An urn contains $N$ cards numbered from 1 to $N$. Two are drawn at random. What is the expected value of the ratio of the smaller to the larger number drawn?
I. solution. We can draw as many number pairs as there are second-class combinations (the order is indeed secondary) that can be formed from $N$ elements; thus, the number of all possible draws is $C_{N}^{2}=\binom{N}{2}=\frac{N(N-1)}{2}$, and thus the probability of drawing a specific number pair is $\frac{1}{\binom{N}{2}}=\frac{2}{N(N-1)}$.

The sought expected value:

$$
M=v_{1} x_{1}+v_{2} x_{2}+\ldots+x_{i}+\ldots+v_{\binom{N}{2}} x_{\binom{N}{2}}
$$

where $v_{1}=v_{2}=\ldots=v_{\binom{N}{2}}=\frac{2}{N(N-1)}$ and $x_{i}\left(i=1,2, \ldots,\binom{N}{2}\right)$ are all the fractions $\frac{p}{q}<1$, with $p<N, q \leq N$.

Thus, if we group the fractions $x_{i}$ by the size of their denominators:

$$
\begin{gathered}
M=\frac{2}{N(N-1)}\left[\frac{1}{2}+\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)+\ldots+\right. \\
\left.+\left(\frac{1}{N}+\frac{2}{N}+\ldots+\frac{N-1}{N}\right)\right]
\end{gathered}
$$

Since (using the formula for the sum of an arithmetic sequence)

$$
\frac{1}{k}+\frac{2}{k}+\ldots+\frac{(k-1)}{k}=\frac{1}{k} \cdot \frac{(k-1) k}{2}=\frac{k-1}{2}
$$

therefore

$$
\begin{gathered}
M=\frac{2}{N(N-1)} \cdot \frac{1}{2}[1+2+3+\ldots+(N-1)]= \\
=\frac{2}{N(N-1)} \cdot \frac{1}{2} \cdot \frac{(N-1) N}{2}=\frac{1}{2}
\end{gathered}
$$

Géza Biczó (Bp. II., Rákóczi g. IV. o. t.)

II. solution: The probability of the formation of any fraction that meets the conditions is equal. The expected value is therefore the arithmetic mean of all possible fractions, that is,

$$
M=\frac{t_{1}+t_{2}+\ldots+t_{k}}{k}
$$

It is easy to see that the

$$
\left(1-t_{1}\right),\left(1-t_{2}\right), \ldots,\left(1-t_{k}\right)
$$

sequence also contains all the fractions that meet our conditions, so

$$
M=\frac{\left(1-t_{1}\right)+\left(1-t_{2}\right)+\ldots+\left(1-t_{k}\right)}{k}
$$

The sum of (1) and (2)

$$
2 M=\frac{1+1+\ldots+1}{k}=\frac{k}{k}=1
$$

and thus

$$
M=\frac{1}{2}
$$

Pál Bártfai (Bp. I., Petőfi g. IV. o. t.)