This problem comes from Feller's Introduction to probability. and it goes like:

"Seven balls are distributed randomly in seven cells. Given that two cells are empty, show that the (conditional) probability of a triple occupancy of some cells equals 1/4"

So far i have found out that all of the distributions that form a triple are given by: ${7 \choose 3}(4!) = \frac{7!}{3!}$.

Yet, i am not sure of how to go on, because there are $7^7$ total distributions, but clearly if you divide these two quantities the result is not even close to $1/4$.

I have also tried to go on by considering that the only occupancy numbers in the distribution can be $0,1,3$ so the total number of distributions that coincide with the ocuppancy numbers $3,1,1,1,1,0,0$ in some order are given by: $\frac{7!}{3!}\frac{7!}{3!4!}$. But if you divide this by $7^7$ the result is still no close to $1/4$.

And also when you consider the fact of conditional probability the results only get worse. So what can i do?

  • 1
    $\begingroup$ Are exactly two cells empty, or at least two? Is a triple occupancy exactly three balls in one cell, or at least three? Is "a triple occupancy" exactly one triple occupancy, or at least one? $\endgroup$ – Karl Mar 1 at 4:16
  • $\begingroup$ There are exactly 2 cells empty and exactly one triple occupancy i think I will update with the exact statement from the problem $\endgroup$ – Sam Haze Mar 1 at 4:22

Let $T$ be the event that there is a triple occupancy, and let $E$ be the event that exactly two cells are empty. We are asked for $$\Pr(T|E)=\frac{\Pr(T\cap E)}{\Pr(E)}$$

The number of distributions with exactly two empty cells and a triple occupancy is $$7\binom73\binom624!=88,200\tag1$$ There are $7$ ways to choose the cell with three balls, $\binom73$ ways to choose the balls to go into it, $\binom62$ ways to choose the two empty cells, and $4!$ to distribute the remaining balls into the remaining cells.

To compute the number of distributions with exactly two empty cells, we use the principle of inclusion and exclusion to get $$\binom 72\left(5^7-\binom514^4+\binom523^7-\binom532^7+\binom541^7\right)=352,800\tag2$$

There are $\binom72$ choices for the two empty cells, and $5^7$ ways to distribute the balls into the other $5$ cells. Now, we must subtract the distributions that use only $4$ of the cells, that is $\binom514^7$. We have subtracted distributions that use only $3$ of the cells twice, so we must add them back in, and so on.

Dividing $(1)$ by $(2)$ does indeed give $\frac14$.


To me, it looks like throwing a $7$ sided die $7$ times, with the the desired conditional probability being

n(1 cell has triple occupancy with 2 cells unoccupied) $\div$ n(2 cells unoccupied)

Using the multinomial coefficient, we get

= $\dfrac{\dbinom{7}{3,1,1,1,1,0,0}\dbinom{7}{1,4,2}}{\dbinom{7}{3,1,1,1,1,0,0}\dbinom{7}{1,4,2} + \dbinom{7}{2,2,1,1,1,0,0}\dbinom{7}{2,3,2}} = 1/4$

  • $\begingroup$ Using Stirling numbers of the second kind simplifies computations a lot, if you have recourse to the tables. If not, I find using the multinomial coefficient the most routine (hence less error prone) way of computing the results of dice throws or analogous problems. $\endgroup$ – true blue anil Mar 1 at 7:21

Using Stirling Number of the second kind -

If we take any $5$ cells, number of ways to distribute $7$ balls such that none of the cells are empty is given by $StirlingS2[7,5] = 140$, without distinguishing between cells.

Now we consider cases where one of the cells has $3$ balls, which is $\displaystyle {7 \choose 3} = 35$ ways of choosing $3$ balls and rest $4$ balls go one each in remaining $4$ cells.

So the desired probability $\displaystyle = \frac{35}{140} = \frac{1}{4}$.


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