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?
 A: If you want to know the probability of finding the fixed number $x$ in $n_{th}$ position you should proceed by first finding the total number of favorable cases and dividing by the total number of possible cases. The number of favorable cases can be obtained by fixing $x$ in the $n_{th}$ position and considering all permutations of the remaining $k-1$ numbers.
$P(x \ is \ in \ n_{th} \ position) = \frac{(k-1)!}{k!}=\frac{1}{k}$ 
Also note that you may consider $n$ to be any position, the result will remain the same.
A: Well, no. The probability that the second number is $x$ is
$(1 - \frac{1}{k})\frac{1}{k-1} = \frac{k-1}{k}\frac{1}{k-1} = \frac{1}{k}$.
You forgot that, once you know that the first number is not $x$, you only have $k-1$ remaining, not $k$, so the probability that the first of the rest is $x$ is $\frac{1}{k-1}$.
It's also not a coincidence that the answer is $\frac{1}{k}$. The situation is symmetric, so the probability that $x$ is in any one position is the same.
A: Unless I am misunderstanding what you're trying to figure out, that's not correct.
It sounds like you're trying to answer the question:
First, fix some number for $x$, say $17$. Now you want to know:
What is the probability that the first number in the set is $17$?
What is the probability that the second number in the set is $17$?
etc.
Well, they all have a probability of $\frac{1}{k}$ of being $17$
This should be immediately intutive, since the probability should be the same for each of the positions. But, if you use your approach of calculating it, you find it as well:
$$P(x_2 = 17)= P(x_2=17|x_1\not = 17)\cdot P(x_1 \not = 17) = \frac{1}{k\color{red}{-1}}\cdot \frac{k-1}{k}=\frac{1}{k}$$
A: 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?
