probability of extracting consecutively an even numbered ball and the following numbered one An urn contains 2n numbered balls that are extracted one by one until no one remains. Which is the probability that for some integer $k \geq 1$ the $(2k-1)$-numbered ball is extracted immediately before the $(2k)$-numbered one? 

I tried using the inclusion-exclusion principle. For one single consecutive couple the probability is $\frac{1}{(2n)!} \cdot (2n-1) \cdot (2n-2)!=\frac{1}{2n}$. For two couples I can use the conditional probability and calculate that:

*

*if the first ball of the first consecutive couple is extracted in one of the n possible odd places, then the chance of extracting a second consecutive couple is $ \frac{ (2n-2) \cdot (2n-4)!}{(2n-2)!}=\frac{1}{2n-3} $;


*if the first ball of the first consecutive couple is extracted in one of the $n-1$ possible even places, then the chance of extracting a second consecutive couple is $ \frac{ (2n-3) \cdot (2n-4)!}{(2n-2)!}=\frac{1}{2n-2} $; 

therefore the chance of extracting two couples of consecutive (odd then even) balls is $\frac{1}
{2n} \cdot \left(\frac{n}{2n-3}+\frac{n-1}{2n-2} \right)$. 
Then I tried generalizing to the extraction of more consecutive couples and using the inclusion-exclusion principle, but I got stuck trying to write down the formula that looks quite long with multiple summations and products. I'm convinced that there's a simpler way to solve the problem but this looked the most promising way to me, considering the complementary seemed worse.
 A: The inclusion-exclusion setup itself isn’t too bad if you think in terms of the $(2n)!$ permutations of the balls (i.e., the possible sequences of draws).
For $k\in[n]$ let $A_k$ be the set of permutations in which $2k-1$ immediately precedes $2k$. Then the inclusion-exclusion principle says that
$$\begin{align*}
\left|\bigcup_{k\in[n]}A_k\right|&=\sum_{\varnothing\ne I\subseteq[n]}(-1)^{|I|+1}\left|\bigcap_{k\in I}A_k\right|\\
&=\sum_{k=1}^n(-1)^{k+1}\binom{n}k(2n-k)!\,.
\end{align*}$$
To see that last step, note that there are $\binom{n}k$ $k$-element subsets $I$ of $[n]$, and if we treat each pair $\{2k-1,2k\}$ for $k\in I$ as a single element, there are $2n-k$ elements to be permuted: the pairs $\{2k-1,2k\}$ for $k\in I$, and the remaining $2n-k$ single elements. The desired probability is then
$$\frac1{(2n)!}\sum_{k=1}^n(-1)^{k+1}\binom{n}k(2n-k)!\,,$$
which can be rewritten
$$\sum_{k=1}^n(-1)^{k+1}\frac{\binom{n}k}{k!\binom{2n}k}\,,$$
for what it’s worth. Unfortunately, I do not at the moment see how to evaluate this in a closed form.
