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Take for example a case where we have to find the probability that one of the boxes contains exactly $3$ balls in a case where $12$ identical balls are to be placed in $3$ identical boxes. My teacher told me that you can take balls and boxes distinct whenever you need to find probability.

Can anyone explain why with an example solved from two cases; $1$) taking distinct balls and boxes and, $2)$ taking identical balls and boxes?

Tyrone
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Abhishek Bansal
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1 Answers1

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What the teacher means is this:

The question asks for the probability that "$1$ of the boxes contains exactly $3$ balls". This question clearly does not distinguish the boxes or the balls.

OK, so what if you do distinguish the boxes and the balls? Well, then you could compute the probability that, say, balls $2,5$, and $7$ end up in box $2$. OK, but obviously this is only one way to get $3$ balls in one of the boxes, so you have to compute it for all combinations of balls and boxes.

Now, that seems more work, but in fact could make things easier: by symmetry, for whatever combination of balls and boxes you have, the probability is the same as for any other combination. Hence, you only have to multiple it by the number of combinations. And once, you do that, you get the probability for getting any $3$ indistinguishable balls in one of the indistingsuihed boxes. Thus, in effect, thinking about this problem in terms of distinguished balls and boxes breaks down the problem for indistinguishable balls and boxes, and can thus make it more easy to solve.

Added Note:

It is clear that the probability of getting balls $2,5$, and $7$ ending up in box $2$ is smaller than the probability of exactly any $3$ balls ending up in one of the boxes. So it is not that the teacher meant that these probabilities are the same, if that is maybe what you thought the teacher meant. No, again, the teacher meant that computing the indistinguishable case is sometimes easier when done via the distinguishable case.

Bram28
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  • https://math.stackexchange.com/questions/1219692/12-identical-balls-can-be-placed-into-3-identical-boxes – Abhishek Bansal Apr 01 '18 at 13:38
  • I still cant judge whose answer is correct in the above. In the comments i got to see that in daniel V answer someone commented that the cases calculated by him are not equally likely. Isnt getting 12 0 0 in 3 boxes equally likely to getting 3 4 5 in 3 boxes(boxes and balls identical) ??? I couldnt find a satsifactory answer in comment section of the above – Abhishek Bansal Apr 01 '18 at 13:39
  • 12 0 0 means 12 balls in 1 box and rest 2 empty – Abhishek Bansal Apr 01 '18 at 13:56
  • @AbhishekBansal DanielV's answer there was not correct: it is more likely to get something like 3 4 5 than to get 12 0 0. Take a simpler example: 3 balls and 3 boxes: if we distinguish the balls and boxes, then there are $3^3$ different distributions. Of those, $3$ have all three balls in one box, $6$ have one ball in each box, and $18$ have two balls in one box and one ball in another. So, 2 1 0 is more likely than 1 1 1 which is more likely than 3 0 0 – Bram28 Apr 01 '18 at 14:06
  • You again took the case of distinct balls and boxes. I got that but can you prove 2 1 0 is not equally likely to 3 0 0 when balls and boxes identical – Abhishek Bansal Apr 01 '18 at 15:18
  • @AbhishekBansal actually, I did prove the indistinct case ... via the distinct case. thus, I showed that with balls and boxes indistinct, the chance of 3 0 0 is $\frac{3}{27}=\frac{1}{9}$ ... which makes sense: the first ball can go anywhere, but the second and third then must go in the same box, and that has chance of $\frac{1}{3}\cdot \frac{1}{3}$ – Bram28 Apr 01 '18 at 15:34
  • I got u.. But i got new confusion too that why is the probability of indistinct case always coming equal to distinct case. Is there any logic behind it? – Abhishek Bansal Apr 01 '18 at 19:41
  • @AbhishekBansal As I explained in the Answer, we are calcualing the same probability, namely that of indistinct cases ... it's just that one way to do that is by considering the distinct cases, and then multiplying those by the number of distinct cases. – Bram28 Apr 01 '18 at 19:43