Monty Hall Problem - what is the probability space for the problem, and a question on $\sigma$-algebras generated by events? So we have the following problem where we have 3 doors, 1 has a car behind it and 2 don't:
i) We first choose door 1
ii) Monty opens one of the other doors which he knows for certain doesn't have a car behind it
iii) We wither keep door 1 or choose the remaining unopened door
So in my mind the sample space is $\Omega=\{\{1,2,1\},\{1,2,3\},\{1,3,1\},\{1,3,2\}\}$ (1st element is our pick, 2nd is Monty's pick, 3rd is our final pick). Is this correct?
And so $\mathcal{F}=\mathcal{P}(\Omega)$.
I'm now unsure exactly how we would define $\mathbb{P}$? Of course, the probability a  car is behind a given door is 1/3, and if door 1 does have the car behind it then Monty can pick either door 2 or 3 with probability 1/2 each. Or if door 1 doesn't have a car behind it then Monty picks either door 2 or 3 with certainty. Is there a neater way in which to define this probability measure?
Now my final issue is how we would define:
(a) The event that door 2 has a no car behind it (call A)
(b) The $\sigma$-algebra generated by A, $\sigma(A)$
For (a) we know that the outcomes $\{1,2,1\},\{1,2,3\}$ both indicate that door 2 has no car behind it. So is event A just the event we pick either one of these outcomes? (however door 1 or 3 may still not have a car behind it, so am I missing outcomes here? If so, this may help me determine $\sigma(A)$?)
 A: Really interesting question! Tricky and fiddly stuff indeed! Hopefully I can help!
How about alternatively we define our sample space, $\Omega := \{D_{1,2}, D_{1,3}, D_{2,3} , D_{3,2} \} $ Where each $D_{i,j} $ is the event that the car is behind Door i and monty opens j. Notice we are always guessing door 1. You should not include the third argument, our choice of final selection, as this is non random.
One could define $P(D_{i,j})$ =
\begin{array}{ll}
      \frac{1}{6} & i =1 \\
      \frac{1}{3} & i \not = 1\
\end{array}
Onto your second part: The event A that there is no car behind door 2. We have 2 ways to write this. Firstly: A = $\Omega \setminus D_{2,3} $ = $\Omega \cap D_{2,3}^c$.
Alternatively:  A = $D_{1,2} \cup  D_{1,3} \cup D_{3,2} $
Now onto the sigma algebra $\sigma(A)$
Firstly: $\Omega , \emptyset \in \sigma(A) $ By definition of a sigma algebra. A is in our sigma algebra obviously, so we have $\Omega, \emptyset $ and A = $D_{1,2} \cup  D_{1,3} \cup D_{3,2} $  are in $\sigma(A)$. Now we can take any interesection, compliment or union to find more sets in our sigma algebra. Infact the only one left is A$^c$. So $\sigma(A) = \{\emptyset , \Omega , A , A^c \}$
Hope this helped!
