9.6: Monty Hall problem (The Bayesian approach)
9.6.1: The review of Problem5.14
(
Monty Hall problem in pure measurement)
$\quad$ 
You are on a game show and you are given the choice of three doors.
Behind one door is a car, and behind the other two are goats.
You choose, say, door 1, and the host, who knows where the car is,
opens another door,
behind which is a goat.
For example,
the host says that

Answer by Fisher's maximum likelihood method in $\S$5.5
Put $\Omega = \{ \omega_1 , \omega_2 , \omega_3 \}$ with the discrete topology $d_D$ and the counting measure $\nu$. Thus consider the classical basic structure:
\begin{align} [ C_0 (\Omega ) \subseteq L^\infty(\Omega, \nu ) \subseteq B(L^2(\Omega, \nu )) ] \end{align}Assume that each state $\delta_{\omega_m} (\in {\frak S}^p (C_0 (\Omega)^* ))$ means
\begin{align} \delta_{\omega_m} \Leftrightarrow \mbox{ the state that the car is } \mbox{behind the door 1} \quad (m=1,2,3) \end{align}Define the observable ${\mathsf O}_1$ $\equiv$ $(\{ 1, 2,3 \}, 2^{\{1, 2 ,3\}}, F_1)$ in $L^\infty (\Omega)$ such that
\begin{align} & [F_1(\{ 1 \})](\omega_1)= 0.0,\qquad [F_1(\{ 2 \})](\omega_1)= 0.5, \qquad [F_1(\{ 3 \})](\omega_1)= 0.5, \nonumber \\ & [F_1(\{ 1 \})](\omega_2)= 0.0, \qquad [F_1(\{ 2 \})](\omega_2)= 0.0, \qquad [F_1(\{ 3 \})](\omega_2)= 1.0, \nonumber \\ & [F_1(\{ 1 \})](\omega_3)= 0.0,\qquad [F_1(\{ 2 \})](\omega_3)= 1.0, \qquad [F_1(\{ 3 \})](\omega_3)= 0.0, \tag{9.24} \end{align}where it is also possible to assume that $F_1(\{ 2 \})(\omega_1)=\alpha$, $F_1(\{ 3 \})(\omega_1) =1 \alpha$ $ (0 < \alpha < 1)$. The fact that you say "the door 1" means that we have a measurement ${\mathsf M}_{L^\infty (\Omega)} ({\mathsf O}_1, S_{[{}\ast{}]})$.
Here, we assume that
$(a):$  "a measured value $1$ is obtained" $ \Leftrightarrow \mbox{The host says "Door 1 has a goat}$ 
(b):  "measured value $2$ is obtained" $ \Leftrightarrow \mbox{The host says "Door 2 has a goat}$ 
(c):  "measured value $3$ is obtained" $ \Leftrightarrow \mbox{The host says "Door 3 has a goat}$ 
Since the host said "Door 3 has a goat"$\!\!\!,\;$ this implies that you get the measured value "3" by the measurement ${\mathsf M}_{L^\infty (\Omega)} ({\mathsf O}_1, S_{[\ast]})$. Therefore, Theorem 5.6 (Fisher's maximum likelihood method) says that
That is because we see that
\begin{align} \max \{ [F_1(\{3\})] (\omega_1), [F_1(\{3 \}){}](\omega_2), [F_1(\{3 \})] (\omega_3) \} & =\max \{ 0.5, \; \; 1.0 , \; \; 0.0 \} \\ & =1.0 = [F_1(\{3\})] (\omega_2) \end{align}
and thus, there is a reason to infer that $[\ast]$ $=$ $\delta_{\omega_2}$. Thus, you should switch to door 2. This is the first answer to MontyHall problem.
Next, let us study Monty Hall problem in mixed measurement theory (particularly, Bayesian statistics).
