What is the probability that the special ball is chosen? I know this question is already posted here, but I doubting my own solution which is I know is wrong. It is quite basic, but I want to learn probability from scratch, so I am posting my question.
Question:

An urn contains $n$ balls, one of which is special. If $k$ of these balls are withdrawn one at a time, with each selection being equally likely to be any of the balls that remain at the time, what is the probability that the special ball is chosen?

My approach:
The number of ways in which we can select $k$ balls from a set of $n$ balls equals:
$$\binom{n}{k}$$
Out of these $ \binom{n}{k}$, we have a single special ball.

As such, the required probability equals: $$\frac {1} {\binom{n}{k}}$$

What am I doing wrong?
 A: It is not certain that your selection contains the special ball. As you said, the amount of ways in which we can pick $k$ balls out of $n$ equals:
$${n \choose k}$$
The amount of ways in which we can pick $k-1$ normal balls and the special ball, equals:
$${n-1 \choose k-1}{1 \choose 1} = {n-1 \choose k-1}$$
As such, the probability of having drawn the special ball in $k$ draws equals:
$$\frac{n-1 \choose k-1}{n \choose k} = \frac{k}{n}$$
A: You need to consider the number of ways of selecting $k$ balls such that the selected set of balls always contain the special ball. This number will be equal to the number of ways of selecting $k-1$ balls from $n-1$ balls.
A: You have to count the number of selections of $k$ balls that contains the special ball : there are $\binom{n-1}{k-1}$, so probability is 
$$\frac{\binom{n-1}{k-1}}{\binom nk} = \frac kn$$
which can be found some simpler way.
A: The probability $\frac{1}{{n\choose k}}$ is the probability of one particular selection of $k$ balls out of an urn with $n$ balls.
But there is more than one selection which includes the special ball.

For example, take $n=3, k=2$. Let's say the balls are marked $\{1,2,3\}$ and that $1$ is the special ball.
Then $\frac{1}{{3\choose 2}}$ is the probability that you will draw $\{1,2\}$, for example. However, it's also possible to draw $\{1,3\}$, and that is also a positive outcome!
A: Say we number the balls $1,2,...,n$, with ball $1$ being the special ball. For most $n$ and $k$, many out of the $n\choose k$ ways to select the balls will have the special ball; two of which are $1,2,...,k$ and $1,3,4,...,k+1$. So, there is more than one outcome in which the special ball is chosen, and so the probability is usually greater than $1/{n\choose k}$. 
In order to solve this problem, try looking at the number of possibilities where the special ball is not chosen. There are $n-1$ non-special balls to choose and therefore ${n-1\choose k}$ ways to choose non-special balls, and so the probability of not choosing the special ball is ${n-1\choose k}/{n\choose k}$. Therefore, the probability of choosing the special ball is
$$1-\frac{n-1\choose k}{n\choose k}$$
which can then be simplified. 
A: 
Out of this $\binom{n}{k}$ [balls], we have a single special ball.

This is your mistake.  You do not have $\binom{n}{k}$ balls.  You have $n$ balls, of which one is special.  $\binom{n}{k}$ is the number of ways of choosing $k$ balls, which is a very different thing from the number of balls.
Imagine the $n$ balls are lined up in random order, like this:
$$
* * * * * * * \cdots * *
$$
Then we draw a line, with $k$ balls to the left of it, and $n-k$ balls to the right.  The $k$ leftmost balls are selected.*
For the first step (random ordering), the special ball is equally likely to be in any position, and we don't actually care about the placement of any of the other balls.  So we can ignore the other balls, and just say that we're picking a position for the special ball.  There are $n$ ways of doing that.
Of those $n$ choices, $k$ of them will be to the left of the line.  So we have a $\frac{k}{n}$ chance of selecting the special ball.

* Incidentally, this line also selects the rightmost $n-k$ balls, which generalizes into a nice proof that $\binom{n}{k} = \binom{n}{n-k}$.
A: Alternative thinking.
The answer is $\frac{k}{n}$.
To understand identify yourself with the special ball. What is the chance that you will be chosen if $k$ balls are selected out of $n$?...
You can think of $n$ spots of which  $k$ are marked as "selected spots", and the special ball is randomly placed in one of these spots. So there are $k$ "good" outcomes among $n$.
The same probability to win a price  in a lottery if you have one lot of $n$ lots in total and on $k$ of them will fall a price.
