Probability: $n$ pairs of shoes You have $n$ pairs of shoes, for a total of $2n$ shoes, which you randomly place in a circle.  Let $E_i$ denote the event that shoes of the $i^\textrm{th}$ pair are next to each other for $i= 1,...,n.$
a) Find $P(E_i)$.
b) For $i \ne j$, find $P(E_j|E_i)$.
c) Approximate the probability, for $n$ large, that no two shoes from the same pair are next to each other.
For part (a), I figured that if you fix one shoe in spot $1$, its pair can be on the left or right for $2$ total arrangements. The first shoe can be in any of $2n$ locations, so the probability should be $P(E_i)=2\cdot \frac{2n}{(2n)!}.$ Is this correct? 
Don't know how to even approach (b) or (c).
 A: Let's change the question to a straight line. So we just need to remember that shoes at first and last place are also considered together.

A) total cases =2n!,
Favourable case : (i) normally say you want to keep 2 particuar  shoes together in a line. Which is 2!(2n-1)!
(ii) one shoe of the pair at first and one shoe at second.
 Which is 2!(2n-2)!
Answer is  $\frac{2}{2n-1}$.

Lets do it for if we want 'r' particular pairs of shoes together
That is 
(i) for normal line $2^r$(2n-r)!
(ii) for atleast one of them is first and last position and other are altogether (remember we have to choose the pair which comes at first and last position) = $2^r$(r)(2n-r-1)!
Total ways (i) + (ii) = $2^r$(2n-r-1)![ 2n-r + r ] = $P_r$
${n \choose r}\ 2^r $ (2n-r-1)!= $T_r$

$P_1$ = 4n(2n-2)!
$P_2$ = 8n(2n-3)!
So the answer to b) part = $\frac{P_2}{P_1}$ = $\frac{2}{2n-3}$

C) by inclusion and exclusion the probability that atleast one shoe is together is 
$S= \frac{1}{(2n-1)!} \sum_{r=1}^n (-1)^{r+1} T_r $ .
$\frac{T_r}{T_{r+1}} = \frac{r+1}{2} \times \frac{2n-r-1}{n-r}$ 
Notice that both the fraction increases as r increase, and 2nd  fraction's rate of increase also increase . (To feel it check out $\frac{9}{5} , \frac{8}{4} ,\frac{7}{3}...$).
So as 'r' increase we can see the terms become more and more neglectable in repect to intial terms . 
$\frac{T_1}{T_2} \approx r+1 = 2$  ($\frac{2n-r-1}{n-r} \approx 2 $ for large n , and when r becomes comparable to n we need not consider the terms because it will beome very less in comparison to first term , you will get the feel in few lines)
$\frac{T_1}{T_3} \approx 2.3 = 6$ and so on...
So $ S \approx \frac{1}{(2n-1)!} T_1(1 - \frac{1}{2} + \frac{1}{6}  - \frac{1}{24} .......) $
(There could be a better approximation  i dont know if there is, i would like to now if there is )
$ S \approx \frac{2n(2n-2)!}{(2n-1)!} (1-\frac{1}{e}) \approx 1-\frac{1}{e}$
Answer is actually 1-S = $\frac{1}{e}$
A: This is really an extended comment on Aryan Bansal's answer, but I'm too bad a typist to type it in a comment box.  Part (c) is called the relaxed ménage problem.  OEIS 3435 lists the number of different solutions for $n=2$ to $n=14$.  Apparently, the number of solutions is the same as the number of directed Hamiltonian cycles on an $n$-cube with a fixed starting node.  So far, I can't see how the two problems are related.   
Simulation agrees well with Aryan's suggestion of a probability of $\frac1e$.  For example, a series of tests with $200$ pairs of shoes any $10,000$ trials gave me results from around $.365$ to about $.372.$
Furthermore, the "Formula" section of the OEIS page cited above gives the asymptotic formula$$
a_n\sim\sqrt{\pi}2^{2n+1}n^{(2n+1)/2}e^{-(2n+1)}
$$
Dividing through by $(2n)!$ and using Stirling's formula in the denominator gives precisely $$\frac{a_n}{(2n)!}\sim\frac1e$$
