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.

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.

Comments:

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Butbutbut! There are two flaws in your argument:

1) You are asking the wrong question. You are asking, "Given these base assumptions about stores, what is the probability that one store scammed JSue twice/two stores scammed her once each?" The question you should be asking is, "Given that JSue has been scammed twice, is it more likely to have been from one store or two?" The former question is simple probability, but the latter question is conditional probability, which generally makes everyone's skin crawl. Your general methodology might be applicable; I'm not sure. Actually...I guess it should work. Except...

2) Your calculations give the probability of a PARTICULAR store (or two particular store) scamming her appropriately. If the question is just "Is it more likely that JSue was scammed by one store or two?" then it doesn't matter WHICH store(s) did the scamming. Suppose JSue has been to n stores.

The probability that a PARTICULAR store scammed her twice, and no other stores scammed her at all, is

P1=xy^2*(1-xy)^(n-1)

The xy^2 you already have; the extra term comes from the probability of every OTHER store NOT scamming her. Now, if we only want the probability that SOME store has scammed her, we add this up over all n stores to get

n*xy^2*(1-xy)^(n-1)

The probability that two PARTICULAR stores scam her is, as you gave above, x^2*y^2. The probability that those two scam her, and no other stores scam her,

is

x^y*y^2*(1-xy)^(n-2)

But instead of summing over n possibilities for a single store, we now have nC2 possibilities for the PAIR of stores. So the probability that SOME pair of stores has scammed her is

P2=(nC2)*x^2*y^2*(1-xy)^(n-2)

=(n/2)(n-1)x^2*y^2*(1-xy)^(n-2)

Comparing P1 and P2, we see

P2/P1 = [(n-1)*x]/[2*(1-xy)]

Assuming xy<<1, the question is largely a matter of comparing n-1 and 2/x. If 2/x is greater (i.e., x is sufficiently small, so dishonest stores are sufficient unlikely) then it is more likely that one store scammed her twice. If n-1 is greater (i.e., JSue has been to sufficiently many stores, basically enough to have an expectation of two or more or them being dishonest) then it is more likely that two separate stores have scammed her.

Now that I've pointed out one of the flaws in your argument, can you spot another one that even I failed to account for?

-T

1) You are asking the wrong question. You are asking, "Given these base assumptions about stores, what is the probability that one store scammed JSue twice/two stores scammed her once each?" The question you should be asking is, "Given that JSue has been scammed twice, is it more likely to have been from one store or two?" The former question is simple probability, but the latter question is conditional probability, which generally makes everyone's skin crawl. Your general methodology might be applicable; I'm not sure. Actually...I guess it should work. Except...

2) Your calculations give the probability of a PARTICULAR store (or two particular store) scamming her appropriately. If the question is just "Is it more likely that JSue was scammed by one store or two?" then it doesn't matter WHICH store(s) did the scamming. Suppose JSue has been to n stores.

The probability that a PARTICULAR store scammed her twice, and no other stores scammed her at all, is

P1=xy^2*(1-xy)^(n-1)

The xy^2 you already have; the extra term comes from the probability of every OTHER store NOT scamming her. Now, if we only want the probability that SOME store has scammed her, we add this up over all n stores to get

n*xy^2*(1-xy)^(n-1)

The probability that two PARTICULAR stores scam her is, as you gave above, x^2*y^2. The probability that those two scam her, and no other stores scam her,

is

x^y*y^2*(1-xy)^(n-2)

But instead of summing over n possibilities for a single store, we now have nC2 possibilities for the PAIR of stores. So the probability that SOME pair of stores has scammed her is

P2=(nC2)*x^2*y^2*(1-xy)^(n-2)

=(n/2)(n-1)x^2*y^2*(1-xy)^(n-2)

Comparing P1 and P2, we see

P2/P1 = [(n-1)*x]/[2*(1-xy)]

Assuming xy<<1, the question is largely a matter of comparing n-1 and 2/x. If 2/x is greater (i.e., x is sufficiently small, so dishonest stores are sufficient unlikely) then it is more likely that one store scammed her twice. If n-1 is greater (i.e., JSue has been to sufficiently many stores, basically enough to have an expectation of two or more or them being dishonest) then it is more likely that two separate stores have scammed her.

Now that I've pointed out one of the flaws in your argument, can you spot another one that even I failed to account for?

-T

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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