Division of Objects into Different Sized Boxes

Question

Suppose you have a set of N distinguishable boxes with lengths $l_1$,$l_2$...$l_N$. Suppose you try to divide x distinguishable objects among them, such that the probability of any object landing in box $i$ is proportional to the length of the box, $l_i$. How would you expect the objects to be distributed among the boxes - i.e. how many objects would a given box $i$ on average contain? What would be the standard deviation of objects in each box?

Answer 1

Let $L=\sum_i l_i$ be the total length of the boxes. The chance of a given object landing in box $i$ is $\frac {l_i}L$. So if you have $x$ objects, you expect $\frac {xl_i}L$ to land in the box by linearity of expectation.

Answer 2

Let $p_i=\frac{l_i}{\sum_1^N l_i}$. Let random variable $X_i$ be the number of balls falling in Box $i$.

Then $(X_1,X_2,\dots, X_N)$ has multinomial distribution.

Each $X_i$ has binomial distribution with parameters $x$, $p_i$.

In particular, $X_i$ has mean $xp_i$, and standard deviation $\sqrt{xp_i(1-p_i)}$.

Note that the $X_i$ are (negatively) correlated. If $i\ne j$, then $\text{Cov}(X_i,X_j)=-xp_ip_j$.