How many pairs of shoes? This is quite standard but I am not sure if this is right. Given $n$ people, each of them 
takes off his/her shoes and puts to a sack. We choose randomly two shoes for each person. What is the chance that each time we get a pair of shoes?
Well, it looks to me that since we have $n$ people, the chance is
$\left( \frac{n}{2n \choose 2}\right)^n = \tfrac{1}{(2n-1)^n}$
Is this right? What if we demand to get exactly the same pair of shoes?
 A: Imagine taking out the shoes one at a time. The probability the second shoe matches the first is $\frac{1}{2n-1}$. Given that this happened, the probability that the fourth matches the third is $\frac{1}{2n-3}$. Given that we got a match on the first two, and the second two, the probability the sixth matches the fifth is $\frac{1}{2n-5}$. And so on. 
So the required probability is 
$$\frac{1}{2n-1}\cdot\frac{1}{2n-3}\cdots \frac{1}{3}\cdot \frac{1}{1}.\tag{1}$$
We can simplify (1) by multiplying top and bottom by $(2n)(2n-2)\cdots (2)$, that is, by $2^n n!$. We get $\frac{2^n n!}{(2n)!}$. 
If we want everyone to get the right pair of shoes, line up the people and let each take out one shoe, then another. The probability the first pick is correct is $\frac{2}{2n}$. Given that it was right, the probability the second was right is $\frac{1}{2n-1}$. If the first person got lucky, the probability the first shoe taken out by the second person is right is $\frac{2}{2n-2}$, and the probability that the second shoe is then right is $\frac{1}{2n-3}$. 
Continue, multiply, and simplify. We get $\frac{2^n}{(2n)!}$. 
Remark: There are "quicker" ways to solve both problems. But I believe that an analysis of the type described above should come first, and then we can look for a more "elegant" solution. 
Added: Here is a quicker way to solve the first problem. There are $(2n)!$ equally likely ways to line up the shoes. Now we count the "favourables," where for every $k$, the $(2k-1)$-th and $(2k)$-th shoes are a pair. 
There are $n!$ ways to line up the pairs, and for each pair the left shoe can be first or second, giving a total of $n!2^n$ favourables. That gives probability $\frac{n!2^n}{(2n)!}$. 
A: For the first part of your question, it appears it is $100$% since no matter what $2$ shoes you choose, it will be a pair of shoes.
For the 2nd part of your question, is this right, the answer is no.
For the 3rd part of your question, what if we demand to get exactly the same pair of shoes... Snowballs chance in hell if n is about $5$ or more.  Actually your question is ambiguous.  By exactly do you mean the same $2$ shoes that the person put "to" the sack or just any matched pair?
