Socks in the dark

The light is out and I must dress in the dark. The drawer holds ten black socks and ten white socks, all mixed. I need two socks of the same colour; I do not mind which.

Dark. The drawer: $10$ black ($B$) and $10$ white ($W$), $|B|+|W|=20$. I need $2$ of the same colour.

$|B|=|W|=10$; least $n$ to be sure of two of one colour.

How many must I take to be sure? Take one: it is black or white. Take a second: the two may differ. Take a third: it must match one of the first two, because there are only two colours and three socks. So three are enough, and two are not.

Take $n=1$: any colour. $n=2$: they may differ. For $n=3$ there are more socks than colours ($3>2$), so two match (the pigeonhole principle). So $n=3$ is enough and $n=2$ is not.

$f:\{1,2,3\}\to \{B,W\}\Rightarrow \exists s\ne t,f(s)=f(t)$.

What if I need two black socks? Now I can be unlucky for a long time: all ten white socks may come first. Only after them is every sock black, and I need two black ones: twelve socks.

For two black ones the worst case is all $10$ white first, then $2$ black: $10+2=12$ socks.

Two black: $n_B=10+2=12$.

The surprise: with three colours of socks I would need four, and with ten colours eleven. Always one more than the number of colours.

For $k$ colours I need $k+1$ socks: $k=3\Rightarrow n=4$ and $k=10\Rightarrow n=11$.

$n_k=k+1,n_3=4,n_{10}=11$.