# Probabilities Involving 4 Married Couples

 P: 1,367 Probabilities Involving 4 Married Couples OK. Here I go. By the inclusion-exclusion identity: \begin{align*}P(E_1 \cup E_2 \cup E_3 \cup E_4) & = P(E_1) + P(E_2) + P(E_3) + P(E_4) \\ & \quad - P(E_1E_2) - P(E_1E_3) - P(E_1E_4) - P(E_2E_3) - P(E_2E_4) - P(E_3E_4) \\ & \quad + P(E_1E_2E_3) + P(E_1E_2E_4) + P(E_1E_3E_4) + P(E_2E_3E_4) \\ & \quad - P(E_1E_2E_3E_4)\end{align*} By symmetry, \begin{align*}P(E_1) &= P(E_2) = P(E_3) = P(E_4) \\ P(E_1E_2) &= P(E_1E_3) = P(E_1E_4) = P(E_2E_3) = P(E_2E_4) = P(E_3E_4) \\ P(E_1E_2E_3) &= P(E_1E_2E_4) = P(E_1E_3E_4) = P(E_2E_3E_4)\end{align*} so the equation reduces down to $$P(E_1 \cup E_2 \cup E_3 \cup E_4) = 4P(E_1) - 6P(E_1E_2) + 4P(E_1E_2E_3) - P(E_1E_2E_3E_4)$$ Define $f(n)$ as the number of ways to position a couple given $n$ row positions. $f(n) = 2(n-1)$ since the couple may occupy positions 1 & 2, 2 & 3, ..., or $n - 1$ & $n$ and the husband or wife may be on the left. $P(E_1)$ is easy to determine: there are $f(8)$ ways to position the first couple and 6! ways to position the rest so $P(E_1) = f(8) \cdot 6! / 8! = 1/4$. Determining $P(E_1E_2)$ is more complicated than $P(E_1)$ since the row positions available to the second couple depend on where the first couple is. In other words, $$P(E_1E_2) = \sum_{i=1}^7 P(E_1E_2|F_i)$$ where $F_i$ is the event that the first couple occupies row positions $i$ and $i+1$. For $i = 1$, the first couple has two available positions: 1 & 2. The second couple has the rest available: 3, ..., 8. The two couples may be positioned in $f(2)f(5)$ = 20 ways. The rest may be positioned in 4! ways. Thus $P(E_1E_2|F_1) = 20\cdot4! / 8! = 1 / 84$. By symmetry, $P(E_1E_2|F_7) = P(E_1E_2|F_1)$. Applying the prior argument for the rest of the values of $i$ yields that \begin{align*}P(E_1E_2|F_2) &= P(E_1E_2|F_6) = 16\cdot4! / 8! = 1/105 \\ P(E_1E_2|F_3) &= P(E_1E_2|F_5) = 1/105 \\ P(E_1E_2|F_4) &= 1/105 \end{align*} Putting it all together yields that $P(E_1E_2) = 1/14$. $P(E_1E_2E_3)$ is found in a similar, but more complicated manner to $P(E_1E_2)$, i.e. the positions available to the third couple depend on where the first and second couple are situated: $$P(E_1E_2E_3) = \sum_i \sum_j P(E_1E_2E_3|F_iG_j)$$ where $G_j$ represents the event that the second couple occupies positions $j$ and $j+1$, $j > i+1$ or $j < i - 1$. Let $N_{ij}$ be the number of ways of situating three couples given $F_i$ and $G_j$. The table below shows some values: $$\begin{tabular}{c|c|l} $$i$$ & $$j$$ & $$N_{1j}$$ \\ \hline 1 & 3 & $$f(2)f(2)f(4) = 24$$ \\ 1 & 4 & $$f(2)f(2)(f(1) + f(3)) = 16$$ \\ 1 & 5 & $$f(2)f(2)(f(2) + f(2)) = 16$$ \\ 1 & 6 & $$f(2)f(2)(f(1) + f(3)) = 16$$ \\ 1 & 7 & $$f(2)f(2)f(4) = 24$$ \\ 2 & 4 & $$f(2)f(2)(f(1) + f(3)) = 16$$ \\ 2 & 5 & $$f(2)f(2)(f(1) + f(1) + f(2)) = 8$$ \\ 2 & 6 & $$f(2)f(2)(f(1) + f(2) + f(1)) = 8$$ \\ 2 & 7 & $$f(2)f(2)(f(1) + f(3)) = 16$$ \\ 3 & 1 & $$f(2)f(2)f(4) = 24$$ \\ 3 & 5 & $$f(2)f(2)(f(2) + f(2)) = 16$$ \\ 3 & 6 & $$f(2)f(2)(f(2) + f(1) + f(1)) = 8$$ \\ 3 & 7 & $$f(2)f(2)(f(2) + f(2)) = 16$$ \\ 4 & 1 & $$f(2)f(2)(f(1) + f(3)) = 16$$ \\ 4 & 2 & $$f(2)f(2)(f(1) + f(3)) = 16$$ \\ 4 & 6 & $$f(2)f(2)(f(3) + f(1)) = 16$$ \\ 4 & 7 & $$f(2)f(2)(f(3) + f(1)) = 16$$ \\ \end{tabular}$$ Let $N_i$ be the number of ways of situating three couples given $F_i$. The table below summarizes the table above for all values of $i$. $$\begin{tabular}{c|l} $$i$$ & $$N_i$$ \\ \hline 1 & 96 \\ 2 & 48 \\ 3 & 64 \\ 4 & 64 \\ 5 & 64 \\ 6 & 48 \\ 7 & 96 \end{tabular}$$ where by symmetry, $N_i$ for $i=5,6,7$ are the same for $i=3,2,1$ respectively. Thus, $$P(E_1E_2E_3) = 2!