peanutman wrote:There are 10 teams eliminated so far. Next set of games will be sent in the next 24 hours for all groups who have settled their first match-ups.
This is a Tournament Players Association Year 3 (TPA3) event, and is governed by its rules, guidelines, and judgments which are detailed at http://www.conquerclub.com/forum/viewtopic.php?f=89&t=175675. All are welcome to join and participate in the association. You need do nothing more than join one or more of the TPA3 events that will be announced each Monday. Check back on the scoreboard to see how you're doing:
There are 14 sets of 3 teams that played their first set of games (one game on each home map plus one random map) between each pairing and in each match there was a team that lost two matches and are now out. Those teams are as follows:
There was one match in which all teams each had a loss after the round robin play (osujacket & Kratos644; zips5000 & jackal31 and seamus & hiddenvariable). These have played a 3-team match-up (1 game on each home map plus 1 random map) with zips5000 & jackal31 winning 3 of 4 games and moving on to the next round.
There was one matchup involving 4 teams, Momo33 & Quiksilver; FreeFalling123 & alvinchek99; reptile & bertbum and silversun6 & benga. The bracket is below:
Re: Peanut's 3X Double-Elimination - Doubles[TPA3] Needs Res
Posted: Tue Apr 02, 2013 2:08 am
by DaveH
Thanks for the responses about the settings being Escalating Spoils rather than No Spoils. All but a couple were OK with the mistake, and I was able to replace one of the sets of games that had not been fully joined where the response was against Escalating.
In fact the original round was also with ES, which I copied erroneously. So, we will complete the present set of games as issued and future rounds will be NS.
With ooge dropping out, jetsetwilly's team go through to the next round. I will assign average rankings in the set of games forfeited by ooge in determining the next set of matchups.
Robespierre__ wrote:I think we received a duplicate set of games (12570373) was against the same opponents that we just received invites to ....
Yes, my interpretation of the original rule is that you have to win 2 Machups. Each matchup is the best of the set of 3 games.
Remember it is a double elimination tournament that has been modified to get multiples of 4 players into the next round, so I think Peanut designed the rule to give a second chance in this preliminary round.
I agree with Robespierre. And I dont see where you read such a weird rule
Double elimination is because when you lose you get a second chance but against another team that also lost the machup. Best of 3, twice, and if one team win once and one team the other time, then we make a best of 3, it gives a best of 3 X 3 ...
EDIT : Is it because we are on round 2?? If yes we have to play 5 match not 6. 2 map home, 2 map away, and 1 random. It is one random too much now. Please confirm
I think that Peanut's original rule for a modified double elimination for three teams is quite clear.....
All teams will play each other once in a match-up. As soon as a team loses 2 match-ups, they are eliminated and the remaining teams will play until one team loses 2 match-ups in the round.
You have played a matchup against each other (ie 3 sets of 3 games) and the loser of the two match ups is eliminated. You now play Machups of three games against each other until one team loses two Machups.
Then we can resume the 4-player double elimination series in the next round.
Round 1 Given that we have 49 teams signed up, we will have 15 groups of 3 and 1 group of 4 all chosen randomly. The groups of 3 will play a modified double-elimination bracket (see spoiler above). The group of 4 will play a standard double-elimination bracket (see diagram above).
hide: Modified 3-team double-elimination bracket
All teams will play each other once in a match-up. As soon as a team loses 2 match-ups, they are eliminated and the remaining teams will play until one team loses 2 match-ups in the round. In the event that all three teams each have a loss after the round robin play, a special 3-team match-up will be played (1 game on each home map plus 1 random map) with the team winning the most games moving on to the next round. In the event that two teams win two match-ups each, they will play a final match-up (home maps plus 1 random map) to determine a winner.
So for the Group of 3 persons :
As far as I understand. I play against everyone in my group once. ( playing against another team is 3 game 1 home 1 away 1 random round 1 and 5 game, 2 home 2 away 1 random round 2). It is group of 3 teams. So it is : Team 1 VS Team 2 Team 1 vs Team3 Team 2 vs team 3
If one one of the 3 team lose twice they are eliminated. This one for Round 1.
Now that we are in round 2 we should be 16 team, splitted into 4 group
Round 2 16 teams grouped in 4s by ranking (see below). Double-elimination brackets for each group (see diagram above). 5 games at a time (2 on each home map, 1 on random map). Match-up winner determined by best of 5. Winner of the bracket moves on to the next round.
