| Revision as of 12:28, 19 June 2006 editCarcharoth (talk | contribs)Administrators73,550 edits →Three-way ties - problem?← Previous edit |
Revision as of 12:38, 19 June 2006 edit undoCarcharoth (talk | contribs)Administrators73,550 edits →Three-way ties - problem?Next edit → |
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::* Switzerland 3-2 Togo; Switzerland 2-2 South Korea; France 4-3 Togo |
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::* Switzerland 3-2 Togo; Switzerland 2-2 South Korea; France 4-3 Togo |
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* Switzerland 3-2 Togo; Switzerland 3-3 South Korea; France 5-4 Togo |
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* Switzerland 3-2 Togo; Switzerland 3-3 South Korea; France 5-4 Togo |
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* Switzerland 3-2 Togo; Switzerland 4-4 South Korea; France 6-5 Togo |
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* Switzerland 3-2 Togo; Switzerland 4-4 South Korea; France 6-5 Togo |
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It is possible that a three-way tie could occur in Group G. If Switzerland beat Togo today, and then France beat Togo while Switzerland and South Korea draw (all three results are quite likely), then three teams will finish tied for 1-3 place in the group.
The problem I foresee, from looking at the FIFA tiebreak criteria, is that they do not say what happens if a three-way tie is only partially resolved when applying criteria d, e, or f (the ones that look only at the points, goal difference and goals scored between the teams in the tie). Eg. One team has inferior points/GD/GS to the other two teams, but the other two teams are still tied on points/GD/GS. Do you then move on, say, from criteria d to e, and still use the values for the three-way tie, or do you recalculate the values for criteria d/e/f for a 2-way tie (effectively reverting to criteria d, as criteria e/f are irrelevant for a 2-way tie)?
The same would apply for a four-way tie, except for the fact that a four-way tie is no longer possible at this World Cup (in Group H, Spain drawing or winning one of its last two games, or losing them both - these conditions cover all possibilities - avoids a 4-way tie). Thankfully. Carcharoth 11:54, 19 June 2006 (UTC)