Paradox of Drawing Samples from Countably Infinite Urn? Consider the following two countable sets of numbers $A = \mathbb{N}\smallsetminus 7 \mathbb{N}$ and $B = 7\mathbb{N}$, i.e., the set of natural numbers that are not multiples of 7 and the set of all multiples of 7. It is easy to prove that $|A| = |B|$.
Given these facts, I have been twisting my brain around the following thought experiment that I came up with - imagine you have an infinitely large urn that contains the elements $A \cup B = \mathbb{N}$. The probability that a random element drawn out of this urn is divisible by 7 = $\frac{1}{7}$ even though, according to cardinality, there are an equal number of elements in from both sets in the urn! In fact this argument would suggest that there are 6 times as many numbers in set $A$ than there are in set $B$!
Going strictly by the sizes of the sets, it seems like it should be equally likely to pick elements from set $A$ and from set $B$ which is even stranger since from empirical evidence we know that the probability should be $\frac{1}{7}$.
How do we resolve this paradox? I feel that I have misunderstood some concept pertaining to cardinality of sets, but then again, $\infty$ is a pretty weird monster on its own and hence I am expecting some interesting responses!
 A: Probability theory (PT) allows you to argue in a consistent way about such questions. If you want to bring in PT you have to set up a space $\Omega$ of elementary events $\omega$ (here $\Omega={\mathbb N_{\geq1}}$), a $\sigma$-algebra ${\cal F}\subset{\cal P}(\Omega)$ of measurable events, and a $\sigma$-additive measure $\mu:\>{\cal F}\to[0,1]$. If you accomplish this – it can be done in several ways, but not in the "intuitive" way – then the paradox will disappear.
E.g., you could fix a large number $N$ and put $$\mu(A):={1\over N}\sum_{1\leq x\leq N} 1_A(x)\ ,$$ or you could put
$$\mu(A):=\sum_{x\in A}2^{-x}\ .$$
A: There is an obvious bijection between the intervals $A = [0,1)$ and $B = [1,4)$, yet if we define a uniform random variable $X$ whose support is on the union of these intervals, the probability of $X \in A$ versus $X \in B$ are not equal, because when we say "uniform distribution," we have imposed a probability measure on the sample space $A \cup B$.
A: Here's my take on this. You can't really talk about probability in the natural numbers. You would have to construct a function $\mu:\mathcal{P}(\mathbb{N}) \rightarrow [0,1]$.
You immediately see that this is not possible taking cardinality into account. If, like in your case, you have an infinite set $A \in \mathcal{P}(\mathbb{N})$ then $$\mu(A) = \frac{|A|}{|\mathbb{N}|} = \frac{\infty}{\infty}$$.
I don't know if it is actually possible to come up with some wacky $\sigma$-additive function that would get you a probability, but for sure it wouldn't be related to the intuitive notion of cardinality.
