If it was a one-game race, meaning the Mets were one game ahead of both the Reds and the Diamondbacks on Sunday morning, there are three games to be played, and eight possible outcomes: 1. Mets win, Reds win, Diamondbacks win 2. Mets win, Reds win, Diamondbacks lose 3. Mets win, Reds lose, Diamondbacks win 4. Mets win, Reds lose, Diamondbacks lose 5. Mets lose, Reds win, Diamondbacks win 6. Mets lose, Reds win, Diamondbacks lose 7. Mets lose, Reds lose, Diamondbacks win 8. Mets lose, Reds lose, Diamondbacks lose Numbers 1, 2, 3, 4, and 8 put the Mets in the playoffs, which makes their chance 62.5 percent. (Five out of eight.) With five games remaining, and 32,768 possibilities, it's seemingly more complicated, but the same rules apply.