Coming from Blitzstein's book, here is a paraphrased version of the question.
Consider a version of Monty Hall where there can be more than one car, and each of the 3 doors independently have probability $p$ of having a car and probability $q = 1-p$ of having a goat behind them. When Monty goes to reveal the contents behind one of the doors, he will always choose a goat when possible. When the contents behind the two other doors are the same, he will choose one of the doors arbitrarily (probability 0.5 for each).
The contestant applied this strategy: They will choose door 1, wait for Monty to reveal the contents of one of the other doors, then they will switch.
a) What is the unconditional probability the contestant wins a car
b) What is the probability the contestant wins a car given that Monty reveals a goat behind door 2.
My approach was the following, and as there are no solutions I wanted to check and see if A) my solution is correct and more importantly B) whether others had different, more efficient or intuitive approaches .
(a)
Let $W$ mean the contestant won a car. Let $C_i$ mean that there is a car behind door $i$.
Below, I just describe all cases with law of total probability.
$P(W) = P(W | C_1, C_2, C_3) P(C_1, C_2, C_3) + P(W | C_1^c, C_2, C_3) P(C_1^c, C_2, C_3) + P(W | C_1, C_2^c, C_3) P(C_1, C_2^c, C_3) + P(W | C_1, C_2, C_3^c) P(C_1, C_2, C_3^c) + P(W | C_1, C_2^c, C_3^c) P(C_1, C_2^c, C_3^c) + P(W | C_1^c, C_2, C_3^c) P(C_1^c, C_2, C_3^c) + P(W | C_1^c, C_2^c, C_3) P(C_1^c, C_2^c, C_3) + P(W | C_1^c, C_2^c, C_3^c) P(C_1^c, C_2^c, C_3^c)$
We can apply some symmetry. We can also substitute the cases where both $C_2$ and $C_3$ as 1 and both $C_2^c$ and $C_3^c$ as 0.
$P(W) = 1P(C_1, C_2, C_3) + 1P(C_1^c, C_2, C_3) + 2P(W | C_1, C_2^c, C_3) P(C_1, C_2^c, C_3) + 0 + 2P(W | C_1^c, C_2^2, C_3) P(C_1^c, C_2^2, C_3)+ 0$
$P(C_1, C_2, C_3) = p^3$
$P(C_1^c, C_2, C_3) = qp^2$
$P(W | C_1, C_2^c, C_3)P(C_1, C_2^c, C_3) = 1(qp^2)$
$P(W | C_1^c, C_2^c, C_3)P(C_1^c, C_2^c, C_3) = 1(q^2p)$
Overall, we have
$P(W) = p^3 + qp^2 + 2(qp^2) + 2(q^2p) $
$= p^3 + 3qp^2 + 2q^2p$
(b)
Let $G_i$ mean that monty opens door $i$ and reveals a goat.
$P(W | G_2) = P(C_3 | G_2)$ as we always switch
$\frac{P(G_2 | C_3)P(C_3)}{P(G_2)}$
$P(G_2 | C_3)P(C_3) = P(G_2 | C_2^c, C_3)P(C_2^c |C_3)P(C_3) = 1(q)(p)$
$P(G_2) = P(G_2 | C_3)P(C_3) + P(G_2 | C_3^c)P(C_3^c)$
$P(G_2 | C_3^c)P(C_3^c) = P(G_2 | C_2^c, C_3^c)P(C_2^c | C_3^c)P(C_3^c) = \frac{1}{2}(q)(q) = \frac{1}{2}q^2$
Overall, we have $P(W | G_2) = \frac{pq}{pq + \frac{1}{2}q^2} = \frac{p}{p + \frac{1}{2}q} = \frac{2p}{2p + q}$.
Thank you in advance.