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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?

Yağız Alp Ersoy
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    https://xkcd.com/221/ – Sandejo Dec 09 '20 at 16:01
  • 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? – Bram28 Dec 09 '20 at 16:01
  • What do you mean by "randomly"? Do you mean that any number has the same probability of being selected? – G. Chiusole Dec 09 '20 at 16:04
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    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. – Henrik supports the community Dec 09 '20 at 16:05
  • @G.Chiusole Yeah, any number has the same probability of being selected. – Yağız Alp Ersoy Dec 09 '20 at 16:17
  • As Henrik points out, such a probability measure does not exist – G. Chiusole Dec 09 '20 at 16:18
  • @Bram28 Well, you have a very good point there. I have no idea. Infinity is itself is like a paradox, it's not yet fully known. – Yağız Alp Ersoy Dec 09 '20 at 16:19

2 Answers2

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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.

Mark S.
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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.

anotherOne
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  • Possible does not necessarily mean "non-zero probability". See Zero probability and impossibility. – Mark S. Dec 09 '20 at 21:39
  • 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. – anotherOne Dec 09 '20 at 21:56
  • "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 $0$. – Mark S. Dec 09 '20 at 22:41
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    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 – anotherOne Dec 09 '20 at 23:09
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    By reading this question: https://math.stackexchange.com/questions/2270193/limit-of-1-x-as-x-approaches-infinity#:~:text=In%20standard%20real%20analysis%2Fcalculus,)%E2%88%92c%7C%3C%CE%B5.&text=So%2C%201%2Fx%20really%20does,gets%20arbitrarily%20close%20to%200. 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. – anotherOne Dec 09 '20 at 23:51