Filling cups and buckets continuously 
There are $n$ cups labeled $1,\dots,n$, each with a water tap that adds water into it at the same rate. There are also $k$ buckets, and $k$ sets $S_1,\dots,S_k\subseteq\{1,\dots,n\}$. At any point, if there is one unit of water in the cups in $S_i$ combined, you pour all that water into bucket $i$. (If the condition is satisfied for two or more $i$ simultaneously, choose the lowest $i$ first.)
As time goes on, what does the ratio between water in the $k$ buckets converge to (and does this limit even always exist)?

As a concrete example, suppose $n = 7$, $k = 3$, $S_1 = \{1,2,3,4\}$, $S_2=\{4,5,6\}$, $S_3=\{6,7\}$, then the ratio between the water in the three buckets converges to $9:4:4$. This can be observed via a computer program and then proved formally, with calculations again based on a computer program. Is there a way to determine an answer based on $S_1,\dots,S_k$ without relying on a computer? Can we set up some kind of "steady-state" equations?
 A: Is it clear that the limit in the general setting exists and the ratios don't oscillate wildly? Anyway, the limit (if it exists) depends on more than just the $S_k$.
If $n=1$ and $k>1$, then it clearly depends on how we break ties. But even if we insist that the $S_k$ are pairwise different, the limit depends on how we break ties.
Breaking ties deterministically can for instance be achieved by filling each cup with a tiny amount of water before we start. In this case, the solution depends on the initial configuration. Consider for instance the case $n=4$, $k=4$ with $S_1 = \{1,2\}$, $S_2 = \{3,4\}$, $S_3 = \{1,3\}$, $S_4 = \{2,4\}$. If the initial content of the cups are $(\epsilon, \epsilon ,0 ,0)$ then only the first two buckets are filled, but if the initial contents are $(\epsilon, 0, \epsilon ,0)$ then only the last two buckets are filled.
We can also break ties randomly. In this case the limit could be a non-constant random variable. Consider the case $n=8$, $k=9$ with  $S_1 = \{1,2\}$, $S_2 = \{3,4\}$, $S_3 = \{1,3\}$, $S_4 = \{2,4\}$,  $S_5 = \{5,6\}$, $S_6 = \{7,8\}$, $S_7 = \{5,7\}$, $S_8 = \{6,8\}$, $S_9 = \{1,5\}$, starting with all cups empty.
At first, cups are always emptied when all of them contain half a unit of water, this continues until at some point bucket 9 is picked - at this point we can't empty all cups and by symmetry w.l.o.g. cups 2 and 6 still contain half a unit. Once another quarter unit has been poured in all cups,$S_1$, $S_4$, $S_5$, $S_8$ hold one unit of water.
If at this point we (randomly) choose buckets 1 and 8, then similarly as in the above example with $\epsilon = \frac 14$ henceforth only buckets 1, 2, 7, 8 are filled. If we select buckets 4 and 5, then henceforth only buckets 3, 4, 5, 6 are filled. Note that in both of these cases we can never pick bucket 9 again since whenever cup 1 is emptied, cup 2 only contains a quarter unit and vice versa.
