There are 3 couples sitting randomly round a 6-seater circular table. What is the probability that all the husbands and wives sit next to each other?

My attempt:

First wife, say, takes any of the six seats. That leaves 2/5 seats where her husband can sit next to her. Second wife, say, can take any of the four remaining seats. There is then only 1 seat out of the remaining 3 where her husband can sit next to her AND leave two empty adjacent seats for the last couple.

So the answer is 2/5 * 1/3 = 2/15.

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    $\begingroup$ Looks reasonable to me. $\endgroup$ – Scott Caldwell Jul 10 '14 at 15:16
  • $\begingroup$ @ScottCaldwell Thank you $\endgroup$ – oks Jul 10 '14 at 15:18
  • $\begingroup$ math.stackexchange.com/search?q=husband+probability many similar questions. $\endgroup$ – Emily Jul 10 '14 at 17:39
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    $\begingroup$ So 13/15 of the time, someone is sleeping on the couch. $\endgroup$ – corsiKa Jul 10 '14 at 22:26
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    $\begingroup$ 100%. "Don't you sit next to that little tramp - get over here next to me where I can keep an eye on you!". $\endgroup$ – Bob Jarvis Jul 10 '14 at 22:52

The general case:

Consider $n$ groups of $k$.

Because the table is round there is a cyclic symmetry.

The total number of permutations is therefore $(nk-1)!$.

Each group has $k!$ permutations.

The total number of permutations for the groups is $(n-1)!$.

So we get

$$ \frac{k!^n (n-1)!}{(nk-1)!}. $$

Case $n=3$ and $k=2$ gives

$$ \frac{2!^3 (3-1)!}{(2\cdot3-1)!} = \frac{2}{15}. $$



6 people can sit around a circular table in $(6-1)! = 5!$ ways

Fix each couple as one. They can be permuted in 2 ways. There are three couples, so each one of them can be seated in $2^3$ ways as couple. Now the three couples are three people and they can be placed in around the circular table in $2!$ ways.

Now the required probability $= \frac{2^3.2}{5!} = \frac{2}{15}$

This is one way you can solve it.




I think you might do too the following:

6 persons can sitting randomly 6! ways.

3 couples (whole) can sitting randomly 3!*2 ways. Each of the pairs may sit in two ways (wife, husband or husband, wife) - that's $2^3=8$ ways.

So I would say that the answer is $\frac{3!\cdot 2\cdot 8}{6!}=\frac{2}{15}$

$\color{red}{So\, the\, solution\, is\, not\, correct - see\, comments.}$

  • $\begingroup$ Except you need to multiply the top by two. 3 couples can sit randomly $3!$ ways, but they could be seats 1-2, 3-4, 5-6 or seats 2-3, 4-5, 6-1. $\endgroup$ – Duncan Jul 10 '14 at 17:14
  • $\begingroup$ @Duncan: Thanks for the warning - fixed. $\endgroup$ – georg Jul 10 '14 at 17:22
  • $\begingroup$ The table is round - so you only have $5!$ $\endgroup$ – johannesvalks Jul 10 '14 at 17:25
  • $\begingroup$ @johannesvalks: I think not. 6 seats --> 6! $\endgroup$ – georg Jul 10 '14 at 17:33
  • $\begingroup$ @georg, Your answer is right, but what johannesvalks said is true, n people can sit in a circular table in (n-1)! ways from a traditional argument to take care of rotation. $\endgroup$ – Satish Ramanathan Jul 10 '14 at 17:38

So 2/15 if they all sit randomly, let’s say like with blind folds on. The actual probability depends on their wants and priorities. Does A actually not want to sit next to his wife A1? Does B1 want sit next to C? And so on. The probability depends entirely on the individuals involved. And yes, this particular answer was partly dependant on that I just could not do the math. :-)


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