I have this following problem in my problem set and I would like to check if my work is in the right directio.
A drunk man with $n$ keys wants to open his door and tries the keys at random. Exactly one key will open the door. Find the mean number of trials if
a) unsuccessful keys are not eliminated from further selections;
b) unsuccsesful keys are eliminated
It seems clear to me that in the first case we have a Geometric distribution with parameter $1/n$, so the expected number of trials is just $n$.
For the second case, my reasoning follows. Let $X$ denote the number of trials until he opens the door.
$P(X = 1) = 1/n$
$P(X = 2) =(n-1)/n \cdot 1/n-1 = 1/n$
$P(X = 3) = (n-1)/n \cdot (n-2)/(n-1) \cdot 1/(n-2) = 1/n$, and so on.
I'm inclined to say that, in the second case, the probabilities over the possible $n$ values of $X$ are uniformly distributed. If this is case, then the average number of trails should be $(n+1)/2$.
Does it seem correct? If not, how to do it? Thanks in advance!!