What a fairer 24 team UEFA Euro could look like

2.2.1Arbitrariness of the placement of the third-placed teams in the bracket

The somewhat strange, asymmetric values of the probabilities p_l and p_r (see Table 3) result from the arbitrary placement of third-placed teams described in the left hand side of Table 1. However, remember that there exists other allocation procedures that respect the group diversity constraint, reported in the right hand side of Table 1. There are 2¹⁶ = 65, 536 deterministic allocation procedures that respect the group diversity constraint (2 acceptable allocation rules for each of the 15 lines in Table 1, except for the 12th line, for which 4 allocation rules are acceptable). Running a simple computer code shows that among those 65,536 allocation rules, exactly 1,000 are such that p_l = p_r for all groups.

A symmetric allocation where p_l = p_r is desirable as it ensures that teams 3B, 3C, 3D, and 3F have a 50% chance of ending in an ‘easy’ quarter of the bracket, i.e., a quarter with 2 runners-up, and a 50% chance of ending up in a ‘tough’ quarter of the bracket, i.e., a quarter with 2 group winners. Team 3A can only end up in a ‘tough’ quarter, while Team 3E can only end up in an ‘easy’ quarter, therefore for these 2 teams the values of p_l and p_r have no impact on W₃ and A₃. Once the effect of the asymmetry of the allocation of the third-placed teams has been removed, we can see more clearly the impact of the global structure of the bracket (see Section 2.2.2).

UEFA happens to have arbitrarily picked one of the 64,536 asymmetric rules. Arbitrarily picking one of the 1,000 symmetric rules would already be an improvement, but it would still be an arbitrary choice. A symmetric, non-arbitrary, more natural allocation procedure would be the following: once the 4 best third-placed teams are known, one of the two (or four) admissible allocation rules would be drawn, uniformly. It is easy to check that in this case the new probabilities p_l and p_r, denoted pl′ and pr′ , are the same for all groups, equal to 0.5, which means that if a third-placed team advances to the round of 16, then it has a 50% chance of being placed in the left half of the bracket, and a 50% chance of being placed in the right half of the bracket, whatever the group it comes from. For instance, looking at the first 10 lines of Table 1, one can compute ℙ(3A plays 1B|3A qualifies) = 110(pABCD + pABCE + pABCF + ⋯+pADEF)=110(0 + 12 + 1 + 12 + 1 + 1 + 0 + 0 + 12 + 12) = 12

where p_X is the probability that 3A plays 1B given that the 4 best third-placed teams come fromGroups X.

The new corresponding values of W₃, W, A₃, and A are denoted with a prime and are also given in Table 2. A comparison of W and W′ (resp. A and A′) is given in Fig. 6. Some symmetry would be restored: Groups B, C, D, and F share the same value of W3′ , and the same value of A3′ . As a consequence, the worst case advantage W′ would order groups asfollows:

Fig.6

Comparison of W and W′ (left) and of A and A′ (right).

As for the teams ordering, it would be the same as under W and A, except that now we would have 3B = 3C = 3D = 3F. Therefore, the fact that W orders C > D > B, and that A orders C > B and F > D, is only due to the arbitrary allocation of third-placed teams, not to the arbitrary global structure of the bracket. Considering the ‘prime’ version allowed us to disentangle between the arbitrary global structure of the bracket, and the arbitrary allocation of third-placed teams. It will also prove useful when we compare various global bracket structures, as we now show.

2.2.2Arbitrariness of the global structure of the bracket

Let us consider the following three advantages for a given group:

There are exactly 5 admissible global structures, which are reported in Table 5. Structure 6 (also reported) is not admissible since no placement of third-placed teams satisfies the group diversity constraint: as proved in the appendix, with Structure 6, we cannot rule out that one quarter of the bracket have two teams from the same group. Structure 6 would become admissible if one slightly relaxed the group diversity constraint by allowing at most one repeated matchup, involving a third-placed team, in quarterfinals.⁴. The global structure that was used by UEFA for the Euro 2016 is Structure 4, in which 1 group gets the three advantages (Group A), 1 group gets AdvW3 and AdvQF only (Group D), 2 groups get AdvW3 and AdvRR only (Groups B and C), 1 group gets AdvRR only (Group F), and 1 group gets no advantage at all(Group E).

Note that for all admissible global structures of the bracket, at least one group gets all 3 advantages. Note also that Structures 3, 5, and 6 leave no group without advantage. However, this implies a lack of win incentive for both Groups Y and Z, since in those structures both their winner and runner-up would play against a runner-up in the round of 16 (see Section 2.3).

