The algorithm for the simplest variant of that problem (take a single random sample from a list of a priori unknown length) can be derived by simple inductive reasoning. If the list has size 1, the problem is trivially solved. Otherwise suppose we have a list (x:xs) and we recursively solve the problem for xs. If all the recursion returns is the random sample from xs, we clearly have insufficient information to take a random sample from (x:xs), so we need to strengthen the induction hypothesis by having the recursion also return the length of xs. Let (r', n') be the random sample and length for xs. Then the result for (x:xs) is (r, n) where n = n' + 1 and r = x with probability 1/n or else r'.

Another way to derive this algorithm is to notice that any solution must implicitly or explicitly compute the length of the list as a byproduct. If you start by writing down the recursive function for computing the length of a list, composed with the straightforward recursive function for taking a random sample from a list of known length, you can apply a standard fusion and deforesting transformation to combine them into a single pass, and you end up with the same algorithm as above.

Here's a fun problem to ponder. Find the most frequently occurring element of a list in O(n) time and O(1) space. You may assume that this element occupies more than half of the list's entries.