Let's consider the following game of two players:
A fair coin is tossed as long as no one has won. Player $A$ wins if the sequence $(Z,K,Z)$ appears and player $B$ wins if $(K,K,K)$ appears. What is the probability that $A$ wins?
(Note: To solve this problem we are only allowed to use the notion of conditional probability, law of total probability and bayes theorem!)
My approach:
Let be $P(A)$ the probability that $A$ wins. I am not sure how to tackle this problem in a rigorous way because I am not able to find an appropriate probability space. What I could do is assume that we are at some stage of the game where nobody has won. Then, we define two sets \begin{align*} &B_{K}:=\{\omega_1=K,\omega_2\in\{Z,K\},\omega_3\in\{Z,K\},\dots\}\\ &B_{Z}:=\{\omega_1=Z,\omega_2\in\{Z,K\},\omega_3\in\{Z,K\},\dots\}. \end{align*}
Those two sets are obviously disjoint and all future sequences are either member of $B_{K}$ or $B_{Z}$. Further, we can easiliy see that there exists a bijection between both sets.
Is it correct if I simply define a sample space $\Omega$ which consists of all future sequences that either start with $K$ or $Z$? This would mean that $P(B_{K})+P(B_{Z})=1$ and $P(B_{K})=P(B_{Z})=0.5$ and allows us to apply the law of total probability $$ P(A)=P(A\mid B_{K}) P(B_{K})+P(A\mid B_{Z}) P(B_{Z})=0.5\cdot(P(A\mid B_{K}) +P(A\mid B_{Z})). $$
Is this the way to go? It doesn't feel rigorous because of two things.
1.) The sample space is only valid for some arbitrary stage of the game - not the whole game.
2.) How do I know that there exists a probability function which I can use? The sample space I have defined is not countable, so I don't know if I can assign a probability to each element of $\Omega$?
Following Lulu's suggestion, let us first define the states on last toss(es):
$S$ = starting toss, $\,A1\equiv$ first $Z$, $A2\equiv$ $ZK$, $\,B1\equiv$ first $K$, $\,B2\equiv KK$,
Then tracing movement step-by-step from one state to another, we have
$S = .5A1+.5B1$
$A1= .5A2 +.5A1$
$A2= .5*1 +.5B2\quad$ [Either we get ZKZ and are done, or go to $B2$]
$B1= .5B2+.5A1$
$B2= .5*0 +.5A1\;$ [We bar $KKK$ as we compute for $A$'s win ]
Note that these equations involve using conditional probabilities
Solving these equations, we get $S = P(A\;wins)=\frac7{12}$
As a countercheck,
if we exchange the multiplier $1$ and $0$ in the equations for $A2$ and $B2$,
we get P(B wins) $= \frac5{12}$