For a given group, the values of the worst case advantages W₁, W₂, and W3′=(W3l+W3r)/2 depend only on the combination of advantages that the group was awarded. The same is true for the average advantages A₁, A₂, and A3′=(A3l+A3r)/2 . Since we aim at measuring the impact of the global structure of the bracket on group advantage, regardless of the allocation of the third-placed teams, we have assumed a symmetric allocation of the third-placed teams, i.e., p_l = p_r = 0.5 (see Section 2.2.1). In Table 6 we report the values of the worst case advantage and average advantage for a given group, as a function of the various possible combinations ofadvantages.

Fig. 7 is then derived from Tables 5 and 6. For each global structure of the bracket (from top to bottom: Structure 1 to Structure 6), it shows how the worst case advantage (left) and the average advantage (right) are distributed over the 6 groups. In Table 7 we report the mean, standard deviation, and range of those 12 distributions.

Fig.7

Distribution of the worst case advantage W′ (left) and the average advantage A′ (right) of the 6 groups, for the 6 global structures of the bracket. From top to bottom: Structure 1 to Structure 6. Structure 6 has the smallest standard deviation of group advantage.

Of course, all structures have the same mean. However, standard deviations and ranges vary. A small standard deviation of group advantage is a desirable feature, as it increases fairness by minimizing the impact of group advantage. The global structures that have the smallest standard deviation and range of group advantage (both for worst case and average advantage) are the structures in which the 2 groups that do not benefit from AdvW3 (Groups Y and Z in Table 5) benefit from AdvRR (Structures 3, 5, and 6; Structure 6 yields the smallest standard deviation and range of group advantage). This means that all groups benefit from at least one advantage. However, this raises an issue of win incentive, as we discuss in the next section, since in Groups Y and Z both the winner and the runner-up play against a runner-up in the round of 16.

Conversely, the global structure that does not suffer from this win incentive issue (Structure 1, in which the 4 groups that benefit from AdvRR also benefit from AdvW3) is the structure that has the largest standard deviation of group advantage (both for worst case and average advantage): in this structure 2 groups benefit from the 3 advantages, and 2 groups benefit from none of those. This illustrates that, if one considers only balanced brackets that respect group diversity and are based only on group ranks, one faces a trade-off between minimizing group advantage and maximizing win incentive.

Among the 2 other global structures of the bracket (Structures 2 and 4), Structure 4, the one chosen by FIFA for the 1986, 1990, and 1994 World Cups⁵ and by UEFA for the Euro 2016, has the smallest standard deviation.

Based on this work, in February 2018, UEFA has adopted Structure 6 for the Euro 2020 (see articles 21.04 and 21.05 in UEFA Euro 2020 regulations (2018)). This means that in 2020 all groups will benefit from at least one advantage and group advantage will be minimized. The small price to pay for this is that for two groups (Groups A and D) both the winner and the runner-up will play against a runner-up in the round of 16, and that there is a 10% chance that one quarterfinal will oppose two teams who were in the same group (see footnote 4).

Finally, when we compare Table 2 with Table 7 (or Fig. 7), we observe that most of the variability in group advantage results from the global structure of the bracket, not from the allocation of third-placed teams.

3.1.1The suggested procedure

A first solution consists of slightly distorting the ideal bracket (Fig. 8) in a deterministic way. Here is what we suggest. The position of each group winner (teams 1 to 6) in the bracket is kept intact. Then a rearrangement of the 6 runner-up positions and a rearrangement of the 4 third-placed team positions are performed as follows:

3.1.2An example

We recall the result of the draw of the group stage of the Euro 2016 in Table 8. Table 9 shows what the team rankings from 1 to 16 would have been at the end of the group stage of the UEFA Euro 2016.

Note that in this example the ideal bracket (reported in Fig. 9) does not satisfy the group diversity constraint:

Fig.9

Ideal bracket if we use the Euro 2016 rankings reported in Table 9. It does not satisfy the group diversity constraint.

Fig. 10 shows what the bracket would be in this example if we follow our suggested deterministic procedure. First, France, Germany, Croatia, Wales, Italy, and Hungary are placed in positions 1, 2, 3, 4, 5, and 6 of the ideal bracket. The three right runners-up come from the groups of France, Wales, and Italy (the left hand side group winners): Switzerland, England, and Belgium. In our example, Switzerland is the lowest ranked runner-up after the group stage among those 3, so it goes to position 11 of the ideal bracket (against Hungary, the lowest ranked group winner). The other two runners-up, England and Belgium, go to positions 7 and 10. Symmetrically, Iceland, the lowest ranked left runner-up, goes to position 12 (against Italy), and Poland and Spain play against each other in positions 8 and 9. The 4 best third-placed teams come from Groups B, C, E, and F. Since the upper right quarter would have contained England (Group B), Germany (Group C), and Belgium (Group E), the only possible opponent for Germany would have been Portugal (third of Group F). This would probably have been described as a strong quarter, even though it would have contained only one group winner. Since the lower left quarter would have contained Wales (Group B) and Italy (Group E), the only possible remaining opponent for Wales would have been Northern Ireland (Group C). Eventually, the third-placed teams from Groups B and E (Slovakia and Ireland) could have been placed equally in the two remaining quarters of the bracket (upper left and lower right). To foster win incentive, Ireland, the lowest ranked of the two remaining third-placed teams, would have been placed against France, the best ranked team among the two remaining group winners, therefore Slovakia would have been placed against Croatia.

