Quote:
Originally Posted by AESP_pres
Did that ever happened in the real Major League history to have more than two teams with the same win / loss record in a league?
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Don't think so, but that doesn't mean that it wasn't a possibility. In, for example, the early 1900's there were lots of rainouts & ties, with many games not made up. So it was common that the pennant contenders - as the season approached conclusion - were going to end up playing a different number of games, meaning it would be unlikely for two teams to have identical records at season's end, and even more unlikely that three or more teams could end up tied.
Since the 1940's, though, there were instances where three-team and four-team ties were a possibility, and schedules were drawn up - just in case - for some of those...
Generally speaking, for a three-team tie the format would depend upon what the format was for a two-team tie:
- If a two-team tie was a one-game playoff, then the league would typically draw lots to determine teams A, B, and C, and then team A would play at team B, with the winner hosting team C.
- If a two-team tie was a best-of-three, then the league would have a round-robin, double-elimination tournament of sorts.
For four-team scenarios, a lot draw would determine teams A, B, C and D, and then A would play at B, and C would play at D, and then the winners meeting. These would either be one-game playoffs or best-of-three, depending upon what the league's current two-team tiebreaker procedure was.
I believe the only time a 5-team tiebreaker wwas a possibility was in the 1973 NL. I'm not sure what you'd call the format, but - presumably after a drawing of lots - the schedule was reported as follows:
Tue 10/2: Pitt @ NYM, Mont @ StL; bye for Chic
Wed 10/3: Chic @ Mont/StL winner
Thu 10/4: NYM/Pitt winner @ winner of Wed game