Ask HN: How to redeem a gift card without risking lock-out?

My young daughter got 2 Amazon Gift cards for her birthday, but they look like they've been tampered with. I would like to check whether they are valid or not without having to go through the redeeming process: I want to avoid being at the mercy of a fraud detection algorithm that may lock me out of my account. There was a HN story not too long ago about someone who got locked out of their iCloud account for similar reasons, and the risk doesn't seem to be worth it to me. Surely, I can't be the only one with this problem. How do people do this "safely"?

Comments

al_borlandJan 25, 2026, 4:21 AM
Does Amazon have a number on the card to call for the balance, or is that only a thing with cards for brick and mortar shops?

For what it’s worth, if the gift card was purchased at a store, part of the back needs to be torn off for it to be scanned and activated, which makes all those cards look tampered with. Not sure if that’s what you’re seeing or not.

B1FIDOJan 24, 2026, 10:50 PM
I've got a crazy bizarre idea. Why don't you contact Amazon customer service and ask them about it?

You can describe the cards and give them the numbers and then tell them why you believe they've been tampered.

I mean it's totally illogical, right?

When I was accused of possessing counterfeit money, I went right to my bank and showed the bills to them (they were, of course, authentic).

toomuchtodoJan 24, 2026, 7:05 PM
Create a new Amazon account just redeem them? TLDR Sacrificial account.
bdangubicJan 24, 2026, 8:45 PM
This will 100% never work, it is exactly the kind of thing Amazon etc will prevent. There is not a single business that accepts gift cards that will not immediately flag this and kill the account. It is the first thing you’d write if you were in charge of this, right?
toomuchtodoJan 24, 2026, 8:46 PM
Have you tried it?
bdangubicJan 24, 2026, 9:29 PM
Amazon ($25) and iCloud (this one was $10, bought from apple.com), both killed the account, did not try others but safe to assume results would be the same