Fig.10

Bracket produced by our suggested deterministic procedure if we use the Euro 2016 rankings reported in Table 9. By construction, the group diversity constraint is satisfied.

3.2.1Case 1: {G₇, G₈} ≠ {G₁, G₄} and {G₇, G₈} ≠ {G₂, G₃}

In this case, let us consider the bracket of Fig. 11. We set X₁ = 1, X₂ = 2, X₃ = 3, and X₄ = 4, and we will draw a combination of teams X₅, …, X₁₆ that satisfies the following constraints:

Using a simple computer code, it is easy to check that for any possible sets of groups {G₇, G₈} and {G₁₃, G₁₄, G₁₅, G₁₆}⁸ such that {G₇, G₈} ≠ {G₁, G₄} and {G₇, G₈} ≠ {G₂, G₃}, the number N of combinations of teams X₅, …, X₁₆ that satisfy all the constraints above belongs to {6, 8, 10, 12, 16, 18, 20, 24}. This means that, when the group stage is over, teams 1 to 16 would be known, corresponding to particular sets {G₇, G₈} and {G₁₃, G₁₄, G₁₅, G₁₆}, and if {G₇, G₈} ≠ {G₁, G₄} and {G₇, G₈} ≠ {G₂, G₃}, then the exhaustive list of the N admissible brackets (i.e., admissible combinations of teams X₅, …, X₁₆) would be published, and one of the N brackets would be randomly drawn, uniformly. Section 3.2.3 below provides an illustrating example.

3.2.2Case 2: {G₇, G₈} = {G₁, G₄} or {G₇, G₈} = {G₂, G₃}

The only situation where N = 0 is when {G₇, G₈} = {G₁, G₄} or {G₇, G₈} = {G₂, G₃}. Indeed, if {G₇, G₈} = {G₁, G₄}, for any permutation (X₇, X₈) of teams (7, 8), the runner-up X₈ must come from the same group as that of a group winner (team 1 or team 4), in the left half of the bracket. Symmetrically, if {G₇, G₈} = {G₂, G₃}, for any permutation (X₇, X₈) of teams (7, 8), the runner-up X₇ must come from the same group as that of a group winner (team 2 or team 3), in the right half of the bracket.

In this situation, we suggest to set X₁ = 1, X₂ = 2, X₃ = 4, X₄ = 3, X₅ = 5 and X₆ = 6, i.e., compared with Fig. 11, we permute teams 3 and 4, and additionally we can impose that X₅ = 5 and X₆ = 6 (see Fig. 12). Then for any possible set of groups {G₁₃, G₁₄, G₁₅, G₁₆}, the number N of combinations of teams X₇, …, X₁₆ that satisfy all the constraints above belongs to {8, 10}. Like in case 1, we would then simply draw one of the N admissible brackets randomly, uniformly, to decide the final bracket.

3.2.3An example

We use the example of the Euro 2016 team rankings after the group stage given in Table 9. In this example, {G₁, G₄} = {A, B}, {G₂, G₃} = {C, D}, and {G₇, G₈} = {C, D}, so we are in Case 2 (Section 3.2.2 and Fig. 12). There are N = 8 admissible brackets, all reported in Table 10. Note that, as Germany is in the right half (in position X₂ = 2) and Croatia is in the left half (in position X₄ = 3), Poland, who is Team 7 and comes from the same group as Germany, can only take position X₈ (in the left half), and cannot take position X₇ (in the right half), and Spain (Team 8) can only take position X₇. As a consequence, the 8 admissible brackets all have X₇ = 8 (Spain) and X₈ = 7 (Poland).

Since France (Group A) and Italy (Group E) are on the left hand side in positions X₁ = 1 and X₅ = 5, Switzerland (Team 12, Group A) and Belgium (Team 9, Group E) can only be on the right hand side, in positions X₁₀ or X₁₁. Symmetrically, as Wales (Group B) and Hungary (Group F) are on the right hand side in positions X₃ = 4 and X₆ = 6, England (Team 11, Group B) and Iceland (Team 10, Group F) can only be on the left hand side, in positions X₉ or X₁₂. For each placement of Switzerland and Belgium (2 possibilities), and each placement of England and Iceland (2 possibilities), there are two acceptable positions for the four third-placed teams. This leads to a total of 8 admissible brackets.

Right after the last matches of the group stage are over, a computer would give this exhaustive list of the 8 possible brackets. This list would be numbered, from 1 to 8, and a random number from 1 to 8 would be drawn to decide the final bracket of the Euro.