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A woman has $n$ keys, of which one will open her door.

$(a)$ If she tries the keys at random, discarding those that do not work, what is the probability that she will open the door on her $k$th try?

$(b)$ What if she does not discard previously tried keys?

I know the method to solve both parts, but what I want to understand is (conceptually) why we get different answers to them.

For example, let the $k$th key be the right one. It is still the right one, whether you discard the first $(k-1)$ keys or not. The $k$th key doesn't care what you do with the other keys. Then why is the probability different for these two cases?

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The woman does not try the keys in a set order. So replacing and not replacing tried keys gives different results. You are right though that if she tried the keys in a fixed order it wouldn’t matter what she did with the previous keys.

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