# Is it possible to randomly select a single member of $\mathbb N$? [duplicate]

The title is my question and the reason for asking it is the following.

Define a set $$\mathbb N (≤ n) \equiv$$ {1,2,3, … , n} and define a “random selection” to be a selection in which each member of $$\mathbb N (≤ n)$$ has an equal probability, $$p$$, of being selected.

As $$\mathbb N (≤ n)$$ is the whole population from which the selection will be made then the probability of selecting some member of $$\mathbb N (≤ n)$$ is $$np = 1$$.

Since $$p \to 0$$ as $$n \to ∞$$ then the probabilty of any member of $$\mathbb N$$ being randomly selected is $$p = 0$$.

Is this correct?

Duplicate Question ?

I agree my question is similar to the other question but it is not similar enough to be called a duplicate. For example, the definition of “random selection” I use is different.

That is mostly correct. What is true is that there is no distribution on $$\mathbb N$$ that associates to each $$n \in \mathbb N$$ an equal probability of being selected. On the other hand, that doesn't mean you can't select a random nonnegative integer "at random", because "at random" doesn't imply the distribution is uniform.

I don’t understand the statement about there being “no distribution that associates to each $$n \in \mathbb N$$ an equal probability of being selected”. Doesn’t the argument in my question show that there is such a distribution, namely $$p = 0$$ for each member of $$\mathbb N$$ ?

Also I don’t understand how a member of $$\mathbb N$$ could be randomly selected unless each member of $$\mathbb N$$ had the same probability of selection?

The conclusion of the argument in my question seems reasonable because it implies it is impossible to make a random selection of a single member of $$\mathbb N$$. This agrees with the practical impossibility of an algorithm that randomly selected a single member of an infinite set because an infinite set as input could not be processed by such an algorithm in a finite time.

User334732 comment

Let $$f(n)=\frac{1} {6}$$ for all $$n \in \mathbb N$$. Let the order be 1,2,3,4,5... Then start with 1 and roll a dice. If you get a 6, stop and you have selected number 1 as your answer. If you don't, then go to $$n$$=2 and roll the dice again. Continue until you roll a 6.

With this procedure the probability of selecting $$n$$ is $$(\frac{5} {6})^{(n-1)} \frac{1}{6}$$ so smaller numbers have a greater probability of being selected than larger numbers. This selection does not seem to be “random”. Specifically it does not satisfy the definition of “random selection” I use in my question which is all $$n \in \mathbb N$$ must have an equal probability of being selected.

Comment from Vincent

Thanks for the clarification. What Snarski is saying is: 1) You are correct that there is no "random selection" according to your definition 2) your definition is somewhat non-standard. Usually (at least among mathematicians, the general population might lean more toward your definition) 'random' (roughly) means that probabilities are involved, not that they are necessary all equal. So yes, a random process for selecting integers that gives each equal, positive probability is impossible, but the next best thing, a random process that gives all numbers at least some positive probability, is.

user334732's comment is an example of that latter, 'next best' thing, and a very beautiful example on top of that because it is very hands on.

It seems we agree on the impossibility of a random selection of a single member of $$\mathbb N$$ given the definition of "random selection" in my question.

You mention that my definition is "somewhat non-standard" and while my definition may be favored by the "general population" usually to mathematicians, " 'random' (roughly) means that probabilities are involved, not that they are necessary all equal".

The definitions of a "random sample", at least the ones I have seen, say a "random sample" has only two defining properties:

P1. each member of the population has an equal chance of selection, and

P2. each selection is independent of the other selections.

My question asks about the random selection of a single member of $$\mathbb N$$, that is, a random sample of size one from the population $$\mathbb N$$ , so property P2 is not applicable and the general definition specializes to property P1 which is the definition in my question.

If you know of a definition of "random selection" that differs from or extends the P1 + P2 definition above then can you supply a reference ? I would be interested to read about it.

I do not share your opinion of the procedure suggested by u334732 although I appreciate u334732 taking the time to attempt to answer my question.

Problem 1 with u334732's suggestion is that the procedure does not give an equal probability of selection for each member of $$\mathbb N$$ since the probability of selecting $$n$$ is $$(\frac{5} {6})^{(n-1)} \frac{1}{6}$$. Hence the procedure does not comply with the definition of "random selection" in my question. I mentioned this problem in a previous reply.

Problem 2 is that the procedure may never select a member of $$\mathbb N$$ because the probability of not selecting a member of $$\mathbb N$$ that is ≤ n is $$(\frac{5} {6})^{n}$$ which is > 0 for all $$n \in \mathbb N$$ hence there is a non-zero probability that no member of $$\mathbb N$$ will be selected by any finite number of applications of the procedure.

I think these problems reflect the impossibility of randomly selecting a single member of $$\mathbb N$$ and there would be some problem with any suggested procedure.

## marked as duplicate by Michael Hoppe, Joel Reyes Noche, Vincent, user334732, Claude LeiboviciDec 12 '18 at 11:28

• Progress through the integers along a wellorder and combine that with a probability $p_x=f(x)$ of stopping on any given element (given that one arrives at it). – user334732 Dec 12 '18 at 10:46
• @user334732 - Can you give more detail about how your procedure actually selects a member of $\mathbb N$. – Xiu LL Dec 12 '18 at 18:27
• Then start with $1$ and roll a dice. If you get a 6, stop and you have selected number $1$ as your answer. If you don't, then go to $n=2$ and roll the dice again. Continue until you roll a $6$. – user334732 Dec 12 '18 at 18:35
That is mostly correct. What is true is that there is no distribution on $$\mathbb{N}$$ that associates to each $$n \in \mathbb{N}$$ an equal probability of being selected. On the other hand, that doesn't mean you can't select a random nonnegative integer "at random", because "at random" doesn't imply the distribution is uniform.