Probability of 2 married couples sitting in a circle

Question

I am not able to figure out where I am going wrong. The question is this -

A total of $2n$ people, consisting of $n$ married couples, are randomly seated (all possible orderings being equally likely) at a round table where men and women are to sit alternately. Let $C_i$ denote the event that the members of couple $i$ are seated next to each other, $i = 1, . . . , n$.

(a) Find $P(C_i)$.

(b) For $j \neq i$, find $P(C_j|C_i)$.

For part (a), my reasoning goes as follows - Pick a man and place him at a particular position. Then, his partner can be choosen in $1$ way and she can be seated in $2$ ways. Then remaining $n-1$ men and $n-1$ women can be seated in $(n-1)!(n-1)!$ ways. The total number of ways of seating $n$ men and $n$ women is $n!(n-1)!$. So, probability is $2/n$.

For part (b), I am starting in same way - Pick a man and place him at a particular position. Then, his partner can be choosen in $1$ way and she can be seated in $2$ ways. Now pick another man, which can be done in $n-1$ ways and his partner, and they can be seated in $n-1$ positions (as there are $n-1$ remaining positions for men and women to sit in alternate positions). Remaining $n-2$ men and $n-2$ women can be seated in $(n-2)!(n-2)!$ ways. So,

$P(C_i\cap C_j) = 2(n-1)^2(n-2)!(n-2)!/(n!(n-1)!) = 2/n$

and $P(C_i) = 2/n$ (from part (a) )

So, $P(C_j|C_i) = P(C_i\cap C_j)/P(C_i) = (2/n)/(2/n) = 1$

I don't understand here where I am going wrong.

Answer 1

(a) Fix one of the partners in the couple. The other one is equally likely to be in any of the remaining $2n-1$ places, of which $2$ are next to the first partner, so $P(C_i)=\frac2{2n-1}$.

(b) If couple $i$ are seated next to each other, that leaves $2n-2$ seats in a row. All selections of two of these seats for couple $j$ are equally likely. There are $\binom{2n-2}2$ such selections, and $2n-3$ of them have adjacent seats, so $P(C_j\mid C_i)=\frac{2n-3}{\binom{2n-2}2}=\frac{2(2n-3)}{(2n-2)(2n-3)}=\frac1{n-1}$.