9.10:Bayesian statistics: Monty Hall problem [three prisoners problem]
This section is extracted from
the following:
$(\sharp):$  S. Ishikawa; The Final Solutions of Monty Hall Problem and Three Prisoners Problem arXiv:1408.0963v1 [stat.OT] 2014 
Bayesian statistics: Monty Hall problem[resp. three prisoners problem]
$\quad$  Suppose you are on a game show, and you are given
the choice of three doors
(i.e., " Door $A_1$"$\!\!\!,\;$ " Door $A_2$"$\!\!\!,\;$ " Door $A_3$"$\!\!)$.
Behind one door is a car, behind the others, goats.
You do not know what's behind the doors.
However, {you pick a door, say "Door $A_1$"}, and the host, who knows what's behind the doors, opens another door, {say " Door $A_3$"$\!\!\!,\;$ which has a goat}. And he adds that

$\quad$ 
Three prisoners, $A_1$, $A_2$, and $A_3$ were in jail.
They knew that one of them was to be set free and
the other two were to be executed.
They did not know who was the one to be spared, but they know that

In the dualism, we have the confrontation \begin{align} \mbox{ "observer$\longleftrightarrow$system" } \end{align} as follows.
Let $\Omega = \{ \omega_{1} , \omega_{2} , \omega_{3} \}$ be a state space with the discrete metric. Each pure state $\delta_{\omega_{{m}}} (\in {\frak S}^p (C(\Omega)^* ))$ means as follows.
\begin{align} & \delta_{\omega_{{m}}} \Leftrightarrow \mbox{The state such that a car is behind the door $A_m$} \nonumber \\ {[\mbox{resp.}} \;\;\; & \color{blue}{ \mbox{ $ \delta_{\omega_{{m}}} \Leftrightarrow $ } \mbox{ the state such that a pisoner $A_m$ is pardoned ] } } \nonumber \\ & \qquad \qquad (m=1,2,3 ) \tag{9.31} \end{align}The observable ${\mathsf O}_1$ $\equiv$ $(\{ 1, 2,3 \}, 2^{\{1, 2 ,3\}}, F_1)$ is defined by
\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{9.32} \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 measurement
${\mathsf M}_{L^\infty (\Omega)} ({\mathsf O}_1, S_{[{}\ast{}]})$"
$ \Leftrightarrow $ $ \left[\begin{array}{ll} \mbox{the host says "Door $A_1$ has a goat" } \\ \color{blue}{ \mbox{the emperor says "Prisoner $A_1$ will be executed" } } \end{array}\right] $ 
(b): 
"measured value $2$ is obtained
by
the measurement
${\mathsf M}_{L^\infty (\Omega)} ({\mathsf O}_1, S_{[{}\ast{}]})$
"
$ \Leftrightarrow $ $ \left[\begin{array}{ll} \mbox{the host says "Door $A_2$ has a goat" } \\ \color{blue}{ \mbox{the emperor says "Prisoner $A_2$ will be executed" } } \end{array}\right] $ 
(c): 
"measured value $3$ is obtained
by
the measurement
${\mathsf M}_{L^\infty (\Omega)} ({\mathsf O}_1, S_{[{}\ast{}]})$
"
$ \Leftrightarrow $ $ \left[\begin{array}{ll} \mbox{the host says "Door $A_3$ has a goat" } \\ \color{blue}{ \mbox{the emperor says "Prisoner $A_3$ will be executed" } } \end{array}\right] $ 
Here, assume that,
by the
statistical
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
$
\left[\begin{array}{ll}
\mbox{the host said
"Door $A_3$ has a goat"
}
\\
\color{blue}{
\mbox{the emperor said
"Prisoner $A_3$ is to be executed"
}
}
\end{array}\right]
$
Then, Bayes' theorem 9.11 says that the posterior state $\nu_{ post}$ $(\in {\cal M}_{+1}^m (\Omega))$ is given by
\begin{align} \nu_{ post} = \frac{F_1(\{3\}) \times \nu_0} {\bigl\langle \nu_0, F_1(\{3\}) \bigr\rangle}. \tag{9.33} \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. \tag{9.34} \end{align} Then,$(I_1):$  In Problem 9.23, \begin{align} \left\{\begin{array}{ll} \mbox{ if $\nu_{ post} (\{ \omega_1 \})$ $<$ $\nu_{ post} (\{ \omega_2 \})$ (i.e., $p_1 < 2 p_2 $), you should pick Door $A_2$} \\ \mbox{ if $\nu_{ post} (\{ \omega_1 \})$ $=$ $\nu_{ post} (\{ \omega_2 \})$ (i.e., $p_1 = 2 p_2 $), you may pick Doors $A_1$ or $A_2$} \\ \mbox{ if $\nu_{ post} (\{ \omega_1 \})$ $>$ $\nu_{ post} (\{ \omega_2 \})$ (i.e., $p_1 > 2 p_2 $), you should not pick Door $A_2$} \end{array}\right. \end{align} 
$(I_2):$  In Problem 9.24, \begin{align} \left\{\begin{array}{ll} \mbox{ if $ \nu_{0} (\{\omega_1\}) < \nu_{\rm post} (\{\omega_1\})$ (i.e., $p_1 < 1 2 p_2$), the prisoner $A_1$'s happiness increases } \\ \mbox{ if $ \nu_{0} (\{\omega_1\}) = \nu_{\rm post} (\{\omega_1\})$ (i.e., $p_1 = 1 2 p_2$), the prisoner $A_1$'s happiness is invariant } \\ \mbox{ if $ \nu_{0} (\{\omega_1\}) > \nu_{\rm post} (\{\omega_1\})$ (i.e., $p_1 > 1 2 p_2$), the prisoner $A_1$'s happiness decreases } \\ \end{array}\right. \end{align} 