I’m not sure how it would work, as after total number of wins, the tie breaking procedure goes with which team “has the higher winning percentage in all games played against all of the other tied Club(s)”. In this scenario, the four teams would have the following records and winning percentages against the other tied teams:
Hamilton - 1-0-0 - 100%
Ottawa - 1-1-0 - 50%
Toronto - 0-1-0 - 0%
Montreal - 0-0-0 - TILT!
With no games against the other tied teams, Montreal will have no winning percentage. So would they be in first? Or last? Maybe, like Schroedinger’s Cat, they’d be in both first AND last?
But no matter what, Toronto would still be behind both Hamilton and Ottawa. :rockin:
The tie-breaking procedure only comes into play when teams are tied in the standings. Toronto at 1-4 is not tied in the standings with the 1-3 clubs. So they are not part of the tie-breaking procedure; they are fourth.
Between Hamilton, Montreal and Ottawa, only Hamilton has a win against the other two clubs. Hamilton is therefore in first.
Between Ottawa and Montreal, Ottawa has the higher net aggregate. Ottawa is therefore in second.