This challenge pertains to algebraic limits, and can be solved through calculus, clever arithmetic, graph theory, or other methods, though is only classified as the foremost.
n \in \mathbb{N} \\[4pt] \operatorname f : \mathbb{R}, \mathbb{Q} \to \mathbb{Q} \\[8pt] \operatorname f {(p, x)} = \begin{cases} p > x : x-\frac1{2n} \\[4pt] p < x : x+\frac1{2n} \\[4pt] p = x : x \end{cases}The function f id defined above. Consider evaluating f recursively, starting at 0.5, with x as the previous result and p a random number from 0 to 1. Eventually an integer would be reached, and the average time to reach one could be found given n. The challenge is to find this average for as many n as possible, then infinitely many, and perhaps all.