Does the limit of proportions of objects in a sequence depend on the order of the objects in that sequence? Motivation : Suppose A is an infinite sequence of some objects, and I want to get some sense of the proportion of objects satisfying some property P. Intuitively, I can define
proportion = $\lim \limits_{n \to \infty} \frac {\sum_{k=1}^{n} f(A_k)}{n}$, where $f(A_k)=1$ if $A_k \in P$, and $0$ otherwise.
Question : Does this limit (or even its existence) depend on the order of the objects in the sequence A?
Yes, quite dramatically. Take $A = \mathbb{N}$ to be the positive integers WLOG and let $S$ be the subset having property $P$. The full group $\text{Aut}(\mathbb{N})$ of permutations of $\mathbb{N}$ acts transitively on the set of all infinite subsets of $\mathbb{N}$ whose complement is also infinite, so if $S$ is infinite with infinite complement then the limit you describe can have any value in $[0, 1]$, or be undefined, depending on the order.
For example, take $S$ to be the odd numbers. In the usual order the limit is $\frac{1}{2}$; this is called the natural density. But if we consider the order
$$1, 2, \\ 3, 5, 4, \\ 7, 9, 11, 6, \\ 13, 15, 17, 19, 8, \dots$$
in which we alternate between one even number and $k$ odd numbers for all positive integers $k$, then the limit is $1$. Variations of this idea can be used to produce any other limit or no limit.
To produce a limit $r \in [0, 1]$ take a sequence $\frac{p_k}{q_k}$ of rational approximations converging to $r$ (which can become eventually constant if $r$ is rational) and alternate between $q_k$ even numbers and $p_k$ odd numbers. To produce no limit, alternate between $2^k$ even numbers and $2^k$ odd numbers; this causes the density to fluctuate between $\frac{1}{2}$ and $\frac{1}{3}$. This is related to the fact that the &quot;Benford subset&quot; of positive integers whose first digit is $1$ has no natural density.
On the other hand, if $S$ is finite then the density is $0$ regardless of the order, and if $S$ is cofinite (has finite complement) then the density is $1$ regardless of the order.
Yes the limit does exist.  Define an event of picking up an object and noting whether it has property P.  Then your limit function is basically the definition of the probability of an event.  Although your function f will flip-flop between 0 and 1, the impact of each new 0 or each new 1 on the total sum will become vanishingly small as n becomes increasingly large.
So if my objects are coins and P is the property (lying heads-up), then as I examine more and more coins the closer your limit will get to 0.5. By the time I've looked at 1,000,000 coins, say, the sum might be at 502,301/1,000,000.  If the next coin is a head, it will become 502,302/1,000,001.  If it is a tail, 502,301/1,000,000, but the difference is negligible.