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[Tournament] Codex Asynchronous suMmer Swiss (CAMS19)

One of the benefits of everyone still playing being eligible to win the whole thing is being able to pair across win records without creating obvious opportunities for bribery.

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I’m unclear what the bribery weakness is with matching by win record as a primary, can you elaborate?

I believe Eric refers to how in normal Swiss, you can lose so many games that you’re mathematically unable to make the cut to a bracket, or to win. This means that sometimes players can be bribed to intentionally throw specific games to bolster one player’s win record.

And that brings problems because how are you going to enforce rules against throwing games? What if they’re just not playing well in that game?

With the battle royale/“elimination after three losses” format, anyone who’s playing is in contention to win.

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Is that a problem with how rewards are usually distributed for Swiss tournaments, then? The rewards and the tournament system’s assumptions being mismatched? n-elimination tournaments would only give rewards to the top n players, because those are the only placings they reliably rank. Since Swiss tournaments reliably rank more players without such a cut-off, the rewards should be more spread out, so having the same reward scheme as for elimination tournaments would encourage collusion.

Right but that’s still true if I prioritize win/loss record for matchmaking over rematches, correct? I was under the impression the “prioritize win/loss record” was being called out as problematic w.r.t. bribes but it still isn’t clear to me why it would it

Unless you have a mostly flat prize distribution (ie each win gives exactly the same prize, so 4-0 is only 4 times better than 1-3 when usually 1-3 is “no prize at all”) there will be a break point where pairing different records produces:
(Player A prize if Player A wins + Player B prize if Player A wins) is greater (sometimes much greater) than (Player A if B wins + Player B if B wins).

Additionally, there is Hype! benefits to having a single clear “Finals” (and semis, and quarters) which pure swiss doesn’t support, and cut-to-top-N is a very sharp prize gradient.


Minimizing re-pairs is more fun for the participants, as they get to experience more different pairings & opponents.

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Oh, would an arithmetic progression not prevent that? I need to go away and work on my intuition for this. EDIT: Never mind, that’s what you meant by a flat distribution. In that case, I’m basically saying that’s a reasonable prize distribution for normal Swiss, in order to avoid collusion. The equivalent for this format would be a flat distribution for the top 3.

The main reason people don’t use that distribution is that it doesn’t reward the “best” players, and isn’t very exciting.

Going X-0 in swiss is more than twice as hard as going X-X, due to the ability filtering that happens with pairing-by-record. Folks generally want to recognize that fact, and that mere recognition (aside from any physical prizes) constitutes a differential payout for 4-0 & 2-2 vs 3-1 & 3-1.

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So not for bribery risk reasons, but just for a general “reward the player on a tear with a ‘easier’ opponent instead of a ‘harder’ rematch”, makes sense from that perspective.

MonoGreen and PseudoBlue in the top 4? What is happening?

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Pretty cool to see :sunglasses:

I almost had him last time, so we will see how this rematch goes. :slight_smile:

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Also folks, I am trying to get more careful about doing matchups, but feel free to call out errors in matchmaking if you see them!

I think Round 7 results might have been recorded incorrect in the spreadsheet.

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DOH! I need to stop doing this so early in the morning! You’re right @bansa I gave Persephone the W against Zhav by mistake.

that means there will be a bye, I’ll fix!

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Sorry to be difficult again but shouldn’t zhavier and I go first since we have the same number of losses?

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I don’t know? You’ve all had a bye and played each other, none of you have a bye, pairing you two means one of you goes to 5 games on P2 with only 3 on P1, this pairing has the possibility of an elimination…

I’m open to input from @zhavier and @EricF but the “how pairings work” algo doesn’t make this situation unambiguously clear to me

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If the number of players is odd, a bye (which counts as a win) will be assigned to the player who:

  1. Has had the fewest byes in the tournament (normally zero)
  2. Has played against all other players, or so many of the other players that a bye is needed to avoid a re-match elsewhere in the tournament (probably won’t happen)
  3. Has the most losses

This gives EricF the bye.

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You know… I should have just kept my mouth shut cause I didn’t mind the bye at all :wink:

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OKAY I apologize for prematurely posting with incorrect participants (again!!!), here comes the ACTUAL ACTUAL pairing. @EricF has been given the bye (as he has the most losses and all other assignment criteria are equal), and Bansa & Zhav get a flipped runback! GL HF to both of you.

CAMS19 Round8

  1. [CAMS19 Round8] zhavier :medal_sports: [Anarchy]/Strength/Growth vs bansa :knockdown: [Law/Peace]/Finesse
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