**
5.5: Manty Hall problem: non-Baysian approach
**

**Problem 5.14 [Monty Hall problem]**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. |

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 $\delta_{\omega_m} (\in {\frak S}^p (C(\Omega)^* ))$ means \begin{align} \delta_{\omega_m} \Leftrightarrow \mbox{ the state that the car is } \mbox{behind the door $m$} \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{5.21} \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" clearly means that you take a measurement ${\mathsf M}_{L^\infty (\Omega)} ({\mathsf O}_1, S_{[{}\ast{}]})$. Here, we assume that

(a): | "a measured value $1$ is obtained by the measurement ${\mathsf M}_{L^\infty (\Omega)} ({\mathsf O}_1, S_{[{}\ast{}]})$" $ \Leftrightarrow \mbox{The host says "Door 1 has a goat"}$ |

(b): | "measured value $2$ is obtained by the measurement ${\mathsf M}_{L^\infty (\Omega)} ({\mathsf O}_1, S_{[{}\ast{}]})$" $ \Leftrightarrow \mbox{The host says "Door 2 has a goat"}$ |

(c): | "measured value $3$ is obtained by the measurement ${\mathsf M}_{L^\infty (\Omega)} ({\mathsf O}_1, S_{[{}\ast{}]})$" $ \Leftrightarrow \mbox{The host says "Door 3 has a goat"}$ |

Recall that, in Problem 5.14, 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

and thus, there is a reason to infer that the unknown state $ [\ast]$ is equal to $\delta_{\omega_2}$. Thus, you should switch to door 2. This is the first answer to Problem 5.14 (Monty-Hall problem).

$\fbox{Note 5.4}$ | Examining the above example, the readers should understand that the problem "What is measurement?" is an unreasonable demand. Thus,
\begin{align}
\mbox{
we have to abandon the realistic approach,
and accept
the metaphysical approach.
}
\end{align}
In other words, we assert that
\begin{align}
\mbox{
the concept of measurement is metaphysical.
}
\end{align}
Also, for a Bayesian approach to Monty Hall problem, see Chapter 9 and Chapter 19. |

**Remark 5.15 [The answer by the moment method]**

In the above, a measured value "3" is obtained by the measurement ${\mathsf M}_{L^\infty (\Omega )} ({\mathsf O} {{=}}(\{ 1, 2,3 \}, 2^{\{1, 2 ,3\}}, F),$ $ S_{[{}\ast{}]})$. Thus, the approximate sample space $(\{1,2,3\}, 2^{\{1,2,3\}}, \nu_{1})$ is obtained such that $\nu_1 (\{1 \}) =0 $, $\nu_1 (\{2 \}) =0 $, $\nu_1 (\{3 \}) =1 $. Therefore,

[when the unknown $[\ast]$ is $\omega_1$] \begin{align} "(5.19) \mbox{ in $\S$5.4"}=|0-0|+ | 0 - 0.5| + | 1- 0.5| = 1, \end{align} [when the unknown $[\ast]$ is $\omega_2$] \begin{align} "(5.19) \mbox{ in $\S$5.4"}=|0-0|+ | 0 - 0| + | 1-1| = 0 \end{align} [when the unknown $[\ast]$ is $\omega_3$] \begin{align} "(5.19) \mbox{ in $\S$5.4"}=|0-0|+ | 0 - 1| + | 1-0| = 2. \end{align}Thus, we can infer that $[\ast]$ $=$ ${\omega_2}$. That is, you should change to the Door 2.