Just ran a crappy little excel simulation using
these football match goal distributions from 0-0 to 3-3. My simulation is therefore based on random results, where all teams have the exact same chance of winning ie. no teams are better or worst than anyone else. These are crappy simulations, but based on those;
Scenario 0: There is a 13-15% chance of team A losing their first two games, leaving team B and team C to fight it out for first place in the final match. 0-3-3 after two games. No collusion is necessary as Team A are already out.
Scenario 1: There is an 18-20% chance of team A drawing one, and losing one game, leaving them with 1 point, whilst team B and C have 1 point and 3 points. 1-1-3 after 2 games.
Team B and C can collude to draw and send Team A home, regardless of score.
Scenario 2: There is a 6-7% chance of team A drawing their first two matches, leaving them with 2 points, whilst team B and C have 1 point each. 2-1-1 after two games.
Team B and C can collude to have a higher scoring draw to send Team A home. i.e. A 1-1 sends Team A home if Team A had two 0-0 draws so far. A 2-2 draw sends Team A home if they had two 1-1 draws so far.
Scenario 3: There is a 27-29% chance of Team A winning one of their first two games, and losing the other. 3-3-0 after two games. Team B and Team C can collude,
only if Team A's loss is greater or equal to their win. I.e. if Team A won 3-0 and lost 1-0, they are topping the group over all, and cant be caught on goal difference by both teams. If the reverse is true, they will be losing the group with -2 GD (compared to -1 for the team they beat), and
the two teams can collude by allowing the team that Team A beat, to beat the other team by anything less than six goals! There is around a 70% chance of this (Team A losing by more goals than they won by, or having the same GD is both matches) occurring in Scenario 3, although that includes the chance that in both games the goal difference was the same (40%) meaning that, only a result with the same goal difference again, but higher scoring, will send both teams through.
Scenario 4: There is an 18-20% chance that Team A wins one and draws one. 4-1-0 after 2 games. Team A can't be caught by both teams, so no collusion is possible.
Scenario 6: There is a 13-15% chance that Team A wins both games. Team A can't be caught by both teams, so no collusion is possible.
Overall, I calculate there is a 45% chance that Team B and Team C
can collude in their final game to cause Team A to come last. This does include some situations where it would unlikely if they could collude in such a manner, but the possibility is there.
For example:
After a 3-3 draw in Match 1, and a 0-0 draw in match 2, a -4 draw will send Team B and C through - but this would seem to be unlikely
After a 2-0 win by Team A in match 1, and a 3-1 loss for Team A in match B, Team B and C can collude in Match C, team B and C can collude by allowing Team B to win 4-2.
tl;dr There is a 45% chance that collusion is possible in the third and final match in any group.