I've been asked by several people, during events ranging from local free-entry tournaments all the way up to the Pro Tour (really!), how to read standings and figure out whether to draw or not, or whether they even have a chance to make it if they win. This article aims to be a tutorial for understanding how to read standings and how to use them to your advantage. I'm going to use some simplified examples in order to demonstrate various scenarios that come up in tournaments.

The Basics

A win is worth 3 points. A loss is worth no points at all. A draw is worth 1 point. This is commonly reported as Win-Loss-Draw, so a player who won 3 rounds, lost two, and drew one would be reported as 3-2-1. With zero draws they're commonly left off, so a player who went 5-2 is a player who won 5 rounds and lost 2. (Whereas 5-0-2 would refer to 5 rounds won, zero lost, and 2 drawn.)

The round count in an event with a cut to the top 8 is set so that a player who loses only one round and has no draws is guaranteed to make top 8, regardless of what his tiebreakers are. Because of this, people tend to simply say they're "X-0" or "X-1". This doesn't always apply to smaller events (which sometimes cut to top 4, or have no cut at all) or to events with byes - Magic Online in particular has had problems with miscalculating the number of rounds to play and X-1s have missed it.

At events like PTQs, this isn't a problem, and X-1 can be safely assumed to make the top 8. Usually, a player who is X-1-1 makes it into the top 8, but sometimes not every person who is X-1-1 can make it. What usually happens is that people figure this out when looking at the standings, and opt not to draw, so that 8th will be an X-1-1, or occasionally an X-2.

At events with a low player count, or a player count which barely exceeds the amount which made the event add a round, a "double draw" is fairly common. For instance, an event with 34 players almost always has exactly 4 people at 4-0 after 3 rounds, and those 4 players will draw in both the 4th and 5th round, taking them to 11 points, guaranteeing their entry to the top 8.

Round Count

I'm going to quote the Magic Tournament Rules, Appendix F here for simplicity, then explain it in more detail:

The following number of Swiss rounds is often required for Premier tournaments. It may be used at the Tournament Organizer’s discretion for non-Premier tournaments. It is included here for reference only.

Players	Rounds
8	3
9-16	4
17-32	5
33-64	6
65-128	7
129-226	8
227-409	9
410+	10

Team tournaments consider each team as a single player for this purpose. Individual or team tournaments that cut to top 4 should be run with one extra round. Individual or team tournaments that cut to top 2 should be run with two extra rounds.

So, why are those the numbers? Well, for the numbers up to 128, they're exactly on the powers of 2. This should make a certain intuitive sense when you consider that exactly half the players will be undefeated after each round.

Remember that players are paired from the top down, with another player of the same record, so long as others exist. If there's one left over, that person gets paired down to someone as close as possible - if there are 5 people at 12 points, then the 5th will be paired with someone at 11 if one exists, or one at 10 if there isn't an 11, or a 9-point player if nobody has 10 or 11. (At the very bottom of the standings, of course, if one person is left over that person receives a bye.)

In short, these round counts make it so that, at most, 1 player can go undefeated. That's important, since this applies for all Swiss events, not just ones with a top 8 cut. It'd be absurd to give a number for an 8-player event otherwise! In addition, these numbers also guarantee that no more than 8 players can have a single loss.

A Simple Example

At a local 7-round PTQ, the standings after Round Six look like this:
1. Adam Anderson 18
2. Bob Bennett 18
3. Chris Campbell 15
4. David Davis 15
5. Edgar Edwards 15
6. Frank Fisher 15
7. Greg Gray 15
8. Hubert Hall 15
9. Isaac Isakson 12
(players continue from there)

The short version of this? 2 people at 18, 6 people at 15, and an arbitrary amount at 12 or below.

As we can see, if everyone plays it out, the people currently at 18 are a lock. This is the way it should be, since 18 points can only be had via being 6-0, and X-1 is supposed to make top 8 no matter what. A 6-0 player losing the final round becomes 6-1 and still makes the top 8.

As for the players currently at 15, the 3 winners will make it to 18 and be guaranteed to make top 8, but what about the losers? They will remain at 15, and since there are already 5 people above 15, only 3 of the people at 15 points will advance - the people advancing from 12 to 15 can pass them via tiebreakers.

What if there were a way to guarantee that they could avoid losing? Let's bring the intentional draw into the equation! As we've discussed, people currently at 12 points cannot ever get past 15 points in a single round, and since the highest person at 12 points is in 9th, we know that making it to 16 or more points means you cannot finish at 9th or below.

