Expected number of pinballs to light up all 3 channels You're playing pinball. When you first shoot a ball it randomly comes down through 1 of 3 gates. When you go through an unlit gate, it lights up. Similarly, a lit gate will go out. What is the expected number of balls you have to throw for all 3 gates to light up?
For example, ball A could go through gate 2, B through gate 3, and C through gate 1. This scenario took 3 rolls and has probability 1/27.
I've put serious thought into this question twice over the last couple of years but my answer gets more and more complicated until my brain explodes.
Follow up
Douglas hit the nail on the head. For kicks, here's the Python script I used as a reality check for both the 2 and 3 gate cases.
from random import randint

def pinball(gates):
    trials = []
    for trial in range(10000):
        state = [False for g in range(gates)]
        balls = 0
        while not all(state):
            gate = randint(0, len(state) - 1)
            state[gate] = not state[gate]
            balls += 1
        trials.append(1.0 * balls)
    print sum(trials) / len(trials)

pinball(2)
pinball(3)

 A: This is the average time it takes for a random walk on the 1-skeleton of a cube to reach the opposite vertex. There are more general theories for such values, but you can determine this particular one with a simple set of linear equations. Let $T_i$ be the expected time from when $i$ lights are lit. You want to determine $T_0$.
$T_0 = 1 + T_1$
$T_1 = 1 + T_0/3 + 2T_2/3$
$T_2 = 1+ 2T_1/3 + 0$ 
which has the solution
$\{T_0=10, T_1=9, T_2=7\}$.
A: After 3 balls you are either at all lit up, or 1 lit up. Hand calculations give these probabilities at 2/9ths and 7/9ths respectively.
If you throw another ball you can't be all lit up, so throw 2 more in.
There are 9 ways these 2 balls can land.
3 of these options will take all lit up to all lit up (namely they both fall in the same hole)
and 2 of these options will take 1 lit up to all lit up (they fall in one whole then the other - both ways round). And as there are either all lit up or 1 lit up at this stage this is all that can happen. 
So if the probability of all lit up at the one possible time is p, then at the next possible time is
3p/9 + 2(1-p)/9 = p+2/9
this value converges on a quarter as (1/4 + 2)/9 = (9/4)/9 = 1/4
