Is there a general rule or formula to figure out how many wins you have to have in a swiss tournament to be able to draw into the top 8?

Take tonight's Premier Event sealed deck tournament on MTGO. 77 entrants call for 7 rounds of Swiss followed by a top 8 draft. How many rounds would I have to win to safely draw the rest of the way into the top 8?

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Top 8 of SCG Invitational, Somerset, NJ, Aug 28-30, 2015
Winner of SCG Worcester Team Sealed Open with Gerard Fabiano and Curtis Sheu, September 28, 2013

Is there a general rule or formula to figure out how many wins you have to have in a swiss tournament to be able to draw into the top 8?

Take tonight's Premier Event sealed deck tournament on MTGO. 77 entrants call for 7 rounds of Swiss followed by a top 8 draft. How many rounds would I have to win to safely draw the rest of the way into the top 8?

It is extremely difficult to make accurate blanket statements about this. There are a lot of variables involved with this: how many people are drawing, tiebreakers, etc.

That said, in a tournament of that size it seems unlikely anybody with more than one loss would proceed.

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It is extremely difficult to make accurate blanket statements about this. There are a lot of variables involved with this: how many people are drawing, tiebreakers, etc.

That said, in a tournament of that size it seems unlikely anybody with more than one loss would proceed.

Yeah, I know there's obviously no way to be precise with it. Was just wondering if there's some general estimation.

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Winner of SCG Worcester Team Sealed Open with Gerard Fabiano and Curtis Sheu, September 28, 2013

just win every round... but general rule of thumb, if u lose 2 rounds, especially early then u cant make it... x-1-1 usually makes it, and x-0-1 always does

You can't draw on Modo, but it's pretty complicated to predict what you need. X-1 will top 8. X-2 will not. If you could draw, it would be a lot more complex.

Since making this thread, I've noticed it's become a go-to resource on Google (top hit!) for people interested in finding out how to make the cut. Unfortunately, the image I linked to was not mine and appears to have been lost on the web, and some people have sent me PMs and e-mails asking for it; rather than re-create it, I've found the "Top Cut Calculator" at http://sixprizes.com/top-cut-calculator/. It appears to exactly implement the formula I described and would be of more use to anyone looking to work out if they'll make the cut at a given tournament.

Mods, I know there's a general prohibition against necro edits and the like; I'm hoping an exception can be made here since otherwise Google will continue to direct people to my original (broken) link and they'll continue to e-mail me unless this edit is permitted.

ORIGINAL MESSAGE:

Generally, you can take a look at the number of players, the numbers of rounds, and the cut-off and guesstimate the number of losses you can absorb before landing out of contention. Experience helps in this regard. Or you can use this plug-and-play formula to really nail down the numbers, which simplifies the model by assuming no draws, and then you can figure in draws just by simple reasoning.

Where n is the number of players in the tournament, r is the number of rounds and P subscript l is the number of players with l losses. The function symbol is the product symbol, so for 64 players in a 4-round tournament, the number of players with 0, 1, 2, 3 and four losses would be:

This means that in this particular tournament, to guarantee yourself top 8, you'll have to either have no losses or be in the top quarter of the players with one loss on breakers. Players attempting to draw into top 8 will make it harder for players with one loss to get in based on breakers, but there's no easy formula to figure out how hard.

More sensibly, a 64-man Swiss tournament would usually have six rounds. So plugging that into the formula:

P0 = (64/2^6) = 1
P1 = (64/2^6)((6)*(5/2)) = 15

At this point, we can see that we have found our top 8 players - the one undefeated player, and seven of the 15 players with one loss. Again, players attempting to draw in to the top 8 makes it more difficult for players with one loss to get into the top 8.

Last example. Let's say it's a 1,436 man tournament with 10 Swiss rounds, cut to the top 128.

P0 = (1436/2^10) = 1.4 (This partial figure means that random pairings can result in a number of undefeated players between 1 and 2)
P1 = (1436/2^10)((10)*(9/2)) = 63.1 (Somewhere between 63 and 64 players)
P2 = (1436/2^10)((10)*(9/2)*(8/3)) = 168.3

At this point, we've found that 0 or 1 loss in this tournament guarantees you top 128. Somewhere between 62 and 64 of the 168 to 169 two-loss players will get into the top 128 cutoff, and whenever a player attempts to go 8-1-1 to get into the top 128, that makes it more difficult for the 8-0-2 players to get in. Incidentally, we can also see that more than one loss puts you out of contention for the top 64, and more than zero losses mean you'd have to fight your way into the top 32.

