begingroup$ You need to specify a distribution telling you with what probability you draw what number. And there is no uniform distribution on the natural numbers. $\endgroup$ – Arthur Apr 12 '17 at 8:28
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    $\begingroup$ That's what I'm saying: making each natural number equally likely is impossible. $\endgroup$ – Arthur Apr 12 '17 at 8:31
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    $\begingroup$ math.stackexchange.com/questions/146844/… $\endgroup$ – Asaf Karagila Apr 12 '17 at 10:33
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    $\begingroup$ "I'm thinking by common sense" - that's going to give you pretty terrible results for anything to do with probability. Human intuition is notoriously awful at probability. $\endgroup$ – user2357112 Apr 12 '17 at 17:33
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    $\begingroup$ @JaideepKhare: The Monty Hall problem, Simpson's paradox, that thing with Bayes' theorem and the medical test, etc... human intuition breaks down all over the place for probability. It also breaks down all over the place for anything involving infinities, but that just means it's terrible at more than one thing. $\endgroup$ – user2357112 Apr 12 '17 at 17:48

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 satisfying 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.

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    $\begingroup$ Can't you just take the limit as the sample space goes to N? $\endgroup$ – Kevin Apr 12 '17 at 15:50
  • $\begingroup$ What would be the probability to pick $1$ for example? $\endgroup$ – Augustin Apr 12 '17 at 15:57
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    $\begingroup$ @Kevin: You can take the limit of the probability of even-ness, and get $\frac{1}{2}$, but then you’ve computed “the limit, as $n \to \infty$, of the chance of getting an even number when picking uniformly from $\{1,…,n\}$”. You can’t reasonably call it “the chance of getting an even number when picking uniformly from $\mathbb{N}$” unless you can find a way to make sense of what “picking uniformly from $\mathbb{N}$” means in itself — and this is rather difficult to make sense of, since taking a limit of probability distributions doesn’t give a probability distribution in general. $\endgroup$ – Peter LeFanu Lumsdaine Apr 12 '17 at 17:40
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    $\begingroup$ @PeterLeFanuLumsdaine: I agree, but I think that limit is more concisely called "the proportion of even numbers to natural numbers." If you use primality instead of evenness, you get "the proportion of primes to naturals is zero" as a trivial corollary of the Prime Number Theorem, or in even simpler terms, "almost all natural numbers are composite." So this isn't totally meaningless, it's just not a "real" probability distribution (in the sense that it doesn't have all of the properties we want in a probability distribution). $\endgroup$ – Kevin Apr 12 '17 at 17:59
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    $\begingroup$ @Kevin: absolutely, agreed — it’s a very meaningful and interesting limit to consider, and is extensively used and studied, mainly under the names natural density or asymptotic density. $\endgroup$ – Peter LeFanu Lumsdaine Apr 12 '17 at 18:14

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 $(1)$.

  • $\begingroup$ So you mean, that may be, selecting $2$ is more or less probable than selecting $1$? $\endgroup$ – Jaideep Khare Apr 12 '17 at 8:46
  • $\begingroup$ @JaideepKhare Note that there is a $\cdots$ behind them. In other words, there is no probability measure such that all natural numbers have the same probability. $\endgroup$ – PSPACEhard Apr 12 '17 at 8:48
  • $\begingroup$ @JaideepKhare It's up to you to choose whatever distribution on the natural numbers you see fit. Any one of these can work. $\endgroup$ – rwols Apr 12 '17 at 10:40
  • $\begingroup$ How is this problem normally resolved? Don't sample from $\mathbb{N}$? Or define a different probability measure? Something else? $\endgroup$ – Burnsba Apr 12 '17 at 12:40
  • $\begingroup$ @Burnsba There are other probability measures defined on $\mathbb{N}$ such as Poisson distribution and Geometric distribution. $\endgroup$ – PSPACEhard Apr 12 '17 at 12:41

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.

  • $\begingroup$ See : en.wikipedia.org/wiki/Cardinality $\endgroup$ – Jaideep Khare Apr 12 '17 at 8:28
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    $\begingroup$ You're right that $n(E)=n(S)$ and that both equal aleph-0. But it doesn't matter that the cardinalities are equal, you still can't divide them. You're using countable infinity as a number, which is a dangerous game. It's more coherently seen as an ever continuing process. After all, $\frac{n(E)+1}{n(S)}=1$ if we used it as a number, but we know that $\frac{a}{b}\neq\frac{a+1}b$ $\endgroup$ – Mitchell Faas Apr 12 '17 at 8:32
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    $\begingroup$ That's what's my question is, why can't we, if we compare them? $\endgroup$ – Jaideep Khare Apr 12 '17 at 8:33
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    $\begingroup$ Because it's not a number, it's a process. $\infty+1=\infty$ might seem logical, but the following also holds: $\infty-\infty=\infty$. The problem is that they never end, so normal arithmetic is impossible. We say that the cardinalities are equal, but that only gets us so far as to say that they continue in the same way. Example: countable infinity is usually proven by making a list and showing how it progresses. We can't use that for uncountable infinities, those progress differently. $\endgroup$ – Mitchell Faas Apr 12 '17 at 8:40
  • $\begingroup$ @JaideepKhare: We can't precisely because trying to leads to contradictions. $\endgroup$ – Will R Jul 6 '17 at 19:25

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 whatever 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.


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 $p=1/2$.

  • $\begingroup$ I've seen that sum used in showing 1/3 of all reduced fractions have even denominators. $\endgroup$ – James Waldby - jwpat7 Apr 13 '17 at 5:21

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:


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.


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.


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 <= n. (Note that s(n) = cardinality of the intersection of S and {1, 2, ..., n}.) Now define the natural density of S: d(S) = limit of s(n)/n as n->infinity (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 = {evens}. It is easy to prove that d(S) = 1/2 here.

Example: S = {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


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.


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