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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?

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    $\begingroup$ The reason for it being unnecessary is that when calculating expected value we do not actually need to know the pdf of the distribution. We can proceed via the linearity of expectation. Here, we found the probability that a specific couple (say, the Jone's family) were seated together and we multiplied that probability by the number of couples total to get the expected value. This because the expected number of couples seated together can be expressed as a summation of much more easily calculated random variables, each describing whether or not a particular couple is seated together. $\endgroup$ – JMoravitz Oct 15 '18 at 17:29
  • $\begingroup$ I am still not sure if I understand. Isn't the probability that my partner will sit at my table $=$ the number of combinations in which my partner does sit at my table divided by the number of total combinations, in which case we should take into account combinations of the other tables as well? $\endgroup$ – agblt Oct 15 '18 at 17:32
  • $\begingroup$ When calculating probabilities via counting, you may set your sample space to be however descriptive or however nondescriptive as you like so long as the outcomes in the sample space are equally likely and the event you are interested in calculating probability of is a subset of the sample space. Here, to be efficient we only considered the sample space as being the different ways in which the table where Mr. Jones is seated is populated. In doing so, we don't bother ourselves with the other tables muddying up our calculations. $\endgroup$ – JMoravitz Oct 15 '18 at 17:36
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    $\begingroup$ Consider an experiment where we flip a coin, throw a die, and buy a lottery ticket and we want to find the probability we flipped a head. While we could have if we wanted considered our sample space as all ways in which a coinflip, diceroll, and lottery ticket could have been purchased, the information about the dice roll and lottery ticket are entirely irrelevant. As such, when forming our sample space we need only concern ourselves with the coin and we can describe our sample space here as having only two equally likely outcomes, simplifying counting considerably. $\endgroup$ – JMoravitz Oct 15 '18 at 17:38
  • $\begingroup$ Ah, that does make sense now... thank you! $\endgroup$ – agblt Oct 15 '18 at 17:39
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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 $10\cdot{3\over19}=1.579$.

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  • $\begingroup$ I see, this a way to arrive at the probability while avoiding the whole discussion of all the combinations of people at the table. $\endgroup$ – agblt Oct 15 '18 at 18:55

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