So all you have to do is plug in the appropriate numbers and do a bit of quick math on a calculator until you've found enough players with each bracket of losses to fill your cutoff and you can make your decisions from there.

You can't draw on Modo, but it's pretty complicated to predict what you need. X-1 will top 8. X-2 will not. If you could draw, it would be a lot more complex.

Just want to clarify this for the op, X-2 is not guaranteed to top 8, but it is not that unusual for some X-2's to top 8 on Magic Online.

X-2's almost never t8 in large events in real life. The reason is mathematically complicated, but stems from the fact that is in real life, you can take draws into account to determine when to "lock" in t8. On magic online you can't.

Okay, thanks guys. I was unaware that draws weren't possible on MTGO. So, basically, for a 7 round, 77 person tournament...obviously 7-0 is guaranteed in, as well as 6-1? Or will there be more than 7 6-1's, meaning tiebreakers would be used?

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Top 8 of SCG Invitational, Las Vegas, NV, Dec 13-15, 2013
Top 8 of SCG Invitational, Somerset, NJ, Aug 28-30, 2015
Winner of SCG Worcester Team Sealed Open with Gerard Fabiano and Curtis Sheu, September 28, 2013

Okay, thanks guys. I was unaware that draws weren't possible on MTGO. So, basically, for a 7 round, 77 person tournament...obviously 7-0 is guaranteed in, as well as 6-1? Or will there be more than 7 6-1's, meaning tiebreakers would be used?

There would be 4 to 5 6-1s:

P0 = (77/2^7) = 0.6 (It is possible for an incomplete tableau of 128 to produce no undefeated players)
P1 = (77/2^7)((7)) = 4.2.

This means that all of the X-0 and X-1 players will make it into the top 8, guaranteed; it is the X-2s that need to go to tiebreakers.

Another way of figuring this out is to use the tool located here.

In any properly run tournament, the number of rounds will be such that X-1-0 (X wins, 1 loss, 0 draws) and better will always make it into the top 8. This goes for both cardboard and online play. The tournaments on MtGO will be properly run, so X-1-0 will always make it. It is never guaranteed that any players who are worse than X-1-0 will get in, but it can and does happen from time to time.

In any properly run tournament, the number of rounds will be such that X-1-0 (X wins, 1 loss, 0 draws) and better will always make it into the top 8. This goes for both cardboard and online play. The tournaments on MtGO will be properly run, so X-1-0 will always make it. It is never guaranteed that any players who are worse than X-1-0 will get in, but it can and does happen from time to time.

No; any tournament in excess of roughly 725 participants has the potential to produce more than 8 combined X-0 and X-1 players. It doesn't come up too often, since that is a large number, but at big events like GPs, it's entirely possible to go X-1 and not top 8.

I don't know if any events on MTGO ever get that large, but if they do, the math is the same.

The larger your event, the more combined X-0/X-1 players you'll produce, and the problem is somewhat exacerbated by the 10-round cap on Swiss Rounds; like the short-run (4-round) 64-man tournament example I provided, if you're running fewer than 11 rounds for a tournament in excess of (2^10 + 1) participants, you're going to have more X-0s and X-1s than will fit in the top 8.

More sensibly, a 64-man Swiss tournament would usually have six rounds. So plugging that into the formula:

P0 = (64/2^6) = 1
P1 = (64/2^6)((6)*(5/2)) = 15

At this point, we can see that we have found our top 8 players - the one undefeated player, and seven of the 15 players with one loss. Again, players attempting to draw in to the top 8 makes it more difficult for players with one loss to get into the top 8.

15 is the number of X-2's, the number of X-1's is 6
'P1 = (64/2^6)((6)) = 6' by your scheme.

No; any tournament in excess of roughly 725 participants has the potential to produce more than 8 combined X-0 and X-1 players. It doesn't come up too often, since that is a large number, but at big events like GPs, it's entirely possible to go X-1 and not top 8.

I don't know if any events on MTGO ever get that large, but if they do, the math is the same.

The larger your event, the more combined X-0/X-1 players you'll produce, and the problem is somewhat exacerbated by the 10-round cap on Swiss Rounds; like the short-run (4-round) 64-man tournament example I provided, if you're running fewer than 11 rounds for a tournament in excess of (2^10 + 1) participants, you're going to have more X-0s and X-1s than will fit in the top 8.

GPs tend to have 15-plus rounds overall, meaning X-2 *will* make top 8 guaranteed (because it cuts to X-2 after day 1)

No; any tournament in excess of roughly 725 participants has the potential to produce more than 8 combined X-0 and X-1 players. It doesn't come up too often, since that is a large number, but at big events like GPs, it's entirely possible to go X-1 and not top 8.

