Breaking down the Cavs’ play-in odds for each seed

Breaking down the Cavs’ play-in odds: Claiming Seventh

The most likely spot for the Cavaliers remains seventh place, their current spot…but just barely. According to PlayoffStatus.com, the Cavs have a 42 percent chance of landing at seventh. That’s more than any other single spot, but overall it’s more likely the Cavs drop than stay at the top of the four-team tournament.

The single most important game to determine where the Cavs fall is likely to be Friday night’s showdown in Brooklyn against the Nets, who are just a game back of Cleveland. The winner of that game will secure the tiebreaker. If Cleveland loses they will be tied with the Nets, who play the moribund Indiana Pacers in the last game of the season; essentially, lose Friday night and the seventh seed is gone. If Cleveland wins, they will be 2 games up and have a great chance at hosting a play-in game.

There is also Atlanta to pay attention to in the race for seventh. They are just one game back of the Cavs and own the tiebreaker. If the Cavs beat the Nets but lose to the Bucks, Atlanta could knock them out of seventh by winning both of their remaining games: Friday in Miami, and Sunday in Houston. Charlotte has no functional path to the seventh seed, although it’s technically still possible if someone goes back in time and steps on a butterfly, or something like that.

Win out, and it’s the seventh seed. Beat Brooklyn and lose to Milwaukee and Atlanta has a small chance to catch them. Lose to Brooklyn and beat Milwaukee, and the Cavs would need the Nets and Hawks to lose at least one game.