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If $12$ identical balls are to be placed in $3$ identical boxes, then the probability that one of the boxes contains exactly $3$ balls is?

This question has already been posed on this site, but in the answer it is assumed that the total number of placements is $3^{12}$. But why is this so? Since the boxes and balls are identical what should matter is the relative number of balls in the boxes, right?

Any help to understand this point will be much appreciated.

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  • If the boxes are identical, it would seem placing one ball can be done in only one way. Doesn't there have to be some kind of distinction among the boxes in order to get a count of how many balls end up in each box? – coffeemath Apr 08 '15 at 13:23
  • @coffeemath: No: once the balls have been placed in the boxes, boxes containing different numbers of balls can be distinguished. – Brian M. Scott Apr 08 '15 at 17:02

1 Answers1

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If you’re talking about this question, note that it has two answers. The accepted answer is based on the assumption that there are $3^{12}$ equally likely outcomes and includes an explanation for this assumption. Specifically, it’s derived from a specific model for the experiment: each of the $12$ balls is deposited in a box, and the box is chosen uniformly at random. That is, each ball has probability $\frac13$ of going into each box. This is not in conflict with the hypothesis that the balls and the boxes are indistinguishable: there are still $12$ different balls and $3$ different boxes, even if we have no way to tell which ball or box we’re looking at when we see only one.

If you like, imagine that we throw the balls one at a time into a funnel with three outlets, one leading to each box. The funnel is so designed that a ball is equally likely to exit from each of the outlets, and we cannot see what happens to the ball after it goes into the funnel. After we’ve thrown all $12$ balls, an assistant retrieves the boxes and shows them to us. We can now distinguish only the $19$ outcomes listed in the other answer to the linked question, but that does not make those $19$ outcomes equally likely; in fact they are not equally likely, as there are only $3$ ways to produce the outcome with all $12$ balls in one box, while there are many ways to produce the more balanced outcomes.

The other is based on the assumption that the $19$ distinguishable outcomes are equally likely. It’s hard to come up with a natural version of the experiment that would justify this assumption, but that answer does correctly work out its consequences.

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