As I understand that the "Monty Hall Problem" is well studied, e.g., here or here, etc. [I am relatively new to probability theory. So, please help me to learn something from you experts here.]

In my understanding so far, summary is that "always switching" has a success rate of 2/3 while "always staying" has a success rate of 1/3.

Just to support my questions, let me denote the Car by C, Monty by M, and the user by X. The door number is a subscript to these defined letters accordingly.

ADD: $C_i$ means Car is behind $i$th door (e.g., $C_1$ means Car is behind door 1); $M_i$ means that Monty opens $i$th door (e.g., $M_3$ means Monty opens door 3), and $X_i$ means that the user selects $i$th door (e.g., $X_1$ means that the user picks door 1).

We start with the assumption that the user selects door 1, that is $X_1$. Then, according to the above study problems, the posteriors are

\begin{align} p\left(C_1 \mid M_3, X_1\right) = \frac{1}{3}, \\ p\left(C_2 \mid M_3, X_1\right) = \frac{2}{3}, \\ p\left(C_3 \mid M_3, X_1\right) = 0 \ . \\ \end{align}

My questions are as follows:

  1. Which door will have the maximum a-posteriori (MAP) solution?

    • It should be door 2, clearly for the above case, isn't it (just a clarification)?
  2. What will be the posterior mean?

    • I am confused and not sure. Can you please help me?

Thank you.

Attempt for the posterior mean of the doors:

\begin{align} \mathbb{E}\left( C_i \right) &= \sum_{i=1}^3 i p\left(C_i \mid M_3, X_1\right) \\ &= 1 \underbrace{p\left(C_1 \mid M_3, X_1\right) }_{\frac{1}{3}} + 2 \underbrace{p\left(C_2 \mid M_3, X_1\right) }_{\frac{2}{3}} + 3 \underbrace{p\left(C_3 \mid M_3, X_1\right) }_{0} \\ &= 1 \cdot \frac{1}{3} + 2 \cdot \frac{2}{3} + 0\\ &= \frac{5}{3} \end{align}

Am I correct?

  • $\begingroup$ What's the difference between $M_1$, $M_2$, and $M_3$? $\endgroup$ – Kodiologist Nov 2 '18 at 14:29
  • $\begingroup$ $M_1$ means that the Monty opens door 1. Likewise for $M_2$ and $M_3$. $\endgroup$ – learning Nov 2 '18 at 15:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.