Statistics problem about Possibility principle I have a question about the following problem from the first edition of Introduction to Probability by Joseph K. Blitzstein and Jessica Hwang. This is problem 68 from Chapter 4.

Each of 111 people names his or her 5 favorite movies out of a list of
11 movies.
Show that there are 2 movies such that at least 21 of the people name
both of these movies as favorites.

The possibility principle states that

The possibility principle: Let A be the event that a randomly chosen
object in a collection has a certain property. If P(A) > 0, then there
exists an object with the property.

Is it enough to state that the probability of two specific movies being chosen by at least 21 people is ${111\choose 21}{(2/11)}^{21} $ (where 2/11 is the probability of the two movies being chosen by a specific person) and therefore since this probability is greater than 0 then by the possibility principle there exist two  movies that are chosen by at least 21 people? But then it could also be said that there are 5 movies that are chosen by all 111 people.
 A: The given problem describes a procedure -- which is not supposed to be random -- that produces a list of $111$ (not necessarily distinct) size-$5$ subsets of $\{1,\ldots,11\}.$ Let's denote such a list as $s=[s_1,\ldots,s_{111}],$ and, for any two numbers $(a,b)$, define
$$X(a,b,s):=\text{number of elements in list $s$ that contain both $a$ and $b$}.$$ We're asked to prove the following:
$$\forall s\,\exists(a,b):X(a,b,s)\gt 20\tag{*}$$
where the quantifiers $\forall$ and $\exists$ abbreviate "for all" and "there exists", respectively.
Now we consider how to make $X$ a random variable to give meaning to $P(X>20)$ in order to apply the possibility principle. This can be done by "randomizing" $(a,b);$ i.e., we consider a random pair of numbers $(A,B)$ and nonrandom lists $s$, so the possibility principle gives the following:
$$\begin{align}\forall s\,P(X(A,B,s)>20)>0\\
\implies\forall s\,\exists (a,b): X(a,b,s)>20\end{align}$$
which is just the form needed for solution (*).
NB: It would not be appropriate randomize $s$, i.e. to consider a random sample $S$, not only because the people are not choosing randomly, but also because in that case the possibility principle would only give, for specified $(a,b)$,
$$\begin{align}P(X(a,b,S)>20)>0\\
\implies\exists s: X(a,b,s)>20\end{align}$$
which is not the form needed for solution (*).
Now, we need to assign a distribution to $(A,B)$ for which we can in fact show
$$\begin{align}&\forall s\,P(X(A,B,s)>20)>0\end{align}.$$
By letting $(A,B)$ be a pair of numbers chosen at random (without replacement) from $\{1,\ldots,11\},$ the resulting hypergeometric distribution allows an easy calculation using indicator functions and the linearity of expectation. Thus, defining the indicator functions $$I(a,b,s_i)=\begin{cases}1&\text{if both $a$ and $b$ occur in set $s_i$}\\
0&\text{otherwise.}\end{cases}$$
for any list $s=[s_1,\ldots,s_{111}],$ we have
$$X(a,b,s)=\sum_{i=1}^{111} I(a,b,s_i)$$
hence
$$\begin{align}E\,X(A,B,s)
&=E\,\sum_{i=1}^{111} I(A,B,s_i)\\
&=\sum_{i=1}^{111} E\,I(A,B,s_i)\\
&=111\,E\,I(A,B,s_1)\\
&=111\,P(\text{Both $A$ and $B$ occur in $s_1$})\\
&=111\,{\binom{5}{2}\binom{11-5}{0} \over \binom{11}{2}}\quad\text{(hypergeometric)}\\
&=111\,{2\over 11}\\
&=20.1\ldots
\end{align}$$
Thus we have shown that $\forall s\,EX(A,B,s)>20$, and hence that $\forall s\,P(X(A,B,s)>20)>0$; consequently, $\forall s\,\exists(a,b): X(a,b,s)>20$, as required.

Generalization:
If $n$ people each name $k$ favorites from a list of $m$ movies, then there are $2$ movies such that at least $p$ people name both of these movies as favorites, where $$p = \left\lceil n{k(k-1)\over m(m-1)}\right\rceil.$$
