A man has \(n\) keys on a key ring, one of which opens the door to his
apartment. Having celebrated a bit too much one evening, he returns home only
to find himself unable to distinguish one key from another. Resourceful, he
works out a fiendishly clever plan: He will choose a key at random and try it.
If it fails to open the door, he will discard it and choose at random one of
the remaining \(n-1\) keys, and so on. Clearly, the probability that he gains
entrance with the first key he selects is \(1 / n\). Show that the probability
the door opens with the third key he tries is also \(1 / n\). (Hint: What has to
happen before he even gets to the third key?)