TITLE: Expected number of steps before three counters reach N modulo 2N at the same time
QUESTION [13 upvotes]: We have three counters, $i, j, k$, all initialized to zero. Each step consists of adding or subtracting one from one of the counters, so $(\Delta i, \Delta j, \Delta k)$ is selected among $(\pm1, 0, 0), (0, \pm1, 0), (0, 0, \pm1)$, each with probability 1/6. What is the expected number of steps before all the counters are N modulo 2N at the same time?

REPLY [5 votes]: Here's a sketch of another derivation of the hitting time for one counter (re Mike's answer).
Consider the usual $\pm 1$ random walk on the integer line. It is well known that the expected time it takes to reach distance $N$ is $N^2$. In our case reaching distance $N$ and reaching the value $N$ modulo $2N$ is the same, since we can't get to $\pm 3N$ without moving through $\pm N$.
We move the specified counter in expectation once every three steps, and therefore it's $3N^2$ and not $N^2$ (because the direction of movement and this "waiting for our turn" are independent).
This argument shows that the case of one counter is rather special.