Here is the problem, which comes from the probabilistic method section of Introduction to Probabilty.
A group of 100 people are assigned to 15 committees of size 20,such that each person serves on 3 committees. Show that there exists two committees that have at least 3 people in common.
My solution:
Pick one of the committees and call it $A$. Suppose we randomly select one of the 14 remaining committees, and let $X$ be the random variable that is the number of people in $A$ that are also in the randomly selected committee. Then $X = I_1 + I_2 + \cdots + I_{15}$ where each $I_i$ is an indicator variable that is $1$ if the $i$th person in $A$ is also in the randomly selected committee or $0$ otherwise. Then the expected value each $I_i$ is just the probability that the $i$th person in $A$ is also in the randomly selected committee, which is $\frac{2}{14}$ since each person is in exactly $2$ other committees other than $A$. Then $E[X]$ = $15\times \frac{2}{14} = \frac{15}{7} > 2$.
However, the given solution is slightly different:
Suppose we pick both committees at random, and let $X$ be the number of people in both committees. Then $X = I_1 + \cdots + I_{100}$ where each indicator variable $I_i$ is $1$ if person $i$ is in both committees and $0$ otherwise. Then $E[I_i] = \frac{\binom{3}{2}\binom{12}{0}}{\binom{15}{2}} = \frac{1}{35}$, so $E[X] = \frac{100}{35} = \frac{20}{7} > 2.$
My question is why do we arrive at different values for $E[X]$? It seems nonintuitive that they have would be different values given that in both solutions $X$ just measures the size of the overlap of the two committees, and I can't find any reason that fixing one rather than selecting both at random would decrease the expected overlap.
Best Answer
This comment correctly answers this question.