On Saturday, 256 teams will play 10 matches over 16 rounds. At the beginning of the day, each team will be assigned a card from 1 to 256. For each round, the teams will go to rooms as dictated by their card, where they will play another team with a similar record. After the match, the winning team will get the lower-numbered card and the losing team takes the higher-numbered card. Therefore, one does not necessarily keep the same card throughout the day.

After all the Saturday rounds have been played, all teams that finished with a 6-4 or better record will advance to the playoffs. As with previous years, there will be a modified double-elimination bracket. All 7-3 and better teams will begin in the Winner’s Bracket, with the top teams starting with up to three byes. If a team in the Winner’s Bracket loses, they will enter the Loser’s Bracket, which is where all the 6-4 teams start. If a team from the Loser’s Bracket loses, then they are eliminated from the tournament.

Once there are four teams remaining, the last Winner’s Bracket team (Team 1) will play the lowest-seeded team, while the two remaining teams play each other. If Team 1 loses that match, then the two winners will play each other to play Team 1 in a one-game final. Otherwise, Team 1 will have two chances to beat the winner of the other match in a two-game advantaged final. This ensures that only the winner will have fewer than two losses in the playoffs.

If you wish to see a more visual representation of the playoffs, you can find last year’s bracket here.

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With 256 teams attending, how many teams can/will finish on Saturday with 10-0, 9-1, 8-2, and 7-3 records (and 6-4, I suppose)? Given that 256 is a power of 2 and power-matching gets a lot more regularized with powers of two, will there necessarily be a 10-0 team or do the initial four “seeding games” (in pools of 16 before breaking out into true power matching) mean that there might not be one?

Comment by Matt Jackson — May 21, 2013 @ 1:39 pm |

There will be 43-45 teams that are 7-3 and better, and 50-55 teams that are 6-4. After eight games, there will be one 8-0 team, and they will have to play two further games against teams that have already suffered a loss. Until we reach a 1024-team HSNCT, one cannot guarantee a 10-0 team.

Comment by Harry — May 21, 2013 @ 2:05 pm |

I wrote a python script that would simulate power matching for 256 teams and 10 games, and here are the expected number of teams for each record:

10-0: 0 teams

9-1: 2 teams

8-2: 12 teams

7-3: 30 teams

6-4: 53 teams

5-5: 62 teams

4-6: 53 teams

3-7: 30 teams

2-8: 12 teams

1-9: 2 teams

Of course, I made an assumption that each team in each game have an equal chance of winning, which may not be correct for unbalanced matchups (9-0 vs. 8-1 for example), so there may be more 10-0/9-1/8-2 teams then the simulation suggests.

Comment by Ryan L. — May 21, 2013 @ 3:34 pm |

I think that, just as there will be one (and only one) 8-0 team at the end of 8 rounds, so also there will also be one (and only one) 0-8 team. That team will then play a pair of 1-loss teams in the last two rounds and may or may or may not win one or both of them. There will be either 0 or 1 undefeated teams, and either 0 or 1 winless ones.

Comment by Nathan M. — May 21, 2013 @ 4:53 pm

Will early Sunday be similar to last year, with the top 6 getting a double bye and the next 6 or so getting a single bye? Also, will there be one of those super-cool spreadsheets for each day like in years past?

Comment by Max Schindler — May 21, 2013 @ 4:30 pm |

This year, the top team will get a triple-bye, and teams two through four will get double-byes.

And I do plan on making another spreadsheet for Sunday.

Comment by Harry — May 21, 2013 @ 4:32 pm |

Ryan: That assumption is not correct even for balanced matchups. We use simulations in which the higher seed has a 75% chance of winning. Running 10,000 trials of the tournament, we got the following:

In 16.8% of trials, there was one 10-0 team, two 9-1 teams, eleven 8-2 teams, twenty-nine 7-3 teams, and fifty-four 6-4 teams.

In 11.1% of trials, there was one 10-0 team, two 9-1 teams, ten 8-2 teams, thirty-one 7-3 teams, and fifty-three 6-4 teams.

In 9.4% of trials, there were no 10-0 teams, four 9-1 teams, ten 8-2 teams, twenty-nine 7-3 teams, and fifty-four 6-4 teams.

In 9.4% of trials, there was one 10-0 team, one 9-1 team, twelve 8-2 teams, thirty 7-3 teams, and fifty-three 6-4 teams.

All other scenarios occurred in less than 6% of trials.

I’m not I understand Matt’s question about “four seeding games”. There are no seeding games. It is pure perfect power-matching until everyone has played 8 matches, after which there are some mismatched games.

