'Free-to-play' games can have real costs, in money and privacy

The internet has evolved to the point where if you are getting something for free, then it's you who are the product.

A totally brilliant observation!

South Park did this issue a couple of weeks ago. Seems about the same as a few years back before sensible data packages came along when a cellular service would send someone a bill for $8K after their teenager got addicted to texting and went ape with a couple hundred messages a day.

I actually work for the company mentioned in the article and I'm shocked they associate us with companies using private informations for wrongdoings. And it's the CBC writing those things about a canadian company ?

By the way, the "Connect to Facebook" in the game is for using a common account because the game can be played on Android, iOS and PC. Yes, it can also share your progress if you accept to. But, that's not at all what is implied in this article.

I've played a few 'freemium' games on occasion, but I NEVER link my social networks to them, nor do I download them if they have a long list of permissions.

I don't agree for all cases.

https://www.youtube.com/watch?v=SLMJpHihykI

Some of those freemium games are addictive. Outside of Civilization, or the new Sid Meir game or a Blizzard game. Those I pay for and play intensively. I find that some of the freemium games quite good, so what if they suck up a little data, instead of charging my $17.00 a month after paying $70 for the game. Nowadays, I really don't spend money on very much. The necessities I need. Taxes, the rest I test. Books, music, apps. Every 3 years, I will buy a new tablet/iPhone and computer( to play the latest games I buy), but that's it. Then I recondition all my equipment and give it away (I don't even give it to my wife, I also buy here new equipment every 3 years).

All this to say, that its all ready very expensive to be able to access the internet with equipment powerful enough to be usable. So that fact that there is free games is sort of normal.