Is there a way to buy individual cards?

By gandalfrockman, in X-Wing

I need more veteran instincts!

I need engine upgrades!

But there is no way in hell I am going to buy 2 more firesprays ( I already have 2), or a millennium falcon (I only play empire).

Is there a way I can just buy the cards I need, or just the cards from a set?

I haven't done it yet, but I'm gonna start printing on glossy paper scanned copies of these cards and then print as many as needed.

For casual games, just do that or use proxies. If you're doing tournaments you'll need the real thing, but for pick-up games a printed version would work fine.

Im not interested in printed copy's, is there anyway to buy just cards?

I might be selling my extra cards that I get from each pack. Both the mini cards and regular side cards. Probably later down the road.

There is no official Print-On-Demand or Print-At-Home product(s) from FFG for X-Wing.

You could look at the used/second hand market. Put up a WTB thread on here or BGG.

Edited by Hexis

No way to do it outside of official channels other than back-alley trading and the black market.

Just like getting a copy of the Star Wars Holiday Special.

I have seen blocks of cards (multiple cards) being sold on ebay.

I have seen blocks of cards (multiple cards) being sold on ebay.

I was gonna say the 'Bay has single cards for sale. Beat me to it.

Someone but $50 for 1 TIE Advanced and 1 X-Wing????

I would have sold them for less than that and included shipping...

Edited by Ken at Sunrise

Someone but $50 for 1 TIE Advanced and 1 X-Wing????

There's a sucker born every minute.

Ebay is pretty good proof of that. I remember once seeing someone selling instructions on how to buy a iPad for $150, and I believe there was at least a couple bids. He made it quite clear what he was selling, so no one could really say it was fraud or anything.

But in this case I think you may also see people who don't know what they are really buying, putting bids on stuff that they think is collectable, and will go up in value.