19.2: The principle of equal odds weight
From the above arguments, we see that
Proclaim 19.4 [The principle of equal weight]
Consider a finite state space $\Omega$ with the discrete metric, that is, $\Omega=\{\omega_1,\omega_2,\ldots,$ $\omega_n\}$. Let ${\mathsf O}=(X, {\cal F}, F)$ be an observable in $C_{} (\Omega)$. Consider a measurement $ {\mathsf M}_{C_{} (\Omega )}({\mathsf O} , S_{[ \ast]} ) $. If the observer has no information for the unknown state $[\ast]$, there is a reason to assume that this measurement is also represented by the mixed measurement $ {\mathsf M}_{C_{} (\Omega )}({\mathsf O} , S_{[ \ast]}(\rho_{\scriptsize{\mbox{prior}}}) ) $, where
\begin{align} \rho_{\scriptsize{\mbox{prior}}} = \frac{1}{n} \sum_{k=1}^n \delta_{\omega_k} \tag{19.9} \end{align}Explanation In betting, it is certain that everybody wants to choose an unpopular $\omega_k$. Thus, I believe that nobody disagrees with Proclaim 19.4. Also, it should be noted that
$(J):$  the term "probability" can be freely used within the rule of Axiom 1 or Axiom$^{(m)}$ 1. 
$\fbox{Note 19.1}$  In this book, we deal with three
"the principle of equal weight" as follows:

Problem 19.5 [Monty Hall problem; Problem 5.14 ;The principle of equal weight]
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
And further,
he now gives you the choice of
sticking with door 1 or switching to door 2?
What should you do?
Figure 19.6:
Monty Hall problem
$(\flat):$  the door 3 has a goat. 
Proof It should be noted that the above is completely the same as Problem 5.14. However, the proof is different. That is, it suffices to use Proclaim 19.4 and Bayes theorem (B$_2$). That is, the proof is similar to Problem 9.16.
$\square \quad$