Given a bunch of boxes and balls, what is the chance that randomly dropping the balls will fill all the boxes?

By Math Vault | Combinatorics

Lots of Colorful Balls

Carl asked the following discrete-math-style probability question:

$N$ balls are randomly dropped into $k$ boxes $ (k \le N)$. What is the probability that all boxes are filled?

Want to know the answer? Here it is!

Warming-Up: The Case With 2 Boxes

Of course, when there is only 1 box, the probability of filling the box with $N$ balls is just $1$. No mystery here.

(Remember, we are assuming that each ball needs to go into exactly one of the boxes — nothing more, nothing less)

So what if $k=2$ (i.e., we have 2 boxes) instead? Well, the probability of filling all the boxes seems intractable, so let’s calculate the probability of not filling all the boxes instead! Let $E_i$ be the event that the ith box is empty, then we have that:

$P(\text{not all boxes are filled}) = P(E_1 \cup E_2)$

Fortunately, due to the simplicity of the case, we also have that:

$P(E_1 \cup E_2)$ = $P(E_1) + P(E_2)$

Now, how can $E_1$ happen? Exactly when all $N$ balls go into box 2 of course! So $P(E_1)=(\frac{1}{2})^N = \frac{1}{2^N}$.

Similarly, $P(E_2)=\frac{1}{2^N}$ as well, which means that:

$P$ (not all boxes are filled) = $P(E_1) + P(E_2) = \frac{2}{2^N} =\frac{1}{2^{N-1}}$

So to wrap up, we have that:

$P$ (all boxes are filled) = $ 1 – P$(not all boxes are filled) = $1 – \frac{1}{2^{N-1}}$

In practice, what that means is that when $N=2$ (i.e., when we have 2 balls):

$P$ (all boxes are filled) = $1 – \frac{1}{2^{2-1}} = \frac{1}{2}$

With 3 balls, the same probability becomes $\frac{3}{4}$. 4 balls? $\frac{7}{8}$. 5 balls? $\frac{15}{16}$.

So as is to be expected, the chance of filling all the boxes increases exponentially towards $1$ as the number of balls increases.

Generalizing: The Case With 3 Boxes

Let’s say that we have $3$ boxes and that $E_i$ denotes the event that the ith box is empty, then, proceeding as before, we have that:

$P(\text{not all boxes are filled}) = P(E_1 \cup E_2 \cup E_3 )$

Except that there is no cheating now — meaning that we have to appeal to the so-called inclusion-exclusion principle:

\begin{multline*} P(E_1 \cup E_2 \cup E_3) = P(E_1) + P(E_2) + P(E_3)-P(E_1 \cap E_2)-P(E_1 \cap E_3) \\ -P(E_2 \cap E_3) + P( E_1 \cap E_2 \cap E_3) \end{multline*}

Now, $E_1$ happens precisely when all balls go into box 2 or box 3, so $P(E_1)=(\frac{2}{3})^N = \frac{2^N}{3^N}$.

All other probabilities can be calculated fairly easily, for example, $E_1 \cap E_2$ happens when all balls go to box 3, so the probability of it would be $(\frac{1}{3})^N = \frac{1}{3^N}$.

Putting everything together, we now have that:

$P(E_1 \cup E_2 \cup E_3)$= $\frac{2^N}{3^N}+\frac{2^N}{3^N}+\frac{2^N}{3^N}-\frac{1}{3^N}-\frac{1}{3^N}-\frac{1}{3^N}$ = $\frac {2^N – 1}{3^{N-1}}$

And now to wrap up:

$P$ (all boxes are filled) = $1 – P$ (not all boxes are filled) = $1-\frac {2^N – 1}{3^{N-1}}$

All right. Let’s put that to test again! When $N=3$ (e.g., we have 3 balls):

$P$ (all boxes are filled) = $1-\frac {2^3 – 1}{3^{3-1}} = \frac{2}{9}$

5 balls? $\frac{50}{81}$. 7 balls? $\frac{602}{729}$. 10 balls? $\frac{18660}{19683}$.

See the same trend? While one might have a hard time filling the boxes with only 3 balls, increasing the number of balls (by perhaps just a bit more) will greatly facilitate this process.

For the Geeks: The General Case With k Boxes

The probability of getting balls into boxes via random throwing

A dorky party idea heh?

Surprise! For the general case, the formula will look a bit nastier with some binomial coefficient and some $(-1)^{i+1}$. All right. Enough said. Here we go!

$P$ (not all boxes are filled) = $P(E_1 \cup \ldots \cup E_k)$ =

(invoking the inclusion-exclusion principle…)

\begin{multline*}P(E_1)+\dots+P(E_k)-P(E_1 \cap E_2) -\ldots-P(E_{k-1} \cap E_k) \\ + P(E_1 \cap E_2 \cap E_3)+ \ldots+P(E_{k-2} \cap E_{k-1} \cap E_k) \\ + \ldots + (-1)^{k+1}P(E_1 \cap \ldots \cap E_k) \end{multline*}

(and after some lengthy number crunching…)

(and some intense parsing…)

$\displaystyle k(\frac{k-1}{k})^N-\binom{k}{2}(\frac{k-2}{k})^N+\binom{k}{3}(\frac{k-3}{k})^N-\ldots+(-1) ^k \binom{k}{k-1}(\frac{1}{k})^N$

(converting into sigma notation now…)

\[ \sum_{i=1}^{k-1} (-1)^{i+1}\binom{k}{i}(\frac{k-i}{k})^N \]

Lastly, putting everything together, we get that:

\begin{align*} P (\text{all boxes are filled})  & = 1-P(\text{not all boxes are filled}) \\ & = \displaystyle 1-\sum_{i=1}^{k-1} (-1)^{i+1}\binom{k}{i}(\frac{k-i}{k})^N \end{align*}

And that’s about it! Finito! 🙂


About the Author

Math Vault and its Redditbots has the singular goal of advocating for education in higher mathematics through digital publishing and the uncanny use of technologies. Head to the Vault for more math cookies. :)

(2) comments

Add Your Reply