If an urn is divided into two parts and one part contains $v_{1}$ red and $f_{1}$ white balls, and the other part contains $v_{2}$ red and $f_{2}$ white balls, what is the probability of drawing a red ball blindly? What would the probability be if there were no partition in the urn, and under what conditions would the two probabilities be equal?
$1^{\circ}$ Anna's probability of drawing a red ball from one part of the urn is $\frac{v_{1}}{2\left(v_{1}+f_{1}\right)}$; similarly, the probability of drawing a red ball from the other part is $\frac{v_{2}}{2\left(v_{2}+f_{2}\right)}$. Thus, the total probability in the first case is:

$$
V_{1}=\frac{v_{1}}{2\left(v_{1}+f_{1}\right)}+\frac{v_{2}}{2\left(v_{2}+f_{2}\right)}=\frac{v_{1}\left(v_{2}+f_{2}\right)+v_{2}\left(v_{1}+f_{1}\right)}{2\left(v_{1}+f_{1}\right)\left(v_{2}+f_{2}\right)}
$$

$2^{\circ}$ If there is no partition in the urn, the number of favorable outcomes is $v_{1}+v_{2}$, and the total number of outcomes is $v_{1}+v_{2}+f_{1}+f_{2}$, so the probability is:

$$
V_{2}=\frac{v_{1}+v_{2}}{\left(v_{1}+f_{1}\right)+\left(v_{2}+f_{2}\right)}
$$

$3^{\circ}$ If we subtract $v_{2}$ from $v_{1}$, then:

$$
V_{1}-V_{2}=\frac{\left[\left(v_{1}+f_{1}\right)-\left(v_{2}+f_{2}\right)\right]\left[v_{1} f_{2}-v_{2} f_{1}\right]}{2\left(v_{1}+f_{1}\right)\left(v_{2}+f_{2}\right)\left[\left(v_{1}+f_{1}\right)+\left(v_{2}+f_{2}\right)\right]}
$$

The two probabilities are equal if the numerator of the fraction is 0; this happens if first:

$$
v_{1}+f_{1}=v_{2}+f_{2}
$$

which means the number of balls in the two compartments is equal; or if secondly

$$
v_{1} f_{1}=v_{2} f_{1}
$$

or

$$
v_{1}: f_{1}=v_{2}: f_{2}
$$

which means the ratios of the number of red to white balls in the two compartments are equal.

(Jenö Scharff, Budapest.)

The problem was also solved by: Bayer B., Filkorn J., Hein I., Holzmann M., Keesz J., Kerekes T., Lukhaub Gy., Lupsa Gy., Póka Gy., Rosenberg Á.