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Old 07 Feb 2003, 02:57 PM   #1
IASocFan
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Default 2006 Qualification Process Announced

CONCACAF has announced the proposed qualification process. It looks like a combination of the Asia method and previous CONCACAF groups.

12 initial groups (11 w/ 3 teams; one with 2 teams)
Winners qualify for semis: 3 groups of 4
Top two qualify for final Hexagonal:
Top 3 qualify for WC; 4th will play Asia 5th.

See: http://www.concacaf.com/news/article.page/1202
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Old 07 Feb 2003, 03:04 PM   #2
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It's proposed only in that FIFA has to bless it. I would assume this will be final.

It looks like the sharks - like US, Mexico, and Costa Rica will get a couple of minnows to devour before the semis.

Maybe Costa Rica will draw Barbados again.
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Old 07 Feb 2003, 03:07 PM   #3
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My question is how the teams get divided into that first round (groups of three, etc).

Pot A: 3
Pot B: 3
Pot C: 6
Pot D: 23

How does that get split into those first groups?
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Old 07 Feb 2003, 03:09 PM   #4
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Quote:
Originally posted by BuffloSoldier
My question is how the teams get divided into that first round (groups of three, etc).

Pot A: 3
Pot B: 3
Pot C: 6
Pot D: 23

How does that get split into those first groups?
I believe pot A/B/C can't play each other and thus can only draw from pot D in the first round.
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Old 07 Feb 2003, 03:12 PM   #5
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Quote:
Originally posted by Canadian_Supporter
11 groups of three and one group of two, which equals 34 teams in the first round.
I realize that--but how do those teams get drawn into said 12 groups?

There's not an even amount from the pots to be thrown into those groups. How many from A or B or C or D can be drawn into a three team group?
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Old 07 Feb 2003, 03:12 PM   #6
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Default 2006 Qualification Process Announced

Quote:
Originally posted by BuffloSoldier
How does that get split into those first groups?
I'm guessing draw 1 from Pot A and 2 from D until Pot A is empty, then 1 from B and 2 from D, then C...

The reason for having 4 pots instead of only 2 must have something to do with the semi-final groupings.

More explaination is obviously needed.

I'm just very glad they kept the final stage as a single group format.
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Old 07 Feb 2003, 03:14 PM   #7
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Pots A, B, and C will be drawn into groups 1 through 12, then Pot D will be drawn. You'll have one team from pots A, B, and C in each group, and two teams from pot D, except for group 12 which will have only one team from pot D.
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Old 07 Feb 2003, 03:15 PM   #8
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My guess is that one team from Pots A, B, and C are put in a group with 2 from pot D (with a bye for the odd group).

The average group might include a seeded team, Surinam, and Aruba.

Sounds like a good time for a Caribbean trip.
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Old 07 Feb 2003, 03:16 PM   #9
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I recall the last time, the three teams that had qualified in for France in 98 were given byes into the semis. Doesn't look like that is happening this time around.

I would assume that the split will be like this:

Pot A: 3 Qualifiers (US, Mexico and CR)
Pot B: The other 3 finalsts (Jamaica, T&T and Honduras)
Pot C: The other 6 semifinalists (Can't remember them all but here goes: Guatamala, Barbados, Canada, El Salvador, ?, ?)
Pot D: Everybody else
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Old 07 Feb 2003, 03:17 PM   #10
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Default 2006 Qualification Process Announced

Quote:
Originally posted by IASocFan
It looks like the sharks - like US, Mexico, and Costa Rica will get a couple of minnows to devour before the semis.
I just did a quick random draw based on how I anticipate this will work...

Group A: Mexico, Netherlands Antilles, Turks & Caicos

OUCH!

Although I did get one group with Haiti, St. Kitts, and Surinam which could be intersting.
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