One of the most common questions in playing a swiss tournament in MTGO is that, with number of rounds n, how many losses can I have and still possibly make the top 8 draft? The answer varies depending on the number of people in the tournament, and the math is not easy to do in one's head.
Does anyone know of a calculator or table that can easily address this question, or could anyone make one? I've searched online to no avail. Basically a chart that says with x number of people, n rounds, you have so many 10-0s, so many 9-1s, etc.
One final caveat: the numbers MTGO uses for # of rounds is slightly different:
# of Players # of Rounds
8 3
9-16 4
17-32 5
33-64 6
65-128 7
129-226 8
227-409 9
410 or more 10
Say you have 9 people who went undefeated in the tournament. Only 4 to 8 (depending on number of participants) of those people would go on to enter the final rounds of the tournament. The way it is decided which of those 9 undefeated people go into the final rounds is determined is based on whom they played during the tournament and how well those other players did aside from when they played the top-9 players.
Simply put, not only do you have to play your best, but you also have to play against the best players in the tournament in order to be assured a spot in the top-4 or top-8 of that tournament.
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"As the size of an explosion increases, the number of social situations it is incapable of solving approaches zero." -- Varsuvius, Order of the Stick
I doubt that is as much of an issue on MTGO as it is in Paper Magic. I mean in MTGO is there even really a system in place to allow people to Intentionally Draw? Cause last I heard Intentionally Drawing was impossible on MTGO.
Byes on the other hand, yeah those can be a problem.
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"As the size of an explosion increases, the number of social situations it is incapable of solving approaches zero." -- Varsuvius, Order of the Stick
There is so much variance due to things like pair-downs and drops from potential top finishers that it's very difficult to be accurate for something like that.
I will say this, though. As long as you aren't too far beyond that 410 person limit (I don't know what the threshold player count is for this, though, probably in the 600s), if you finish X-1 after swiss rounds, you are guaranteed in the top 8.
This is the reason that the thresholds for 8/9 rounds aren't at the powers of 2 like the rounds before. Assuming the best possible records, if you do 256 players and 8 rounds, you have 1 8-0 and 8 7-1s; at 226, you won't get more than 7 7-1s.
Does anyone know of a calculator or table that can easily address this question, or could anyone make one? I've searched online to no avail. Basically a chart that says with x number of people, n rounds, you have so many 10-0s, so many 9-1s, etc.
One final caveat: the numbers MTGO uses for # of rounds is slightly different:
Simply put, not only do you have to play your best, but you also have to play against the best players in the tournament in order to be assured a spot in the top-4 or top-8 of that tournament.
People ID as well.
I doubt that is as much of an issue on MTGO as it is in Paper Magic. I mean in MTGO is there even really a system in place to allow people to Intentionally Draw? Cause last I heard Intentionally Drawing was impossible on MTGO.
Byes on the other hand, yeah those can be a problem.
I will say this, though. As long as you aren't too far beyond that 410 person limit (I don't know what the threshold player count is for this, though, probably in the 600s), if you finish X-1 after swiss rounds, you are guaranteed in the top 8.
This is the reason that the thresholds for 8/9 rounds aren't at the powers of 2 like the rounds before. Assuming the best possible records, if you do 256 players and 8 rounds, you have 1 8-0 and 8 7-1s; at 226, you won't get more than 7 7-1s.
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Swiss Tournament Formula
Here is the punchline: