Expected number of times the mode occurs in a collection of finite sets

Question

Let $A$ be the set of all subsets of $[100]:=\{ 1,2,\ldots, 100 \}$ of size $10,$ i.e. $x\in A\iff x\subset [100]$ AND $ \vert x \vert = 10.$ Note then that $\vert A \vert = \binom{100}{10}.$

For any (sub-)collection, E, of members of $A,$ define the maximal order of $E$ to be the number of times that the number most common to all members of $E$ occurs.

Let $k\in\mathbb{N}.$ Across all (sub-)collections $E,$ of $A,$ with $\vert E\vert = k,$ each (sub-)collection (of size $k$) being equally likely chosen, what is the expected maximal order of $E$ in terms of $k?$

This is even interesting for $k=2,$ with me not making much progress, other than going through all the options like, finding the probability that two members of $E$ have no numbers in common, finding the probability that two members of $E$ have one number in common, and so on. But this is tedious. I would be more interested in a good approximation for the expected value, if the maths is easier.

Or maybe there is a similarly-formulated problem that more well-known or easier to tackle?

@BrianTung No; I mean that $\vert E\vert = k.$ I will edit this in. — Adam Rubinson, Jun 19 '24 at 15:26
Oh, OK. So the size of each subset is fixed at $10$, but the number of subsets is $k$. The "of size $k$" in the question was a little confusing to me. — Brian Tung, Jun 19 '24 at 15:28
Yes that is correct. I have just updated the question. If you feel like you could make it read better than it currently does, please feel free to do so. — Adam Rubinson, Jun 19 '24 at 15:29
I applied the "tedious" method you describe for $k=2$, with the help of Wolfram Alpha, where I needed to put the probability of no intersection inside the sum and divided by $10$ to make Wolfram Alpha compute it, and obtained about $1.6695 = 0.83475k$. — Fabius Wiesner, Jun 19 '24 at 18:32

Expected number of times the mode occurs in a collection of finite sets

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