9.6: Monty Hall problem (The Bayesian approach)
9.6.1: The review of Problem5.14 ( Monty Hall problem in pure measurement)

Problem 9.15 [Monty Hall problem(The answer by Fisher's maximum likelihood method)]
$\quad$ 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
$(\flat):$ the door 3 has a goat.
And further, He now gives you the choice of sticking with door 1 or switching to door 2?
What should you do? 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

you should pick door number 2.

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 Monty-Hall problem.

$\square \quad$

9.6.2:Monty Hall problem in mixed measurement

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
 $(\sharp_1):$ the car was set behind the door decided by the cast of the distorted dice. That is, the host set the car behind the $k$-th door (i.e., "number k"$\!$) with probability $p_k$ (or, weight such that $p_1 + p_2 + p_3 =1$, $0 \le p_1 , p_2 , p_3 \le 1$ $)$.

And further, the host says, for example,

 $(\flat):$ the door 3 has a goat.
He says to you, "Do you want to pick door number 2?" Is it to your advantage to switch your choice of doors?

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.
$\square \quad$
 $\fbox{Note 9.3}$ 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