1

Can there be a lottery of the natural numbers, so that every natural number is chosen equally likely?

The standard answer would be "No" because: If we define a measure $\mathbf{P}$ on $\mathbb{N}$ so that $\mathbf{P}(n) = r \in (0,1] \; \forall \, \mathbb{N}$, then $\mathbf{P}(\mathbb{N}) = \infty$. If we define a measure so that $\mathbf{P}(n) = 0$, then $\mathbf{P}(\mathbb{N}) = 0$.

But why can we conclude from that, that a lottery of the natural numbers (with every natural number equally likely) is impossible?

Note: the question is not if there can be a uniform probability distribution (satisfying all axioms of probability, including countable additivity) over the natural numbers but if there can be a lottery of the natural numbers so that every number is chosen equally likely!

R. Neville
  • 127
  • You could make $p(n)=1/2^n$. – Gregory Grant Nov 25 '15 at 14:42
  • 1
    I don't understand what you mean by "If we define a measure so that $P(n)=0$ then $P(\Bbb N)=1$.". – Gregory Grant Nov 25 '15 at 14:44
  • @GregoryGrant I suspect that was a typo. I repaired it in an edit. Let the OP check it. – drhab Nov 25 '15 at 14:46
  • 6
    No, there is no way around it. There is no uniform (probability) measure on a countably infinite set. When one needs something like this, a typical approach is to obtain a result for a uniform probability on a finite set ${1,\ldots, N}$ and then pass to a limit as $N \to \infty$, depending on what actually is to be shown. For example, one might ask: what is the probability that "two natural numbers chosen at random" are coprime? A sensible interpretation of this question can be made using the limit approach, although strictly speaking there is no way to uniformly select a natural number. – hardmath Nov 25 '15 at 14:47
  • 1
    If you want each integer to have an identical probability such that the probabilities sum to $1$, then at the very least you'll need to work over a non-archimedian field. Maybe a solution to this exists in non-standard analysis. – Ben Grossmann Nov 25 '15 at 14:50
  • There is a problem in your point $1$: the probability that you obtain a rational number when uniformly drawing a real number in $[0,1)$ is $0$. The average number of trials until picking a rational number is $\infty$. – Taladris Nov 30 '15 at 06:10
  • @Taladris: So if I could chose infinitely often, it would be possible? (as I said, this is a philosophical question) – R. Neville Nov 30 '15 at 06:16
  • "Why does the following not work" Choosing repeatedly numbers in (0,1) with uniform distribution will never produce a rational number. That is, not only the average number of trials is infinite but with full probability no trial will result in a rational number ever. – Did Nov 30 '15 at 07:30
  • @Did: How can you know that? Say, I chose a number $r \in [0, 1)$ with equal distribution. But of course $\mathbb{P}({r}) = 0$, yet it did that I chose $r$, so it is not true that I will never chose $r$. – R. Neville Nov 30 '15 at 07:37
  • Basic probability stuff: each time you draw a number, the probability that it is irrational is 1 hence, with probability 1, none of them is irrational. – Did Nov 30 '15 at 08:02
  • @Did: How do you know that an event with probability zero will never happen? – R. Neville Nov 30 '15 at 08:04
  • @R.Neville: I was going fast by saying that the average number of trials is $\infty$. Actually, the zero probability of drawing a rational means that you cannot define a geometric law: the probability of drawing a rational at the nth trial is also $0$. The probability of drawing a rational after an infinity number of trial has no meaning. – Taladris Nov 30 '15 at 08:19
  • While the answers to the linked question do sort of answer this question, a better choice to close this as a duplicate of would probably be this question. – Ilmari Karonen Feb 20 '16 at 14:41

1 Answers1

3

The probability that some number is drawn cannot be positive because the sum of the probabilities would be infinite.

Probability $0$ for an event not impossible is possible, if the number of events is uncountable. But for countably many events, $P(X)=0$ is equivalent to $X$ is the impossible event.

Peter
  • 84,454
  • Where is the flaw in the following: 1.) I choose randomly on $[0, 1)$ with uniform distribution until I get a result in $A = \mathbb{Q} \cap [0, 1)$ (rationals are spreaded equally in the reals) 2.) I use a bijection $\phi\colon A \rightarrow \mathbb{N} $, voilà, I have chosen randomly a natural number (with all of them equally likely)! – R. Neville Nov 30 '15 at 05:59
  • The probability that you ever choose a rational number in the interval $[0,1)$ is $0$ due the the uncountability of the reals – ASKASK Nov 30 '15 at 06:07
  • So the whole "choose randomly until I get a rational result" will have you never reaching an answer – ASKASK Nov 30 '15 at 06:08
  • @ASKASK: Yes, I know $\mathbb{P}(A) = 0$, but it is still possible! After all, if I chose a real number $r$ in $[0,1)$ it was $\mathbb{P}(r) = 0$, but still it happened! – R. Neville Nov 30 '15 at 06:14
  • To be completely honest, I have very little knowledge on this subject so feel free to ignore my answer as it may be wrong, but from how I see it, you can't simply rely on the notion of "inevitably landing on a rational" because there is nothing that guarentees that it will actually ever happen (in fact, odds are it just never happens) – ASKASK Nov 30 '15 at 06:19
  • @ASKASK: So, maybe there is an infinite lottery of the natural numbers, which sadly just fails to give me a natural number with probability 1, but in the cases it gives me a natural number then all of them are equally likely? :-D – R. Neville Nov 30 '15 at 06:26
  • Well sure... But if you are willing to stoop to that, then why not just pick a random number in $(0,1]$ and then take its reciprocal? Every natural number will appear with equal probability, it's just there is a probability of 0 that any of them will appear. – ASKASK Nov 30 '15 at 07:22
  • @ASKASK: Because that way they are not "spreaded" equally which I believe is needed as an assumption so that every natural number is equally likely. In $[0, 0.1)$ there are infinitely many reciprocals of natural numbers, in $[0.1,1)$ there are only the reciprocals of $2, \ldots, 10$. – R. Neville Nov 30 '15 at 07:47
  • Interesting. Now I'm inclined to ask, how do you know that your given bijection from Q to N preserves the "equally spread" property? – ASKASK Nov 30 '15 at 07:49
  • @ASKASK: It's nothing that can be preserved. I just assume that it is important that the the countably infinite set to which we construct a bijection to $\mathbb{N}$ is equally "spread" on $[0,1)$. – R. Neville Nov 30 '15 at 07:55
  • Interesting. So basically you just want a bijection between $[0,1)$ and $[0,\infty)$ such that the natural numbers are all hit with equal probability? – ASKASK Nov 30 '15 at 07:57