# What is the proper way to split uneven draft pods?

The optimal number for a draft pod is 8 people, but it can be often hard to get a even multiple of 8.

If there are an odd number of drafters how should the pods be split? Also if there is any justification as to why it's done that way that would be nice to know.

I am aware of the Wizards Event Reporter, and that it can split players into pods automatically. I want to know how to split pods by hand, in case I ever need to run an event with pencil and paper.

• I thought there would have to be some kind of official rules on this, but apparently, the only thing the tournament rules say is that "players assemble into random drafting circles (called pods) of roughly equal size at the direction of the Head Judge."(!) Apr 15 '16 at 15:03

In my LGS it's solved this way.

6 players => 1 x 6

7 players => 1 x 7

8 players => 1 x 8

9 players => 1 x 9

10 players => 1 x 10

11 players => 1 x 11

12 players => 2 x 6

13 players => 1 x 6 + 1 x 7

14 players => 1 x 6 + 1 x 8

15 players => 1 x 7 + 1 x 8

16 players => 2 x 8

17 players => 1 x 8 + 1 x 9

18 players => 3 x 6

19 players => 2 x 6 + 1 x 7

20 players => 2 x 6 + 1 x 8

21 players => 1 x 6 + 1 x 7 + 1 x 8

22 players => 1 x 6 + 2 x 8

23 players => 1 x 7 + 2 x 8

24 players => 3 x 8

The idea behind is to not have a draftpod with less than 6 people, which yields in just too less cards in the entire cardpool. 9 players is still quite doable, 10 and especially 11 player pods are really bad (you basically won't see any card coming back to you).

To summarize:

• Don't go below 6

• Avoid going above 9

• Try to fill as many 8 pods as possible

• This is basically what my store does, so it would seem to be a common practice at least but I was hoping for a more definitively correct answer. Particularly since there is always a debate on numbers like 14, where some argue that two 7's is better because it's closer to 8 for the actual draft and others argue that preventing a bye takes higher priority. Apr 15 '16 at 17:18
• AFAIK one should have the minimum number of draft pods, so 18 players should result in 2 pods an 8-player, and a 10-player. 8 is ideal, but one should err on the 9-11 side than the 6-7 side. Apr 15 '16 at 21:32
• `1 x 7`, `2 x 8` etc - what are these numbers meant to represent? A single 7-player multiplayer game? A 7-player something or other? Apr 17 '16 at 4:06
• @doppelgreener: it means 1 pod à 7 players, 2 pods à 8 players, etc.. Nothing to do with multiplayer, we are still talking draft here Apr 17 '16 at 6:36

Realistically, the simplest way to resolve having X players where X is odd, would be to treat it as if you were running an X+1 man pod and then someone dropped suddenly.

So you draw up the rankings with "Mr Missing" as the 8th players, and then the player that gets matched up with him gets a "bye", where you automatically win the round.

Obviously, this still means that one player is having to sit out every round (this is unavoidable unless you're doing a Conspiracy-style multiplayer draft), but because they count as winning, they'll automatically be facing someone next match, whereas one of the losers of the match will now get the bye for the next round, will win, and will then get another match.

This allows you to constantly rotate through who's sitting out the next game, and gives a real benefit to winning, because you get to play every round.

EDIT: Did some research, and the official rules for Swiss Pairings can be found here; basically confirming the above.

• There's a difference between a bye and an opponent that drops. A bye does not affect tiebreakers. If your opponent drops, that opponent will still contribute to your tiebreakers. You can read more about tiebreakers here, but ultimately the best advice you can give someone is to not worry about all of that and just use the Wizards Event Reporter software to run the event properly. Apr 15 '16 at 15:35
• This just tells is what to do in Swiss rounds when a single pod does not have an even number of players. I am concerned with how to place players into pods when the number of players is not divisible by 8. Apr 15 '16 at 17:15
• Ah, I see. Okay, then @t.rathjen's answer is the best, but for the pods, I would apply the same philosophy. If you have 7 people, form the pods you would for 8 people and then cut one down. Apr 15 '16 at 17:18