I don't know if any events on MTGO ever get that large, but if they do, the math is the same.

The larger your event, the more combined X-0/X-1 players you'll produce, and the problem is somewhat exacerbated by the 10-round cap on Swiss Rounds; like the short-run (4-round) 64-man tournament example I provided, if you're running fewer than 11 rounds for a tournament in excess of (2^10 + 1) participants, you're going to have more X-0s and X-1s than will fit in the top 8.

With 700+ players you should have run more rounds... the reason the table in the back of the MTR doesnt factor in those high numbers is that its so unlikely to happen that you will have a 700+ non-GP event, and if you do... you likely have the required skills to figure out the correct number of rounds...

No; any tournament in excess of roughly 725 participants has the potential to produce more than 8 combined X-0 and X-1 players. It doesn't come up too often, since that is a large number, but at big events like GPs, it's entirely possible to go X-1 and not top 8.

I don't know if any events on MTGO ever get that large, but if they do, the math is the same.

The larger your event, the more combined X-0/X-1 players you'll produce, and the problem is somewhat exacerbated by the 10-round cap on Swiss Rounds; like the short-run (4-round) 64-man tournament example I provided, if you're running fewer than 11 rounds for a tournament in excess of (2^10 + 1) participants, you're going to have more X-0s and X-1s than will fit in the top 8.

Please read my post properly before commenting. I said "properly run". I did not say "you have to always have X rounds when you have 2^X players". "Properly run" means that the number of rounds is such that X-1-0 always makes it. That can mean having more rounds than the table at the back of the Tournament Rules indicates.

In GPs and other large tournaments, the number of rounds will even be such that X-2-0 always makes it, as some other posters have already said.

You play until someone has two wins or a player's clock runs out. In your scenario no win would be recorded for either player and you would go to the next game.

So if we are both on 10 life and I Earthquake for 10, one of us wins and the other loses? How does the game decide who gets the win? Is it me since I cast the spell?

I have only ever played the free trial a couple of years ago, but I had never heard of this...

~ Tim

It's possible to have the game be declared a draw depending on the game state (life totals, etc.). However, intentional drawing is unavailable as a feature on MODO as of now.

So if we are both on 10 life and I Earthquake for 10, one of us wins and the other loses? How does the game decide who gets the win? Is it me since I cast the spell?

I have only ever played the free trial a couple of years ago, but I had never heard of this...

~ Tim

You can draw games (unintentionally), but never matches...

FWIW, Swiss style tournaments are designed to be essentially double elimination events, except they allow for eliminated players to continue to participate in the event through its entirety - except in the situation where you make a cut based on record or standing to lessen the number of players, like at a GP day 2. In an event with X players, if you have the correct number of rounds {log(base2) X = rounds [rounded up]}, it takes an exorbitant amount of draws in awkward places to allow X-2 to make top 8.

FWIW, Swiss style tournaments are designed to be essentially double elimination events, except they allow for eliminated players to continue to participate in the event through its entirety - except in the situation where you make a cut based on record or standing to lessen the number of players, like at a GP day 2. In an event with X players, if you have the correct number of rounds {log(base2) X = rounds [rounded up]}, it takes an exorbitant amount of draws in awkward places to allow X-2 to make top 8.

This isnt actually true, magic events are only double elimination IF people are drawing, since on magic online at least a x-2 almost always makes it in (times when one doesnt, when you have an exact number for that number of rounds). Draws are actually what knocks x-2s out of contention.

This isnt actually true, magic events are only double elimination IF people are drawing, since on magic online at least a x-2 almost always makes it in (times when one doesnt, when you have an exact number for that number of rounds). Draws are actually what knocks x-2s out of contention.

Also depends if you are just over the count for an exstra round, or almost at the next threshold (70 players vs 120 player... equal amount of rounds)

Also depends if you are just over the count for an exstra round, or almost at the next threshold (70 players vs 120 player... equal amount of rounds)

Yeah I think this is a point that has been glossed over...

For example, from 33-64 players, the tournament will be 6 Swiss rounds, cut to top 8.

Assuming online (so no draws), in a 64 person tournament, you will usually have:
One 6-0
Six 5-1's
Leaving one spot for a 4-2 player, which will almost always be someone who didn't lose until the later rounds.

However, in a 33 player tournament (still 6 rounds) you will usually have:
Zero or One 6-0
Three to Four 5-1's
Leaving four to five spots for 4-2 players.

So, the closer the tournament is to the low end of the "# of rounds" range, the more likely you will be able to get in with a record worse than X-1.

