Whoa. Since the order of victories doesn't matter, there are effectively 64 possible outcomes. Each of the three teams has four possible outcomes - they win 0, 1, 2, or 3 games. 4³= 64. Of those, There are 16 combinations where the Mets win 3, and the Mets clinch in all There are 16 combinations where the Mets win 2. The only scenarios where the Mets don't clinch are the four where the Reds win 3 (and the DBacks win 0, 1, 2, or 3). That leaves 12 where the Mets clinch. 16 combinations where the Mets win 1. Of those, the Mets clinch where the Reds win 1 or 0 and the DBacks win 0, 1, or 2. That's 6 combinations. Where the Mets win 0, they only win the WC if the Reds also win 0 and the DBacks win 0 or 1. That's 2 combinations. So, out of 64, 16+12+6+2 = 36 where the Mets win the WC. So, if we assume 50-50 for each win, that's a 56.25% probability. If we assume that each team has a slightly higher probability of winning because they'll be more willing to use rather than rest better players, or simply that in the Mets' case, they're playing a weaker team, that favors the status quo and increases the probability that the Mets win the WC. Not showing work on that, but picture the extreme cases where the win probabilities were 0 or 100%. Those would guarantee the status quo.