Classic $2n$ people around a table problem A total of $2n$ people, consisting of $n$ married couples, are randomly seated (all possible orderings being equally likely) at a round table. Let $C_i$ denote the event that the members of couple $i$ are seated next to each other, $i = 1,...,n$
a) Find $P(C_i)$
Attempt: So this is the prob that one particular couple sit next to each other. There are $2n$ seats for the first person. Given that the table is round, there are 2 seats for the wife/husband, (either to the left or right) and $(2n-2)!$ rearrangements of the remaining people.  Since each of the orderings are equally likely, $|s| = (2n)!$ and the prob we want is $(2n) \cdot 2 \cdot (2n-2)! / (2n)! = 2/(2n-1) $ Is it a good argument?
b) For $j \neq i,\, \text{find} \, P(C_i|C_j)$ 
Attempt: This is the prob that given some couple $j$ already sitting next to one another, what is the probability that some other couple $i$ are sitting next to each other. By definition, this is equal to $P(C_i, C_j)/P(C_j)$. I already computed the denominator in a), so I need only worry about the numerator. 
For the numerator: If couple $i$ and $j$ are to sit next to each other, there are $(2n)$ places for the first person and $2$ choices for the next person. For the other couple, I am not really sure what to say since if one member of couple $j$ sits next to a member of couple $i$, then there is only one place for the other member of couple $j$. But, if the couple $j$ do not sit any where near couple $i$ then there is more than one place for the other member. It seems reasonable to compute therefore, $$P(C_i,C_j) = P(C_i,C_j| \text{one member of j next to i})P(\text{one member of j next to i}) + P(C_i,C_j|\text{member of j not next to i})P(\text{member of j not next to i})$$ Does this make sense and is my approach good or not?
c) Approximate the probability, for $n$ large, that there are no married couples who are seated next to each other. 
Attempt: $P(\text{no married couples next to each other}) = 1-P(\text{at least one couple sit next to each other})$I know the approximation will be Poisson since n is large, but I am not sure where to go from here. Thanks!
 A: a) The probability that a particular wife has her husband sitting next to her is $\dfrac{2}{2n-1}$ since she has two neighbours.
b) You can regard couple $i$ together as breaking the circle so that the question now involves a row of $2n-2$ people.  These can be arranged in $(2n-2)!$ ways.  But the number of ways they can be arranged if couple $j$ sit together is $2(2n-3)!$ since we could treat couple $j$ as a single person, but doubling the number as they can sit either way round.  So the probability is $\dfrac{2}{2n-2}$.
c) Going back to (a), the expected number of couples sitting together is $\tfrac{2n}{2n-1}$ which for large $n$ approaches $1$.  Using your Poisson approximation [which also uses the almost independence between couples illustrated by the answer to (b) being close to the answer for (a)] with an expectation of $1$, the limit of the probability of no couples together is $e^{-1}\approx 0.3678794$.  For a similar question (couples in a row rather than a circle) see Showing probability no husband next to wife converges to $e^{-1}$
A: This solution covers parts a) and b).
To do so, let us consider a simpler question: A total of $2n$ people, consisting of $n$ married couples, are randomly divided into $n$ pairs. Let $W_{i}$ denote the event in which the members of couple $i$ are paired together, where $i=1,\dots,n.$
Now, the question involves a linear arrangement instead of a circular arrangement. As such, we have that
$$P(W_{i}) = \frac{(2n-2)! \times 2 \times n}{(2n)!}=\frac{1}{2n-1},$$
where $(2n-2)!$ describes the linear arrangement of the $2n-2$ people not in couple $i,$ $2$ describes the ways in which to pair the people in couple $i,$ and $n$ describes the ways in which to slot the couple to form the linear arrangement.
What about $P(W_{i} \cap W_{j})?$ Similarly, we have that
$$P(W_{i} \cap W_{j}) = \frac{(2n-4)! \times 2 \times n \times 2 \times (n-1)}{(2n)!}=\frac{1}{(2n-1)(2n-3)}.$$
Hence,
$$P(W_{i}|W_{j}) = \frac{P(W_{i} \cap W_{j})}{P(W_{j})} = \frac{1}{2n-3}.$$
But this only covers the simpler linear arrangement question. What about the original circular arrangement question? For that I follow the circular arrangement solution from n-f-taussig.
That is,
$$P(C_{i}) = \frac{2 \times \frac{(2n-1)!}{2n-1}}{\frac{(2n)!}{2n}}=\frac{2}{2n-1},$$
where $2$ describes the ways in which to fuse the people from couple $i$ into a single "person," $\frac{(2n-1)!}{2n-1}$ describes the circular arrangement of the $2n-1$ "people," and $\frac{(2n)!}{2n}$ describes the circular arrangement of all of the $2n$ people.
What about $P(C_{i} \cap C_{j})?$ Similarly, we have that
$$P(C_{i} \cap C_{j}) = \frac{2 \times 2 \times \frac{(2n-2)!}{2n-2}}{\frac{(2n)!}{2n}}=\frac{4}{(2n-1)(2n-2)}.$$
Hence,
$$P(C_{i}|C_{j}) = \frac{P(C_{i} \cap C_{j})}{P(C_{j})} = \frac{2}{2n-2}.$$
