In a Swiss tournament, if I recall correctly you need
- Roundup(Ln2(X)) round to determine the winner with X players
- So for 50 players, you would need (Ln2(50), rounded up which is 6 rounds.
- Roundup(Ln2(Y)) rounds
- So if you want the top 4, you add an extra Ln2(4)=2 rounds.
- Roundup(Ln2(x)+Ln2(Y)) rounds or
- Roundup(Ln2(X)) + Roundup(Ln2(Y) )
For me, without mathematical proof, it would appear that splitting qualifiers in n tournaments makes qualification more variable and uncertain, and therefore less fair: even for stronger players, it is difficult to constantly hit 90+% (which is usually needed if Y/n is between 1-3), while they are often almost certain to be part of Y qualifiers in a tournament that has enough rounds.
If we take the recent British which had 3 qualifiers for a total of 12 places (3 of which were on cumulative score, which I'll ignore here for simplicity), which is best. Some qualifiers had 80-90 participants. For simplicity, let's assume 100 unique players
- 3 qualifiers of 7 rounds as per the tournament
- 3 qualifiers of Ln2(100)+Ln2(4)= 9 rounds
- a single qualifier of Ln2(100)+Ln(12)= 11 rounds