Netflix May Lose ‘Friends’ in 2019

     October 17, 2018


The streaming wars are in full swing. As consumers move away from cable and towards streaming channels, they’re faced with a new conundrum: How do you gather all of the content you want? What sacrifices will you have to make as the shows and movies you want to see fragment across the streaming landscape? Is it feasible to pay for 10 or 15 streaming services?

In the beginning, when Netflix was leading the way in streaming, it was able to score huge deals like landing the entire series of Friends. But with WarnerMedia planning to launch its own streaming service next year, it looks like that licensing deal with Netflix is likely to expire and Friends would move to WarnerMedia’s streaming service since Warner Bros. Television owns the property. That would be a serious blow to Netflix as Deadline reports that it’s the third most popular sitcom in the U.S. behind Brooklyn Nine-Nine and The Big Bang Theory, and certainly not too shabby when you consider that Friends went off the air 14 years ago.

If you’re wondering why Netflix is churning out so much content, this is a large reason why. The streaming service was able to build its name by licensing content from other providers. However, those companies are now figuring out they can just have their own streaming service and they’re taking their toys with them. Disney will be doing it with their movies and TV shows, and now it looks like WarnerMedia is following suit. Netflix is making sure that when that happens, it will have a large library of its own content to woo consumers, who will then be faced with the decision if they want to spend money on additional streaming services.

In the meantime, if you don’t like the thought of being batted around by which streaming service has which title you want to watch at a particular moment, you could always just buy physical media and not worry about it. Friends: The Complete Series is currently only $110 on Blu-ray. Buy it and discover that Ross Gellar is a total sociopath.