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By random seating, we mean the following:

  1. Take the first couple. The husband has twenty choices in his seat, and the wife has nineteen.
  2. Now, take the second couple. The husband has eighteen choices, and the wife has seventeen.
  3. We continue with the other couples in the same manner.

The answer to the question is $\tfrac{30}{19}$ which suggests that the probability that any particular couple is seated together is $\tfrac{3}{19}$. For each $i \in \{ \, 1, 2 \ldots 10 \, \}$, and define: $$X_i = \begin{cases} \ 1 \ \ & \text{if the } i \text{-th couple is seated together} \\ \ 0 & \text{otherwise} \end{cases}$$ If it is true that $E(X_i) = \tfrac{3}{19}$, then the expected number of married couples sitting together equals: $$E(X_1 + X_2 \ldots + X_{10}) \ \ = \ \ \tfrac{30}{19}$$ Is there an easy way confirm that $E(X_i) = P(X_i = 1) = \tfrac{3}{19}$? The case with $X_1$ is simple, but the conditional probabilities soon get complicated ....

  1. For the first couple, the husband may sit any where. Then, the wife has $3$ out of $19$ chances to be seated at the same table. Thus, $P(X_1 = 1) = \tfrac{3}{19}$.
  2. For the second couple. With conditioning, we have: $$\tfrac{3}{19} \left[ \, \tfrac{2}{18} \tfrac{1}{17} + \tfrac{16}{18} \tfrac{3}{17} \, \right] + \tfrac{16}{19} \left[ \, \tfrac{3}{18} \tfrac{2}{17} + \tfrac{3}{18} \tfrac{2}{17} + \tfrac{12}{18} \tfrac{3}{17} \, \right] \ \ = \ \ \tfrac{3}{19}$$
  • 2
    $\begingroup$ "the case with $X_1$ is simple, but . . ." You seriously think the odds of Mr. and Mrs. Jones sitting at the same table depends on whether you choose to call them "couple number one" or "couple number three"? $\endgroup$ – bof May 21 '17 at 3:53
  • $\begingroup$ So the order in which they are seated is irrelevant? No, I did not try to think of it that way. $\endgroup$ – Andy Tam May 29 '17 at 5:36
  • $\begingroup$ Nothing in the problem statement suggests to me that people are seated sequentially, one by one. I understood "random seating" to mean that all possible seating arrangements are equally likely. $\endgroup$ – bof May 29 '17 at 5:42
  • $\begingroup$ Ok, fair enough .... Thanks. $\endgroup$ – Andy Tam May 29 '17 at 6:03

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