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.
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.