The matchematical traps
A.) As long
as there is no draw in the group matches, the following three special
cases are theoretically possible (it depends on the particular results
who beats who):
-1) A team with 6 points (2
wins, 1 loss) could finish only third because there are three teams with
6 points.
The
famous A beats B, B beats C, and C beats A while all three beat D.
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-2) A team with 3 points (1
win, 2 losses) could make it to the second round because three teams finish
with 3 points.
A beats B, B beats C, and
C beats A while all three lose to D.
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-3) The Group is already decided
after two matches played, when two teams that meet themselves in the third
encounter already have collected six points from their first two matches.
(decided refers to who advances)
To avoid a disappointing
situation, the African Cup Of Nations has tried a new mode: The succession
of the matches had not been determined by the draw but it had been fixed
that the two top seeds would meet in the first match. This created an exciting
discourse with a favorite behind from the start and an outsider from the
other opening match with a good position and indeed kept all four groups
open.
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-4) The 3-point-regulation
implies a new problem: A team can not only be qualified after their first
two matches but also can have clinched the first place for sure. This happens
if the other two results have been draws. This might lead to a let-down
in the third match and might give an advantage to the team who meet this
favorite in the third match.
This situation in reverse
to the one before seems less likely to be avoided by the mode of the last
African Cup Of Nations: Here it is the 3 point rule to blame although it
seems to do good work elsewhere.
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B.) In reverse
- if there is a draw the group will not be
completely decided before the third match as a team that has gained one
point has always a theoretical possibility to advance. Also it can be said
that a team with two wins here in all cases will advance, a team with two
losses in any case will have to go home.
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C.) Unbeaten
but out When there have been draws some special
cases are possible in which an undefeated team has to walk home or a winningless
team may go through:
-1) Three teams finsish with
5 points (1 wins, 2 draws)
A
draws with B, B with C, and C with A while all three beat D.
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-2) Three teams finsish with
2 points (1 loss, 2 draws)
A
draws with B, B with C, and C with A while all three lose to D.
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-3) One team draws with all
the other three. This leaves the following options for the remaining three
matches between the other teams:
-3a) The most spectacular case:
All matches end with a draw - all teams finish with 3 points, 2 go home,
2 advance, all unbeaten, all winningless.
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-3b) The other matches see
2 draws and one sole win. This leaves 2 teams with 3 draws=3points. The
other 2 teams have 2 and 5 points. So of the unbeaten teams with 3 points
one advance, the other one goes home.
example:
A draws with B, C with D. A draws with D, B and C draw too. A draws with
C, and B beats D. This results as follows: B 5pts., A 3pts., C 3pts., D
2pts.
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-3c) One team draws with all
three opponents while the other matches see exactly one more draw: Two
cases are possible, in the first one the team with the three draws advances
in the other they have to go home.
i)
A draws with B, C, D. B and C draw also. B and C both beat D. B and C advance.
ii) A draws with B, C, D. B and C draw also. B and C both lose to D. A
and D advance.
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-3d) One team draws with all
three, but the other matches all have a winner. This in any case means
the drawing team is out.
The
final table is either 7-4-3-1 or 4-4-4-3.
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D.) The closest
table of all other options is when each team
has one win, one draw, and one loss.
This is
only possible if the 2 draws happen on the same gameday (means in both
1st, or both 2nd, or both 3rd matches) while the other two gamedays see
only wins/losses.
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