Comment by Jonah Greenthal from NAQT — May 21, 2013 @ 4:31 pm |

Oh, and overall, there is a 63% chance (by the aforedescribed simulation method) that there will be a 10-0 team. There cannot be more than one such team.

Comment by Jonah Greenthal from NAQT — May 21, 2013 @ 4:42 pm |

Won’t there be one 8-0 team, which has to play two games at 3/4 a chance, for 9/16=56%?

Comment by Max Schindler — May 21, 2013 @ 4:46 pm

Max: I misremembered the parameters of the simulation; we actually use an 80% chance of the higher-seeded team winning, so since there are two one-off matches, one would expect .80² = a 64% chance of there being a 10-0 team. The simulation obviously didn’t get that exactly (we came up with 62.64%, if you want more decimal places), but it’s about right.

Comment by jonahgreenthalnaqt — May 21, 2013 @ 4:54 pm

My mistake, Jonah. I just wanted to write a short, simple script just so I could see an approximate distribution for the records.

If you don’t mind answering, what is the rationale for having the higher seed have an 80% change of winning for balanced matchups? In other words, if two teams are both 2-2, why would the one who happens to have the lower number have a much better chance of winning?

Comment by Ryan L. — May 21, 2013 @ 5:23 pm |

Well, the tournament is seeded, so it’s indeed true that higher seeds have (we think) a better chance of winning. (Now, that doesn’t mean that, say, the 1 v. 256 match in round 1 is actually the 1 seed versus the 256 seed, but the way we scramble the seeds makes the simulation valid.) I believe 80% was just arrived at by feel, but it’s given us pretty good projections for previous tournaments.

Comment by jonahgreenthalnaqt — May 21, 2013 @ 5:28 pm

This is very situational, and the simulations that have been run so far are simplistic.

Cards are assigned somewhat randomly, so for the first two rounds, the probabilities should be 50%. When a 1-1 team plays another 1-1 team, however, one of them is a team that won its first match and is 1-1 because they lost to a team that is now guaranteed to be 2-0, and the other is a team that lost its first match and is 1-1 because they beat a team that is now guaranteed to be 0-2. In this case, there is a reason to give the team with the better card, which will be the team that won its first match, a better than 50% chance of winning. It is difficult to know what that percentage will be, and of course saying that a team has a better than 50% chance of winning based on limited information is a far cry from that team actually winning.

For the matches at the end that determine whether or not a team will be undefeated, keep in mind that they feature an 8-0 team vs a 7-1 team, so there is at least one reason to believe that the 8-0 team is better. If the 8-0 team wins that match, then you have a 9-0 team vs an 8-1 team. I haven’t done any research to determine whether the win probability for the undefeated team should be 50%, 75%, 80%, or something else, but I think it is fair to say that nobody should be shocked either way whether 0 or 1 team finishes undefeated.

Comment by Reinstein — May 21, 2013 @ 7:02 pm

Jonah: I was told that the first four matches take place in pre-determined smaller “pools” of 16, and open up so teams of the same record (or off-by-one) play each other outside those initial sets of 16 from then on out. Is this no longer true?

Comment by Matt Jackson — May 21, 2013 @ 6:16 pm |

Not exactly. It’s true that there’s perfect power-matching among groups of 16 for the first four matches — but then such groups are paired up to create groups of 32, then those are paired up to groups of 64, etc., until it’s just one big pool of 256. That post was based on last year’s format, for which it was true because 16 is the largest power of 2 dividing 240 (last year’s field size). I’ll ensure that that thread on the boards is clarified.

Comment by jonahgreenthalnaqt — May 21, 2013 @ 6:18 pm |

I just clarified this on hsquizbowl. Jonah is correct.

Comment by Reinstein — May 21, 2013 @ 6:52 pm |

If I had to bet, that “misconception” comes from it being 240 teams last year, so, since 16 is the power of 2 that maximally divides 240, it worked like that then.

Comment by Max Schindler — May 21, 2013 @ 7:22 pm

Thanks for the clarifications, Jonah.

Comment by Matt Jackson — May 21, 2013 @ 8:02 pm |

Sorry to continue being terrible at bracketology, but: Do we know yet what the numerical places will be for ties below 4th place? Will it once again be 5th, 8th, 13th, 21st, and 33rd, or will those numbers shift?

Comment by Matt Jackson — May 22, 2013 @ 1:00 pm |

There will be three 5th-place teams, five 8th-place teams, eight 13th-place teams, thirteen 21st-place teams, etc.

Comment by Jonah Greenthal — May 22, 2013 @ 1:02 pm |

Unless I’m reading it wrong, the places will be as you mentioned. The first difference is that it will be 48th place instead of 49th place.

Comment by Harry — May 22, 2013 @ 1:05 pm |