TITLE: Probability of picking a random natural number
QUESTION [15 upvotes]: I randomly pick a natural number n. Assuming that I would have picked each number with the same probability, what was the probability for me to pick n before I did it?

REPLY [3 votes]: Hi :) I've always had a nagging suspicion about drawing (uniform) random numbers from an infinite set. I'm not convinced it's possible: here's an intuition about why I'm sceptical: it would be nice if someone could explain where I go wrong. My counterargument is structured as follows: if it's possible to draw random natural numbers (i.e. from the whole infinite set) then there must be a probability of selecting an EVEN number: I argue that $Pr(even) = \frac{1}{2}$ AND $Pr(even) = 1$, causing a contradiction. 
First we need to define:
URD: A "uniform random draw" for a set $S$, ($S = \mathbb{N} = \{1, 2, 3, ... \}$ in this case) is a random selection $x$ from $S$ such that every $y$ in $S$ has an equal chance of being picked. This is what we're doing in the video, and it's what we usually mean when we just say "random". 
Consider the statement '$Pr(\texttt{n even}) = 0.5$', where $n$ is a random element from $\mathbb{N}$  -- this is what I assume most people would agree with. I disagree that this is true, I think it depends on arbitrary selection rules. 
Firstly the intuition we normally have in mine: if you represent a natural number as a rounded scaler from 1 to infinity or a finite string of bits defining a binary number (like your computer does), then it's true that $Pr(\texttt{n even}) = 0.5$ : because it boils down to whether the last bit is 0 (even) or 1 (odd), and since both 0 and 1 bits are equally likely, it follows that Pr(even) = Pr(not even), which must sum to 1 so Pr(even) = 1/2. You can jiggle this for $Pr(\texttt{n is a multiple of m})$ to get $\frac{1}{m}$. It's easy to see that this way of drawing number satisfies URD. 
There is another way to uniquely represent numbers, not just as a line or scale: we can represent n as a product of primes via the fundamental theorem of arithmetic. 
Consider $n = 2^{a1} * 3^{a2}  * 5^{a3} * 7^{a4} * 11^{a5} * ..... $
If we start by drawing random URD numbers a1, a2, a3, ..., over $\{0, 1, 2, ....\}$ and then defining $n$ as above, we have drawn a random $n$. In fact, since every different sequence $\{a1, a2, a3 ... \}$ defines one, and EXACTLY one, natural number $n$, it also follows that this method of drawing $n$ should also satisfy URD. 
But under this last method, $prob(\texttt{n is even}) = prob(\texttt{n not odd}) = 1 - prob(a1 = 0) = 1 - 0 = 1$
If you want to be more formal, you can note the following observation: suppose you have $N$ (non-empty) sets $S_1, S_2, ...., S_N$. We want to choose a random element $X$ from the cartesian product set $S = S_1 \times S_2 \times S_3 .... \times S_N$. Then drawing $X$ is equivalent to drawing $N$ elements $s_i$ from $S_i$, and then setting $X = (s_1, s_2, ..., s_N)$. 
In effect, I've used this result in my argument above that $Pr(even) = 1$, because I've quoted the fundamental theorem of arithmetic and defined $S_i = \{\texttt{powers of the }$i$\texttt{th prime}\}$, e.g. $S_2 = \{1, 3, 9, 27, 81, ....\}$. 
Since I've shown that Pr(even) = 1/2 and shown that Pr(even) = 1, it follows that it makes no sense to even assign a probability to drawing random numbers from the infinite set of natural numbers.
I assume I'm wrong somewhere, but I don't know where I've gone wrong?