Probability of picking a number from a set of unique integers Suppose I have a set of $k$ integers such that every number is unique.
Let $A = \{1,2,3,4,...,k\}$ 
Now suppose that we rearrange these numbers to a random permutation in the set. I want to find the probability of finding a fixed number $x$ at any position of the set.  Here's my understanding:
The probability of the first number in the set being $x$ is $\frac{1}{k}$ 
The probability of the second number in the set being $x$ is $(1-\frac{1}{k})\frac{1}{k}$
The probability of the third number being $x$ is $(1-\frac{1}{k})^2 \frac{1}{k}$ 
That leads to the probability of the $n^{th}$ term being $x$ to be $(1-\frac{1}{k})^{n-1} \frac{1}{k}$ 
Is this correct?

Hint:
You are wrong at the second place in the set.
$P(x=2)=P(x\neq1 )\cdot P(x=2|x\neq1)=(1-\frac{1}{k})\cdot \frac{1}{k-1}=\frac{k-1}{k}\cdot \frac{1}{k-1}=\frac{1}{k} $
This is because you already know it is not in the first place.
Now try to look for the third one, and then the fourth..
And you can ask yourself, why should the first place will have more chances the the others to find x in it?
