In a box are n red socks and n+1 blue socks. I choose 2 socks. Use a proof to show that I am twice as likely to pick one of each rather than 2 blue. 
In a box are $n$ red socks and $n+1$ blue socks. I choose $2$ socks. Use a proof to show that I am twice as likely to pick one of each rather than $2$ blue.

I am unsure which combinatorial proof to use. This looks like pigeonhole but I don't know how I would use it.
 A: There is no need to find the actual probabilities:

*

*If you choose a blue sock first (without replacement) then you are then equally likely to choose a blue or a red sock second since there remain $n$ of each. So blue followed by blue is as likely as blue followed by red.


*But the socks are exchangeable, so the total probability of a mixed pair is twice the probability of a blue sock followed by a red sock,  and so (by 1) twice the probability of an all blue pair.
A: The denominator will remain the same in both cases, so just comparing numerators is enough

*

*For blue pair, numerator $= (n+1)n$


*For mixed socks which can come as $BR\; or\; RB,\;\; 2*(n+1)n \;\;...[QED]$
A: I would say this is what a combinatorial argument would look like.
Number the red and blue socks: $r_1,\cdots,r_n,b_1,\cdots,b_{n+1}$ (respectively). So when we take two socks out we will have one of the possible pairs $(r_i, r_j)$, $(r_i, b_k)$, or $(b_i, b_j)$, where $i≠ j$ and the possible values are in the obvious ranges in each case. Now count how many combinations of each there are:
For two reds we have $\frac{n(n-1)}{2}$, for two blues $\frac{(n+1)n}{2}$, and for one blue, one red we have $n(n+1)$.
From now it's clear what to do.
A: One red one blue is
$\dfrac{\binom{n}{1}\binom{n+1}{1}}{\binom{2n+1}{2}}$ which is
$\dfrac{n+1}{2n+1}$.
Two blue is $\dfrac{\binom{n+1}{2}}{\binom{2n+1}{2}}$ which is $\dfrac{n+1}{2(2n+1)}$
and we proved that $P(1 $and$ 1)$=$2P(2 red)$!
