A box contains $n$ white and $n$ black pebbles. A fair sided die with $n$ sides(numbers from $1$ to $n$) is thrown and the number you get, let's call it $r$, is how many pebbles are taken out randomly.
One pebble is chosen randomly from the box, what is the probability that the chosen pebble is black?
(preferably using as simple probabilty as possible, no binomial distribution and stuff like that, it's sadly the basic stuff that's bothering me.)
My attempt:
So I assume we need $\sum_{r=1}^{n} P(r)\cdot \sum_{k=0}^{r} P($k black pebbles are taken out$)\cdot P($black pebble chosen from the remaining$)$
$P(r) = 1/n$
And then, for a fixed $r \in \{1,...,n\}$:
$P($k black pebbles are taken out$) = \frac{\binom{n}{k}\cdot \binom{n}{r-k}}{\binom{2n}{r}}$
$P($black pebble chosen from the remaining $2n - r$ $) = \frac{n-k}{n-r}$
But I'm really not sure how to make anything useful out of this. Thank you in advance!
EDIT: I know the answer is $1/2$, but I want to know how to reach it!
The summation you need to simplify is $$ \sum_{r=1}^n\frac1n\sum_{k=0}^r\frac{\binom{n}{k}\binom{n}{r-k}}{\binom{2n}{r}}\cdot \frac{n-k}{n-r}=\frac1n\sum_{r=1}^n\frac{1}{\binom{2n}{r}(n-r)}\sum_{k=0}^r\binom{n}{k}\cdot (n-k)\cdot \binom{n}{r-k} $$ Since $\binom{n}{k}(n-k)=\binom{n-1}{k}n$, this reduces to $$ \sum_{r=1}^n\frac{1}{\binom{2n}{r}(n-r)}\sum_{k=0}^r\binom{n-1}{k}\binom{n}{r-k} $$ Using Vandermonde's$^*$ identity, the inner summation is equal to $\binom{2n-1}r$. Combined with the fact that $\binom{2n}{r}(n-r)=\binom{2n-1}{r}2n$, this becomes $$ \sum_{r=1}^n\frac{\binom{2n-1}r}{2n\binom{2n-1}r}=\sum_{r=1}^n\frac1{2n}=\frac12. $$
$^*$ The Wikipedia link gives several proofs of Vandermonde's identity, but here is one with a probabilistic flavor. The probability that a hypergeometric random variable with parameters $2n-1,n-1$ and $r$ is equal to $k$ is $\binom{n-1}k\binom{n}{r-k}/\binom{2n-1}r$. Summing over all possible values of $k$, the result must be one.