Therefore, the people at 15 points should all choose to take the intentional draw to make top 8, meaning in this event everyone at X-1-1 makes top 8 and nobody at X-2 does. That's great news for Hubert, but terrible news for Isaac, and the reason why "0-2 drop" is a common phrase. Even in the cases where X-2 players can make top 8, the way tiebreakers are calculated (a subject for another article) means that people who lose early in the event have far worse tiebreakers than people who lose late in the event.

The Double Draw

In smaller events, it's often common that a person at X-0-2 can make it. Consider a 16-player event (thus, 4 rounds) with a cut to top 8, with no unintentional draws in the first two rounds.

After round one:
8 players at 3 points
8 players at 0 points

After round two:
4 players at 6 points
8 players at 3 points
4 players at 0 points

Now, if people know what they're doing, you'll see the 4 people at 6 points choose to draw. Why? Well, let's consider what happens if they don't:

After round three:
2 players at 9 points
6 players at 6 points
6 players at 3 points
2 players at 0 points

If you learned the lesson in the simple example well, you'll notice that in this case the people at 9 and 6 can all draw to make 7 points and be assured of making top 8, rendering the last round largely irrelevant.

However, there is an edge case where if there are some unintentional draws it's possible that 7 points puts people in a position where their tiebreakers matter. Consider what happens if two people who are at 3 points draw:

After round three:
2 players at 9 points
5 players at 6 points
2 players at 4 points
5 players at 3 points
2 players at 0 points

Not all of the people at 6 points can draw in! One of them will be paired against a 4-point player, and the 4-point player will be unable to agree to a draw. This is not necessarily the person who has the worst tiebreakers! As a result, it is better for the people who go into round 3 at six points to take an intentional draw, resulting in the following standings:
After round two:
4 players at 6 points
8 players at 3 points
4 players at 0 points

After round three:
4 players at 7 points
4 players at 6 points
6 players at 3 points
2 players at 0 points

And even if there is an unintentional draw between 3-point players, all the people who go into round 3 at 7 points are going to be able to draw to 8.
After round three:
4 players at 7 points
3 players at 6 points
2 players at 4 points
5 players at 3 points
2 players at 0 points

Here the pairings are 7-7, 7-7, 6-6, 6-4, 4-3, 3-3, 3-3, 0-0. The first two of these will continue on the "double draw" plan, the 6-6 players can draw to 7 and safely make it, the 6-4 people have to play it out since the 4 drawing to 5 can't make it. 7 points (2-1-1) is guaranteed to make it, but depending on what happens in the 4-3 matchup, a single person at 6 points can make the top 8 - but it comes down to tiebreakers, and it's not necessarily going to be the person who beat the 4-point player.

In practice you don't have to evaluate a whole tournament because you only care about a particular person - yourself. The simplest possible rule of thumb for the double draw in small events is as follows: if there are exactly 4 people at X-0, and nobody at X-0-1, you can safely double draw.

Complex Double Draw

Round 5 of a 7-round PTQ has just finished, and we have the following standings due to some undefeated players going to time in this round and the last:
1. Adam Anderson 15
2. Bob Bennett 15
3. Chris Campbell 15
4. David Davis 13
5. Edgar Edwards 13
6. Frank Fisher 13
7. Greg Gray 12
8. Hubert Hall 12
9. Isaac Isakson 12
...more 12-point players follow, for a total of 18...
Nobody is at 11 points, but there is one person at 10 points.

When standings are posted, Adam and Bob are paired against each other. Chris is "paired down" against David, and Edgar and Frank are paired against each other. (The 10-point player is paired against a 9-point player.) Can Adam and Bob plan on double-drawing?

First, let's observe that a double draw puts them on 17 points. Is 17 a lock for the top 8? The important thing is how many people there are at 12 points. With 22 other people at or above 12 points, 11 can win the first round and at most 6 can win the next. That works out perfectly for Adam and Bob- except for one small problem. They've got to have other people to draw with.

There is a "pair-down" involved, which means there are multiple scenarios that could play out.

Let's see what happens if David wins (We'll say Edgar wins his match against Frank, but it doesn't matter which one it is):

Adam 16
Bob 16
David 16
Edgar 16
10 players at 15 points, including Chris
Frank 13

The players at 16 clearly can't draw the next round, since there will be 5 18-point players after this round. One of them will lose to the other 3 on tiebreakers.

Let's see what happens if Chris wins (We'll say Edgar wins his match against Frank, but it doesn't matter which one it is):

Chris 18
Adam 16
Bob 16
Edgar 16
9 players at 15 points
David 13
Frank 13

One of them will be paired up against Chris (who can clearly agree to draw) and the other will be paired against Edgar. Is 17 points safe here? You'll have one player at 19, three at 17, and either 4 or 5 people at 18. If it's 4, then they're all in just fine, but if the person who gets paired down against a 13-point player wins, then it comes down to tiebreakers, and one of the people at 17 points will lose out.

