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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!

CAF
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  • (a) is correct but too complicated. Pretend that wife #$i$ chose her seat first among all $2n$ people and husband #$i$ chose next. It doesn't matter where she sits; hubby has $2n-1$ places to choose from at random, and only two of them put him next to his wife. – Dilip Sarwate Dec 08 '12 at 15:55
  • That's nice. Can you help me at all with parts b), c)? – CAF Dec 08 '12 at 16:29
  • Try developing it in your own. Couple #$j$ is already seated next to each other. Now let wife #$i$ choose next, and break up the problem into two parts: she sits next to the already seated couple, and she sits apart from the already seated couple. – Dilip Sarwate Dec 08 '12 at 17:07
  • So what I wrote was good for b)? I wanted to verify that the equation I wrote was right before I started computing the terms. Thanks. – CAF Dec 08 '12 at 17:19
  • For (b), if couple $j$ are neighbours, the rest of the table is essentially a line. There are $\binom{n-2}{2}$ equally likely ways to choose the (unordered) pair of chairs occupied by couple $i$, and $n-2-1$ of these choices are "good." – André Nicolas Dec 08 '12 at 18:06
  • What do you mean by '$n-2-1 $ choices are 'good''. Is the eqn correct that I wrote down in my question? – CAF Dec 08 '12 at 18:37
  • I have the correct answer for b) now. Any hints on how I may get c)? Thanks! – CAF Dec 09 '12 at 11:00

2 Answers2

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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 $e^{-1}$

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Henry
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  • I'm a bit confused on how you get the expected value for part c). – Vedaad Shakib Sep 07 '16 at 07:26
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    @VedaadShakib If from (a) you have the probability that a particular wife has her husband sitting next to her is $\dfrac{2}{2n−1}$ and there are $n$ couples, then by linearity of expectation the expected number of couples sitting together is $n \times \dfrac{2}{2n−1}$ – Henry Sep 07 '16 at 07:31
  • Thanks for the speedy response. So do you assume that the values of $P(C_i)$ are independent? – Vedaad Shakib Sep 07 '16 at 07:37
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    @VedaadShakib You do not need independence for that expectation. You do need to assume it for CAF's Poisson approximation – Henry Sep 07 '16 at 07:43
  • I think that this argument that $EX=2n/(2n-1)$ is at least obscure. IMO, the correct way to argue the poisson aproximation with $\lambda=1$ is to observe that, by b) we have a kind of almost independence, so we can see the random variable X=number of couples seated next each other as a poisson variable with parameters $(n,p)=(n, 1/n)$. – Eduardo Mar 14 '22 at 00:14
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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, $$P(C_{i}|C_{j}) = \frac{P(C_{i} \cap C_{j})}{P(C_{j})} = \frac{2}{2n-2}.$$

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