Ok my suggestion is the following.
You surely know how to use the hint: If you denote the probability of not having a matching term by $P$ then the probability of having at least one matching pair is ... (I'll leave it to you to think about this one)
Now you might want to use the following formula.
Let $\Omega$ be a (finite) probability space and $A\subset \Omega$ an event, then for the probability of $A$ holds$$
P(A)=\frac{|A|}{|\Omega|}
$$
That means the probability of $A$ is the number of all results such that $A$ has happened divided by the total number of possible results.
Now you have to compute those cardinalities: $\binom{n}{k}$ denotes the number of possibilities to draw $k$ items out of $n$ items when you do not care about the order in which the items were drawn (which is the case in your problem).
So $|\Omega|=\binom{16}{6}$ in your case, because you draw 6 socks out of 16 socks in total.
Furthermore there are $16 \cdot 14 \cdot 12 \cdot 10 \cdot 8 \cdot 6$ possibilities to draw $6$ socks without having a matching pair when taking into account the order (why?). Furthermore you have $6!$ possibilities to arrange $6$ objects on $6$ slots, so you have to divide by that number. I'll leave the rest to you. I hope this was helpful (and I didn't make a mistake somewhere).
EDIT: I just read your comment asking to explain the $16 \cdot 14 \cdot 12 \cdot 10 \cdot 8 \cdot 6$ part. Well this is just counting socks. I'll call the socks that result in the event "no matching pairs" as good socks and any sock preventing that result a bad sock.
- In the first round you cannot get a matching pair obviuosly (as you'd
need $2$ socks for a pair) so the number of good socks is 16.
- In the second round you have only 15 socks left, however one of those is of the same color as the one you already have and therefore a bad sock. Makes in total 15-1=14 good socks
- round three: you have 14 socks left. Among those are 2 bad socks (the corresponding sock to your first and second pick) so you have 14-2=12 socks left.
- and so on.
You also asked for a way to write this with binomials. Look at the following$$
\frac{\binom{2}{1} \cdot \binom{2}{1} \cdot \binom{2}{1} \cdot \binom{2}{1} \cdot \binom{2}{1} \cdot \binom{2}{1} \cdot \binom{2}{0} \cdot \binom{2}{0} \cdot \binom{8}{2}}{\binom{16}{6}}
$$
What do I want to tell you with that fraction? Well we now consider pairs. One pair consists of $2$ socks and we draw $6$ socks in total. That means we want exactly one sock out of $6$ pairs (therefore $6$ times $\binom{2}{1}$) and no sock out of the resulting $2$ pairs (therefore the two $\binom{2}{0}$). Of course you could just omit the $\binom{2}{0}$ term as it is $1$ anyway, I just put it there for illustration. Ok and now you have $\binom{8}{2}$ possibilities to choose "which pair you don't get any sock from" ( I hope that sounds remotely logical).
If you type it in a calculator you'll find that the two results are the same.
Best Answer
First of all, your probability of drawing socks of the same color is not quite right. After we draw one sock, there are $n/2-1$ socks of the same color remaining, and $n-1$ total socks remaining, so the probability of picking a matching sock is $$\frac{n/2-1}{n-1}=\frac{n-2}{2n-2}$$ not $\frac{n}{2(n-1)}$. Now we may write the expected value of the number of times you need to pick socks as the following infinite series: $$\sum_{i=1}^\infty \bigg(1-\frac{n-2}{2n-2}\bigg)^{i-1}\cdot\frac{n-2}{2n-2}\cdot i$$ which is not divergent, since $1-\frac{n-2}{2n-2}\in (0,1)$. Can you compute the value of this sum?