Also, as a general rule of thumb, if you've lost 2 rounds before the final 2 rounds have started (e.g. by the end of round 5 in a 7-round tournament), you're probably not going to make the cut because of bad tiebreakers.

x-1-1 and if you have good tie breakers, you'd make it in. This past weekend I played in a MWS and went 5-1-1 and just got in the top 8. There was 2 other players with the same record as me but due to tie breakers, i got in.

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Format: Standard - NPH
Location: 2011 MMS Qualifier - Minneapolis, MN (6/4)
Players: 99
Finished: 2nd Place

Take tonight's Premier Event sealed deck tournament on MTGO. 77 entrants call for 7 rounds of Swiss followed by a top 8 draft. How many rounds would I have to win to safely draw the rest of the way into the top 8?

Top 8 of SCG Invitational, Las Vegas, NV, Dec 13-15, 2013

Top 8 of SCG Invitational, Somerset, NJ, Aug 28-30, 2015

Winner of SCG Worcester Team Sealed Open with Gerard Fabiano and Curtis Sheu, September 28, 2013

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It is extremely difficult to make accurate blanket statements about this. There are a lot of variables involved with this: how many people are drawing, tiebreakers, etc.

That said, in a tournament of that size it seems unlikely anybody with more than one loss would proceed.

Yeah, I know there's obviously no way to be precise with it. Was just wondering if there's some general estimation.

Top 8 of SCG Invitational, Las Vegas, NV, Dec 13-15, 2013

Top 8 of SCG Invitational, Somerset, NJ, Aug 28-30, 2015

Winner of SCG Worcester Team Sealed Open with Gerard Fabiano and Curtis Sheu, September 28, 2013

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EDH:

UBGThe MimeoplasmUBG

UPDATE 09/17/2015:Since making this thread, I've noticed it's become a go-to resource on Google (top hit!) for people interested in finding out how to make the cut. Unfortunately, the image I linked to was not mine and appears to have been lost on the web, and some people have sent me PMs and e-mails asking for it; rather than re-create it, I've found the "Top Cut Calculator" at http://sixprizes.com/top-cut-calculator/. It appears to exactly implement the formula I described and would be of more use to anyone looking to work out if they'll make the cut at a given tournament.

Mods, I know there's a general prohibition against necro edits and the like; I'm hoping an exception can be made here since otherwise Google will continue to direct people to my original (broken) link and they'll continue to e-mail me unless this edit is permitted.

ORIGINAL MESSAGE:Generally, you can take a look at the number of players, the numbers of rounds, and the cut-off and guesstimate the number of losses you can absorb before landing out of contention. Experience helps in this regard. Or you can use this plug-and-play formula to really nail down the numbers, which simplifies the model by assuming no draws, and then you can figure in draws just by simple reasoning.

Where n is the number of players in the tournament, r is the number of rounds and P subscript l is the number of players with l losses. The function symbol is the product symbol, so for 64 players in a 4-round tournament, the number of players with 0, 1, 2, 3 and four losses would be:

P0 = (64/2^4) = 4

P1 = (64/2^4)(4) = 16

P2 = (64/2^4)((4)*(3/2)) = 24

P3 = (64/2^4)((4)*(3/2)*(2/3)) = 16

P4 = ((64/2^4)((4)*(3/2)*(2/3)*(1/4)) = 4

This means that in this particular tournament, to guarantee yourself top 8, you'll have to either have no losses or be in the top quarter of the players with one loss on breakers. Players attempting to draw into top 8 will make it harder for players with one loss to get in based on breakers, but there's no easy formula to figure out how hard.

More sensibly, a 64-man Swiss tournament would usually have six rounds. So plugging that into the formula:

P0 = (64/2^6) = 1

P1 = (64/2^6)((6)*(5/2)) = 15

At this point, we can see that we have found our top 8 players - the one undefeated player, and seven of the 15 players with one loss. Again, players attempting to draw in to the top 8 makes it more difficult for players with one loss to get into the top 8.

Last example. Let's say it's a 1,436 man tournament with 10 Swiss rounds, cut to the top 128.

P0 = (1436/2^10) = 1.4

(This partial figure means that random pairings can result in a number of undefeated players between 1 and 2)P1 = (1436/2^10)((10)*(9/2)) = 63.1

(Somewhere between 63 and 64 players)P2 = (1436/2^10)((10)*(9/2)*(8/3)) = 168.3

At this point, we've found that 0 or 1 loss in this tournament guarantees you top 128. Somewhere between 62 and 64 of the 168 to 169 two-loss players will get into the top 128 cutoff, and whenever a player attempts to go 8-1-1 to get into the top 128, that makes it more difficult for the 8-0-2 players to get in. Incidentally, we can also see that more than one loss puts you out of contention for the top 64, and more than zero losses mean you'd have to fight your way into the top 32.

