4 ms·
In my case it's not, and you also need my billing address to use my card, which isn't printed on it. Still, if you manage to collect 5000 credit card numbers, t
by Asooka 9y ago
In my case it's not, and you also need my billing address to use my card, which isn't printed on it. Still, if you manage to collect 5000 credit card numbers, the CVV is 3 digits, which gives you 1000 possibilities and 1 - 0.999 * * 5000 = 0.99, so statistically you should hit one in 5000 cards if you guess CVV randomly.
- fnl 9y agoWell, as you point out, the probability of not hitting a valid CVV this way is non-zero, or 0.672% - if using R: choose(5000,0)*0.999^(5000-0)*0.001^0 But that means, by brute forcing, while you should crack at least one card, there is still a non-zero chance that this will be your unlucky day... (Cracking all cards with this methods, is - essentially - a zero-chance game, but even for that an infinitesimally small chance is left, that my computer cannot reproduce.)
- goldenkey 9y agoNothing is certain when combining uncertain independent events. Didnt need a calculator for basic measure theory..