9.7:Monty Hall problem ( The principle of equal weight )

9.7.1: The principle of equal weight--- The most famous unsolved problem

Let us reconsider Monty Hall problem (Problem 9.14, Problem9.15) in what follows. We think that the following is one of the most reasonable answers (also, see Problem 19.5).

Problem 9.17 [Monty Hall problem(The principle of equal weight)]

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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. $(\sharp):$ You choose a door by the cast of the fair dice, i.e., with probability $1/3$. According to the rule ($\sharp$), you pick a door, say number 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. He says to you, "Do you want to pick door number 2?" Is it to your advantage to switch your choice of doors?

Answer By the same way of Problem9.15 and Problem9.16 (Monty Hall problem), define the state space $\Omega = \{ \omega_1 , \omega_2 , \omega_3 \}$ and the observable ${\mathsf O}=(X, {\cal F}, F)$. And the observable ${\mathsf O}=(X, {\cal F}, F)$ is defined by the formula (9.11). The map $\phi:\Omega \to \Omega $ is defined by

we get a causal operator $\Phi:L^\infty (\Omega) \to L^\infty(\Omega)$ by $[\Phi(f)](\omega) = f(\phi(\omega))$ $(\forall f \in L^\infty(\Omega), \;\forall \omega \in \Omega )$. Assume that a car is behind the door $k$ $(k=1,2,3)$. Then, we say that

We, by the argument in Chapter 11 (cf. the formula (11.7)) see the following identifications:

Thus, the above (a) is equal to

Here, note that $\frac{1}{3}(\delta_{\omega_k}+\delta_{\phi(\omega_k)} +\delta_{\phi^2 (\omega_k)})$ $=$ $\frac{1}{3}(\delta_{\omega_1}+\delta_{\omega_2} +\delta_{\omega_3})$ $(\forall k=1,2,3)$. Thus, this (b) is identified with the mixed measurement ${\mathsf M}_{L^\infty (\Omega)} ({\mathsf O}, S_{[{}\ast{}]}(\nu_e ))$ , where

Therefore, Problem 9.17 is the same as Problem 9.16. Hence, you should choose the door 2.

$\fbox{Note 9.4}$

The above argument is easy. That is, since you have no information, we choose the door by a fair dice throwing. In this sense, the principle of equal weight --- unless we have sufficient reason to regard one possible case as more probable than another, we treat them as equally probable --- is clear in measurement theory. However, it should be noted that the above argument is based on dualism.

From the above argument, we have the following theorem.

Theorem 9.18 [The principle of equal weight] Consider a finite state space $\Omega$, that is, $\Omega=\{\omega_1,\omega_2,\ldots,\omega_n\}$. Let ${\mathsf O}=(X, {\cal F}, F)$ be an observable in $L^\infty (\Omega, \nu)$, where $\nu$ is the counting measure. Consider a measurement $ {\mathsf M}_{L^\infty (\Omega )}({\mathsf O} , S_{[ \ast]} ) $. If the observer has no information for the state $[\ast]$, there is a reason to that this measurement is identified with the mixed measurement $ {\mathsf M}_{L^\infty (\Omega )}({\mathsf O} , {\overline S}_{[ \ast]}(w_e) ) $ $\Big($ or, $ {\mathsf M}_{L^\infty (\Omega )}({\mathsf O} , S_{[ \ast]}(\nu_e) ) $ $\Big)$, where

Proof. The proof is a easy consequence of the above Monty Hall problem (or, see