Can there be a lottery of the natural numbers? [duplicate]

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!

marked as duplicate by Did, Charles, jameselmore, Rory Daulton, Alex ProvostDec 30 '15 at 20:25

• You could make $p(n)=1/2^n$. – Gregory Grant Nov 25 '15 at 14:42
• 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
• 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
• 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. – Omnomnomnom Nov 25 '15 at 14:50

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.
• 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 5:59
• 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 6:07
• @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 6:14