Suppose that 3 indistinguishable balls are placed at random into 3 distinguishable cells. What is the probability that exactly one cell remains empty?

The book's answer is $$\frac{3(3-1)}{3+3-1\choose{3}}=\frac{3}{5}$$.

In words, the number of ways of distributing the 3 balls so that there is exactly one cell that is empty divided by the number of ways to distributing the 3 balls.

My question is, that if I interpret "randomly" to mean the probability that any ball falls into a cell is $ \frac{1}{3} $. I think the answer comes out different. The probability that the first cell is empty is $\frac{6}{27}$. So the probability that one of them is empty is

$$3 *\frac{6}{27}=\frac{2}{3}$$.

Therefore is it right to say that there are two interpretations to this question and based on the wording we should infer the first interpretation?

  • 3
    $\begingroup$ Yes. The problem does not adequately specify the probability space. I think your interpretation is more natural, but others may disagree. $\endgroup$ – saulspatz Jul 13 '19 at 15:11


It is not only a question of interpretation.

Your first answer is based on the interpretation that all the distinguishable arrangement of the three undistiguishable balls in the three distinguishable boxes are equally likely.

Below there are two arrangements of distinguishable balls which look different:

$$\boxed 1\ \boxed 2 \ \boxed 3$$ $$\boxed 2\ \boxed 1 \ \boxed 3$$

These arrangements fuse if you cannot distinguish the balls:

$$\boxed *\ \boxed * \ \boxed *.$$

Your first answer is correct if the interpretation is acceptable.

With real physical marbles (you can touch them) it is hard to imagine that this interpretation is correct. How would you put the in concreto distinguishable balls into the boxes so that the interpretation works? So this interpretation is not really acceptable in the case of distinguishable marbles whose "marks" get deleted. Since they are touchable, fingerprints remain.

Regarding the second, more acceptable interpretation the distinguishable arrangements of distinguishable matbles get equally likely and the probability we seek is indeed:


Then, however you don't get the same probability for the arrangements that remain distinguishable after removing the fingerprints.

Now, the moral of the example is that the interpretation is a tough thing. Most importantly the first interpretation cannot be realized by throwing balls into the boxes.

(Don't get mislead by the Bose-Einstein distribution in the case of which the number of arrangements of the indistinguishable balls is given as a definition and not as a result of a calculation.)

  • $\begingroup$ Is my second answer wrong then? $\endgroup$ – johnson Jul 13 '19 at 16:04
  • $\begingroup$ I hit the submit button accidentally. I am not done yet. $\endgroup$ – zoli Jul 13 '19 at 16:06
  • $\begingroup$ I am done now , finally . $\endgroup$ – zoli Jul 13 '19 at 16:33
  • $\begingroup$ I need to know why my calculation is wrong. Let's calculate the probability the first cell is the only one empty. In each subset, let the first set represent the balls in the second cell and the second set the balls in the third cell. The outcomes are {{1},{2,3}} {{2},{1,3}} {{3},{1,2}} {{1,2},{3}} {{1,3},{2}}, {{2,3},{1}} each of these happens with probability 1/27 $\endgroup$ – johnson Jul 13 '19 at 16:42
  • $\begingroup$ I am terribly sorry. You are right regarding the second interpretation. $\endgroup$ – zoli Jul 13 '19 at 17:06

From the solution given and from the fact that the balls are indistinguishable, the question seems to be suggesting that you are to consider each configuration of balls in cells as equally likely, rather than your interpretation to consider each placement of an individual ball as equally likely. This changes the relative probabilities of the final configurations. I'll illustrate.

With indistinguishable balls there are $\binom{5}{2}=10$ configurations, each equally likely. So $[3,0,0]$ is as likely as $[2,1,0]$ is as likely as $[1,1,1]$, $\frac{1}{10}$ chance for any of these.

With each ball having a $\frac{1}{3}$ chance of being in a given cell, there is now a $(\frac{1}{3})^3=\frac{1}{27}$ chance of $[3,0,0]$, but a $3\cdot\frac{1}{27}=\frac{1}{9}$ chance of $[2,1,0]$, as well as a $\frac{1}{9}$ chance of $[1,1,1]$. This is because you have to account for the different ways these last two configurations can occur (i.e. with $[2,1,0]$ you can have the first and second ball in cell 1, first and third, or second and third), whereas there's only one way for $[3,0,0]$ to occur.

