A tough tournament!

The tournament involves $\binom{5}{2} = 10$ matches, each of which has two possible outcomes, thus there are $2^{10} = 1024$ possible ways the tournament can play out in general.

Now it will be easier to calculate the complement here, i.e., to determine the number of tournament scenarios in which either one of the teams wins all its matches or one of the teams loses all its matches, or both. (Note that at most one team can win all of its matches, as all the other teams would then have at least one loss. Similarly, at most one team can lose all of its matches.)

Let $A$ be the set of all tournament scenarios in which one team wins all of its matches and $B$ be the set of tournament scenarios in which one team loses all of its matches. We then wish to find

$|A \cup B| = |A| + |B| - |A \cap B|$ .

By symmetry $|A| = |B|$ . Now if one team wins all of its 4 matches, then the remaining 6 matches can each still have 2 outcomes. As this is the case if any of the 5 teams wins all of its matches, we have $|A| = 5 \times 2^{6} = 320$ .

If one team wins all of its matches and one team loses all of its matches, then the outcomes of 7 of the matches have been determined, (not 8 since one of the matches involves both these two teams), so the remaining 3 matches have 2 outcomes each. We can choose the all-wins team in 5 ways and then the all-losses team in 4 ways, so $|A \cap B| = 5 \times 4 \times 2^{3} = 160$ .

Thus $|A \cup B| = 320 + 320 - 160 = 480$ , and so there are $1024 - 480 = 544$ tournament scenarios in which no team either wins or loses all of its matches. The desired probability is then

$\dfrac{544}{1024} = \dfrac{17}{32}$ , and so $a + b = 17 + 32 = \boxed{49}$ .