By using this site, you agree to our Privacy Policy and our Terms of Use. Close
NextGen_Gamer on 29 July 2020
Edit Thread
theDX said:

I don't believe in a $100 price difference between the regular and digital PS5.

The base storage capacity is more important to the digital PS5 than the regular PS5 so it should actually be higher for that model. And a 1TB SSD costs more than a Blu-ray drive.


Edit: even if both models have the same storage capacity it still means Sony can't cut costs on the digital PS5 by reducing storage.

There is a chance that there could be a $100 difference, if Sony wants to take a hit on the cost upfront and make it up later. Before I get into that though, I wanted to say that it is confirmed (in interviews with Sony execs) that the *only* difference between the standard and digital editions is the removal of the optical drive - both have the same 825GB SSD capacity.

Now, the 4K Blu-ray drive itself is probably only around $30 to Sony. So making the Digital Edition $50 cheaper right off the bat is making it less profitable (you are removing a $30 part but lowering the price by $50). However, I could see Sony making it a full $100 cheaper because of one simple fact: they make FAR more money off digital games then retail copies. Sony and MS both have a 70/30 split on their digital stores. So, the publisher takes 70% of a $60 game, and Sony takes the other 30%. For retail games, it's closer to a flat fee - but that flat fee is FAR less than 30% of $60 because there are also actual manufacturing costs, shipping costs and of course the retailer cut. I think I read somewhere that Walmart for example keeps $20 of the $60 standard game price. Right away that only leaves $40 for for Sony/publisher to split.

In other words, Sony may make as much as 3 to 5 times off a single digital game sale versus a retail copy. And since you HAVE to buy digital on the digital edition (duh), Sony could rely on that fact to price it lower than most are predicting since they know they can make up the costs long term.