As @MikeEarnest wrote, the usual rules of the game mean everyone will be called infinitely often (with prob $1$), and then the details of the random model / adversarial behavior of the jailer will merely affect the finish time of the algorithm.
Instead, you're asking if the problem can be solved assuming some prisoner only appears a finite no. of times, right? I think the answer is No. First, lets be more precise.
Let $P = \{1, 2, \dots, 100\}$ denote the prisoners.
Let $x_t \in P$ be the prisoner called on day $t$.
Let $X_t = \{x_n: n \le t\}$ be the set of prisoners who has been called (at least once) by day $t$.
Define an infinite sequence $X = (x_1, x_2, \dots)$ to be legal by your rules if there exists $T \in \mathbb{N}$ s.t. by day $T$ everyone has appeared at least once, i.e. $X_T = P$.
Then, to solve your version of the problem means to have some deterministic distributed algorithm $A$ s.t. for any legal sequence $X$ as input, the prisoners go free at a finite time $T'(X)$. It is allowed that $T'$ be a function of $X$ (i.e. of $T$, or any other aspect of $X$).
Here is one key point: if algorithm $A$ must work for any legal input $X$, that means an adversary can first look at $A$, then decide on $X$, and $A$ must still work.
It probably takes a lot of work to formalize an algorithm, so I will (sadly) keep that part vague. Subject to that caveat, here is proof that an algorithm (that works for any legal input) cannot exist.
I will prove that an "easier" version of this problem is still unsolvable: the jailer told all the prisoners, truthfully, that a specific prisoner S (for Spartacus!) will be called infinitely often (each of the others will still be called at least once, i.e. $X$ is legal).
Now suppose S first get called at $t=1000$. He can see the lightbulb on or off.
Case $1$: everyone has been called already, i.e. $X_{1000} = P$. Without loss of generality, lets say for such an input the lightbulb will be on at $t=1000$. (How the others contrive it, I don't know, but lets assume they can do it.)
In this case S must eventually call for freedom, say at time $T'$, which is a constant hard-coded into the algorithm for this class of inputs, because otherwise the algorithm would fail for the legal input where $\forall t > 1000: x_t = S$ i.e. he will be called every day from now on.
So we have established that:
- If S gets called first on $t=1000$, sees the lightbulb on, and gets called from $t=1001$ to $T'$, then he calls for freedom on day $T'$.
This immediately also means, if $X_{1000} \neq P$, the lightbulb needs to be off. (Again, how the others contrive it, I have no idea.) Because if the lightbulb were on while $X_{1000} \neq P$, then for the legal input where S gets called between $t=1001$ to $T'$ while all remaining prisoners get called beyond $T'$, S would erroneously call for freedom on $T'$.
Case $2$: $X_{1000} \neq P$ and prisoner $V \notin X_{1000}$ i.e. V (for Valjean) has not yet been called.
We have established that in Case $2$, S sees the lightbulb off at $t=1000$. Lets say he leaves the lightbulb in state $E \in \{on, off\}$. Now lets construct the rest of $X$: lets say $x_{1001} = V$ and $x_{1002} = S$, at which time S can see the lightbulb in state $F \in \{on, off\}$. With this extra bit he needs to distinguish between these cases:
Case $2.1$: $X_{1001} = P$, i.e. V, the new guy, is also the last guy and now everyone has been called.
Case $2.2$: $X_{1001} \neq P$, i.e. V, the new guy, is not the last guy and someone is still not called.
In 2.1, S needs to call for freedom at some fixed time $T''$ (for the algorithm to work in case the legal input is s.t. he gets called forever from now on) whereas in 2.2, S must not call for freedom at $T''$ (even if he gets called between $t=1002$ to $T''$). That's why S needs to distinguish 2.1 vs 2.2, and he can only do so based on the bulb status $F$ when he sees it again at $t=1002$.
However, as far as V is concerned, he has never been called until $t=1001$, he sees the bulb in state $E$, and he has no idea if the legal input so far is case 2.1 or 2.2. So in a deterministic algorithm he must make the same choice for what to do with the bulb and leave it in the same state $F$ for both 2.1 and 2.2. Therefore, S cannot distinguish between 2.1 and 2.2. I.e. no matter how the algorithm is designed, S must either do the wrong thing in case 2.1 (never calls for freedom) or in case 2.2 (calls for freedom erroneously).
In conclusion: no deterministic algorithm can work for all legal inputs.
