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In many mathematical problems given as (natural language) text we are making assumptions. One example:

We roll two dices. What is the probability that a sum of 3 is thrown ?

By answering "$\frac{1}{18}$" we have assumed that two "fair" dices are used - although the text does not say if the dices are "fair" or "weighted". Using two "weighted" dices the result could be any number in the range $[0,1]$.

However we would definitely not assume that the probability of rolling the combination 1+2 (which can be rolled as 1+2 and as 2+1) is the same probability as the one of rolling the combination 1+1.

We would also not assume that all 11 possible sums from 2 to 12 have the same probability of $\frac{1}{11}$.

Against this background I have a problem understanding the assumptions made in questions with "indistinct objects" like this one:

There are 8 numbered cells and 12 indistinct balls. All 12 balls are randomly divided between all of the 8 cells. What is the probability ...

The answer is ... and it was confirmed by the official homework solution of the university.

If I understood correctly, the formula used in the official homework solution assumes that all "possible (distinguishable) results" in the process of dividing the balls between the cells have the same probability:

When two "indistinct" balls are divided between two (distinct) boxes the formula seems to result in the probability of $\frac{1}{3}$ for each of the three results (2+0, 1+1 and 0+2).

Comparing to the problem with the two dices however, I would expect that probabilities of $\frac{1}{4}$ are assumed for 2+0 and 0+2 and a probability of $\frac{1}{2}$ is assumed for 1+1.

Searching the internet for documents about the "indistinct balls in distinct boxes" problem I found out that the formula assumed by the homework solution is used by many universities.

In some of the answers and comments to the question linked there are comments like "I wonder if it is possible to simulate an experiment with this probability distribution at all".

My questions:

  • Is there a reason why this probability distribution is assumed by different universities?
  • Are there "real-life" examples (e.g. in science or in gambling) with "indistinct" objects that (approximately) have this probability distribution?
Martin Rosenau
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  • It means that each of the 12 balls has 1/8 chance to end up in cell $j$, for each $j$, and the locations of the 12 balls are mutually independent. So there are 8 to the 12th power equally likely outcomes if we imagine the balls being distinguishable. However, with the balls identical to each other, the distinguishable outcomes are no longer equally likely, for example "all 12 balls in cell 1" is much less likely than "3 balls each in cells 1,2,3,4". It's NOT the case that all possible distinguishable results are equally likely -- just like with the dice. – Ned Nov 02 '18 at 01:25
  • @Ned I edited my question: My problem in understanding is that the official solution of that university obviously DOES assume that the distinguishable results have equal probabilities. And my question is why. – Martin Rosenau Nov 02 '18 at 05:53
  • Ok I see. The solution they offer to the linked problem does assume that each distinguishable outcome is equally likely. That does not correspond to any natural "balls in boxes" situation. If they want to test enumeration by counting the number of distinguishable outcomes, they should have done just that with two questions for the numerator and denominator. Stars and Bars outcomes are not naturally equally likely in any reasonable context of "balls in boxes." You could "pick 7 bar locations between 20 stars" with each equally likely in a natural way, but balls in boxes I just don't see it. – Ned Nov 02 '18 at 10:42
  • The counting method is used by many universities; it is reasonable to suppose that almost all regular academic universities teach this method, since it is a standard method of combinatorics. The use of this counting method as a basis of an equiprobable distribution seems much less standard. We have an example (via the linked MSE question) in which one course at one university assumed the states of "indistinguisable balls in numbered boxes" are equiprobable; did you find any other examples? (I tried the search you suggested and have not yet found any.) – David K Nov 03 '18 at 20:56
  • @DavidK It's much worse than "less standard" it's completely ridiculous. It's a mistake that should be corrected and I doubt you'll find other instances of counting distinguishable arrangements of identical balls in boxes as being equally likely. It's just wrong. – Ned Nov 05 '18 at 13:09

2 Answers2

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When two "indistinct" balls are divided between two (distinct) boxes the formula seems to result in the probability of 1/3 for each of the three results (2+0, 1+1 and 0+2).

This is very strange. Could you give the precise problem text? Probabilities in this case should be 1/4, 1/2 and 1/4.

Here is the trick I use when reasoning about problems of this kind.

We have indistinguishable balls. One person makes experiments with these balls and cells. He puts balls randomly into cells, repeats the experiment many times and measures the frequency of different outcomes. He can't distinguish the balls.

Now suppose that there is another person sitting next to him. He has secretly attached very small lables on the balls, so he CAN distinguish the balls. And he also measures frequencies of different outcomes.

The second experimentator can see 4 different outcomes. First ball and second ball in cell 1, first ball in cell 1, second ball in cell 2, etc. Probabilities of these 4 outcomes are equal and it is 1/4.

This is not because of some convention!

Suppose 1000 experiments are taken simultaneously. First they put ball 1 into random cell. In about 500 experiments ball 1 is now in cell 1. Now we pay attention to these 500 experiments only. All them put ball 2 in a random cell. Again, in about half of experiments ball 2 will go into cell 1, and in the other half - into cell 2. "Half" - because placing ball 2 doesn't depend on what happened earlier. So we have that in about 250 of all the experiments ball 1 is in cell 1, and ball 2 is in cell 2. Situation with other possible outcomes is the same.

This may be not a rigorous prove that probabilities of all outcomes are equal. I wanted not to prove it, but to make it obvious.

Now back to the first experimentator - the one who can't distinguish the balls. It can't happen that he would see 1/3 probability of the 3 outcomes he can see. Because his results wouldn't agree with results of the second experimentator. Both of them have records of the same set of experiments. Second one would see that in about 1/4 of all experiments both balls are in cell 1, let's say that happened in experiments number 1, 8, 9, 19, 20,.... And the first experimentator would agree that in exactly these experiments both balls are in cell 1, so the probability of this outcome is 1/4.

Consider the following as offtopic:

Elementary particles ARE indistinguishable. It's not possible to put marks on them. Even theoretically. And if one puts two such indistinguishable particles into two boxes the probabilities of outcomes wouldn't be 1/4, 1/2, 1/4. This note was not about math, but about strange properties of the world we are living in.

lesnik
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  • The linked problem does count the distinguishable outcomes with identical balls as equally likely (analogous to getting 1/3, 1/3, 1/3 here). It's ridiculous IMO and the only thing I can think of is "in our probability class, when these words are used, it means we should count by stars-and-bars and then treat those outcomes as equally likely" even though that makes no sense whatsoever for balls in boxes. Otherwise I don't get it at all. – Ned Nov 02 '18 at 14:53
  • @Ned Ok, I see. I checked the linked problem and I totally agree with you. It's very strange solution. Probability of the outcome "there re no empty cells" doesn't depend on if cells are distinguishable or not. – lesnik Nov 02 '18 at 16:11
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Is there a reason why this probability distribution is assumed by different universities?

  • Those so-called universities are teaching alternative facts.

Are there "real-life" examples (e.g. in science or in gambling) with "indistinct" objects that (approximately) have this probability distribution?

  • No. It doesn't matter if the balls are distinguishable or not. The easiest way to debunk this is to think about an experiment with colored balls observed by two people, one being color blind. If those universities' ideas were correct, the two persons would experience different results, which is obviously nonsense.
ThomasR
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