General Question

nettodo's avatar

What happens to a used but unregistered AMEX gift card?

Asked by nettodo (468points) June 20th, 2014

(disclaimer: I accept all liabilities that stem from this question and the actions taken)

So, while cleaning out backstage at my local theatre, a person I was helping clean up with spotted a $50 American Express gift card (you know, the one that you give when you don’t want to “restrict” someone but don’t feel like cash). The card was not signed and appeared to be not registered.

However, after checking the balance on the American Express website, I found it was used once and had less than the $50 starting balance. We’re trying to find the owner and will give it back if requested.

My question comes from the “if” of “we couldn’t locate the owner because the card was not registered nor signed”. Would the card be treated similar to lost cash or as a lost credit/debit card?

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3 Answers

Response moderated (Unhelpful)
SavoirFaire's avatar

It would be treated like lost cash. Most of these cards have a message to that effect somewhere on them. There is one important difference, however. If the previous owner knows the card number and security code, they can deactivate the original card and get a replacement issued to them.

JLeslie's avatar

I don’t know about AmEx cards specifically, but some of the visa cards like that lost money every month. There was a fee taken out and if you didn’t use it you lost all your money. I hope that doesn’t exist anymore, because it was so unscrupulous in my mind. If there are cards like that still out there you might want to research if that card is like that.

If there is a lost and found at the theatre you could turn it in and if no one claims it in a week or so I assume they will let you have it back. Or, you can just wait a few weeks and give the owner a chance to deactivate it, and if it doesn’t get deactivated go ahead and use it. It is like lost cash to me. I would feel good if I could get it back to its rightful owner, but it sounds unlikely in this case.

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