There are two situations in which three people each win a game:

The first type: the probability that A wins B wins C wins A is 2/3 *1/2 *1/3 =1/9.

Second, the probability that B wins A C and B A C is (1-2/3) * (1-kloc-0//2) * (1-kloc-0//3) =1/9.

So the total probability of the first question is 1/9+ 1/9 =2/9.

The second question is, think again. C acts as a bystander for at most 2 times and at least 1 time. It is not difficult to calculate the exhaustion probability of various situations.