Private Practice Season 6 Casts Alfre Woodard and Matt Long

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Private Practice has a busy season ahead of it, what with the question of whom Addison (Kate Walsh) is going to choose to be with still up in thee air – will it be Sam (Taye Diggs) or Jake (Benjamin Bratt)? Will she say yes to Sam’s proposal, or is Jake the man for her? Season 6 will hopefully give us some kind of answer in its Sept. 25 premiere, even if it happens she chooses neither of them. Personally, I’m leaning toward thinking she chooses neither; based purely on the fact Kate Walsh is leaving Private Practice this season after 13 episodes. Still, she might choose one of them, and some casting announcements may shed some light on the ultimate direction Addison will choose.

According to TVLine, Sam’s mother has been cast, with four-time Emmy Award winner Alfre Woodard taking on the part. She’ll make her appearance in a November episode, suggesting she might be around for Sam’s engagement. To be honest, I just don’t buy that. Unless Sam and Addison are getting married within a few short months after their engagement, why should his mother be hanging around? No, I think she’ll be introduced for another reason. I’m still leaning toward the idea Addison will leave the show having chosen neither Sam nor Jake, and here’s why.

Now, this might be a long shot but TV Guide reports Private Practice has tapped Matt Long to recur as a new doctor at St. Ambrose. Not much is known about him except he’s playing a young, attractive doctor entering as Addison is leaving. Because this is a hospital drama, love will almost certainly be in the air for Long’s new character, so why not have him enter into a relationship with Addison? I know it sounds a little crazy – he’s so new after all – but how can Addison be in a relationship with any of the main characters if she’s just going to be leaving? It would make more sense to involve her with a new recurring character that could then leave with her. Addison could leave happy and Private Practice wouldn’t lose another cast member.