# Probability of picking a random natural number

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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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/… –  Dan Piponi Jan 18 '11 at 23:06

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)+... = 1. So c+c+c+... = 1. But if c > 0 you get c+c+c+... = $\infty$. And if c=0 you get c+c+c+... = 0. There's no value of c for which this sum equals 1.

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So, are you saying that whenever I pick a natural number at random, some are more likely to get picked by me than others? –  Thomas Dec 13 '10 at 11:20
Yes. Here's an example: Choose n with probability $1/(2^{n})$. This is a valid distribution because $1/2 + 1/4 + 1/8 + ... = 1$. –  user3533 Dec 13 '10 at 11:22
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. –  user3533 Dec 13 '10 at 11:27
@Thomas: Not "whenever I pick a natural number at random", but "in general, when I pick a natural number at random". –  Shai Covo Dec 13 '10 at 11:48

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).

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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 $p_1 = p_2 = \cdots = p_N = p$.

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