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Its time to go back to the business-model drawing board. Get rid of the ads and start charging for the service? It would be interesting to see how many social n
by LostInTheWoods 18y ago
Its time to go back to the business-model drawing board. Get rid of the ads and start charging for the service? It would be interesting to see how many social networks could justify their existence based on utility alone.
- shutter 18y agoIf the charge was _minimal_, I'd be willing to pay a few bucks a year for Facebook. If you use it properly (read: non-obsessively), it's a great way to keep connected with people you've known in the past several years. Without Facebook, I would have lost touch with a lot of people, whom I could now communicate with easily when I need to. I gave up the day-to-day "social" aspect of it a while ago, but for keeping connections with past high-school and college friends, there isn't a better way to keep the doors open. That being said, if I was paying for the service, I'd hope they would open up their data (for export) and give a few more bells and whistles for the money.
- soundsop 18y agoThey could do a flickr-type model and charge only heavy users (say, more than 1000 friends) or charge for premium services (high-res photos or similar features). EDIT: Even if social ads don't work out, I think Facebook will probably find a way to make lots of money. Although, maybe not enough money to justify the highest valuations we've heard.
- dangoldin 18y agoThat's a very interesting point. How much would people be willing to pay for social networks. It seems that people would stick with one general site(Facebook) and potentially a few more specialized ones. Unfortunately, if the price ends up being too high for most people, they will leave and go on to somewhere else, thus making it less worth it for you to have it so you'll end up leaving too. If a major social network did start charging, a lot of people would probably move to a competitor unless the offering was too good to leave.. Hmm. Very interesting point though.