If you have multiple American Express cards stored in your Amazon account, or at other online sites, you may have been stymied by a user interface that only displays the last four digits of your American Express cards – two of which happen to be the same.
Why is that? And what’s the probability that you’ll have two American Express cards with the same last four digits? The answer lies in the rather unique way American Express credit and charge card numbers are issued, and has an interesting link to the birthday problem from probability theory.
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How Do the Last Four Digits Work?
American Express credit card numbers are 15 digits, and those issued by American Express have the last four digits determined as follows:
- (12th digit): number of cards that have been issued on a revolving credit line, starting at 1 and increasing for each replacement card (usually due to theft or loss)
- (13th – 14th digit): number of additional cardmember, starting at 00 for the primary cardmember and increasing by 1 for each additional cardmember
- (15th digit): checksum digit
If your card has never been replaced and you are the primary cardmember, your last four digits are always 100x, where x is the checksum digit. Since the checksum digit can be 0–9, there is a 10% chance that any two American Express credit or charge cards fitting those two conditions share the same last four digits—compared to other issuers, a very high probability! This is why you must confirm the last five digits when talking to American Express.
How to Avoid Having the Same Last Four Digits
If having the same last four digits on two American Express cards is an issue, there are a few workarounds.
My solution is to memorize the expiration date (or rather, remember when I applied for the card) because Amazon displays the expiration date. (And I generally don’t store my card information on other sites, preferring to use LastPass.)
An Amazon-specific solution mentioned by a few readers is to put identifying information about the card in the “name” field of any cards you may be confused about. This currently works because Amazon does not verify the name on your credit card when charging it. (In other words, it’s YMMV and could change without notice.)
If that isn’t working for you, requesting a replacement card will increase the twelfth digit and also change the checksum digit. You can do this online or by phone; it’s my understanding that calling and mentioning the last four digit issue is an acceptable reason for customer care professionals.
Probability Theory
Now, for the exciting part! It’s easy enough to calculate the probability that any two American Express cards will share the same last four digits (10%). But, you have to go a bit deeper into basic probability theory to find the probability that two of a given number of American Express credit cards share the same last four digits. (It is important to note that the probabilities we are about to calculate are for cards with the same 12th thru 14th digits. In other words, the probabilities can be for “never replaced” & “primary cardmember”; “never replaced” & “first additional cardmember”; “replaced once” & “primary cardmember”; etc.—but never a mix of the categories.)
If you have eleven American Express cards, the answer is 100% because there are only ten possible checksum digits (the pigeonhole principle).
For two to ten American Express cards, we can find the probability by asking the opposite question: what’s the probability that no two cards have the same last four digits? Because the probability that no two cards have the same last four digits and the probability that two cards do have the same last four digits must be 100%, we can subtract our answer from 100% to answer our original question.
This is the same way the birthday problem is solved, which you may recognize by the formulation that (discarding February 29), there is a 50% probability that two people in a room share a birthday when there are 23 people in the room. (Astute readers may recognize that this assumes the distribution of birthdays is uniformly random, which it is not, but the effect on the probability is negligible enough to ignore. Likewise, I make that assumption about the checksum digit.)
We can calculate the probability that no two cards share the same last four digits by calculating the probability that each subsequent card beyond the first card does not have the same checksum digit. The probability that a second card does not have the same last four digits is 90% – if the checksum digit of the first card is x, then there are 9 out of 10 digits the second card can have without having the same last four digits. The probability that a third card does not have the same last four digits as the first or second card is 80%, which is then multiplied by 90% to obtain the probability that both a second and third card do not have the same last four digits as the first card. And so on. What you wind up with is probabilities that escalate quickly:
| # Cards | Not the Same Last 4 Probability |
Same Last 4 Probability |
| 2 | 90% | 10% |
| 3 | 72% | 28% |
| 4 | 50.4% | 49.6% |
| 5 | 30.24% | 69.76% |
| 6 | 15.12% | 84.88% |
| 7 | 6.048% | 93.952% |
| 8 | 1.814% | 98.186% |
| 9 | 0.363% | 99.637% |
| 10 | 0.036% | 99.964% |
With just four American Express cards, you have a near-50% probability of having two cards with the same last four digits!
For fun, we can also calculate the probability that two of your American Express cards share the same last five digits:
| # Cards | Not the Same Last 5 Probability |
Same Last 5 Probability |
| 2 | 99% | 1% |
| 3 | 97.02% | 2.98% |
| 4 | 94.109% | 5.891% |
| 5 | 90.345% | 9.655% |
| 6 | 85.828% | 14.172% |
| 7 | 80.678% | 19.322% |
| 8 | 75.031% | 24.969% |
| 9 | 69.028% | 30.972% |
| 10 | 62.816% | 37.184% |
Questions? Comments? Other mathematical funsies? Drop them below!
