Four cards, four pins. What's probability to block one of them? Do not beat me much... I've studied the probability theory more than 15 years ago in the university and don't remember much.
So the problem is. We have 4 credit cards and 4 pin-codes. We know that all pin-codes are correct and they are for these 4 cards. But we do not know which pin-code is for which card. A card becomes blocked if one tried 3 incorrect pins. 
What's the probability of blocking one of the cards during our attempts to match all the pins to all the cards?
I've tried to solve the problem directly and here what I've got. Let's suppose we started from the card #1 and a correct pin is the pin #1.
Outcomes which do not block the card (because the pin #1 is within the first 3 attempts) are:


*

*1,2,3,4 

*2,1,3,4

*3,1,2,4 

*1,3,2,4

*2,3,1,4

*3,2,1,4

*4,1,3,2

*1,4,3,2

*3,4,1,2

*4,3,1,2

*1,3,4,2

*3,1,4,2

*2,1,4,3

*1,2,4,3

*4,2,1,3 

*2,4,1,3

*1,4,2,3

*4,1,2,3


The rest outcomes which do block the card:


*

*3,2,4... blocked,1

*2,3,4... blocked,1

*4,3,2... blocked,1

*3,4,2... blocked,1

*2,4,3... blocked,1

*4,2,3... blocked,1


So we have in total 24 outcomes (==#of permutations in array of [1,2,3,4]) and 6 of them lead to the card blockage.
Therefore, probability of having card blocked is exactly 25%.
It seems that we can block only one card in the worst case, so the answer is 25%. Am I right or wrong?
 A: There's a better method to solve your problem. Instead of taking the cards one by one and trying pins on it, pick a pin and try to find the card that belongs to it, by testing that pin on all unmatched cards one by one. But when choosing the next card to match against the current pin, always choose one of the cards with the least amount of failed retries so far.
That way, the chance you'll fail is only $1$ in $24$, which is the case that you pick the wrong card at every chance.
Example scenario:


*

*Try pin 1 on card 1: Failure.

*Try pin 1 on card 2: Failure.

*Try pin 1 on card 3: Success.

*Now you know that pin 1 belongs to card 3.

*Try pin 2 on card 4: Failure.

*Try pin 2 on card 1: Failure.

*Now you know that pin 2 belongs to card 2.

*Try pin 3 on card 4: Failure.

*Now you know that pin 3 belongs to card 1, and pin 4 belongs to card 4.
UPDATE: According to a quick program I wrote, it is impossible to succeed in all 24 cases. If you check every card against at most two pins, you can never completely determine which card belongs to which pin, and checking any card against a third pin will either risk blocking the card or not help because you already knew it was the correct pin.
