n = number of matches before the ace match (4 in a Bo5, 6 in a Bo7)
p = probability of ace match between 2 evenly matched teams
p = 0.5^n * nc(n/2)
where c is the combinatorial. My thoughts are that there are nc(n/2) ways of getting to an ace match, and that the probability of each way happening is the probabilities of each player winning multiplied by each other. (for evenly matched teams, this is 0.5^n no matter who wins each game)
Results: (again, for evenly matched teams)
31.25% chance of an ace match in a Bo7
37.5% chance of an ace match in a Bo5
Given that most teams are not evenly matched, the percentage should not exceed this in reality.
I dunno, something feels wrong here, help?