An ideal gas in equilibrium consists of $N$ molecules in a container of volume $V$. What is the probability that a volume $V^{*}$ of the container is empty? What is this probability if $V^{*}=\frac{V}{N}$, and $N \gg 1$?
Since no part of the container is distinguished for the gas molecules, the probability that a given molecule is in a volume $v$ of a container of volume $V$ is:

$$
p=\frac{v}{V}
$$

For the volume $V^{*}$ of the container to be empty, all $N$ molecules must be in the volume $V-V^{*}$. The probability of this is:

$$
p_{1}=\left(\frac{V-V^{*}}{V}\right)^{N}=\left(1-\frac{V^{*}}{V}\right)^{N}
$$

For $V^{*}=\frac{V}{N}$, then

$$
p_{1}=\left(1-\frac{1}{N}\right)^{N}
$$

We know that $\lim _{n \rightarrow \infty}\left(1+\frac{k}{n}\right)^{n}=e^{k}$. Considering the case $k=-1$, as $N$ becomes very large, the probability arbitrarily approaches $e^{-1}$.

Thus, our result is that for a large number of molecules, the volume $V^{*}=\frac{V}{N}$ of the container is empty with a probability of $e^{-1} \approx 0.368$.

István Németh (Budapest, Fazekas M. Gymnasium, 4th year)