axiomatic approach to 3 prisoner problem? I've decided to start working through Casella and Berger's Statistical Inference. The book introduces probability theory from an axiomatic approach and defines a probability function according to Kolmogorov's Axioms. I would like to prove the well-known solution to the Three Prisoner Problem from an axiomatic approach and I'm a bit confused about how we actually define the probability functions and sample space.
The text doesn't go into much detail in the example given, so I'm hoping I can get some clarity here.
Disclaimers: This is my first time thinking about probability this way. This is the first time I've posted to Stack Exchange. If I'm posting in the wrong place, please let me know :)
The authors give a very short solution to the known Three Prisoner Problem, paraphrased here:

Three prisoners A, B, and C are on death row. One prisoner will be pardoned tomorrow. The warden knows who will be pardoned. Prisoner A asks the warden who will be pardoned, but the warden refuses to tell. So A then asks which of B or C will die tomorrow. The warden says "B will die". The warden thinks he's given no additional information to A about A's own likelihood of being pardoned. Prisoner A thinks their chance of being pardoned has risen to 1/2, given this new information. Who's thinking is correct, the warden or the prisoner?

My thinking and questions:
The known solution is that the warden is correct and I feel I understand this on a logical level. But I can't figure out how to "prove" it on an axiomatic level.
The text starts by stating $p(A)=p(B)=p(C)=1/3$ where A = event A is pardoned, B = event B is pardoned, etc. This makes sense to me, as we can define the function $p$ on the sample space $ S = \{A, B, C\}$.
Then the text says, "Let $W$ denote the event the warden says $B$ will die. A can update his probability of being pardoned to $p(A|W)=p(A\cap W)/p(W)$." This is where I started to get confused.
I understand this equation for $p(A|W)$ is given by definition of conditional probability. But the book doesn't given any indication of how to update the sample space. Am I correct in thinking $p(\cdot)$ and $p(\cdot|W)$ are two different functions, one defined on $S$ and one defined on a new sample space?
It confuses me that $p(A\cap W)$ has any meaning, since $p$ is a function defined on the very simple sample space $ S = \{A, B, C\}$.
Logically, I understand the solution to this problem. But, if I'm going to learn probability from the axioms, I want to understand the specifics of how we define the sample space (well, the sigma-algebra of the sample space) and probability functions for this problem.
My tentative solution would be to define a new sample space $S^* = \{AW_B, AW_C, BW_C, CW_B \}$ where $AW_B$, for example, represents the event that $A$ is pardoned and the warden says $B$ will die.
Then, to define a new probability function $p^*$ on $S^*$, where $p^*(AW_B)=p^*(AW_C)$
By the definition of a probability function: $p^*(AW_B \cup AW_C) = p^*(AW_B) + p^*(AW_C)$
Then, if we could assume $p(A) = p^*(AW_B \cup AW_C)$ we could show $p^*(AW_B) = p^*(AW_C)=1/6$.
But honestly, I'm very confused on how to connect this back to the definition of the conditional probability $P(A|W)$. I guess primarily because I don't understand what initial sample space is being considered for the original definition of $p$. Any guidance on this problem would be greatly appreciated.
 A: You are right. You have to use a larger probability space to define a $\mathbb{P}\left(A\middle|W\right)$. Because for $W$, the randomness comes additionally from the wardon, which means $W$ doesn't belong to any $\sigma$-algebra generated by subset of $S$ hence not measurable.
To connect your construction to $\mathbb{P}\left(A\middle|W\right)$. Note that $\left(S^{\ast},\mathscr{P}\left(S^{\ast}\right),p^{\ast}\right)$ is a well-defined probability space, where $\mathscr{P}\left(\cdot\right)$ denotes powerset. In addition, using your notation, we have $W=AW_{B}\cup CW_{B}$, $A=AW_{B}\cup AW_{C}$, and $A\cap W=AW_{B}$, hence $A$, $W$, $A\cap W$ are all measurable in this space. Therefore, we can define a valid conditional probability like this
$$p^{\ast}\left(A\middle |W\right)=\frac{p^{\ast}\left(A\cap W\right)}{p^{\ast}\left(W\right)}=\frac{p^{\ast}\left(AW_{B}\right)}{p^{\ast}\left(AW_{B}\right)+p^{\ast}\left(CW_{B}\right)}=\frac{\frac{1}{6}}{\frac{1}{6}+\frac{1}{3}}=\frac{1}{3}=p^{\ast}\left(A\right). $$
