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We have $16$ red identical balls and $4$ black identical balls. We have $4$ bins and put all balls randomly into the bins such that each bin contains $5$ balls at the end. Each arrangement has the same probability.

How many ways are there to distribute the balls into the bins? What is the probability that all black balls are in different bins?

My idea was to realize that the red balls have no influence, so the number of all arrangements is simply ${4+4-1\choose 4}$. As all black balls are identical there is only one way that all black balls are in different bins, hence the probability is $\frac{1}{{7\choose 4}}$. But this Sound too easy... what do I miss? Or is it correct?

Philipp
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  • The red balls do matter. It would be better to think of putting $20$ balls into the boxes in a way that does not look at their colours until the very end. – Henry Aug 26 '23 at 14:26
  • Imagine we have 1000 balls (996 red + 4 black). Knowing that each bin contains 250 balls does not suggest that each bin will contain 1 black ball. Imagine now that we have 8 balls (4 red + 4 black). Knowing that each bin contains 2 balls seems to be a big 'incentive' to suggest that the 4 black balls are in different bins. So the number of red balls has an impact. – Lourrran Aug 26 '23 at 18:46

4 Answers4

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When we work over statistics, it does not matter whether the balls or bins are identical or not, because all of them are seen as distinct. So, the denominator part is $4^{20}$.

Now, lets think about the numerator. We have $16$ distinct red balls and $4$ distinct black balls to disperse $4$ distinct bins where all black balls are in different bins .Then we can disperse those $4$ blacks into $5$ bins by $P(4,4)$ ways. The rest reds can be dispersed bu $4^{16}$ ways. So $$\frac{24 \times 4^{16}}{4^{20}}$$

With restriction where each bins have exactly $5$ balls:

Now, if each bins contains exactly $5$ balls, then we have actually $20$ distinct balls, so we can separate them into $4$ bins where each have exactly $5$ balls: $$\binom{20}{5}\binom{15}{5}\binom{10}{5}\binom{5}{5}$$

Now, lets disperse those blacks into those $4$ bins by $P(4,4)$.After that, diperse those $16$ reds into $4$ bins where each bins have $4$ reds. So, the numerator is $$P(4,4)\times \binom{16}{4}\binom{12}{4}\binom{8}{4}\binom{4}{4}$$

The answer is $$\frac{P(4,4)\times \binom{16}{4}\binom{12}{4}\binom{8}{4}\binom{4}{4}}{\binom{20}{5}\binom{15}{5}\binom{10}{5}\binom{5}{5}}=0.128999..$$

Not a Salmon Fish
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    Your answer of $\dfrac{125}{969}$ seems plausible by simulation – Henry Aug 26 '23 at 14:14
  • You make the balls distinct which violates the condition and in your approach the arrangement where each bin contains one black ball is more likely than other arrangements which also violates the conditions in the question. – Philipp Aug 26 '23 at 15:09
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    @Philipp no, it is not. I mentioned about my reason in the begining of my post. The identical objects are possible for quantum objects. So, it does not matter whether the balls are identical or distinct. You should think them as distinct. For example, think that we have $2$ identical red balls and $3$ identical black balls. What is the probability that we select red ball ? The answer is $2/5$ instead of $1/2$ – Not a Salmon Fish Aug 26 '23 at 15:14
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    Your example is different from the original problem. We do not choose randomly between the bins we just put them (in a an unknown process) into the bins. The problem says "each arrangement" has the same probability. So $4$ black balls in bin $1$ has the same probability like each black ball in a different bin. Your approach doesn't respect this condition. I know it sounds too easy but if my approach is wrong then where is my mistake? I did not violate any of the conditions. – Philipp Aug 26 '23 at 15:22
  • Have written an snswer In support of your result (+1) – true blue anil Aug 26 '23 at 19:52
  • @Philipp Your "4 black balls in bin 1 has the same probability like each black ball in a different bin" would suggest that when throwing six indistinguishable dice, you are as likely to get one of each value as to get them all showing $1$. It is in fact $720$ times more likely. – Henry Aug 27 '23 at 19:45
  • @Henry, the problem says each arrangement has the same probability. So your dice example doesn't apply here. If you want to apply the dice example then you have to assume that each outcome of the six undistinguishable dice has the same probability. So indeed: one of each value is as likely as to get six $1$'s. Maybe this has no real world application but the description of the problem is very clear this regard. – Philipp Aug 28 '23 at 09:15
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A very useful ploy is to fit a given problem into a well recognised framework.

