Just out of curiosity. After 8, 16, 32, 64 and 128, my guess for the next cap was be 256, but it's 226. So why that number exactly? =)

I don't know the exact math, but the concept behind the number is this:
Anyone with only 1 loss (X-1) at the end of Swiss will make the Top 8. If
you look at your swiss triangle (http://swisstriangle.net/), you'll see
that 256 will have 9 people at X-1 or better. That said, I'm not sure
about that math behind it.

I also do not have the exact formulae, but the main reason for the shifts is due to increasing impurity in the Swiss triangle as draws propagate. After 100 players, the probability of draws increases the fallibility of the Swiss triangle's ability to fairly adjudicate fair standings with just the rounds indicated for pure win-loss records.

I was wondering about this same thing. At first I thought it was a safety margin to account for draws or DQs, but I went through the math and it turns out Shawn is exactly right. If we're using the Swiss algorithm to determine a winner, we can do 2^n rounds for n players and there will be at most one undefeated player, but since we want top 8, and since players want an X-1 record to be a guaranteed top 8, that isn't enough for us. As it happens, if we've done n swiss rounds for 2^n players, we can guarantee that there will be at most 1 player at X-0 and n players at X-1. At 128 players, n=7 and we have 8 players at X-1 or better. At fewer than 128 players, we have not enough players at X-1 or better, so some players from X-2 make top 8. But, at 256 players, we have 1 undefeated plus 8 at X-1, and 1+8 is more than 8 and some poor fellow thinks he'll make top 8, but really won't. At the 129-226 player range, we again statistically expect exactly 8 players at X-1 or better.

This isn't perfect though, it depends very significantly on the results of matches where players of different records are paired. 226 is the sweet spot where the odds are 50% but it isn't a true breakpoint like 128 players is. Plus, draws, DQs, and matches between players at different records can all screw you up, no matter how much math you do, and if you're running a 200+ player event without any of those then you must just be incredibly lucky.

By the way, here are the “real” breakpoints - the number of players where, regardless of the results of paired up/down matches, there will be /at most/ 8 players at X-1 or better:
Up to 216 players - 8 rounds
Up to 384 players - 9 rounds
Up to 704 players - 10 rounds
Up to 1280 players - 11 rounds

As for draws - if you're very close to these breakpoints, there's the possibility that a player with an X-1-1 record will not make top 8. A player with X-0-1 will always make top 8. If you're at the lower end of a player range (129-170ish, 217-300ish, etc) then there is an “extra” spot in top 8, and if there are two players at X-1-1, both can make top 8. DQs move everyone up a spot, no way to predict those so there's no point worrying about it, “worst” case is an additional player at X-2 makes top 8. If everyone who gets paired down loses their match we'll have fewer players at X-1, but as long as you have fewer than the numbers I listed in the last paragraph, there's no way that your players can manage to kick someone at X-1 out of the top 8.

I don't even want to think about byes, that has the potential to get messy. It doesn't affect the player who was already going to be undefeated, but if we have a 216 player tournament and offer 20 2-round byes, then we now have two undefeated players and 8 more at X-1. Players with byes have a better chance of making top 8, but the “average” record is also much better because for each round bye, that player is getting a win, but there isn't a corresponding player getting a loss, and so you have more players at X-1 records.

Edited Dan Collins (Jan. 7, 2014 05:55:38 AM)

With 226 players, it's theoretically possible to have a player at 7-1 being 9th. The chances of that happening are really small, since it would require virtually no draws (when determining the minimum record needed for T8, a draw is as bad as a loss: you can afford to have one, but not two), and all players that are paired down at X-0 should lose their matches, while all players paired down at X-1 should win their matches. Besides that, there are players with human behavior throwing a wrench in the mathematical model. People drop at 2-1 because they have a party, or they don't like their pool and are done playing. This last effect is larger at bigger events too, the longer the event takes, the more people drop because they don't want to play on anymore, even when they're winning.

Note that there is a treshhold defined for 9 -> 10 rounds at 410 players, but none for 10 to 11 rounds. I'm sure Dan has mathematically sounds calculations for the mathematical cutoff between 10 and 11 rounds, but there is no official announced number for when to go to 11 rounds. If you plan to organise an event with more than 700 expected players, I recommend running it on 2 days. If you organise a PTQ and so many players show up: Congratulations you broke the previous record by 100% ! Let the TO make a call to WotC and discuss the options (preferably a few days/weeks before the event).

Long ago - more than five years, at least - the formula was included in the rules. (My oldest copy of the Floor Rules - remember them? - is from 2008, and it's not in there.)

Anyway, it was slightly more complex than 2^N, there was a log() function in there somewhere. The point, however, was as has been stated re: Top 8 including all X-1 or better records.

Events that include byes are more complex; each bye equates to some fraction of an additional player, so 120 players with a total of 42 byes might need 8 rounds, instead of 7. Or, not - I don't have that formula, either. (Thankfully, 99.x% of us will never need that - the scorekeepers at GPs and similar events need to know, but I don't.)

d:^D

The calculator (incl. byes) was also built into DCIR, but maybe only the
special version.


On Tue, Jan 7, 2014 at 5:17 PM, Scott Marshall <
forum-7783-563d@apps.magicjudges.org> wrote:

> Long ago - more than five years, at least - the formula was included in
> the rules. (My oldest copy of the Floor Rules - remember them? - is from
> 2008, and it's not in there.)
>
> Anyway, it was slightly more complex than 2^N, there was a log() function
> in there somewhere. The point, however, was as has been stated re: Top 8
> including all X-1 or better records.
>
> Events that include byes are more complex; each bye equates to some
> fraction of an additional player, so 120 players with a total of 42 byes
> might need 8 rounds, instead of 7. Or, not - I don't have that formula,
> either. (Thankfully, 99.x% of us will never need that - the scorekeepers at
> GPs and similar events need to know, but I don't.)
>
> d:^D
>
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–
Adam Cetnerowski
Gdansk, Poland

Special Version? ¬_¬ There is no “special version”! :)

Jasper - obviously the data I listed is 100% theoretical, for comparison with the official numbers. The actual numbers allow for a faster tournament that still allow probably 70-80% confidence of a clean Top 8. It all depends on exactly which players get paired down and lose, and how early that happens. YMMV.

I was under the impression that:
1 bye = 2 players
2 byes = 4 players
3 byes = 8 players




2014/1/7 Adam Cetnerowski <forum-7783-89b8@apps.magicjudges.org>

I think we've answered the original question, a few additional ones, and added a healthy dose of mild speculation as well … in short, this topic has run its course. (Closing for traffic control…)