Got a replacement card and now I have two cards with the last 4 digits that are in sequential order. One from U.S Bank and one from Chase. This is going to make things a little confusing in the future. Yes pics included.

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scrouds

KayK's running mate

posted: Jul. 16, 2016 @ 8:20a

Cool story bro

burgerwars

just a salad for me

posted: Jul. 16, 2016 @ 8:22a

Amazing.

enterkey

In trouble

posted: Jul. 16, 2016 @ 8:37a

Maybe there are different pattern hidden with the rest of numbers. Allow us to help you find out.

dannon35

Thrifty Member

posted: Jul. 16, 2016 @ 6:50p

Happens about 1 in 5000 times.

MilleniumBuc

Scrouds Butch

posted: Jul. 18, 2016 @ 6:52p

dannon35 said: Happens about 1 in 5000 times. Can you provide the mathematical factualization of this?

goku2

Greedy Member

posted: Jul. 19, 2016 @ 1:16a

MilleniumBuc said: dannon35 said: Happens about 1 in 5000 times. Can you provide the mathematical factualization of this? Yeah I think it's actually more frequent than that but I'm not sure.

MilleniumBuc

Scrouds Butch

posted: Jul. 19, 2016 @ 2:31p

So would 3 in a row be 1 in 3333 times?

Or because of the checksum error verification, would the odds actually be smaller?

76hhma

CAUTION: Finance Veteran

posted: Jul. 19, 2016 @ 4:58p

The first 15 digits are somewhat correlated according to Luhn Algorithm; the last digit is a deterministic error-check bit. Another challenge in probability.

BrianGa

Posting Under the Influence

posted: Jul. 20, 2016 @ 10:37a

This thread sucks without your mother's maiden name.

kamalktk

Dwayne Elizondo Mountain Dew Herbert Camacho

posted: Jul. 20, 2016 @ 11:02a

goku2 said: MilleniumBuc said: dannon35 said: Happens about 1 in 5000 times. Can you provide the mathematical factualization of this? Yeah I think it's actually more frequent than that but I'm not sure. if you only care about them being sequential, and don't care what the particular numbers are (01-02 vs 155-156), then the odds are 1 in 10,000. It's a four digit sequence so it has 10,000 possible numbers, and what the first number is would not be relevant. [and assuming there are no limits on what cc numbers can be].

Odds of three sequential (or two sequential if you care what the numbers are) would be 100,000,000.

dannon35

Thrifty Member

posted: Jul. 20, 2016 @ 12:28p

kamalktk said: goku2 said: MilleniumBuc said: dannon35 said: Happens about 1 in 5000 times. Can you provide the mathematical factualization of this? Yeah I think it's actually more frequent than that but I'm not sure. if you only care about them being sequential, and don't care what the particular numbers are (01-02 vs 155-156), then the odds are 1 in 10,000. It's a four digit sequence so it has 10,000 possible numbers, and what the first number is would not be relevant. [and assuming there are no limits on what cc numbers can be].

But... if my first card ended in "5678", the second card could be either "5677" or "5679" for the cards to be sequential, so 2/10,000 or 1/5,000.

kamalktk

Dwayne Elizondo Mountain Dew Herbert Camacho

posted: Jul. 20, 2016 @ 1:05p

dannon35 said: kamalktk said: goku2 said: MilleniumBuc said: dannon35 said: Happens about 1 in 5000 times. Can you provide the mathematical factualization of this? Yeah I think it's actually more frequent than that but I'm not sure. if you only care about them being sequential, and don't care what the particular numbers are (01-02 vs 155-156), then the odds are 1 in 10,000. It's a four digit sequence so it has 10,000 possible numbers, and what the first number is would not be relevant. [and assuming there are no limits on what cc numbers can be].

But... if my first card ended in "5678", the second card could be either "5677" or "5679" for the cards to be sequential, so 2/10,000 or 1/5,000. Good point. I hadnt thought of that.

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