(2 \cdot 96 + 2 \cdot 48 + 3 \cdot 64) / 8! = 1/42$$ Finally, there are 4! ways to order all four couples such that all couples are together. Hence, $P(E_1E_2E_3E_4) = 4!/8! = 1/1680$. And so, $$P(E_1 \cup E_2 \cup E_3 \cup E_4) = 4\cdot\frac{1}{4} - 6\cdot\frac{1}{14} + 4\cdot\frac{1}{42} -\frac{1}{1680} = \frac{1119}{1680}$$ Now obviously I did something wrong somewhere, but where?
 P: 70 You are definately on the right track but it seems you fell into error in computing the very last term in your application of the inclusion-exclusion formula. Otherwise your work seems correct. In any event I'll run through my reasoning of this problem thus we may compare notes. For P(E1) we have the case where one couple sits together. If we think of the couple which sits together as a single entity there are 7! ways they can be permuted among the remaing 6 persons. In other words I am looking at the couple that sits together as a single entity (as 1 person) permuted among 6 others, in other words a permutation of 7 "persons". In other words a permutation of the group C P1 P2 P3 P4 P5 P6 where C represents the couple that sits together... and P1 ... P6 represent the remaining persons who do not sit together. Now bear in mind for any couple that sits together there are 2! = 2 ways they can be permuted among each other. Thus for couple C consisting of a man/wife pair there are 2! permatations of the pair. Therefore we have. $4P(E1) = \frac{(4)(2)(7!)}{8!} = \frac{(4)(2)}{8} = \frac{8}{8} = 1$ which is what you got for the first term . Now for the case where 2 couples sit together we reason the same way. Thinking of each couple that sits together as a single entity we have 6! permutations of the 2 couples and the remaining 4 persons. each of the couples that sit together can be permuted in 2! = 2 ways therefore $6P(E1E2) = \frac{(6)(2^2)(6!)}{8!} = \frac{3}{7}$ We keep applying similar reasoning and get $4P(E1E2E3) = \frac{(4)(2^3)(5!)}{8!} = \frac{2}{21}$ Thus far you are correct Now for that last term If all 4 couples sit together there are 2! = 2 permutations among each couple. There are 4! possible permutations of the couples among each other. i.e. a permuation of the 4 couples: C1 C2 C3 C4 Furthermore each husband/wife pair can be permuted 2! ways and there are 4 couples. Therefore we have.. $P(E1E2E3E4) = \frac{(2^4)(4!)}{8!} = \frac{1}{105}$ Plugging into the inclusion-excusion formula we get $P(E1 \cup E2 \cup E3 \cup E4) = 1 - \frac{3}{7} + \frac{2}{21} - \frac{1}{105} = \frac{23}{35}$ It was that last term that messed things up. Otherwise you seem to have reasoned the problem correctly. Skins
 P: 70 Yes, that does help to think of each couple that sits together as a single entity permuted among the remaining persons. As soon as I considered solving it that way the calculations became a lot simpler. As a matter of fact each of the terms $$4P(E1), 6P(E1E2), 4P(E1E2E3) and P(E1E2E3E4)$$ can be defined via the expression: $C(4,n)\frac{2^n(8-n)!}{8!}$ where n is the number of married couples sitting together. Thus for example: $P(E1E2) = C(4,2)\frac{2^2(8-2)!}{8!}$ Anyway, glad to be of help in solving this problem. I wish there were forums like this around back in the days when i was learning this stuff. Skins