I randomly pick a natural number n. Assuming that I would have picked each number with the same probability, what was the probability for me to pick n before I did it?

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    $\begingroup$ This is more a question for a psychologist than for a mathematician. Humans tend to pick certain numbers more often than other numbers. For example 7, 17, 35 and 37 are quite popular. No amount of mathematics would allow you to predict such a distribution without some facts from psychology. See here for some more details: scienceblogs.com/cognitivedaily/2007/02/… $\endgroup$
    – Dan Piponi
    Jan 18, 2011 at 23:06

6 Answers 6


There is no uniform distribution on the natural numbers. So the question itself is self-contradictory.

Some explanation:

You say that each number is chosen with the same probability. Say probability $c$. Let's denote the probability to choose $n$ by $p(n)$. So you say $p(n)=c$ for any natural $n$. For any distribution, the sum of probabilities always equals 1: $$p(1)+p(2)+p(3)+ \dots = 1.$$ So $c+c+c+\dots = 1.$ But if $c > 0$, you get $$c+c+c+\dots = \infty.$$ And if $c=0$ you get $$c+c+c+\dots = 0.$$ There's no value of $c$ for which this sum equals $1$.

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    $\begingroup$ So, are you saying that whenever I pick a natural number at random, some are more likely to get picked by me than others? $\endgroup$
    – Thomas
    Dec 13, 2010 at 11:20
  • $\begingroup$ Yes. Here's an example: Choose n with probability $1/(2^{n})$. This is a valid distribution because $1/2 + 1/4 + 1/8 + ... = 1$. $\endgroup$
    – user3533
    Dec 13, 2010 at 11:22
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    $\begingroup$ If this seems counterintuitive, think of a person choosing a "random natural number". Any normal person would choose a number with much less than a billion digits (I'm exaggerating on purpose), so he's actually choosing from a finite set of naturals, giving all larger ones 0 probability. $\endgroup$
    – user3533
    Dec 13, 2010 at 11:27
  • $\begingroup$ @Thomas: Not "whenever I pick a natural number at random", but "in general, when I pick a natural number at random". $\endgroup$
    – Shai Covo
    Dec 13, 2010 at 11:48
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    $\begingroup$ The point is that it's not possible to pick a natural number uniformly at random. Not that some are more likely than other, this depends on the distribution used. The point is that there is no uniform distribution. $\endgroup$
    – Benjamin
    Dec 3, 2015 at 18:44

There's an aspect that doesn't seem to have been mentioned in the other answers yet.

The question presupposes that it is possible to pick each natural number with the same probability. Since this is not possible, it raises the question what may have led to the assumption that it should be possible. It's instructive to compare this with typical cases where there is a uniform distribution.

When we roll a die, there's a symmetry between the sides. There's no reason why any of them should be more or less likely than the others to come up, so it's clear that the probability for each of them should be the same (if the die is symmetric and we throw it hard enough).

If we spin a wheel, there's again a symmetry among the angles at which it might come to rest. There's again no reason why any of them should be more or less likely than the others, and it's clear that the probability should be uniform across the range of angles (if the wheel is symmetric and we turn it hard enough).

The idea that "I randomly pick a natural number n. Assuming that I would have picked each number with the same probability" seems to be modeled on such cases. So it's instructive to realize that there is no similar experiment that this corresponds to. As has been pointed out in some of the comments, asking a person for a "random" answer won't do, because there's no reason to except symmetry between small and large numbers. Another idea might be to mark the rational numbers on the wheel, choose some bijection mapping them to the natural numbers, and then choose the natural number that corresponds to the rational number at which the wheel comes to rest. But if you think about it, this doesn't make the sense that it does for real numbers.

So there's a direct correspondence between the impossibility of setting up a situation in which there is the required symmetry between all natural numbers and the impossibility of defining a uniform distribution on the natural numbers (or any countably infinite set).


If you want to choose a natural number from the set $\lbrace 1,2,3,\ldots \rbrace$ of all natural numbers, then user3533 is right. But if you pick a natural number from a finite set, say $A = \lbrace 1,2,\ldots,N \rbrace$, each number with the same probability, then this probability must be the reciprocal of the number of elements in that set. So, if $A = \lbrace 1,2,\ldots,N \rbrace$, then $p=1/N$. This follows immediately from $\sum\nolimits_{i = 1}^N {p_i } = 1$ with $p_1 = p_2 = \cdots = p_N = p$.


Some clarification should be added to the above answers.

