Sister Wives viewers vote: which of Kody Brown’s remaining wives will be the last wife standing?

Sister Wives star Kody Brown and his remaining wives Meri, Janelle, and Robyn
Sister Wives viewers voted on which one of Kody Brown’s remaining wives will be the last one standing. Pic credit: TLC

Since his split from Christine Brown, Kody Brown has three wives remaining, and Sister Wives fans voted for which of the three will remain the last wife standing.

As Sister Wives fans are aware, Kody’s third wife Christine left him and their plural marriage, making it Instagram official in November 2021.

Since Christine’s departure, Kody is left with the task of keeping three wives satisfied now instead of four — his first wife Meri, his second wife Janelle, and his fourth wife, Robyn.

Kody and Meri (who was just 19 at the time) wed in 1990 and welcomed a second wife into their marriage, Janelle, in 1993. Christine entered the plural marriage in 1994 and Kody didn’t take on another wife until 2010 when he spiritually wed Robyn.

Meri and Kody legally divorced in 2014 to allow him to legally wed Robyn in order to adopt her three children from her previous marriage.

Sister Wives fans vote: ‘Who will be the last wife standing?’

Now that Christine has left the polygamous union, Sister Wives viewers are curious about which of Kody’s three wives will stick around the longest.

In a Reddit post titled “Who will be the last wife standing?” Sister Wives fans took part in a poll and voted between Meri, Janelle, and Robyn.

“Now that the tell all is done, who do you think will be the last wife left?” the post read. “Will Meri be his movie and sleep over on the couch buddy for life? Will Sobyn get to keep her Legal Wife title or do tears bounce off of Teflon too?”

sister wives fans voted on reddit for which of kody's wives will be the last one standing
Pic credit: u/SwissCheese4Collagen/Reddit

Sister Wives fans explain their choices: Meri, Janelle, or Robyn?

At the time this story was published, Robyn was in the lead, with 450 of the 654 total votes. Meri came in second place with a respectable 143 votes, and Janelle was voted the least likely to stick around with just 61 votes.

Some of the voters took to the comments section to explain their choice.

“Janelle is my bet,” wrote one Redditor. “Meri will eventually just live full time running the B&B, and Sobyn will self-destruct and leave on her own when the show is finally done.”

During Season 16 of Sister Wives, Janelle revealed to Robyn that she had considered leaving Kody, but ultimately decided it was too easy to give up. If Sister Wives fans had it their way, Janelle would be the next of Kody’s wives to leave him.

sister wives voted on reddit for who will be kody brown's last wife standing
Pic credit: u/SwissCheese4Collagen/Reddit

Another Sister Wives fan voted for Meri after initially considering Robyn and explained why: “If you had said ‘real’ wife I would have voted Robyn. But Meri is in for the long game. She will probably out live them all just to prove her devotion.”

Meri has made it clear that despite Kody’s lack of romance and sex, his mistreatment, and admitting to having a platonic friendship rather than an intimate marriage, she isn’t going anywhere.

“If I quit, if I walk away, then it’s not… it’s not going to get better and I’m not going anywhere y’all,” Meri said last season. “You’re stuck with me, whether you like it or not.”

One Redditor who voted for Robyn commented with their reason for voting for Kody’s fourth wife: “Robyn has no income or a way to earn without TLC. She isn’t going anywhere.”

Robyn admitted that her marriage to Kody is the only one of his that’s “fully functioning.” Robyn is Kody’s only legal wife, and Sister Wives fans have long thought her intentions were to push away Kody’s other wives to become his sole wife, and the viewers’ votes reflected that.

Sister Wives fans have spoken and according to their Reddit poll, Robyn wins — with 450 of the total 654 votes — the title of Kody’s last wife standing.

Sister Wives is currently on hiatus.