Describing the process as "random" is never enough to determine the probability space of a problem like this. Your interpretation isn't wrong per se, but with them mentioning the balls are indistinguishable, in probability courses the expected calculation given that language is to assume that the final configurations have uniform probability of occurring. They definitely could have made that clearer in the question.


$\frac{1}{3}$ is the probability that any one of the three balls falls into a cell. But in the second interpretation, since you've eliminated one cell, one would think that the probability space becomes different. Now you're distributing three balls into two cells. If you were to imagine a sample space for the the second interpretation , then you could create three disjoint spaces within the universal sample space: $A_{1}$, $A_{2}$, and $A_{3}$. For example, $A_{1}$ would be the sample space with the first cell not filled, $A_{2}$ the sample space with the second cell not filled, and $A_{3}$ would be the sample space with the third cell not filled. Let's consider the elements that make up $A_{1}$ for the moment. How were the elements in $A_{1}$ generated? We know that the balls are dropped randomly, but now we've said that the first cell cannot be filled. So even if we're performing a real-world experiment, there is a physical cell that is to remain empty. To make things more concrete we can just throw away the first cell. Thus interpreting this physical experiment mathematically means that we're essentially considering only cells $2$ and $3$ in $A_{1}$. We now see that the elements in $A_{1} \cup A_{2} \cup A_{3}$ are the elements in which two cells are filled by three balls randomly.


To illustrate @Solipsist's answer and make the distinction between the interpretations clearer for anyone confused:

Interpretation 1

Label the cells $0,1,2$, make the balls indistinguishable and mark them by *. When each arrangement is equally likely, there are only $10$ possible outcomes for the number of balls in each cell:

$$\begin{array}{|c|c|c|} \hline 0&1&2\\ \hline &&***\\ \hline &\color{red}{*}&\color{red}{**}\\ \hline &\color{red}{**}&\color{red}{*}\\ \hline &***&\\ \hline \color{red}{*}&&\color{red}{**}\\ \hline &\ldots\\ \hline \color{red}{**}&\color{red}{*}&\\ \hline ***&&\\ \hline \end{array}$$

Permissible outcomes are marked in red. The number of outcomes is given by a stars-and-bars problem, to find the number of 3-tuples of non-negative integers whose sum is 3, which is simply $\binom{3+3-1}{3-1}=10$. There are $4$ impermissible outcomes: $3$ where all balls are in $1$ cell and $1$ where each cell has exactly $1$ ball. So the probability is $\frac{6}{10}=\frac{3}{5}$.

Interpretation 2

As before, label the cells $0,1,2$ but now label the balls $a,b,c$. If each ball is equally likely to fall into each cell, there are $27$ outcomes. The possible outcomes of where the balls end up are the numbers $0$ to $26$ in ternary:

$$\begin{array}{|c|c|c|} \hline a&b&c\\ \hline 0&0&0\\ \hline \color{red}{0}&\color{red}{0}&\color{red}{1}\\ \hline \color{red}{0}&\color{red}{0}&\color{red}{2}\\ \hline \color{red}{0}&\color{red}{1}&\color{red}{0}\\ \hline \color{red}{0}&\color{red}{1}&\color{red}{1}\\ \hline 0&1&2\\ \hline &\ldots\\ \hline \color{red}{2}&\color{red}{2}&\color{red}{1}\\ \hline 2&2&2\\ \hline \end{array}$$

The permissible arrangements have one empty cell, so they are missing exactly one cell-number ($0$, $1$ or $2$). So the probability that cell $2$ is the only empty cell is the probability that $a,b,c$ are in either cell $0$ or $1$, i.e. $\left(\frac23\right)^3$, minus the probability that $a,b,c$ are all in cell $0$ or $1$, i.e. $\frac2{3^3}$. By symmetry this is the same when cell $0$ or $1$ is the only empty cell so the probability, as @johnson showed, is $3\left(\frac{2^3}{3^3}-\frac{2}{3^3}\right)=\frac{18}{27}=\frac23$.

You can easily demonstrate that the two probabilities are different by looking at the full tables of outcomes. The distinction between the interpretations is whether you consider a difference between the cases:

  • $a,b$ are in cell $0$ but $c$ is in cell $1$
  • $a,c$ are in cell $0$ but $b$ is in cell $1$

The book considers these to be identical, so they aren't given separate probabilities. Without specifying the probability space, or what outcomes were equally likely, it's not perfectly clear what the question was intending. But it is slightly implied to be the second.


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