Model
Suppose, we have $k$ urns with $n \cdot k$ balls dropped independently at random in those $k$ urns initially. At each step, we draw balls from each urn, where each ball could be drawn independently at random with probability $p$. After we have picked those balls, they are returned to the urns, with each ball returned independently at random to one of the urns.
UPDATE
- All the above "random" is used in the following sense : each urn have a probability of $\frac{1}{k}$ to be picked for each ball at each step [incl. the initial stage]
- "Independently" is used to emphasize, that each ball's "fate" is decided independently of other balls.
Suggested by @Henry, we can abstract ourselves off when each specific ball changed its location, and concentrate on where it could be now.
My thoughts
Suppose, we can track each individual ball, i.e. they have some id, which is used by us only for tracking ball's location, and not utilized for any other purpose in any "black box" model [see mention below].
if I look at "where the ball is now?", then I don't think the present location is influenced by anything, except the last decision, where to drop the ball.
So, conceptually present state, have the same distribution as initial one: all balls are just randomly distributed between the urns.
I got another opinion, that stated the "game" in a bit different way.
- "Order" the balls in line initially and assign to each ball an urn at random (this is our initial drop)
- "Mark" balls, picked at stage $1$, and move them to the "end of the line". Assign "new" urns to them only.
This "re-ordered" line has the same distribution, as the initial one. And by induction, this proves, that each stage is identically distributed.
Question
If "Mark" balls stage is made a black box, the above claim is still valid. Just "re-order" "marked" balls to the end-of-line and voila! we have the same balls distribution at each stage.
What is wrong with this? I mean I've just made another try for the partitioning of the balls, and this try is still a random assignment of the urns. Why the distribution of the balls now depends on the black-box? Was the initial problem then solved by this method? Or the reasoning is also wrong for it [i.e. when the "black box" in initial problem is a random picking of balls before reassignment]?
[E.g., let another "black-box" be the "pick few balls from the urn number $1$ only" model. Then, eventually, the urn number $1$ will become empty, and the distribution will stuck at some specific point, which is obviously far from being the eventual outcome of the Model [at the top]]