Problem 9.16 [Monty Hall problem (The answer by Bayes' method)]
$\quad$  Suppose you are on a game show, and you are given
the choice of three doors
(i.e., "number 1"$\!\!\!,\;$ "number 2"$\!\!\!,\;$ "number 3"$\!\!)$.
Behind one door is a car, behind the others, goats.
You pick a door, say number 1.
Then,
the host,
who set a car behind a certain door,
says
And further, the host says, for example,

Answer In the same way as we did in Problem 9.15 (Monty Hall problem: the answer by Fisher's maximum likelihood method), consider the state space $\Omega = \{ \omega_1 , \omega_2 , \omega_3 \}$ with the discrete metric $d_D$ and the observable ${\mathsf O}_1$. Under the hypothesis ($\sharp_1)$, define the mixed state $\nu_0$ $(\in {\cal M}_{+1} (\Omega))$ such that
\begin{align} \nu_0= p_1 \delta_{\omega_1} + p_2 \delta_{\omega_2}+p_3 \delta_{\omega_3} \end{align} namely, \begin{align} \nu_0 (\{ \omega_1 \}) = p_1, \quad \nu_0 (\{ \omega_2 \}) = p_2, \quad \nu_0 (\{ \omega_3 \}) = p_3 \end{align}Thus we have a mixed measurement ${\mathsf M}_{L^\infty (\Omega)} ({\mathsf O}_1, S_{[{}\ast{}]} ( \nu_0))$. Note that
$a):$ 
"measured value $1$ is obtained
by the mixed measurement
${\mathsf M}_{L^\infty (\Omega)} ({\mathsf O}_1,
S_{[{}\ast{}]} ( \nu_0))$"
$ \Leftrightarrow \mbox{the host says "Door 1 has a goat"} $ 
b): 
"measured value $2$ is obtained
by the mixed measurement
${\mathsf M}_{L^\infty (\Omega)} ({\mathsf O}_1,
S_{[{}\ast{}]} ( \nu_0))$"
$ \Leftrightarrow \mbox{the host says "Door 2 has a goat"} $ 
c): 
"measured value $3$ is obtained
by the mixed measurement
${\mathsf M}_{L^\infty (\Omega)} ({\mathsf O}_1,
S_{[{}\ast{}]} ( \nu_0))$"
$ \Leftrightarrow $ the host says "Door 3 has a goat" 
Here, assume that, by the mixed measurement ${\mathsf M}_{L^\infty (\Omega)} ({\mathsf O}_1, S_{[{}\ast{}]} (\nu_0))$, you obtain a measured value $3$, which corresponds to the fact that the host said "Door 3 has a goat"$\!\!\!.\;$ Then, Theorem 9.11 (Bayes' theorem) says that the posterior state $\nu_{ post}$ $(\in {\cal M}_{+1} (\Omega))$ is given by
\begin{align} \nu_{ post} = \frac{F_1(\{3\}) \times \nu_0} {\bigl\langle \nu_0, F_1(\{3\}) \bigr\rangle}. \end{align}That is,
\begin{align} & \nu_{ post} (\{ \omega_1 \})= \frac{\frac{p_1}{2}}{ \frac{p_1}{2} + p_2 }, \quad \nu_{ post} (\{ \omega_2 \})= \frac{p_2}{ \frac{p_1}{2} + p_2 }, \quad \nu_{ post} (\{ \omega_3 \}) = 0. \end{align}Particularly, we see that
$(\sharp):$  if $p_1 = p_2 = p_3 = 1/3$, then it holds that $\nu_{ post} (\{ \omega_1 \})=1/3$, $\nu_{ post} (\{ \omega_2 \})=2/3$, $\nu_{ post} (\{ \omega_3 \})=0$, and thus, you should pick Door 2. 
$\fbox{Note 9.3}$  It is not natural to assume the rule ($\sharp_1$) in Problem 9.16. That is because the host may intentionally set the car behind a certain door. Thus we think that Problem 9.16 is temporary. For our formal assertion, see Problem 9.17 latter and $\S$18.2 