So aside from that, like I said there are 27 remaining combinations of outcomes (3 x 3 x 3) - We wind up in the #1 WC slot in 22 of them, 20 of them free and clear of both teams plus two others where we're tied w/the Giants for slot #1 which is the same as winning outright since we hold the tie-breaker. ODDS = 89.07% - In two other cases we wind up with WC #2 all to ourselves: Mets lose twice + Giants win twice + Cards either lose once or lose twice. ODDS = 4.69% - There's the one case mentioned above where we tie with the Cards for WC #1 with SF out of it: ODDS = 1.56% So that's 25 of the 27 cases where we're assured of playing one game for the right to meet the Cubs. TOTAL ODDS = 95.32% That leaves the two we don't want to see where we'll be forced into a Monday play-in or maybe even a Monday/Tuesday play-in: - Mets lose both, Giants win both, Cards win both: This puts us in a play-in vs StL for WC #2. ODDS = 1.56% OR - Mets lose both, Giants win once, Cards win both: that outcome throws us all into a 3-way tie and multiple play-in games which I don't even know how they'd be set up. ODDS = 3.13% Subtle error here in the calc - assuming 50% probability of a win of any given game, each team has a 25% likelihood of winning 0 games, a 50% likelihood of winning 1, and a 25% likelihood of winning 2. The calc above doesn't take into account that a team is twice as likely to win exactly one game than zero or two. There are four possible outcomes for each team - win none (one outcome), win one (two outcomes), or win two (one outcome). So 4^3 = 64 outcomes in total. The only outcomes under which the Mets don't win the wild card are these: Cards win 2, Mets win 0, Giants win 1. 2 out of 64 ways that this can happen (1 way for the Cards to win 2 * 1 way for the Mets to win 0 * 2 ways for the Giants to win 1). If this happens, there's a 3-way tie [crossout]and the Mets still have only a 1/3 probability of not winning a wild card. So the probability of Mets losing the WC this way is (2/64)*(1/3) = 2/192[/crossout] OE - in a 3 way tie, the Mets have a 1/4 probability of not winning a WC, because of the way the tiebreaker works. So prob of Mets losing this way is (2/64)*(1/4) = 2/256 Cards win 2, Mets win 0, Giants win 2. Only one way that this can happen, and it leads to the Mets and Cards tying for WC2, so there's still only a 1/2 probability that the Mets lose the WC even if it were to happen. So the probability of the Mets losing the WC this way is (1/64)*(1/2) = 1/128. The outcomes are mutually exclusive, so you can add them. [crossout]2/192[/crossout] 2/256 + 1/128 = [crossout]7/384[/crossout] 1/64 probability that the Mets don't get the WC. So there's a [crossout]377/384 = 98.2%[/crossout] 63/64 = 98.4% probability that the Mets do get the WC. I'm not motivated enough to figure out the probabilities of WC1 and WC2. The real probability is much higher, of course, because of strength of opponents (also, the Dodgers still have something to play for, and the Pirates and Phillies don't).