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Imagine infinite set of, let's say, natural numbers. I choose one of the infinite numbers randomly. Let's call that number n. If I choose another number too, can it be the same number (n), theoretically?

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    $\begingroup$ xkcd.com/221 $\endgroup$ – Sandejo Dec 9 '20 at 16:01
  • $\begingroup$ What are your own thoughts on this? Intuitive, it would seem yes, sure, why couldn't the second be the same as the first? But since you posted this question, I assume you have some doubts or concerns? What are those? Can you please express those in your post? $\endgroup$ – Bram28 Dec 9 '20 at 16:01
  • $\begingroup$ What do you mean by "randomly"? Do you mean that any number has the same probability of being selected? $\endgroup$ – G. Chiusole Dec 9 '20 at 16:04
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    $\begingroup$ How do choose your number? You need a probability distribution, and note that as probability distributions are countably additive, if you make one on a countable set (such as the natural numbers) not all elements can have the same probability. $\endgroup$ – Henrik supports the community Dec 9 '20 at 16:05
  • $\begingroup$ @G.Chiusole Yeah, any number has the same probability of being selected. $\endgroup$ – Yağız Alp Ersoy Dec 9 '20 at 16:17
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As mentioned in the comments, there is no way that each natural can have the same probability. See this answer to "Can you pick a random natural number?..., for instance.

However, we can still pick a number randomly with unequal probabilities. For instance, suppose I choose a number by flipping a coin, and the number is the number of heads before the first tail. (If you want to include a case of flipping heads forever, let's say that also counts as the number $0$.)

I just got the following sequences of flips: $HHT,T,HT,T$. Which means that the number $0$ was randomly selected twice.

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I'd loved to put this in a comment, but I can't because I am new here.

I just wanted to say that since you picked that particular number the first time, it has a non-zero probability of being picked. Thus it definitely can be picked again.

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  • $\begingroup$ Possible does not necessarily mean "non-zero probability". See Zero probability and impossibility. $\endgroup$ – Mark S. Dec 9 '20 at 21:39
  • $\begingroup$ Oh I see, you are right. That was a bad choice of words from me. I should have said that "it is pickable" instead of "has a non-zero probability of being picked". However I could even argue that the limit of $k/x$ for k > 0 and x tending to infinity it's not exactly 0, but a bit more than that. $\endgroup$ – Sheik Yerbouti Dec 9 '20 at 21:56
  • $\begingroup$ "It is pickable" would be fine. For your other comment, even in nonstandard analysis where there are a variety of infinities and infinitesimals, we would still say the limit is exactly $0$. $\endgroup$ – Mark S. Dec 9 '20 at 22:41
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    $\begingroup$ But take for example 1/x, the function never "touches" the horizontal axis. For any value of x arbitrarily large there always will be an e : 0 < e < 1/x $\endgroup$ – Sheik Yerbouti Dec 9 '20 at 23:09
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    $\begingroup$ By reading this question: math.stackexchange.com/questions/2270193/…. I've understood that yes, we say that the limit is 0, but when we say so we actually mean that it is arbitrarily close to zero. So I may conclude that in a certain way we are both right. $\endgroup$ – Sheik Yerbouti Dec 9 '20 at 23:51

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