For (a variation) of the Monty Hall problem, what permits you to repaint doors 2 and 3? What permits you to replace p with $1  p$?
Please see the text colored in red at the bottom.

Why can you haughtily just — and what legitimizes you to — "[i]magine repainting doors 2 and 3, reversing which is called which"? This is completely unworkable, halfbaked because you can't reverse doors in the real game!

What legitimates you to replace $1  p$ in part (b), with $p$ for part (c)? In general, you can't haphazardly replace variables!
 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 $P(WD_2)$, 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,
$P(C3 \mid D2) = \dfrac{P(D_2 \mid C_3)P(C_3)}{P(D_2)} = \dfrac{P(D_2 \mid C_3)P(C_3)}{P(D_2 \mid C_1)P(C_1) + P(D_2 \mid C_2)P(C_2) + P(D_2 \mid C_3)P(C_3)}
= \dfrac{1 \cdot 1/3}{p \cdot 1/3 + 0 \cdot 1/3 + 1 \cdot 1/3} = \dfrac{1}{ 1 + p}.$(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). $\color{red}{\text{Imagine repainting doors 2 and 3, reversing which is called which. By part (b) with $1p$ in place of $p$}}$, $P(C_2 \mid D_3) = \dfrac{1}{1+(1p)} = \dfrac{1}{2p}.$
Blitzstein, Introduction to Probability (2019 2 edn), Chapter 2, Exercise 40, p 91.
pp 156 in the publicly downloadable PDF of curbed solutions.
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