By partitioning the reals into non-measurable sets, one can devise a way to relate a real selected uniformly at random from an interval such as [0, 1) to a natural number using a method by which all natural numbers have an "equal," however undefined, probability of being mapped to (here, equal simply means that no natural is given preference as to being mapped to over any other natural). The problem is that, because non-measurable sets are used to do this, no meaningful cumulative distribution function can be established (as might be expected). To see an example of such a selection process, see here:

Is this fraction undefined? Infinite Probability Question.


It would require a definition of a probability distribution with an infinitesimal probability mass at every natural number. Dirac delta functions of infinitesimal length.

In (Bayesian) statistics, functions that relax the axiom of "summation to one" are known as improper priors and have an infinite total mass. This loss of the axiom on tautology ($p(T)=1$ for any tautology $T$) comes with severe repercussions. You'll see Dutch book arguments that make this clear.

Note that there are many infinite sequences that people try to tackle with tools from probability theory. Most famous are the attempts to postulate hypotheses about the distribution of subsequent primes. So, this means that questions like the one you have are very good to ask ourselves.

There is also the concept of natural density. The natural density of picking an even natural number is $1/2$.


Hi :) I've always had a nagging suspicion about drawing (uniform) random numbers from an infinite set. I'm not convinced it's possible: here's an intuition about why I'm sceptical: it would be nice if someone could explain where I go wrong. My counterargument is structured as follows: if it's possible to draw random natural numbers (i.e. from the whole infinite set) then there must be a probability of selecting an EVEN number: I argue that $Pr(even) = \frac{1}{2}$ AND $Pr(even) = 1$, causing a contradiction.

First we need to define: URD: A "uniform random draw" for a set $S$, ($S = \mathbb{N} = \{1, 2, 3, ... \}$ in this case) is a random selection $x$ from $S$ such that every $y$ in $S$ has an equal chance of being picked. This is what we're doing in the video, and it's what we usually mean when we just say "random".

Consider the statement '$Pr(\texttt{n even}) = 0.5$', where $n$ is a random element from $\mathbb{N}$ -- this is what I assume most people would agree with. I disagree that this is true, I think it depends on arbitrary selection rules.

Firstly the intuition we normally have in mine: if you represent a natural number as a rounded scaler from 1 to infinity or a finite string of bits defining a binary number (like your computer does), then it's true that $Pr(\texttt{n even}) = 0.5$ : because it boils down to whether the last bit is 0 (even) or 1 (odd), and since both 0 and 1 bits are equally likely, it follows that Pr(even) = Pr(not even), which must sum to 1 so Pr(even) = 1/2. You can jiggle this for $Pr(\texttt{n is a multiple of m})$ to get $\frac{1}{m}$. It's easy to see that this way of drawing number satisfies URD.

There is another way to uniquely represent numbers, not just as a line or scale: we can represent n as a product of primes via the fundamental theorem of arithmetic.

Consider $n = 2^{a1} * 3^{a2} * 5^{a3} * 7^{a4} * 11^{a5} * ..... $

If we start by drawing random URD numbers a1, a2, a3, ..., over $\{0, 1, 2, ....\}$ and then defining $n$ as above, we have drawn a random $n$. In fact, since every different sequence $\{a1, a2, a3 ... \}$ defines one, and EXACTLY one, natural number $n$, it also follows that this method of drawing $n$ should also satisfy URD.

But under this last method, $prob(\texttt{n is even}) = prob(\texttt{n not odd}) = 1 - prob(a1 = 0) = 1 - 0 = 1$

If you want to be more formal, you can note the following observation: suppose you have $N$ (non-empty) sets $S_1, S_2, ...., S_N$. We want to choose a random element $X$ from the cartesian product set $S = S_1 \times S_2 \times S_3 .... \times S_N$. Then drawing $X$ is equivalent to drawing $N$ elements $s_i$ from $S_i$, and then setting $X = (s_1, s_2, ..., s_N)$. In effect, I've used this result in my argument above that $Pr(even) = 1$, because I've quoted the fundamental theorem of arithmetic and defined $S_i = \{\texttt{powers of the }$i$\texttt{th prime}\}$, e.g. $S_2 = \{1, 3, 9, 27, 81, ....\}$.

Since I've shown that Pr(even) = 1/2 and shown that Pr(even) = 1, it follows that it makes no sense to even assign a probability to drawing random numbers from the infinite set of natural numbers.

I assume I'm wrong somewhere, but I don't know where I've gone wrong?


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