Urn problem where you remove as much balls as the r-th side of a fair dice of n faces tells you This is another problem from David Stirzaker Elementary Probability. As context you have $2n$ balls, $n$ of which are tangerine and the other $n$ are heliotrope. The problem goes as follows:
"A fair die with n sides is rolled. If the rth face is shown, r balls are removed from
the urn and placed in a bag. What is the probability that a ball removed at random from the bag is
tangerine"
The books gives as answer $\frac{1}{2}$ but I'm not sure how to get there. What I've been trying to do is consider the procedure from another problem that's almost identical to this except that there is no fair die and you are only taking $n$ balls instead of the $r$ that indicates this problem, the answer to that problem is in this post, it went like this:
$$P(B)=\frac{1}{\binom{2n}{n}}\left(\sum\limits_{k=0}^n\frac{n-k}{n}\binom{n}{n-k}\binom{n}{k}\right)=
\frac{1}{\binom{2n}{n}}\left(\sum\limits_{k=0}^n\color{blue}{\frac{n-k}{n}\cdot\frac{n!}{(n-k)!k!}}\cdot\binom{n}{k}\right)=\\
\frac{1}{\binom{2n}{n}}\left(\sum\limits_{k=0}^n\color{blue}{\frac{(n-1)!}{(n-k-1)!k!}}\cdot\binom{n}{k}\right)=
\frac{1}{\binom{2n}{n}}\left(\sum\limits_{k=0}^n\color{blue}{\binom{n-1}{k}}\cdot\color{red}{\binom{n}{k}}\right)=\\
\frac{1}{\binom{2n}{n}}\left(\sum\limits_{k=0}^n\binom{n-1}{k}\cdot\color{red}{\binom{n}{n-k}}\right)=\frac{1}{2}$$
With this in mind what I tried to do is consider that every face in the die has a probability of $\frac{1}{n}$ to show up and to consider every posible value that $r$ can have as follows:
$$P(B)=\sum\limits_{r=1}^n\frac{1}{n}\frac{1}{\binom{2n}{r}}\left(\sum\limits_{k=0}^r\frac{n-k}{n}\binom{r}{r-k}\binom{r}{k}\right)$$
Where you are considering that you choose $r$ balls from the total of $2n$ to remove and you consider each possible $k$ such that in those $r$ balls you removed exactly $k$ tangerine balls. But I'm not sure to how can i get to the $\frac{1}{2}$ answer, so I would appreciate a lot the help
 A: First, let us find the conditional probability of choosing a tangerine ball given that the roll of the die was $r$. I think you will find the proof here is quite similar to the proof you provided in your post.
$$
\begin{align}
P(\text{choose tangerine}\mid \text{roll} = r)
&=\frac1{\binom{2n}r}\sum_{k=0}^r \frac{r-k}{r}\binom{n}{r-k}\binom{n}{k}
\\&=\frac1{r\binom{2n}{r}}\sum_{k=0}^{r-1} \color{blue}{(r-k)\binom{n}{r-k}}\binom{n}{k}
\\&=\frac1{r\binom{2n}{r}}\sum_{k=0}^{r-1} \color{blue}{n\binom{n-1}{r-k-1}}\binom{n}{k}
\\&=\frac{n}{r\binom{2n}{r}}\binom{2n-1}{r-1}=\frac12
\end{align}
$$
Since the conditional probability of drawing a tangerine is $1/2$ for every possible roll of the die, it follows that the overall probability is $1/2$. In detail,
$$
P(\text{choose tangerine})=\sum_{r=1}^n P(\text{choose tangerine}\mid \text{roll = r})P(\text{roll }=r)=\sum_{r=1}^n \frac12\times \frac1n=\frac12.
$$
A: There is no preference for any particular ball to be chosen. Each of the $2n$ balls in the urn is equally likely to come out of the bag at the end of the process. So since half the balls are tangerine, there is a probability of $\frac12$ that a tangerine ball is chosen.
