Nationwide also to introduce "Card Reader Security"

Your insurer, hopefully.

You have.

Mark

Insert card RESPOND, SIGN OR IDENTIFY Press Identify PLEASE WAIT ENTER PIN: Enter wrong PIN

**** PRESS ENTER? Press Enter WRONG PIN, TRY AGAIN Doesn't respond to any more button presses

Remove card, repeat with right PIN. PIN CORRECT YOUR CODE IS 1234 5678

Theo

"Mark Goodge" wrote

Hmmm. What's the difference between:-

(A) Mugger steals your card under threat of violence, uses it fraudulently to purchase goods - bank pays cost of the fraud;

... and ...

(B) Mugger steals your card+PIN under threat of violence, uses them fraudulently at an ATM to get cash - bank doesn't pay cost of the fraud???

? [Ie, BOTH are: "Mugger steals XXX under threat of violence, then uses XXX fraudulently...", so why the different liability?]

determine

What insurance?

Er, on what basis exactly? The withdrawal wasn't made by the account holder, the only argument the bank could use not to refund was that the cardholder was negligent with the PIN. Given a choice of being stabbed or telling a mugger the PIN won't count as being negligent.

In any case, negligence is irrelavent if the withdrawal is used to borrow money (eg it's on a credit card or an overdrawn current account). You can keep the PIN notification wrapped around the card and lose it, and you are not liable.

Sorry, should have said except for the first 50.

"Andy Pandy" wrote

That's what I thought!

"Andy Pandy" wrote

What's the basis for the above, Andy?

The Consumer Credit Act 1974.

"Andy Pandy" wrote

That's interesting, thanks. Looks like carte-blanche for everyone with a credit card, and a gripe against their bank, to be grossly negligent ... !!

Do you know what section of CCA1974 it's in?

>

Sections 83 & 84.

Sorry, I misread that as being forced to make a cash withdrawal at the cashpoint which is then stolen from you. In that case, it is you who has made the withdrawal and you who stands the loss.

Mark

Which appears to be different from the Barclays Instructions.

What is it you have quoted above?

Yes - he didn't say: (From the instructions)

The card reader says: ?PIN LOCKED?. If you enter your card PIN into the card reader three times incorrectly, the PIN will lock and you won?t be able to use your card. To unlock the PIN, you will need to visit any Barclays ATM and follow the on-screen instructions.

Only if you have used the same PIN for each card - which would be a silly thing to do.

No, you still get 9 guesses regardless.

If I've used the same pin on all the cards then the probability of getting it right in 9 guesses is (assuming all 10000 possible pins are allowed) is about 1 in 1111.1. If I've used a different pin on all three cards the probability is about 1 in 1111.8 (1111.4).

If we assume 6 guesses per card, total 18 guesses then if the pins are the same it's about 1 in 555.6. If the pins are all different it's about 1 in 556.2 (555.9).

I'm astonished that the odds are as close as that. I'd expected it to be a bigger difference. I may have done something wrong with the calculations but I cannot see what. (The numbers in brackets are the probability if the thief knows all the pins are different and therefore picks three random numbers to try on each card with the small possibility of trying the same number on more than one card)

Tim.

Anything in ALL CAPS is a message displayed on the screen. Anything in Mixed Case is an action that you do.

Tim.

That much was obvious. I think you'll find that the question "What is it you have quoted" was trying to ask *whose* instructions were quoted, given that the question followed an indication that those instructions were not the same as Barclay's.

At 23:48:05 on 30/01/2008, Tim Woodall delighted uk.finance by announcing:

That doesn't look right at all.

Why are you assuming 6 guesses per card?

"Tim Woodall" wrote

Agreed, assuming the thief doesn't repeat any guesses on different cards. [For 9,991 out of the possible 10,000 PINs, the thief won't crack any card -- ie 99.91% chance of not getting the PIN, leaving 0.09% chance (or 1 in 1111.111...) of getting it.]

"Tim Woodall" wrote

I'd be interested to see your calculations here - I make it 1 in 1111.4445 either way (as long as you've used a different PIN on each card).