So, in a perfect world, here's how the NOVA's format works ...
80 or 96 attendees ... let's go with 96 b/c I think we'll actually get there.
96 turns into, after 1 Round
48 x 1-0
48 x 0-1
After Round 2 this turns into
24 x 2-0
48 x 1-1
24 x 0-2
After Round 3 this turns into
12 x 3-0
36 x 2-1
36 x 1-2
12 x 0-3
After Round 4 this turns into
6 x 4-0
24 x 3-1
36 x 2-2
24 x 1-3
6 x 0-4
The 6 x 4-0 have all played undefeated opponents, and all advance to day 2, along with the TWO HIGHEST RATED 3-1 PLAYERS, where in Round 5
#3 plays #6
#4 plays #5
#'s 1 and 2 wait
Round 6 =
#1 vs. lowest 5 winner
#2 vs. highest 5 winner
Meanwhile the highest round 5 loser plays #8, and the lowest round 5 loser plays #7 ... you can figure where that'll go from there, to Round 7's finals, and the single loss guys battling for 3rd place
But what happens if 3 people don't show, and we end up with 94 instead of 96 (subbing our single ringer in to get even #'s)?
94 after Round 1 pares down to ...
47 x 1-0
47 x 0-1
Ruh roh!
Either #1 needs to play #48 (0-1), or #47 needs to play #48 ... but who is the "right" one? Do you drop the lowest rated winner to the bottom bracket for that pairing, or do you give the highest winner the highest loser? Which is the proper call?
Let's extrapolate this into another dastardly issue ... 9 people no show out of 96, and you sub in your ringer to get to 88
Round 1 pares to
44
Round 2 pares to
22
Round 3 pares to
11
So going into Round 4, which will determine who gets to play on Day 2, SOMEBODY among the 11 3-0 players gets to / MUST play against a 2-1 opponent. Who gets the "easy" match? #1 or #11?
I'm curious about peoples' inputs here ... it's almost inevitable that people won't show; whether the "on hand" ringers / wait list will be robust enough to manage that and get back to either 80 or 96 will not be known until the day of ... so what's the best course of action here? Should the #1 play the highest from the next lower bracket, or should the lowest rated undefeated get that match?
- Mike
I'm wondering, thanks to this post, what the 'correct' ratio of Ringers to Players is. I'm thinking along the lines of 1 to 12? Surely people are reliable enough fro that to work? Hrmm.
ReplyDeleteAs regards the issue - I'm perfectly positioned, having no vested interest...but neither do I have an easy answer.
For the last example, my instinct is to pair 11 with the 12th player, because they are closest matched, in theory. Whoever is top doesn't need the boost.
I suppose the same applies for the first issue...but in these circumstances I think ideally you would avoid repeat games too. Sorry I can offer no more than a half-formed thought.
As far as the first situation goes I would suggest pairing the lowest rated player in the 1-0 bracket with the highest rated player in the 0-1 bracket, seems to work for magic.
ReplyDeleteI'd say the top 1-0 person should play the second to bottom 1-0 player, and the bottom 1-0 player should play the top 0-1 player. That way you never have the 1st place person playing anyone but an undefeated opponent
ReplyDeleteI guess I disagree with everyone here. People are looking at this like its swiss pairings. But its not. The best ranked person should be playing the lowest ranked person in the 'winners' bracket, even if that means he gets an 'easier' game. Anything else is altering the way you have setup the tournament for no logical reason.
ReplyDeleteLook at it more like earning your way into the top half, quarter, eigth, etc, rather then simply being undefeated.
Regardless of how you do it, what may get tricky is seeding after you've gone down the path of adding defeated players into the 'winners' bracket (e.g. what happens when that 2-1 player wins) You may have to rethink your seeding mechanism to account for it.
Use the same scoring and pairing method you use for everyone else.
ReplyDeleteIf that means the 'top' player gets to play the "easiest" loser, and could get upended and that player then becomes a defeated 'winner', so what?
At the end of the tournament, there will be a clear winner, right? Even if it means you end up with something like a defeated player 'winning' everything--in the end, there would be no more undefeated opponents and the players who essentially end up 1 and 2 won't feel gipped--doesn't it fall to strength of schedule at that point, so wouldn't the 'defeated' player still not win due to that(?), because even if by some crazy happenstance his 'winning' opponent was the OTHER defeated seed moving on and they did not play but both managed to win...you get the idea. lol
I'm not sure what happens if:
Defeated player beats undefeated player and then beats another undefeated player but is seeded in game #1, while undefeated player goes undefeated in game #2. Is that possible? Does it matter at that point? I don't think which match you are in (1 or 2) but who has undefeated records then if none it's strength of schedule? Could this happen?
Two defeated players advance to round 6.
Both players defeat their opponents.
So there are four 5-1 players? Who wins?
lol
once round 2 is over basically 3/4 -2 of the players are eliminated from winning and advancing to day 2..thats alot of people. Is there concern of people just saying fuck it at that point mike ? Ive never played a tournament using this style so im curious to see how it works out.
ReplyDeleteThere are some whiskey40k articles addressing just that issue throughout the blog.
ReplyDeleteI'll go over it tomorrow in more depth ... suffice to say Day 2 has open gaming either way, to play games you'd like with people from around the country, and there's only a single prize given out. Everything else occurs on Day 1, including half the Vegas quals.