I have noticed your question not so long ago while looking for an answer to a question similar to yours. I was interested in how to devise an optimal strategy for picking floors for dropping eggs by maximizing the entropy. Funny how a similar thought crystallized in separate places at almost the same time. I did not find the answer though and see that there were no responses here, so let me share my thoughts on this for a lack of anything better.
I found a way of computing the entropy, which can be seen as a useful test for a proper solution if it ever comes by. It also shows that there a solution for the entropy exists, but possibly in terms of special functions. This solution also showed me that the problem is far from trivial and will require more thought than I can invest for now.
In this problem, the entropy can be thought of as the smallest number of yes/no questions that reveal the number of the first floor where the egg doesn't break if we are given $N$ eggs. The problem can be solved by recursion in the spirit of this post.
Lets denote the distance we can explore with the said number of eggs in $t$ trials as $d_N(t)$. If we have $F$ floors, then the ceiling integer of the positive root of $d_N(t_H) = F$ is the answer. It is not evident that there will be only one of those, but lets roll with it. I have also added an index $H$ in the last expression to indicate that this $t_H$ is the entropy for given $N$ and $F$.
For the case of $N=1$, all is pretty simple. We have 1 egg, so we have to test 1 floor at a time and $d_1(t) = t$, which is one of terminating relations to the recursion I am about to introduce. The second one is $d_N(t) = d_t(t)$ when $N > t$, i.e. a situation occurs that we have more eggs than we could possibly need.
For $N$ eggs, we have $d_N(t) = 1 + d_{N-1}(t-1) + d_{N}(t-1)$. Please read the blog post referenced above for the intuition behind it.
Let me say as a small digression that in the $N > t$ case, one naturally gets the logarithmic search: $$d_N(t) = d_t(t) =\\ 1 + d_{t-1}(t-1) + d_{t}(t-1) =\\ 1 + d_{t-1}(t-1) + d_{t-1}(t-1) =\\ 1 + 2d_{t-1}(t-1) =\\ 1 + 2(1 + 2d_{t-2}(t-2)) = ...= 2^t -1.$$
Solving $2^{t_H} -1 = F$ for $t_H$, we get $t_H = \log_2{(F-1)}$.
Getting back to $N < t$, the best thing I came up with now was unpack the recursion relations for a few more eggs until a pattern emerged. Being very slow to catch on, I had to go up until 5... I will not test your patience and will just give the final answer. It is expressed as
$$d_N(t) = \frac{1}{N!}(t)_N + \sum_{i=1}^{N-1}\frac{2^{i-1}}{(N-i)!}(t-i)_{N-i} + 2^{N-1} - 1,$$
where $(t)_N$ is the falling factorial. Solving $d_N(t_H) = F$ gives the entropy for arbitrary $N$ and $F$. The resulting polynomial can be expressed in terms of Bernoulli polynomials or Hurwitz zeta functions. I tested this numerically and the answers match the tabulated answers in this post.
I realize that I got very wordy and did not answer your question precisely. I hope though that I showed that the entropy can be found for this problem in terms of special functions and that it will be a hairy expression.
