Probability that no team in a tournament wins all games or loses all games.

Solution 1:

Yes, you are right.

Moreover, the similar method can be used to solve a more general problem:

$N$ teams play a tournament in which every team plays every other team exactly once. No ties occur, and each team has a $\frac{1}{2}$ probability of winning any game it plays. Find the probability that no team wins/loses all the games.

Any team has the probability of $\frac{1}{2^{N - 1}}$ to win all games, or (symmetrically) to lose all games. Because there are $N$ teams and only one of them at a time can be the complete winner (or loser), then the probability that one team wins (or loses) all the games is $\frac{N}{2^{N - 1}} \cdot 2$. Now, the probability, that there are the complete winner and the complete loser at the same time is $\frac{N(N-1)}{2^{2N - 3}}$ (which we find by applying the same method to the pairs of teams). Thus by inclusion-exclusion principle, the final probability is $1 - \frac{N}{2^{N-2}} + \frac{N(N - 1)}{2^{2N - 3}}$, which, indeed, does collapse into $\frac{17}{32}$ when $N = 5$.