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Intuitively, why does $p$ vary inversely with $P(C_3 \mid D_2)$? But directly with $P(C_2 \mid D_3)$?

Please see the fractions colored in red and orange at the bottom.

1. $P(C_3 \mid D_2) = \color{red}{\dfrac1{
1 + p}}$  means that $p$ is inversely related with $P(C_3 \mid D_2)$. But this is COUNTERintuitive! The question postulates that "you choose [...] door 1" and "Monty enjoys opening door 2 more than [...] door 3". But if $D_3$ hides the car, then Monty is forbidden from opening $D_3$ by the rules of the game, and must open $D_2$. So it doesn't matter whether Monty enjoys opening $D_2$ more than $D_3$! 

N.B. $1/2 \le p \le 1 \iff 3/2 \le 1 + p \le 2 + p \iff 2/3 \ge {\color{red}{\dfrac{1}{
1 + p}}}  \ge \dfrac2{2 + p}$.

2. $P(C_2 \mid D_3) = \dfrac{1}{1+(1-p)} = \color{darkorange}{\dfrac1{2-p}}.$ Now, $p ∝ P(C_2 \mid D_3)$. Why this about-face?

N.B. $1/2 \le p \le 1 \iff -1/2 \ge \-p \ge -1 \iff 3/2 \ge 2 - p \ge 1 \iff 2/3 \le {\color{darkorange}{\dfrac1{2-p}}} \le 1$.




>40. Consider the Monty Hall problem, except that Monty enjoys opening door 2 more
than he enjoys opening door 3, and if he has a choice between opening these two doors, he opens door 2 with probability p, where $1/2 \le  p \le 1$.
>
>To recap: there are three doors, behind one of which there is a car (which you want),
and behind the other two of which there are goats (which you don't want). Initially,
all possibilities are equally likely for where the car is. You choose a door, which for concreteness we assume is door 1. Monty Hall then opens a door to reveal a goat, and offers you the option of switching. Assume that Monty Hall knows which door has the
car, will always open a goat door and offer the option of switching, and as above assume that if Monty Hall has a choice between opening door 2 and door 3, he chooses door 2 with probability p (with $1/2 \le  p \le 1$).
>
>(a) Find the unconditional probability that the strategy of always switching succeeds
(unconditional in the sense that we do not condition on which of doors 2 or 3 Monty
opens).
>
>(b) Find the probability that the strategy of always switching succeeds, given that Monty
opens door 2.
>
>(c) Find the probability that the strategy of always switching succeeds, given that Monty
opens door 3.
>
>### Solution:
>
>(a) Let $C_j$ be the event that the car is hidden behind door j and let W be the event
that we win using the switching strategy. Using the law of total probability, we can find the unconditional probability of winning:
$P(W) = P(W \mid C_1)P(C_1) + P(W \mid C_2)P(C_2) + P(W \mid C_3)P(C_3)
= 0 \cdot 1/3 + 1 \cdot 1/3 + 1 \cdot 1/3 = 2/3.$
>
>(b) A tree method works well here (delete the paths which are no longer relevant after
the conditioning, and reweight the remaining values by dividing by their sum), or we
can use Bayes' rule and the law of total probability (as below).
>
>Let $D_i$ be the event that Monty opens Door i. Note that we are looking for $P(W|D_2)$,
which is the same as $P(C_3 \mid D_2)$ as we first choose Door 1 and then switch to Door 3. By Bayes' rule and the law of total probability,
>
>$P(C_3 \mid D_2) = \dfrac{P(D_2 \mid C_3)P(C_3)}{P(D_2)} = \dfrac{P(D_2 \mid C_3)P(C_3)}{P(D_2 \mid C_1)P(C_1) + P(D_2 \mid C_2)P(C_2) + P(D_2 \mid C_3)P(C_3)}    
= \dfrac{1 \cdot 1/3}{p \cdot 1/3 + 0 \cdot 1/3 + 1 \cdot 1/3} = \color{red}{\dfrac{1}{
1 + p}}.$
>
>(c) The structure of the problem is the same as part (b) (except for the condition that
$p \ge 1/2$, which was not needed above). Imagine repainting doors 2 and 3, reversing which is called which. By part (b) with $1-p$ in place of $p$, $P(C_2 \mid D_3) = \dfrac{1}{1+(1-p)} = \color{darkorange}{\dfrac{1}{2-p}}.$

Blitzstein, *Introduction to Probability* (2019 2 edn), Chapter 2, Exercise 40, p 91.    
pp 15-6 in the publicly downloadable PDF of curbed solutions.