# Tie breaking in association football championships

I'm pretty sure the algorithm established by FIFA in order to rank teams during 2014 World Cup group stages is ambiguous for the case of three or more teams having equal number of points.

The rule is the following:1

a) greatest number of points obtained in all group matches;

b) goal difference in all group matches;

c) greatest number of goals scored in all group matches.

If two or more teams are equal on the basis of the above three criteria, their rankings shall be determined as follows:

d) greatest number of points obtained in the group matches between the teams concerned;

e) goal difference resulting from the group matches between the teams concerned;

f) greater number of goals scored in all group matches between the teams concerned;

g) drawing of lots by the FIFA Organizing Committee

Say three teams, call them X,Y and Z, are equal based on criteria a), b), c) and d). Then assume step e) puts Z above the other two who still need breaking apart. In other words, counting in only the games involving X, Y and Z, all three teams have equal number of points but Z has a higher goal difference than X and Y, who are equal from that perspective.

This is where it becomes unclear to me. Imagine that Y beat X the one time they played each other. Then, although step d) does not help in ranking the three teams when considering the games involving them, it does separate X and Y when ignoring Z's games. In simpler words :

Z=Y=X for step d) (i.e. when based on all games featuring any two of them)

Z>Y=X for step e)

Y>X for step d) (i.e. when based only on Y vs X, hence Y is better off when ignoring the outcome of Y vs Z)

Then what happens ?

Options are :

1. Having already applied steps d) and e) to all three teams without obtaining a full determination of their rank, we move forwards to step f)
2. Since step e) reduced the number of teams involved in this process, the algorithm is reinitiated at step d)

If we move to step e), do we still count goal difference for X and Y in games involving all three teams since they initiated the algorithm or only the game involving X and Y.

If you are about to comment that steps d) and e) yield the same result when only taking into account the match between X and Y, then imagine that it's a 2 legged championship and that they both win a game with a nonzero aggregate or else that we have four teams of which only one was moved above the other three by step d) and that we are in a group with 5 or 6 teams (e.g. qualifiers).

I can't find the answer anywhere, but this must have happened once in history surely. This is just 2014 World Cup but I've followed football for a while and the rules have always been similar to these ones.

• Y=X for step d. Then how would it be possible for Y>X for step d after factoring in step e? Apr 5, 2017 at 7:36
• You can't go back to step d) after applying step e), instead after applying each step you have to move to the next step, i.e step f) in this case.
– gdrt
Apr 5, 2017 at 11:01
• @CodeNewbie X=Y only when their matches with Z are counted, otherwise Y>X. So presumably X did better against Z than Y did. This is how it can be. Maybe Z should be considered a bunch of teams rather than a single team, because I'm not sure my scenario is possible with only four teams but it definitely is with 5 or 6. Apr 5, 2017 at 15:28
• According to the Wikipedia article History of the FIFA World Cup: "As of the 2014 World Cup, lots have only been drawn once in tournament history. However, they were used to separate second and third place in a group (Republic of Ireland and the Netherlands in 1990) where both were already assured of qualification. Thus, a team has never been eliminated based upon drawn lots." The use of drawing lost is also mentioned in articles about 1990 World Cup. Apr 5, 2017 at 20:09
• Please consider blockquoting, especially in long questions containing a large amount of your own text. Apr 7, 2017 at 2:32

Section 50.5 (a) - (c) states that all teams are first ranked by

• points, then
• goal difference, then
• goals scored.

If any teams are level after applying all three criteria in this order, section 50.5 (d) - (f) states that the matches played between the equal teams are then ranked by (once again):

• points, then
• goal difference, then
• goals scored.

The simplest way to do this would be to construct a league table from the matches played between the level teams. As with a full league table used in (a) - (c), this reduced table would be ordered on points, then goal difference, then goals scored.

Z=Y=X for step d) (i.e. when based on all games featuring any two of them)

Z>Y=X for step e)

Y>X for step d) (i.e. when based only on Y vs X, hence Y is better off when ignoring the outcome of Y vs Z)

the third step is the error. The procedure states that (d) - (g) are applied to all teams that are

equal on the basis of the above three criteria [(a) - (c)]

It does not state that a level team's matches are excluded with the process restarting if they are ahead on one of the criteria. By restarting the process with just two of the teams at (d), you would not be including all teams level on (a) - (c), and this leads to an erroneous ambiguity.

In this example (if it is even mathematically possible), goals scored in matches between X, Y, and Z would be used to separate X and Y, since X, Y and Z are all level on the basis of (a) - (c).

As an aside, if all criteria are exhausted, in accordance with section 50.5 (g), lots will be drawn to separate the teams.