So, can Adam and Bob count on a double draw? The answer is clearly no. There are too many people at 12 points in the initial situation to make it a safe play.

Complex Double Draw Okay

After 5 rounds of a 7-round PTQ, the standings are as follows:

1. Adam Anderson 15
2. Bob Bennett 15
3. Chris Campbell 15
4. David Davis 13
5. Edgar Edwards 13
6. Frank Fisher 13
7. Greg Gray 12
8. Hubert Hall 12
9. Isaac Isakson 12
...more 12-point players follow, for a total of 10...
Nobody is at 11 points, but there are seven people at 10 points.

This is the same situation as before, but a bunch of the 3-1 players picked up an unintentional draw, so there are way fewer people at 4-1 than normal.

Let's rework Adam and Bob's double draw math with these new numbers :

If David wins (We'll say Edgar wins his match against Frank, but again it doesn't matter which one it is):
Adam 16
Bob 16
David 16
Edgar 16
6 players at 15 points, including Chris
Frank 13

Notice how this works out to be identical to the conventional 4-player double draw? That's what Adam and Bob are dependent on - the right amount of X-0-1 players "catching up" to them so they can be assured that they'll have a draw partner in the next round. This is exactly what they want, and is the ideal situation for the double draw in a complex scenario.

But what happens if David loses and Chris wins? Chris won't be X-0-1, but is that a problem?
Chris 18
Adam 16
Bob 16
Edgar 16
5 players at 15 points
David 13
Frank 13

Because David was paired up, Chris ends up ahead of them and can agree to the final-round draw, it works out just as well for Adam and Bob. However, Chris is in with a loss, and may try to get a friend into top 8 by knocking out the person he gets paired against.

In this situation I would overall consider Adam and Bob safe to draw with the goal of double drawing, because the only way one of them gets knocked out is if Chris wins and decides to go for the dreamcrush. The chances of that are pretty low, but not utterly nonexistent.

Breaking Down the Double Draw

As we can see, the number of X-1s makes a big difference when doing double-draw math. Since a double draw requires 4 people to be involved, there can be no more than 4 other people who make the top 8. As a result, this means a maximum of 16 other people can be in contention leading into the second-to-last round. When there were 18 people at X-1 and 6 people above it, the double draw was not viable. Thus, the simple rule of thumb needs a bit more detail:

Exactly 4 people at X-0
Maximum of 16 people at X-1

If there are 3 people at X-0, then having exactly 3 people at X-0-1 will enable the paired X-0 players to double draw, but as a result of the additional players in contention, this only works when there are 10 or fewer people at X-1. (There is a small risk of the paired-down X-0 winning, then deciding to dreamcrush rather than draw, but even if he does that, there's only a 1 in 3 chance that you get paired against that person.)

If there are 2 people at X-0, then having exactly 4 people at X-0-1 enables the X-0 players to double draw. This works with as many as 16 people at X-1.

Drawing to Other Cutoffs

If you're in an event with point cutoffs or other critical thresholds, such as an SCG Open, Grand Prix, or even the Pro Tour, you can apply the single-round math to the final round to ensure meeting the threshold. This is tricky, since you have to account for people above you or at the same record opting to draw, but let's look at the last Pro Tour.

At PT Nagoya, anyone with 30 points after Round 15 could take a draw to guarantee top 50 and win an invite to Pro Tour: Philadelphia. Quite a few people took advantage of that: everyone at 31 except for Carrie Oliver and Kai Fielder, who made it to 31 from 28, took a draw in the final round to secure their invite. Of course, not everyone took that option: people who were already qualified for Philadelphia opted to play the final round to try to make top 32 and get more Pro Points, resulting in their opponents not getting the draw. This is important to remember - once outside of top 8 contention, not everyone has the same goals when choosing to draw!

Application

When handling draws, it's best to avoid all mention of the prize and just focus on explaining to your opponent that taking the draw allows you both to make the relevant cutoff in order to avoid giving the appearance that you are performing bribery or collusion. Needless to say, actual bribery or collusion will get you disqualified and possibly banned by the DCI. Just ask your opponent if they are willing to draw, and if they are unsure, explain that a draw ensures that you both make the cutoff. If they refuse even when you both are locked for top 8, there's nothing you can do but re-offer the draw after the first game is over. If you're below them on the standings, it's often helpful to point this out, since you can simply point out that if a draw were going to screw them over it'd screw you over just as well.

Furthermore, outside assistance will earn you a Match Loss penalty, so once you've left the pairings board and are at the table, don't tell people at adjacent tables whether they're safe to draw or not.

Joshua Justice

@JoshJMTG on Twitter