Here we can consider it as distributing $16$ red and $4$ black cards into $4$ hands, and the answers easily flow out

How many ways are there to distribute the balls into the bins?

$$\binom{20}5\binom{15}5\binom{10}5\binom55 $$

What is the probability that all black balls are in different bins?

$$\frac{\binom{16}4\binom41\;\binom{12}4\binom31\;\binom84\binom21\;\binom44\binom11}{\binom{20}5\binom{15}5\binom{10}5\binom55 }=\frac{125}{969}$$

true blue anil
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  • What I don't understand is that if your answer is right mine must be wrong. So I must have violated some condition. But I don't see which one. Also, if you map the original problem into your framework the arrangements might attain different probabilities which is not legit. Maybe there is no real world situation that applies to such an experiment but that is how I understand the problem. – Philipp Aug 26 '23 at 20:59
  • @Philipp: It is not a very well written question, but you can't ignore the division between red-black. It parallels the division between aces and non-aces in the question I am linking. You may find it instructive to see the variety of ways that question has been solved, but the dichotomy between aces/non-aces has been maintained (and has to) in all answers. https://math.stackexchange.com/questions/1345413/is-there-an-alternative-intuition-for-solving-the-probability-of-having-one-ace/1345436#1345436 – true blue anil Aug 26 '23 at 21:25
  • What exactly do I violate if I "ignore" the red balls? Why is this problem comparable to the aces/non aces problem? The fundamental difference is that in the aces/non aces situation we have a different probability measure. The probability that one player has all aces is different from the probability that each player has one ace. In the above stated problem this doesn't apply. – Philipp Aug 26 '23 at 22:25
  • @Philipp: The linked problem also asks for the probability that each player has an ace, not that one player has all the aces. – true blue anil Aug 27 '23 at 03:53
  • @Philipp: An even simpler analogy is forming teams where some are boys and some are girls. I used the cards analogy only because of red and black. – true blue anil Aug 27 '23 at 04:59
  • I believe the source of confusion is the word "identical" in the problem statement. The example with cards is good because cards are evidently distinguishable. I guess now that the problem's author had no idea how confusing the word could be. As for me, it made me think about Quantum Mechanics and ignore Classical intuition. – kludg Aug 27 '23 at 08:50
  • @kludg, but what if we find other "real" world situations that fit. E.g. we can imagine a bathtub filled with red and black balls. Then we dip four bins into the bathtub as often as each bin contains $5$ balls. In this setting we can't make the balls distinguishable and it is reasonable to assume that each arrangement has the same probability. – Philipp Aug 27 '23 at 10:53
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    @Philipp I prefer the solution that that corresponds to our real world, but the point is: the problem statement is ambiguous and allows different interpretations. It does not make sense to ask, where you did a mistake, you did not, you just interpreted the problem differently, as I did also. – kludg Aug 27 '23 at 12:32
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I consider as distinct the following cases:

  • all 4 black balls in the same bin : 4 cases

  • 3 black balls in the same bin : $4\times 3 = 12$ cases

  • 2 black balls in one bin, 2 black balls in the other bin : $\binom{4}{2} = 6$ cases

  • 2 black balls in the same bin, and other 2 black balls in different bins: $4\times 3 = 12$ cases (choose a bin for 2 black balls, then choose a bin without black balls)

  • all black balls in 4 different bins: 1 case

Total number of cases : $4+ 12 +6+12+1=35$

Probability: $P=\frac{1}{35}$

PS: I am not happy with my solution. It assumes that all distinct cases are equally likely, which is hardly true in a real world. I just followed the problem statement. Maybe the problem statement is intentionally confusing.

kludg
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  • If the bins were also indistinguishable (the question does not say whether they are or are not), would you say there only $5$ possible outcomes and they were all equally likely so $P=\frac15=0.2$? This would also not be right in the real world. – Henry Aug 27 '23 at 19:49
  • @Henry Considering bins indistinguishable is logically possible, but I thought that it kills the problem, and did not consider it more. – kludg Aug 28 '23 at 01:26
  • @Henry Physics considers distributions of identical particles on energy levels (bins) which differ only by the number of particles in each bin, like in my solution. The case of identical bins is beyond Physics. – kludg Aug 28 '23 at 04:36
  • Physically each ball is not in the same place and so are distinct, whether or not you distinguish them when you look at them – Henry Aug 28 '23 at 09:16
  • @Henry Real balls are of course always distinguishable, the question is: if the problem's author meant real distinguishable balls, why he called them identical? Identical are electrons, for example. – kludg Aug 28 '23 at 09:50
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As far as I am concerned, I cannot agree with the previous answer.

For convenience, let's name the four bins A, B, C, D. The first approach supposes that each bin is identical (4 balls in A are considered the same as 4 balls in D, this is ridiculous), we can quickly know that $C_{4+4-1}^{4}$ is not a great solution for the following two reasons:

  1. $3$ black balls in A and $1$ in B is not considered the same distribution as $1$ in A and $3$ in B.
  2. $4$ black balls in A is not considered the same as $4$ black balls in D.

Thus, we can just simply enumerate the distributions, namely $$ 4, 3+1, 2+2, 2+1+1, 1+1+1+1,$$ So there are just 5 possibilities. Which is ridiculous.

The second explanation is that the bins are different. Well, in this case, you are absolutely right. I can enumerate all possibilities to prove it:

  1. $4$ balls all in the same bin, $4$ distributions.
  2. $3+1$, $A_{4}^{2}=12$ distributions.
  3. $2+2$, $C_{4}^{2}=6$ distributions.
  4. $2+1+1,$ $C_{4}^{3}*C_{3}^{1}=12$ distributions.
  5. $1+1+1+1$ 1 possibility.

All in all, $4+12+6+12+1=35=C_{7}^{4}$.

Leo Ji
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