I am working on the below problem (38) from Joe Blitzsteins Introduction to Probability and I accidently answered a slighty different question, which it would be great to check the answer to:
Question:
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 $\frac{1}{2} ≤ p ≤ 1.$
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 $\frac{1}{2} ≤ p ≤ 1.$)
(b) Find the probability that the strategy of always switching succeeds, given that Monty opens door 2
Answer:
I originally missed the line 'You choose a door, which for concreteness we assume is door 1' and have answered a variant on this question, which I would like to check is correct.
Let $S$ be the event you switch doors and win, $C$ be the event that the car is behind the door you initially chose, and let $M2$ be the event that Monty opens door 2, and $D1$ be the event you choose door 1 initially. The above question asks us to find $P(S | M2 , D1)$. I however want to find the unconditional probability $P(S | M2)$.
I have opted to use $C$ as the outcome of interest, and to find $S$ as $1 - C$, assuming the first choice of door is chosen with 1/3 probabiltiy for each door.
We want to know $P(C | M2) = \frac{P(M2 | C) P(C)}{P(M2)}$.
$P(M2 | C) = \frac{1}{3}p + \frac{1}{3}\frac{1}{2} + \frac{1}{3}0$
i.e. if you chose door 1, Monty has a free choice of door 2, 3 and will open with probability $p$. If you choose door 3, Monty has a free choice and will open with probability $\frac{1}{2}$. If you choose door 2, Monty will not open door 2.
$P(C) = \frac{1}{3}$
$P(M2) = \frac{1}{9}p + \frac{2}{9}\frac{1}{2} + \frac{1}{9}\frac{1}{2} + \frac{2}{9}\frac{1}{2}$
I go this from working from a tree diagram, with D1, D2, D3 having $\frac{1}{3}$ probability of being chosen, and within each of these the $P(C) = \frac{1}{3}$ and $P(C^c) = \frac{2}{3}$. For door 1 & $C$, there is $p$ probability Monty will open the door (given free choice). For door 1 and $C^c$, there is $\frac{1}{2}$ probability the car is under 3 and he will open door 2. For door 2, the probabilities are 0 and for door 3 the same logic applies except Monty will always open door 1 vs. door 2 at probability $\frac{1}{2}$.
Putting this all together, $P(C | M2) = \frac{\frac{1}{9}p + \frac{1}{18}}{\frac{1}{9}p +\frac{5}{18}}$. This seems reasonable, giving $P(C | M2) = \frac{1}{3}$ when $p$ is $\frac{1}{2}$, $0.42$ when $p = 1$ and $0.2$ when $p = 0$, which makes sense as Monty opening door 2 should increase the overall chance of the car been under you as he is more likely to open D2 when has a free choice and you are on door 1. To get the answer in the original question you can take the complement of this result. Is this correct?
I don't believe your analysis is correct.
If you choose door $1$ initially, there are four possible subsequent outcomes, $\ M2\,\& \,C$$,M3\,\&\ C$$, M2\,\&\,C3\ $, and$ M3\,\&\,C2\ ,$ with the following probabilities \begin{align} P(\, M2\,\&\,C\,)&= \frac{p}{3}\\ P(\, M3\,\&\ C\,)&=\frac{1-p}{3}\\ P(\, M3\,\&\ C2\,)&= \frac{1}{3}\\ P(\, M2\,\&\ C3\,)&= \frac{1}{3}\ . \end{align} Thus \begin{align} P(\,M2\,|\,C)&=\frac{P(\, M2\,\&\,C\,)}{P(\,C\,)}=p\ ,\\ P(\,M2\,)&=P(M2\,\&\,C\,)+ P(\, M2\,\&\ C3\,)=\frac{1+p}{3}\ , \end{align} and $$ P(\,C\,|\,M2)= \frac{P(\, M2\,\&\,C\,)}{P(\,M2\,)}=\frac{p}{1+p}\ . $$ For $\ p<1\ $ this is less than $\ \frac{1}{2}\ $, so your best option is still to switch to door $3$, when you will win the car with probability $\ P(S\,|\,M2\,)=$$\frac{1}{1+p}\ $.
Likewise, $$ P(\,C\,|\,M3\,)= \frac{P(\, M3\,\&\,C\,)}{P(\,M3\,)}=\frac{1-p}{2-p}\ $$ which is again less than $\ \frac{1}{2}\ $ whenever $\ 0<p\ $, so you should switch to door $2$, when you will win the car with probability $\ P(S\,|\,M3\,)=$$\frac{1}{2-p}\ $.
If you play optimally, your overall probability of winning the car is still \begin{align} P(S)&= P(S\,|\,M2\,)P(\,M2\,)+P(S\,|\,M3\,)P(\,M3\,)\\ &=\left(\frac{1}{1+p}\right)\left(\frac{1+p}{3}\right)+\left(\frac{1}{2-p}\right)\left(\frac{2-p}{3}\right)\\ &=\frac{2}{3}\ , \end{align} independent of the value of $\ p\ $.