Ricsi, Dénes, and Attila often play ping pong against each other, with two of them standing on one side. Dénes and Attila win against Ricsi three times as often as they lose; Dénes wins against Ricsi and Attila as often as he loses; finally, Attila wins against Ricsi and Dénes twice as often as he loses. Recently, they spent an entire afternoon playing, during which they played six matches, with two players on each side. What is the probability that Ricsi won at least once, either alone or as part of a pair?
Solution. If Ricsi wins at least once out of the six matches, he can win once, twice, ..., or six times. We could calculate all these possibilities, but it is simpler to calculate the probability that he never wins, and then subtract the obtained result from 1 to get the answer to the question. (In other words, we calculate the probability of the complementary event.)

If Ricsi plays against Dénes and Attila, then he has a 3/4 probability of losing. If Attila plays against Dénes and Ricsi, then Ricsi has a 1/2 probability of losing, and finally, if Dénes plays against Attila and Ricsi, then Ricsi has a 2/3 probability of losing.

Since the events are independent, the probability that Ricsi loses all his matches is given by the product of the above probabilities:

$$
P(\text { Ricsi always loses })=\frac{3}{4} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3} \cdot \frac{2}{3}=\frac{1}{16}
$$

Therefore, the probability that Ricsi wins at least once is:

$$
P(\text { Ricsi wins at least once })=1-\frac{1}{16}=\frac{15}{16}
$$