How many tickets in the lottery will have an ambo probability of $\frac{5}{473}$, when drawing five numbers?
It is known in probability theory that the ratio of the number of favorable cases to the total number of possible cases is called probability. If the number of lottery tickets is $x$, then the number of favorable cases is $\binom{5}{2}=\frac{5 \cdot 4}{1 \cdot 2}=10$; the total number of cases is $\binom{x}{2}=\frac{x \cdot (x-1)}{1 \cdot 2}$; thus,

$$
\frac{10 \cdot 2}{x \cdot (x-1)}=\frac{5}{473}
$$

from which

$$
x^2 - x = 1892
$$

which has a positive root $x=44$.

(Gyula Póka, Losoncz.)

The problem was also solved by: Baumann J., Bayer B., Beck I., Czank K., Faith F., Filkorn J., Hein I., Holzmann M., Keesz J., Kerekes T., König D., Krausz B., Gy. Krisztián, Kürth A., Lakatos K., Gy. Lupsa, Á. Rosenberg, M. Russo, G. Sasvári, J. Scharff, R. Scheuer, E. Tézner, T. Ulmer, A. Weisz.