This spring, in the women's handball EHF Cup, three Hungarian teams made it to the top eight. The eight teams were paired by a draw. At the draw, all three Hungarian teams were matched against foreign opponents. What was the probability of this happening?

 

The spring saw three Hungarian teams advance to the top eight in the women's handball EHF Cup. The eight teams were paired through a draw. In this draw, all three Hungarian teams were paired with foreign opponents. What was the probability of this occurring?
I. solution. Let's calculate the number of all possible draws and the number of favorable cases.

Number the teams from 1 to 8, with the Hungarian teams receiving the numbers 1 to 3. First, let's choose the opponent for the 1st team; this can happen in 7 ways. Then, from the remaining 6 teams, select the one with the smallest number, and choose its opponent; this can happen in 5 ways. For the remaining 4 teams, the opponent of the one with the smallest number can be chosen in 3 ways, and the last two teams will be each other's opponents. Therefore, the number of all possible draws is $7 \cdot 5 \cdot 3=105$.

If the opponents of the Hungarian teams are all foreign, then the 1st team can have 5 different opponents. The 2nd team can get an opponent from the remaining 4 foreign teams - this can happen in 4 ways - and the 3rd Hungarian team can have 3 different opponents, and finally, the two remaining foreign teams will be each other's opponents. The number of favorable cases is therefore $5 \cdot 4 \cdot 3=60$.

In $\frac{60}{105}=\frac{4}{7}$ of the cases, there is no Hungarian-Hungarian match.

II. solution. The probability that the first Hungarian team gets a foreign opponent is $\frac{5}{7}$, since there are 7 teams besides it, and 5 of them are foreign. The conditional probability that the second Hungarian team also gets a foreign opponent is $\frac{4}{5}$, because out of the remaining 5 teams, 4 are foreign. The third Hungarian team will definitely get a foreign opponent, as the Hungarian teams have already been drawn.

Thus, the probability that there is no Hungarian-Hungarian pair is

$$
\frac{5}{7} \cdot \frac{4}{5}=\frac{4}{7}
$$

III. solution. Consider the following events:

A: all three Hungarian teams get foreign opponents;

B: the 1st Hungarian team's opponent is the 2nd Hungarian team;

C: the 1st Hungarian team's opponent is the 3rd Hungarian team;

D: the 2nd Hungarian team's opponent is the 3rd Hungarian team.

Exactly one of the four events will occur, so the sum of their probabilities is 1. A selected team can get 7 different opponents with equal probability, so the probabilities of events $B, C, D$ are all $\frac{1}{7}$. The sought probability is therefore

$$
P(A)=1-P(B)-P(C)-P(D)=1-3 \cdot \frac{1}{7}=\frac{4}{7}
$$