Probability to open a door using all the keys I have 1 door and 10 keys. What is the probablity to open the door trying all the keys? I will discard every single key time to time.
 A: If I understand you correctly, you want the probability of unlocking the door on the $10$th try, if you discard keys whenever they fail, and assuming exactly one of the keys is correct. If so, just keep multiplying by the probability of getting a wrong key from those that remain, until finally you are down to one key. Notice a "collapsing" pattern as you go along?
A: Let us assume that precisely one of the keys opens the door. We may want the probability that it is the $k$-th key that does it. Since all orderings of the keys are equally likely, the right key is just as likely to be in any position as in any other. So the probability we open the door on the $i$-th trial is $\dfrac{1}{10}$. That in particular applies to $i=10$. 
A: Solution 1:
Hint: How many permutations are there for the order of the 10 keys?
Hint: How many permutations are there, where the last key is the correct key?

Solution 2:
Hint: The probability that the last key is the correct key, is the same as the probability that the nth key is the correct key. Hence ...
