The solution to this problem,
Ten married couples are to be seated at five different tables, with four people
at each table. Assume random seating, what is the expected number of
married couples that are seated at the same table?
says that the probability of any individual married couple sitting at a table together is a Bernoulli random variable and can be calculated with
$$P(X_i=1) = {{{18}\choose2}\over{{19}\choose3}}$$
where ${19}\choose3$ is all the combinations of people that can join one husband from couple $i$ in the three remaining places at the table and ${{18}\choose2}$ is the all the combinations of people that can join a couple $i$ in the two remaining places at the table.
My question is, why are there not more possibilities to consider because of the many combinations of people sitting at the other tables? For example, although there is only ${19}\choose3$ combinations of people that can join one husband from couple $i$ in the three remaining places at the table, is there not ${{16}\choose4}{{12}\choose4}{{8}\choose4}{{4}\choose4}$ combinations of people who can sit at the other tables, multiplying the possibilities?
Is this because these possibilities cancel each other out?
Do we say
$${{{18}\choose2}{{16}\choose4}{{12}\choose4}{{8}\choose4}{{4}\choose4}\over{{19}\choose3}{{16}\choose4}{{12}\choose4}{{8}\choose4}{{4}\choose4}}= {{{18}\choose2}\over{{19}\choose3}}$$
or is there a reason why that cannot be done? Or is there a reason it is unnecessary?
Best Answer
Consider Mrs. Whitespoon. She is sitting at a certain table. Her husband was allocated one of the $19$ other seats with each of them equally probable. Three of these seats are favorable. Therefore Mr. Whitespoon sits at the same table as his wife with probability ${3\over19}$. Since there are $10$ such husbands, by linearity of expectation, the expected number of husbands sitting at the same table as their wives is $10\cdot{3\over19}=1.579$.