**Solution 1. **It is easy to check that the number of all possible matchings of the 8 teams is 7^{.}5^{.}3^{.}1=105.

The number of those cases when the opponents of the Hungarian teams are all foreign is 5^{.}4^{.}3=60.

The requested probability is .

**Solution 2. **Denote the three Hungarian teams by *A*, *B* and *C*. Team *A* has 7 possible opponents. Therefore, the probablility that the opponent of team *A* is team *B*, is exactly 1/7.

The same holds for the pairs *A*,*C* and *B*,*C*.

The three wrong cases are pairwise distinct and all remaining cases are good. Therefore the requested probablity is