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Can anybody explain wild card . . . ?
The paper this morning said that if the Mets had won yesterday, they would then have had two ways to make the postseason: (1) defeat the Phillies in the one-game playoff for the division title, or (2) enter as the wild card, since they would have ended tied with San Diego and Colorado for the wild card spot, all with 73 losses.
But if the Mets had won yesterday and then lost to the Phillies in the one-game playoff, they would have 74 losses. Either Colorado or San Diego, after their one-game playoff, would end with only 73. So how could the Mets have entered via the wild card route? Or is the paper wrong? Or does the one-game playoff not count in the final standings? |
Right: the playoff would not have counted as a regular season win or loss, and so there would have been a 3-way tie for wild card.
In that case, Bud Selig plays "eenie-meenie" to pick the wild card. It's actually in the rules. |
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Thanks. Not counting the one-game playoff game explains it. I wonder how they would have dealt with 3 teams in a tie.
Those extra games used to count. The final National League standings of 1951 show the Giants 98-59 and the Dodgers 97-60, with the 157-game totals reflecting the 3-game playoff. Thomson's home run counted on his regular-season records. Does Selig actually just select a team, mbyron? (Incidentally, you should know that an airline was sued because a flight attendant said something like, "There are plenty of open seats. You can simply do eenie-meenie-miney-moe." So prepare for a demonstration outside your house.) |
Those extra games counted as early as 3 years ago. The rule that you don't count it is brand new.
And yes - if we'd had two ties for the division, they'd have each played for their respective division titles, with the losers playing the next day for the wildcard. As recently as Friday, we had the possibility of a 5-way tie, which would have taken 3 days to resolve. |
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