Suppose we have 4 couples and 8 seats. The seats are oriented such that it is a row. The people take their seats in a random fashion.

What is the probability that 2 couples sit next to each other?

My work: Let us label the people as $A_1, A_2, A_3, A_4, B_1, B_2, B_3, B_4$. $A_i$ and $B_i$ are couples. Then, we want to know the probability of $P(C_i \cap C_j), i \neq j$ and $C_i$ means couple $i$ are seated next to each other.

Then, suppose we fix $C_i$ and $C_j$. Then, the row arrangement could look something like: $\{ C_i, C_j, a,b,c,d\}$, where $a,b,c,d$ are the remaining 4 people. Then, there are $\binom{6}{2}$ ways of picking positions for $C_i$ and $C_j$, $2!$ ways of arranging the elements in $C_i$, $2!$ ways of arranging the elements in $C_j$, $2!$ ways of arranging $C_i$ and $C_j$ relative to each other, and $4!$ ways of arranging the remaining 4 people. There are $8!$ ways to arrange the 8 people in a row. Also, there are $\binom{4}{2}$ ways to pick 2 couples out of 4 couples.

Putting all this together, $$ P(C_i \cap C_j) = \frac{\binom{6}{2} \binom{4}{2} 2! 2! 2! 4! }{8!}$$

I am curious because this differs from the solution given here: cached solution from google

Note that on the bottom of page 5, the difference is the google solution is missing a $\binom{6}{2}$ term, which means we can choose 2 positions for $C_i$ and $C_j$ out of the available 6 positions. (Please observe the quantity $p_2$ in the linked document). The factor $\binom{4}{2}$ is accounted for later on the top of page 6, where the author sums up all the probabilities corresponding to possible indices $i$ and $j$.



I would agree with you that the cached answer looks wrong and that there is a missing factor of 15 in $p_2$ (and of 20 in $p_3$).

When you ask "What is the probability that 2 couples sit next to each other?" you are looking at what was originally the $p_2$ part of the "What is the probability that at least one of the wives ends up sitting next to her husband?" question.

$p_2$ is the probability that two particular couples sit with husband next to wife. As you say, the ${4 \choose 2} = 6$ for which two couples these are is included in the $4p_1-6p_2+4p_3-p_4$ formula.

I would then say that the two particular couples have each become one unit (with $(2!)^2$ ways of arranging the individuals within the couple), reducing the number of orderable units from $8$ to $6$ so we should have $$p_2 = \frac{(2!)^2 \; 6!}{8!} \approx 0.071428571 $$ and similarly $p_1 = \frac{(2!)^1 \; 7!}{8!} = 0.25$, $p_3 = \frac{(2!)^3 \; 5!}{8!} \approx 0.023810$, and $p_4 = \frac{(2!)^4 \; 4!}{8!} \approx 0.00952381$.

For $p_2$ and $p_3$, these are not the same as the cached answer, though the $p_2$ is the same as yours. Using the inclusion-exclusion formula this gives me a probability that at at least one of the wives ends up sitting next to her husband of about $0.657142857$, not the $0.9666667$ of the cached answer.

Some simulation convinces me this looks plausible.

  • $\begingroup$ Hi Henry: Thanks, this is the exact answer, I understand it, and I can also verify this in MATLAB. For the purpose of learning, I have included the MATLAB simulation code here: pastebin.com/UXjPEDaE . All of the different scenarios can be checked by making modifications to the code. $\endgroup$ – jrand Jun 5 '11 at 17:15

Then, there are $\binom{6}{2}$ ways of picking positions for $C_i$ and $C_j$

I think that this is where you're wrong. In fact, there are only 5 (as in $\binom{5}{1}$) ways of placing them (considering that you account for order later):

  • $(C_iC_j,a,b,c,d)$
  • $(a,C_iC_j,b,c,d)$
  • $(a,b,C_iC_j,c,d)$
  • $(a,b,c,C_iC_j,d)$
  • $(a,b,c,d,C_iC_j)$
  • 1
    $\begingroup$ Hi trutheality, you have shown 5 possible arrangements. Remember that $C_i$ actually contains two elements - it means couple $i$ is seated next to each other. Then, this is a possible arrangement: $(C_i, a,b,c,d,C_j)$. To be more clear, $(C_i, a,b,c,d,C_j) = (A_i, B_i, X_a, X_b, X_c, X_d, A_j, B_j)$. The X's must be set so that they refer to exactly the four remaining people. $\endgroup$ – jrand Jun 4 '11 at 22:30
  • $\begingroup$ I think that we understand the question differently: I took "two couples seated next to each other" to mean that not only do we have $A_i$ next to $B_i$ and $A_j$ next to $B_j$, but also $C_i$ next to $C_j$. If you drop the last requirement then your answer is correct. $\endgroup$ – trutheality Jun 4 '11 at 22:34
  • $\begingroup$ In any case, if we use your interpretation, where pairs of couples must be seated next to each other, the probability should be: $P(C_i \cap C_j) = \frac{5 \binom{4}{2} 2! 2! 2! 4! }{8!}$. I have replaced $\binom{6}{2}$ by $5$. This is still different from the claimed answer in the link. $\endgroup$ – jrand Jun 4 '11 at 22:36
  • $\begingroup$ I think you're right then, the answer in the link doesn't seem to account for the different ways of arranging couples and non-couples. $\endgroup$ – trutheality Jun 4 '11 at 22:48

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