This is the problem statement (First Course in Prob., Sheldon Ross, 8th Ed., Q 4.67):
A total of $2n$ people, consisting of $n$ married couples, subject to the constraint that the men and women alternate, 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 $P(C_i)$.
(b) For $j \neq i$, find $P(C_j | C_i)$.
(c) Approximate the probability, for $n$ large, that there are no married couples who are seated next to each other.
I am assuming couples are man-woman here.
For part $(a)$, I let either of the two gender sit first (whom to choose first does not matter in circular arrangement) in $(n-1)!$ and the remaining people in $n$ places in $n!$ ways. Thus total ways is $n!(n-1)!$. Now, let women sit first in $(n-1)!$, followed by husband of $C_i$ in $2$ ways, followed by $(n-1)$ men-spaces in $(n-1)!$ ways. Thus, $P(C_i) = \frac{2 \cdot ((n-1)!)^2}{(n-1)! \cdot n!} = \frac{2}{n}$
For $(b)$, I have seen solution for Problem 4.67 as well as in the solution manual:
Conditioning on whether the man of couple j sits next to the woman of couple i gives the result: $\frac{1}{n-1} \cdot \frac{1}{n-1} + \frac{n-2}{n-1} \cdot \frac{2}{n-1} = \frac{2n-3}{(n-1)^2} $
I do not understand either of the 2 solutions for part $(b)$ . Regarding how they are conditioning on the couple $C_i$ and $C_j$ sitting together, or how is that permutation working. This part $(b)$ is what I need help for.
Kindly explain in the absolute basic terms. Thanks you! {{{(＞﹏＜)}}}
More Intuitive Part A
Two women sits next to husband $i$: one on the left and one on the right. Out of $n$ women, what's the probability that the one sitting on his left is his wife? $\frac{1}{n}$ right? How about on his right? Also $\frac{1}{n}$. Therefore the probability for part $A$ is
$$ \frac{1}{n}+\frac{1}{n}=\frac{2}{n} $$
Part B
The couple $i$ sits together. Husband $j$ cannot sit next to husband $i$, so he either $(1)$ sits next to wife $i$ or $(2)$ he sits away from the couple $i$.
Scenario $(1)$
Husband $i$ sits on one side of wife $i$. From the remaining $n-1$ men, what's the probability that the one sitting on the other side of wife $i$ is husband $j$? $\frac{1}{n-1}$ right?
Now, wife $i$ sits on one side of husband $j$. From the remaining $n-1$ women, what's the probability that the one sitting on the other side of husband $j$ is his own wife? $\frac{1}{n-1}$ right?
This is the origin of the first product in the answer: $\frac{1}{n-1}\cdot\frac{1}{n-1}$
Scenario $(2)$
Husband $i$ sits on one side of wife $i$. From the remaining $n-1$ men, what's the probability that the one sitting on the other side of wife $i$ is NOT husband $j$? $\frac{n-2}{n-1}$ right?
Now husband $j$ sits away from the couple $i$. On the left and right of him sit one woman each. From the $n-1$ remaining women (excluding wife $i$), what's the probability that the one sitting on his left is his wife? $\frac{1}{n-1}$ right? How about on his right? Also $\frac{1}{n-1}$.
This is the origin of the second product in the answer: $\frac{n-2}{n-1}\cdot\left(\frac{1}{n-1}+\frac{1}{n-1}\right)$