I often use this system to determine prizes: http://swisstriangle.net/

However, I'd like to know whether the swiss triangle (with “paired down wins”) can actually guarantee a maximum number of players at each level.

For example, if the triangle says
6:0 - 1 player
5:1 - 5 players
4:2 - 10 players
etc.

I wish to know whether I can be 100% certain (even accounting for any byes, drops, double-match-losses and draws) that at most 1 player will score 6 wins, at most 6 players five wins, at most 16 players 4 wins etc.

For the purpose of this, I do NOT consider 3 draws equivalent to 1 win.

100% certain? No. WER may decide to pair a round with three pair-downs at a given point total, completely negating the tenets of the triangle.

I would also argue that the algorithm that said site uses is slightly flawed, which can seen with 6 players entered: If the pair-down always won, the results would be 1-2-2-1, not 1-3-2-0. It's a minor bug, though, and over-estimates players at top rankings.

That being said, it's accurate enough that trusting it for prizes will give you decent worst-case-scenario numbers; that being said, the number of times that everything will fall into place like that at an actual decent-sized event is pretty small.

(slight plug: I keep a personal implementation of Swiss triangles here)

You should be careful relying too closely on swisstriangle.net. It uses a slightly buggy algorithm, and as a result a N-player event will sometimes end up with more than N/2 match wins per round.

At first glance, the site to which Jordan refers does not have this problem.

It'd be great if either of those sites allowed for an extra round or
two, as it can be nice to see what happens when you run enough rounds
for cut to top4 :)

I just updated mine - goes until there's only one X-0 or X-1, and colors in the cut rounds for Top 8 and Top 4. Note that, because I'm making a “paired down always wins” assumption, I don't believe there's a case where Top 4 won't occur one round beyond Top 8.

Gareth - check out Eli Shiffrin's triangle. He has it where you can enter the number of players and the number of rounds:

http://elishiffrin.org/triangle.html

Jordan your calculator is disagreeing with Eli's and I figure out who is more correct. Someone isn't counting the loss from a pair down correctly. If you get the triangle for 76 players you'll find that round 4 has:
5 - 19 - 28 - 19 - 5 (and Eli has 29 at x-2's and only 4 0-x's)
Which you then go on to have:
3 - 11 - 24 - 24 - 11 - 3
But Eli has:
3 - 12 - 24 - 24 - 11 - 2

5 4-0s generates 3 5-0s and 2 4-1s
19 3-1s has one losing to a 4-0 and the remaining 18 provide 9 more 4-1s

So that should be a total of 11 4-1s?

Eli's and the swisstriangle.net ones both give the same answer; a previous post went into that discrepancy (look at it with 6 players and you'll see the same issue pop up) and I think the issue is the same.

One thing that would be awesome would be the possibility to include players with byes to those pages. It matters for some WPN Premium events.

Updated this one for:

- Byes
- Top 8 cutoff @ X-2.

Note that, at really high numbers, this algorithm starts breaking down a bit, as it doesn't factor in things like “players can't receive a bye twice”.

Not getting a bye twice isn't really a concern, as when it would
impact the system chances are the player would get a second bye
(players can get a second bye when all other players on the same
points have also already had a bye)

Where things would get less accurate is that at the top after many
rounds many of the x-1 players may have already played each other.
That causes some interesting pairings.

Remember that as a HJ/TO you can just announce that you will run
rounds until there are less than Y players at X-1 or better.

(players can get a second bye when all other players on the same points have also already had a bye)

Clarification: Players can only get a bye twice in a tournament under very specific circumstances, which include (but are not limited to) the following:

- In an event which includes drafts, a player can receive a bye within each draft done, as well as once outside the draft rounds of the format. (think limited GPs and 7-man draft pods in Day 2)
- A player who in a bracketed tournament may receive (effectively) a second bye if the match which determines his opponent ends in a double match loss or double-DQ.
- It is theoretically possible for a player in a 2000+-player Grand Prix (or other tournament with a non-top-8 cut) tournament to receive a bye in both Day 1 and Day 2.
- WER randomly decides that a person should get a second bye. (which I have seen…)

The bye is awarded randomly to a player who has not yet received a bye, and is among the lowest point total of players for which at least one player in that point total has not yet received a bye. (to sanity-check this, as the full WotC implementation of Swiss pairings is not public, I did double-check that both DCIR and WER follow this rule)