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Post by Rycchus on Sept 10, 2007 16:03:13 GMT -5
In theory. But generally the pairings go Winner Pod X vs. Runner-up Pod Y, so you ought to end up with the advantage if you win your group, and the disadvantage if you come runner up.
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Post by nawglan on Sept 10, 2007 17:45:59 GMT -5
So, Randomize the winners into their brackets, then randomize the runners-up. I'll be the one warming the bench for the 2nd round, so doesn't really matter to me. 8)
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Post by xade on Sept 10, 2007 19:20:34 GMT -5
but... but... if I make it to the finals, I would *hate* to have the possiblity of running into Toyo again... At least until the grand final...
Can we at least ensure that the people from the same pod *can't* play each other again until at least the semi's... (preferably the grand?)
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Post by Rycchus on Sept 10, 2007 20:06:13 GMT -5
Oh yeah, that's a good point. Well then, divide the pods into two halves. I'm not good at words, so here's a picture: Put the winners of the groups into the eight bottom places (A, B, C, D, E, F , *) randomly. Take the runners up of the corresponding groups from one "half" (LHS) (let's say a, b, c, d) and randomly assign them to the other half (E, F, G, *). Then take the corresponding runners up of the right hand side (e, f, g, *) and assign them randomly to the groups on the left hand side (A, B, C, D). It's not perfect, because the wildcards might end up playing people from their own groups. But it ensures that the top two people from each group won't meet each other until the final if at all. (Note, in the description I've treated the wildcards as a separate group, and grouping one with the winners and one with the runners up. Obviously they're equal to each other (toss a coin for who gets lumped with the winners) and come from one of the other groups.)
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Post by awall on Sept 10, 2007 21:35:23 GMT -5
What Rycchus suggested is technically the "right" way to do it (it's what World Cup Soccer uses, I believe). However, we have to account for the fact that we have 7 winners, 7 runners up, and 2 wild cards. In this case, I believe the two best winners should play the wild cards, the two best runners up should play each other, and the remaining 5 winners play the remaining 5 runners up. The "best" winners/runners up would be the ones with the hardest pod (probably highest average elo, not counting your own). Assuming that A is the hardest pod and and B is second hardest...
Round 1 1.1) Winner A vs. Wild Card 2 1.2) Winner C vs. Runner Up D 1.3) Winner E vs. Runner Up F 1.4) Runner Up A vs. Runner Up B 1.5) Winner B vs. Wild Card 1 1.6) Winner D vs. Runner Up G 1.7) Winner F vs. Runner Up C 1.8) Winner G vs. Runner Up E
Round 2 2.1) Winner 1.1 vs. Winner 1.2 2.2) Winner 1.3 vs. Winner 1.4 2.3) Winner 1.5 vs. Winner 1.6 2.4) Winner 1.7 vs. Winner 1.8
Round 3 3.1) Winner 2.1 vs. Winner 2.2 3.2) Winner 2.3 vs. Winner 2.4
Final Winner 3.1 vs. Winner 3.2
Other than the wild card matches, there is no chance of a rematch until the semifinals, and the only possible rematches here are between the winner/runner up of A and of G. Also, the players who had the hardest opponents in the round-robin get the easiest opponents in the first elimination round.
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Post by Slartucker on Sept 10, 2007 22:08:58 GMT -5
You guys have done a nice job with this. However, I'm pretty set on doing it randomly, without regard to results of round one. There are two reasons for this.
First, day one results are an uneven representation of skill or even predictions of success and therefore not a great basis for seeding. This is due to the unevenness of the pods.
Second, while I originally wanted to avoid the same players playing each other more than once, I have started to think of round one as "Day One" and the other rounds as "Day Two" of a multi-day championship. In that context playing someone repeatedly actually just enhances the flavour of the match for spectators. Grudge matches are good! And besides, round one is just one game whereas later rounds are at least two, so it isn't really a big inflation in terms of actual games played.
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Post by Rycchus on Sept 11, 2007 8:16:18 GMT -5
Only problem with that is that it's theoretically possible to win the tournament just by playing the same few people. You might end up playing only two people that weren't in your group, and I don't think that's much fun.
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Post by Slartucker on Sept 11, 2007 11:32:42 GMT -5
That would still mean you played six different people, which is more than ANYONE played in last year's tournament.
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Post by Rycchus on Sept 11, 2007 12:16:12 GMT -5
I guess. But you could get as far as the quarter-finals with only playing people in your group.
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Post by Slartucker on Sept 11, 2007 12:34:14 GMT -5
Yes, playing at least eight games with four people, compared to four games with three people last year.
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Post by Rycchus on Sept 11, 2007 12:51:58 GMT -5
I wasn't here last year Okay, okay. It's your deal, dude. I was just giving you my opinion
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Post by xade on Sept 11, 2007 19:04:28 GMT -5
I wasn't here last year Okay, okay. It's your deal, dude. I was just giving you my opinion You can't like, own, the tournament man... this is the peoples tournament... simplify man, simplify...
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Post by Slartucker on Sept 11, 2007 22:15:30 GMT -5
Dude, where's my tournament?
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Post by freesoul on Sept 12, 2007 8:14:27 GMT -5
I suppose a way to randomize the participants, and avoid immediate rematches is to take a sort of "shuffle the deck" approach.
Split the pods in half, randomize both sides, and then splice them together.
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Post by Rycchus on Sept 12, 2007 19:48:06 GMT -5
^ Yeah, that's what I said.
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