Monty Hall Uneven Door Probabilities - Conditional Door Opening Probability

Question

I've been reading about this variation of the Monty Hall problem described in "The Monty Hall Problem" by Jason Rosenhouse. Specifically "3.10 - What If the Car Is Not Placed Randomly".

Suppose the car is placed behind door 1, door 2, and door 3 with probabilities $(p_{1}, p_{2}, p_{3})$, respectively, with $p_{1} + p_{2} + p_{3} = 1$. Suppose the contestant chooses door $i$ and Monty opens door $j$, with $i \neq j$. The conditional probability is given by: \begin{equation} \mathbb{P}(C_{i} | M_{j}) = \frac{\mathbb{P}(C_{i}) \times \mathbb{P}(M_{j}|C_{i})}{\mathbb{P}(M_{j})} \end{equation} with $C_{i}$ describing the event that the car is behind door $i$, and $M_{j}$ describing the event that Monty opens door $j$.

The author continues to decompose the unconditional probability in the denominator using the Law of Total Probability. \begin{equation} \mathbb{P}(M_{j}) = \mathbb{P}(C_{i}) \times \mathbb{P}(M_{j}|C_{i}) + \mathbb{P}(\tilde{C}_{i}) \times \mathbb{P}(M_{j}|\tilde{C}_{i}) \end{equation} with $\tilde{C}_{i}$ describing the "car not behind door $i$ event". Clearly $\mathbb{P}(\tilde{C}_{i}) = 1 - p_{i}$. I am having trouble determining the conditional probability $\mathbb{P}(M_{j}|\tilde{C}_{i})$. The author claim's it is $\frac{1}{2}$, because "where our initial choice [i.e. $C_{i}$] is incorrect Monty will open each of the other doors in roughly half of the cases." (p. 79). However I sooner think the conditional probability should sooner be $\frac{p_{-j}}{p_{j}+p_{-j}}$, as the realisation of the actual car's position still matters.

I also ran the following simulation code in R, and it confirms my insight.

library(tidyverse)

car = sample(
  x = c("A", "B", "C"),
  size = 1000000,
  replace=TRUE,
  prob = c(6/10, 3/10, 1/10)

)

test <- data.frame(
  outcome = car
)

test <- test %>% mutate(
  door = case_when(
    outcome == 'A' ~ 'Doesnt matter',
    outcome == 'B' ~ 'C',
    outcome == 'C' ~ 'B'
  )
)

test %>% filter(outcome != 'A') %>% with(table(door) / nrow(.))

Given the author is a full professor in mathematics, I wanted to ask you if you can help me in finding where my intuition is off?

Answer 1

Your intuition is right and the full professor in mathematics is wrong. The problem description explicitly states that “Monty will open a door he knows to conceal a goat”. Since there are only two doors left and they conceal a car with different probabilities, they conceal a goat with different probabilities, and thus Monty will open them with different probabilities.

You could have told that a full professor in mathematics can make basic mistakes by the fact that he writes “What If the Car Is Not Placed Randomly” when he actually means “What If the Car Is Placed Randomly But Not Uniformly Randomly”.

Of course, people with lots of points on math.SE are at least as likely to make basic mistakes as full professors in mathematics, so who knows, perhaps someone will explain why I’m wrong...