# solution by iterated expectation

Considering the problem

There were n couples attending a ceremony. Out of the 2n people, exactly m of them ate pizza, chosen randomly over all subsets of 2n people. Let X be the number of couples that didn’t eat pizza. Find an explicit expression for the mean of X.

There seems to an easy and correct solution by taking an indicator RV for each couple (expectation of which easily calculated), and summing them all for n couples.

However I took a different approach by using the law of iterated expectation as follows:

Let X be the number of couples that didn't eat pizza, and Y be the number of the men that had pizza, noting that $Y$ is a binomial random variable with $p=\frac m{2n}$. We thus are going to use the law of iterated expectations to calculate mean of X: \begin{align*} \mathbf E[X] &= \mathbf E[\:\mathbf E[X|Y]\:] \\ \mathbf E[Y] &= m/2 \\\mathbf E[Y^2] &= \mathbf {var}[Y]+\mathbf E[Y]^2 = m/2(1-m/2n)+m^2/4\end{align*} To find $\mathbf E[X|Y]$, we need to determine the expected number of women that didn't eat pizza while their husbands are among the presumed group of men that didn't have pizza.\ We assume $y$ men had pizza. There are $n$ women, each with probability of $\frac{n-y}{n}$ to be the wife of a man that didn't eat pizza. Also, independently, this woman won't have a pizza with a probability of $1 -\frac{m-y}{n}$. Thus the probability of a couple not having pizza equals: $$p=(1-\frac{y}{n})\cdot (1-\frac{m-y}{n})$$ Thus, we have a binomial random variable with the calculated p, hence the mean is: $$\mathbf E[X|Y] = np = \frac{(n-Y)(n-m+Y)}{n}$$

\begin{align*} \mathbf E[X] &= \mathbf E[\:\mathbf E[X|Y]\:] > \\&=\mathbf E\left[\:\frac{(n-Y)(n-m+Y)}{n}\:\right] \\&= > \frac 1n\left(n^2-mn+m\mathbf E[Y]-\mathbf E[Y^2]\right) > \\&=n-m+m^2/(4n)+m^2/(4n^2)-m/(2n) \end{align*}

which does not lead to the correct answer, if there's no mistake in calculation parts I hope. :) Are the steps taken correct, apart from calculation parts? Would you please tell me where the fallacy is, if you see any?

PS. I believe that my assumption for considering X|Y as a binomial with the calculated p is not correct, since it allows it in probability for no women to be wives of the presumed group of men!

I believe that my assumption for considering X|Y as a binomial with the calculated p is not correct, since it allows it in probability for no women to be wives of the presumed group of men!

So can any alternative assumptions be made to yield the correct expectation?