Combinatorics. Picking items with and without replacement. Am I reasoning in a right way? We have n keys and we want to choose right one to open the door.
a. We choose keys w/o replacement.
What is probability to choose right one: one the first try, second try, k-th try?
b. What if we do it with replacement?
Ok, so when we consider case w/o replacement.
P(1st try) = 1/n
P(2nd try) = (n-1)/n * 1/(n-1) <- not sure if I am right here. But I think for 2nd try we need to include probability that we did not pick right one on the first try. Correct?
P(k-th try) = ( (n-k)/n )^k * 1/(n-k) <- again, not sure if it is correct. But I feel like it should be
b. with replacement (ok, here I am a bit confused)
I think our first probability must be the same
P(1st try) = 1/n
P(2nd try) ??? I am not sure here. Most probably it is same as P(1st try)?
Meaning, P(1st try) = P(2nd try) = P(3rd try) + ...+ P(k-th try)?
From the first glance it seems right, but I believe it is not how it really works..
I will be really thankful for a help!
 A: For without replacement, there is a simple solution. Line up the keys in the order you will try them. The "right" key is equally likely to be in any of the $n$ positions. So for $1\le k\le n$, the probability that we use $k$ trials is $\frac{1}{n}$. 
For with replacement, on any trial the probability of success is $\frac{1}{n}$, and the probability of failure is $\frac{n-1}{n}$.
So the probability that $k=1$ is $\frac{1}{n}$.
We use $2$ trials if we get failure, then success. The probability of this is $\frac{n-1}{n}\cdot\frac{1}{n}$.
We use $3$ trials if we have $2$ consecutive failures, and then success. The probability of this is $\left(\frac{n-1}{n}\right)^2\cdot\frac{1}{n}$.
Continue. In the replacement case, the probability it takes exactly $t$ trials is
$\left(\frac{n-1}{n}\right)^{t-1}\cdot\frac{1}{n}$. Here $t$ can be any positive integer.  
Remark: We can also do the no replacement calculation a harder way. The first success happens on the second trial if we get failure, then success. The probability of failure on the first trial is $\frac{n-1}{n}$. Given that we had failure on the first trial, there are $n-1$ keys left, of which $1$ is the right one, so the conditional probability of success on the second trial, given that we had failure on the first, is $\frac{1}{n-1}$. Thus the probability we get the right key on the second trial is $\frac{n-1}{n}\cdot \frac{1}{n-1}$, which is $\frac{1}{n}$.
For success on the third trial, we must have failure, failure, then success. The associated probability is $\frac{n-1}{n}\cdot \frac{n-2}{n-1}\cdot\frac{1}{n-2}$, which again is $\frac{1}{n}$. 
We could continue, but it is easier to use the argument in the answer above. 
A: B)
$$P(\text{1st try})=\frac{1}{n}$$
$$P(\text{2nd try})=\frac{n-1}{n}\cdot\frac{1}{n}$$
$$P(\text{k:th try})=(\frac{n-1}{n})^{k-1}\cdot\frac{1}{n}$$
