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OK, the Mets have to win their last seven games. It happens. Doesn't happen much, though. Putting aside hard to measure items like home-field advantage and momentum, your average team (playing your average opponent) has a 0.78% of winning their next seven games at any given time.
But this isn't your average team; nor are these your average opponents. When taking into account the Mets winning percentage and their opponents' losing percentages, the chances of the Mets taking the next seven over your Philadelphia Phillies and your Colorado, the chances of the Mets sweeping to the finish is up to 0.99101%.
Wow! Almost 1%!
But, you know, we need a little help from our friends, as we need the Astros to lose their last six. Met fans can take some small comfort in knowing that, if we win tonight, our best friends on Tuesday and Wednesday will be the St. Louis Cardinals, and they are good friends to have. After that, they take on the Cubs, who just stole a victory from them yesterday, and, if the Astros come in on the rebound from a small sweep by St. Louis, will definitely have the 'Stros on their heels. Taking into account the current losing percentage of the Astros and the winning percentages of St. Louis and Chicago —but not home-field advantage, momentum, or who's pitching — this occurrence has a 1.45910% chance of coming about. Whoo-hoo!
Now, these two occurrences con-curring — now we're in tricky territory. Multiply those two percentages by each other and we are at 0.01446%. Um, OK, yeah, we can do that. I used to bullseye womprats in my T-16 back home, and they were no bigger than that.
Of course, there's more. There's often more. Grrr. The Phillies can only win one more game of their remaining six. The good news there is that, in winning out, we'll have already knocked off three of those games. Beyond that, they'll have to come into RFK Stadium and lose two of three to close the year. And that's going to be in adverse circumstances, for I will be in attendance. A sweep would be nice. That would have an 11.12846% chance of occurring. But I don't want to ask for too much. Taking, as usual, current winning and losing percentages into account, the chances Nats taking at least two of three is thrice that — 33.38539%. Of course that still makes the chances of this whole scenario working out two thirds less likely. All this taken into account, we currently have a 0.00483% chance of surviving the season. That's one tiny womprat.
Lastly are Jack McKeon's pesky (but reeling) Marlins. They retain six games on their schedule, and we can't afford them winning more than five. Not bloody likely, but a likelihood we have to account for in that it's small possibility affect our highly improbable march. The chances of them losing at least one game against Washington and Atlanta are a strong 98.22078%.
That whittles our chances of pulling this whole act off down to 0.00474%, only a little less rosy than the pre-Florida calculation. I’m seeing this as about a 20,703-to-1 shot.
Let's go Mets.
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