I have worked out a few successive derivations of Bayes' Theorem...
$\Pr(W) = \Pr(W)$
$\Pr(W\ |\ C_1)\ =\ \frac{\Pr(C_1\ |\ W)\Pr(W)}{\Pr(C_1)}$
$\Pr(W\ |\ (C_1\ \cap\ C_2))\ =\ \frac{\Pr(C_2\ |\ (W\ \cap\ C_1))\Pr(W\ |\ C_1)}{\Pr(C_2\ |\ C_1)}$
$\Pr(W\ |\ (C_1\ \cap\ C_2\ \cap\ C_3)) = \frac{\Pr(C_3\ |\ (W\ \cap\ C_1\ \cap\ C_2))\Pr(W\ |\ (C_1\ \cap\ C_2))}{\Pr(C_3\ |\ (C_1\ \cap\ C_2))}$
I think the pattern is $\Pr(W\ |\ \bigcap\limits_{i = 1}^{n - 1} C_i)\ =\ \frac{\Pr(C_{n - 1}\ |\ (W\ \cap\ \bigcap\limits_{i = 1}^{n - 2} C_i))\Pr(W\ |\ \bigcap\limits_{i = 1}^{n - 2} C_i)}{\Pr(C_{n - 1}\ |\ \bigcap\limits_{i = 1}^{n - 2} C_i)}$. I have intuition that this should be interpretable as a recurrence relation. As far as I can tell, $\Pr(W\ |\ \bigcap\limits_{i = 1}^{n - 1} C_i)$ is playing the role of $a_n$, and $\Pr(W\ |\ \bigcap\limits_{i = 1}^{n - 2} C_i)$ is playing the role of $a_{n - 1}$. However, it is less clear what to do with the other factors. Should I leave them as probabilities, yielding $a_n =\ \frac{\Pr(C_{n - 1}\ |\ (W\ \cap\ \bigcap\limits_{i = 1}^{n - 2} C_i))}{\Pr(C_{n - 1}\ |\ \bigcap\limits_{i = 1}^{n - 2} C_i)} a_{n - 1}$, or is there some transformation I need to apply to the coefficient as well? Or should I leave everything in probability notation? The coefficient is certainly not constant in this form, yet I believe the solution should be simple (probably related to Bayesean updating formulas). How can I massage this into an analytically solvable form?
The Bayes' theorem says: $$\mathbb{P}(W |\bigcap_{i=1}^n C_i ) = \frac{\mathbb{P}(W \cap\bigcap_{i=1}^n C_i )}{\mathbb{P}(\bigcap_{i=1}^n C_i )} \tag{1}$$ Given the sets $(C_i)_{i\in\mathbb{N}_+} \subset \Omega$ , we define the functions $(a_n)_{n\in \mathbb N_+}$ as follows $$ \begin{align} a_n\colon \Omega & \mapsto [0,1] \hspace{5cm} \text{for }n \in \mathbb{N_+}\\[-1ex] A & \longmapsto a_n(A) :=\mathbb{P}(A |\bigcap_{i=1}^n C_i ) \end{align} $$
Then from $(1)$, we have $$\begin{align} a_n(W)&= \frac{\mathbb{P}(W \cap\bigcap_{i=1}^n C_i )}{\mathbb{P}(\bigcap_{i=1}^n C_i )}\\ &= \frac{\mathbb{P}(W\cap C_n \cap\bigcap_{i=1}^{n-1} C_i )}{\mathbb{P}(C_n \cap \bigcap_{i=1}^{n-1} C_i )} \\ &= \frac{\mathbb{P}(W\cap C_n |\bigcap_{i=1}^{n-1} C_i )\cdot\mathbb{P}( \bigcap_{i=1}^{n-1} C_i )}{\mathbb{P}(C_n |\bigcap_{i=1}^{n-1} C_i )\cdot\mathbb{P}( \bigcap_{i=1}^{n-1} C_i )} \\ &= \frac{\mathbb{P}(W\cap C_n |\bigcap_{i=1}^{n-1} C_i )}{\mathbb{P}(C_n |\bigcap_{i=1}^{n-1} C_i )} \\ \color{red}{a_n(W)}&\color{red}{= \frac{a_{n-1}(W \cap C_n)}{a_{n-1}(C_n)}} \tag{2} \end{align}$$
I think that $(2)$ is the best recurrence relation that we can have.