So we should have a standart double elimination brakets since we are 4. for now. And not play again against same team.
donelladan wrote:Seems not clear at all. This is what I read
Round 1 Given that we have 49 teams signed up, we will have 15 groups of 3 and 1 group of 4 all chosen randomly. The groups of 3 will play a modified double-elimination bracket (see spoiler above). The group of 4 will play a standard double-elimination bracket (see diagram above).
hide: Modified 3-team double-elimination bracket
All teams will play each other once in a match-up. As soon as a team loses 2 match-ups, they are eliminated and the remaining teams will play until one team loses 2 match-ups in the round. In the event that all three teams each have a loss after the round robin play, a special 3-team match-up will be played (1 game on each home map plus 1 random map) with the team winning the most games moving on to the next round. In the event that two teams win two match-ups each, they will play a final match-up (home maps plus 1 random map) to determine a winner.
So for the Group of 3 persons :
As far as I understand. I play against everyone in my group once. ( playing against another team is 3 game 1 home 1 away 1 random round 1 and 5 game, 2 home 2 away 1 random round 2). It is group of 3 teams. So it is : Team 1 VS Team 2 Team 1 vs Team3 Team 2 vs team 3
If one one of the 3 team lose twice they are eliminated. This one for Round 1.
Now that we are in round 2 we should be 16 team, splitted into 4 group
Round 2 16 teams grouped in 4s by ranking (see below). Double-elimination brackets for each group (see diagram above). 5 games at a time (2 on each home map, 1 on random map). Match-up winner determined by best of 5. Winner of the bracket moves on to the next round.
So we should have a standart double elimination brakets since we are 4. for now. And not play again against same team.
I'm not sure what invites were sent or games played, but I believe this interpretation is correct. It is essentially, 3 rounds of small double elimination brackets. There were not enough sign ups to run the tournament as intended and I allowed peanut to modify the play slightly in Round 1. In rounds 2 and 3, the double elimination bracket it what was intended to be used.
Thanks for the inputs, though I read Peanut's instructions slightly differently.
Firstly, the one team of 4 is playing the correct double elimination format - no problem.
The remainder have played in groups of 3 a matchup (which is a home game, an away game and a random game) against each of the other two teams i.e. 9 games.
This is where I picked the tournament up and there is one group that have won one matchup each, so they ahave played a three- team game as specified. - no problem
The remainder have had one team that lost twice and so are out, leaving the other two teams to play each other until one team wins TWO matchups.
One matchup has been decided and so I have sent out the second set of the same games to determine which team wins the second match in which case there may then be a third set of the same games to determine which team goes through.
Note that we are still in the first round, though I may have confused over the use of "set" as a label.
After the conclusion of this Round 1 there will be 16 teams organised and ranked as 4 groups of 4 teams that can play a standard double elimination tournament. - no problem
Note that I am not trying to be difficult and do not care too much what system we play, but I am just trying to keep to the original rules. Had Peanut written "All teams will play each other once in a match-up. As soon as a team loses 2 match-ups, they are eliminated and the remaining teams will play until one team loses ONE match-up in the round." than I would agree that we are now ready to move on to the 4 x 4 format, but he put "... win TWO matchups in the round"
Perhaps I need to take into account the earlier matchup between the two teams and so avoid a second set of games other than if a decider is required?
donelladan wrote:Now that we are in round 2 we should be 16 team, splitted into 4 group
For clarification, we are still in Round 1; the games sent out are the second set of games to form matchup 2 as in "..the remaining teams will play until one team loses TWO match-ups in the round"
As it says "Round" though, I perhaps need to take account of the matchup between the two teams in the three team set of games and only if there was 1 win each have the deciding set?
I'm happy to go wiuth whatever Chap think is the original interpretation of course.
We now have 15 of the 16 teams into the final sets of games and awaiting a joiner into the 4-team bracket final. Teams have reached the final stage through 3 different methods due to the different number of entrants than planned, so I can't see a fair way of ranking teams ahead of distributing them according to the Serpentine system (which would form 4 groups viz. 1, 8, 9, 16 & 2, 7, 10, 15 & 3, 6, 11,14 & 4,5,13,12) so I intend to randomise the groups if that is acceptable.