The probability of a sequence of conditional probabilities with a strict ordering Let's assume that we are observing a sequence of choices by an individual. Each choice situation is indexed by $t = 1,..., T$ and ordered such that a choice made in $t = 1$ precedes a choice made in $t = 2$. Furthermore, we cannot observe a choice in $t = 2$ unless a particular choice was made in $t = 1$. As an observer, I am interested in calculating the probability of observing a particular sequence of choices made by an individual.
The problem can be structured as a decision tree like this:

It is obvious from the structure of the problem that observing a choice in $t = 2$ is conditional on choosing $S_1$ and that observing a choice in $t = 3$ is conditional on choosing $S_1$ and $S_2$ in $t = 1$ and $t = 2$ respectively. To solve this I thought of using Bayes' theorem, which states that the conditional probability of $A$ given $B$ is:
$$
P(A|B) = \frac{P(B|A)P(A)}{P(B)}
$$
To put this in the context of the current problem, let $A$ be the probability of observing a choice in $t$ and $B$ be the probability that you chose $S$ in $t-1$. Now, $P(B|A) = 1$ because the probability that you chose $S$ in $t-1$ conditional on us observing a choice in period $t$ is known with certainty given the strict ordering on $t$. This means that $P(A|B)$, i.e. the probability of observing a choice conditional on choosing $S$ in the previous period, reduces to the ratio $P(A)/P(B)$. However, there is no guarantee that $P(A) < P(B)$ which means that $P(A|B)$ is no longer bound by the unit interval. Now this creates obvious problems.
Let us look at a numerical example to illustrate the practical implications of this. In $t=1$ the probability of choosing $S$ is .57, in $t=2$ it's 0.34 and in $t=3$ it is .73. The probabilities of the others vary, which can happen, but they are specifically chosen extreme here to illustrate the problem.

Now, applying Bayes' theorem above, I get the following:

Where $P(A|B)$ in period $t-1$ is $P(B)$ in period $t$. Finally, the probability of observing the sequence of choices is the product over the conditional probabilities. However, given the likelihood of getting $P(A|B) > 1$, I am concerned that this may not be the correct application of the theorem or even if it is possible in sequence like this. In practice, the sequence of observed choices can be very long.
 A: You're incorrect when you state that "there is no guarantee that () < ()". I think the problem boils down to using Bayes' rule, when you really just need conditional probability.
When you mention (), it's important to remember that it is the overall probability of being in the state  at a given time, t. That is, it's the probability of being in , out of all states at time t, even those that didn't originate from . Why? Because () isn't conditional on anything. We don't know anything about what preceded it. (It could have been , or it could have been ~.) Intuitively, I can only reach state  after reaching state  (by the strict ordering). So, the overall probability for  will be less than or equal to the overall probability for .
As an example, when I know that an event strictly precedes another, we can calculate that using conditional probability. For example, Let's say I have a state, , which is sometimes followed by a state . There's no way to reach , except through . Let's further state that  is always preceded by some state, C. C happens at t=0,  at t=1, and  at t=2. C -> B -> A
C occurs as the first step, so there's no conditional probability, we just have (C). Then, () = (|C)*(C). Now, by definition, (|C)*(C) = ( & C). But because we know C already happened if we reached , ( & C) = (). We follow a similar process to find () = (|)*(). You'll note that when we chain these together, we get () = (|)*(|C)*(C). We can continue this chaining for higher values of t.
A: Interesting problem. You have made small mistakes in defining probabilities. For example:

To put this in the context of the current problem, let $A$ be the probability of observing a choice in $t$ and $B$ be the probability that.....

$A$ and $B$ are events not probabilities. So first lets define events that you are interested in.
Most importantly, we need to make a distinction between the event of choosing a certain action and observing a certain action. In your example you seem to have blurred this distinction. Event $A$ is observing that a particular choice is made but in calculation you are using the probability of the event of choosing that action.
Based on your question, there are in total $2\times3\times3=18$ sequences that are possible. Let's call this set $E$ of possible events. However, the observer can only see $4$ possible events. $3$ events are the ones that have $S_1$ and $S_2$ and are therefore observable, and the fourth event is the invisible sequence. We call this set $V \cup {0}$, where we define ${0}$ as the fourth event described above.
Depending on what is the probability distribution at each time point, you can calculate the probability of any event in set $V$.
Whether you need bayes theorem or not depends on the data you have. Some of the relationships that hold true are following:
$$Pr(S_1|e \in V)=Pr(S_2|e \in V)=Pr(S_1,S_2|e \in V)=1$$
So, $$Pr(e \in V | S_1)= \frac{Pr(S_1|e \in V)\cdot Pr(e \in V)}{Pr(S_1)}$$
Now it is easy to show that $Pr(e \in V) \leq Pr(S_1)$. This is because $$Pr(e \in V) < Pr(V) = Pr(S_1)\cdot Pr(S_2)$$.
