Thursday, June 16, 2005
1:58 AM — Swipe your card
A month or two ago my poor friend JSue discovered someone had ripped off her debit card info. Apparently some store (or crooked employee) she visited had stored her card data, used it to create a fake card, and then went on a shopping spree at a 7-11. Luckily, her bank noticed this, stopped the charge, notified her, and she wasn't liable for a single TicTac. Well, just a few days ago it happened again, this time with a credit card and a $4,000 jewelry purchase.
I immediately assumed that it was probably one location doing the card info theft, and that J should cross-reference her statements to determine where it was. In explaining it to T$, I said that I thought it was more likely that it'd be one place than two places, and even if it wasn't it's easier to pursue the possibility of it being a single place so she should act upon that assumption until proven wrong. Which made me doubt my thought that it was a single place. I started musing it statistically, and most of the content from this post was emailed to my former roommate, Jethereal, whom teaches statistics, for verification.
Assume the chance of one store being crooked (or having a crooked employee, same difference) is x (x is a fraction likely <<1). The chance of any OTHER store also being crooked is also x, as I'll assume every store is indistinguishable. Assuming you already have a crooked store, assume the chance of them ripping YOU off is y (y is a fraction <1, but maybe not <<1).
(I believe the second in the "one store twice" x turns into a 1 b/c if it was crooked before, it's still crooked, but each time you have a certain chance they'll rip you off. The same store being still crooked is a dependent event.)
Assuming x<1, y<1, implies xy^2 < x^2*y^2, so it's more likely that J had her data ripped off at a single store than two separate stores. ¶ permalink | 1 musings
I immediately assumed that it was probably one location doing the card info theft, and that J should cross-reference her statements to determine where it was. In explaining it to T$, I said that I thought it was more likely that it'd be one place than two places, and even if it wasn't it's easier to pursue the possibility of it being a single place so she should act upon that assumption until proven wrong. Which made me doubt my thought that it was a single place. I started musing it statistically, and most of the content from this post was emailed to my former roommate, Jethereal, whom teaches statistics, for verification.
Assume the chance of one store being crooked (or having a crooked employee, same difference) is x (x is a fraction likely <<1). The chance of any OTHER store also being crooked is also x, as I'll assume every store is indistinguishable. Assuming you already have a crooked store, assume the chance of them ripping YOU off is y (y is a fraction <1, but maybe not <<1).
- If you walk into any random store, they chance they'll rip you off is x*y.
- If you walk into a different random store, they chance they'll rip you off is also x*y.
- If you walk into two random stores, the chance BOTH will rip you off is (x*y)(x*y) = x^2*y^2. (independent events)
- If you walk into ONE random store TWICE, the chance it will rip you off twice is (x*y)(1*y) = xy^2. (dependent events)
(I believe the second in the "one store twice" x turns into a 1 b/c if it was crooked before, it's still crooked, but each time you have a certain chance they'll rip you off. The same store being still crooked is a dependent event.)
Assuming x<1, y<1, implies xy^2 < x^2*y^2, so it's more likely that J had her data ripped off at a single store than two separate stores. ¶ permalink | 1 musings
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All this may be true, however, the probability that I would like to kick some slimy bastard's ass is
P = 1,
where there is no dependence on number of stores visited, amount spent, phases of the moon, or anything else. :)
I used my credit card at relatively few stores in between when my debit card was stolen and my credit card was stolen. So once I get the list of merchants visited, I'll hopefully be able to narrow it down. And THEN the ass-kicking will begin!
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P = 1,
where there is no dependence on number of stores visited, amount spent, phases of the moon, or anything else. :)
I used my credit card at relatively few stores in between when my debit card was stolen and my credit card was stolen. So once I get the list of merchants visited, I'll hopefully be able to narrow it down. And THEN the ass-kicking will begin!
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