So all you have to do is plug in the appropriate numbers and do a bit of quick math on a calculator until you've found enough players with each bracket of losses to fill your cutoff and you can make your decisions from there.

Hope this helps!

Just want to clarify this for the op, X-2 is not guaranteed to top 8, but it is not that unusual for some X-2's to top 8 on Magic Online.

X-2's almost never t8 in large events in real life. The reason is mathematically complicated, but stems from the fact that is in real life, you can take draws into account to determine when to "lock" in t8. On magic online you can't.

Top 8 of SCG Invitational, Las Vegas, NV, Dec 13-15, 2013

Top 8 of SCG Invitational, Somerset, NJ, Aug 28-30, 2015

Winner of SCG Worcester Team Sealed Open with Gerard Fabiano and Curtis Sheu, September 28, 2013

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There would be 4 to 5 6-1s:

P0 = (77/2^7) = 0.6

(It is possible for an incomplete tableau of 128 to produce no undefeated players)P1 = (77/2^7)((7)) = 4.2.

This means that all of the X-0 and X-1 players will make it into the top 8, guaranteed; it is the X-2s that need to go to tiebreakers.

Another way of figuring this out is to use the tool located here.

No; any tournament in excess of roughly 725 participants has the potential to produce more than 8 combined X-0 and X-1 players. It doesn't come up too often, since that is a large number, but at big events like GPs, it's entirely possible to go X-1 and not top 8.

I don't know if any events on MTGO ever get that large, but if they do, the math is the same.

The larger your event, the more combined X-0/X-1 players you'll produce, and the problem is somewhat exacerbated by the 10-round cap on Swiss Rounds; like the short-run (4-round) 64-man tournament example I provided, if you're running fewer than 11 rounds for a tournament in excess of (2^10 + 1) participants, you're going to have more X-0s and X-1s than will fit in the top 8.

15 is the number of X-2's, the number of X-1's is 6

'P1 = (64/2^6)((6)) = 6' by your scheme.

476.GPs tend to have 15-plus rounds overall, meaning X-2 *will* make top 8 guaranteed (because it cuts to X-2 after day 1)

With 700+ players you should have run more rounds... the reason the table in the back of the MTR doesnt factor in those high numbers is that its so unlikely to happen that you will have a 700+ non-GP event, and if you do... you likely have the required skills to figure out the correct number of rounds...

DCI level 3 JudgePlease read my post properly before commenting. I said "properly run". I did not say "you have to always have X rounds when you have 2^X players". "Properly run" means that the number of rounds is such that X-1-0 always makes it. That can mean having more rounds than the table at the back of the Tournament Rules indicates.

In GPs and other large tournaments, the number of rounds will even be such that X-2-0 always makes it, as some other posters have already said.

It's possible to have the game be declared a draw depending on the game state (life totals, etc.). However, intentional drawing is unavailable as a feature on MODO as of now.

Ruin's

TradingPostYou can draw games (unintentionally), but never matches...

DCI level 3 JudgeThis isnt actually true, magic events are only double elimination IF people are drawing, since on magic online at least a x-2 almost always makes it in (times when one doesnt, when you have an exact number for that number of rounds). Draws are actually what knocks x-2s out of contention.

Also depends if you are just over the count for an exstra round, or almost at the next threshold (70 players vs 120 player... equal amount of rounds)

DCI level 3 JudgeYeah I think this is a point that has been glossed over...

For example, from 33-64 players, the tournament will be 6 Swiss rounds, cut to top 8.

Assuming online (so no draws), in a 64 person tournament, you will usually have:

One 6-0

Six 5-1's

Leaving one spot for a 4-2 player, which will almost always be someone who didn't lose until the later rounds.

However, in a 33 player tournament (still 6 rounds) you will usually have:

Zero or One 6-0

Three to Four 5-1's

Leaving four to five spots for 4-2 players.

So, the closer the tournament is to the low end of the "# of rounds" range, the more likely you will be able to get in with a record worse than X-1.

Also, as a general rule of thumb, if you've lost 2 rounds before the final 2 rounds have started (e.g. by the end of round 5 in a 7-round tournament), you're probably not going to make the cut because of bad tiebreakers.

Format: Standard - NPH

Location: 2011 MMS Qualifier - Minneapolis, MN (6/4)

Players: 99

Finished: 2nd Place

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http://forums.mtgsalvation.com/showthread.php?t=347708