How does this Markov process involving balls and bins behave? I have some set $S_1,\ldots,S_k$ ($k \geq 3$) of bins, each initially with $N_0(S_i)$ balls ($N_t(S_i)$ denotes the number of balls in $S_i$ at time $t$). A bin can contain a negative number of balls. Now I apply the following rule: at time $t$, I choose 3 bins uniformly at random, say $S_{i_1}, S_{i_2}, S_{i_3}$, and
$$N_{t+1}(S_{i_1}) = N_t(S_{i_1}) - 1,$$
$$N_{t+1}(S_{i_2}) = N_t(S_{i_2}) + N_{t}(S_{i_3}) + 1,$$
$$N_{t+1}(S_{i_3}) = 0.$$
Can anything reasonably be said about the distribution of balls as $t\to \infty$?
Edit: I should also say, the 3 bins are chosen among the $\binom{k}{3}$ 3-sets of bins; they are always distinct.
 A: Let's consider an easy case: $k=3$ where the total number of balls (which is a conserved quantity) is $1$.   At each positive time there will be one bin with $0$ balls, so the others will have $x$ and $1-x$ for some integer $x$.  Let $Y_t = \max(x, 1 - x)$.  With probability $1/3$, 
the bin with $0$ will be chosen as $S_{i_1}$ and $Y_{t+1} = 2$. 
With probability $1/3$, the bin with $1 - Y_t$ will be chosen as $S_{i-1}$, so $Y_{t+1} = Y_t + 1$.   With probability $1/3$, the bin with $Y_t$ will be chosen as $S_{i_1}$, so $Y_{t+1} = Y_t - 1$ unless $Y_t = 1$, in which case $Y_{t+1} = 1$.  The transition matrix for the (infinite) Markov chain $Y_t$ looks like this:
$$ \pmatrix{ 1/3 & 2/3 & 0   & 0    &    0 & \ldots \cr
             1/3 & 1/3 & 1/3 & 0    &    0 & \ldots \cr
             0   & 2/3 & 0   & 1/3  &   0 & \ldots \cr
             0   & 1/3 & 1/3 & 0    & 1/3 & \ldots \cr
             0   & 1/3 & 0   & 1/3  & 0   & \ldots \cr
             0   & 1/3 & 0   &  0   & 1/3 & \ldots\cr
            \ldots & \ldots & \ldots & \ldots & \ldots & \ldots}$$
Equilibrium probabilities $\mu_j$ for this Markov chain should satisfy
$\mu_1 = \mu_2/2$, $(5/6) \mu_2 = 1/3 + \mu_3/3$, and $\mu_j = (\mu_{j-1} + \mu_{j+1})/3$ for $j \ge 3$.  The general solution to that recurrence is $\mu_j = A r_1^j + B r_2^j$ where $r_1 = (3 - \sqrt{5})/2$ and $r_2 = (3 + \sqrt{5})/2$, but in order for $\mu_j$ to be summable we must have $B=0$, i.e. $\mu_j = r_1^{j-2} \mu_2$ for $j \ge 2$.  That gives us
$$ 1 = \sum_{j=1}^\infty \mu_j = \frac{1}{2} \mu_2 + \sum_{j \ge 2} r_1^{j-2} \mu_2 = \left(\frac{1}{2} + \frac{2}{\sqrt{5}-1} \right) \mu_2 $$
i.e. $\mu_2 = 2 \sqrt{5} - 4$.  We do get a stationary probability distribution with
$\mu_1 = \sqrt{5} -2$, $\mu_j = (2 \sqrt{5}-4) ((3 - \sqrt{5})/2)^{j-2}$ for $j \ge 2$.
