# Monty Hall Variant a priori probability of winning by whiching when Monty has preference for door 2

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

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?

• The problem statement is not clear. Is the preference for Door $2$ independent of which door the contestant chooses? So that if the contestant chooses door $2$ then the preference is irrelevant? If so, then of course the answers are the same if the contest chooses another door ($1$ or $3$) but if the contestant chooses $2$ then you just have the usual Monty Hall problem. Note that, given the statement in $3$, it appears that we are to assume that $2$ was not selected by the contestant, so the $\frac 13$ that you assumed is not correct.
– lulu
Jan 7, 2022 at 12:03
• But your variant is not clear. In order to exploit the host's preference, the contestant must choose $1$ or $3$. Choosing door $2$ is sub-optimal, as it wastes the competitive advantage. It makes no difference if the contestant chooses $1$ or $3$, but that's a $\frac 12$ matter, not $\frac 13$. If you had a different variant in mind you ought to state it clearly and explictly.
– lulu
Jan 7, 2022 at 17:33
• I don't understand what variant you are solving. I suggest: delete all references to the other problem, the one you read incorrectly. None of that has any relevance to the problem at hand. Start your post with a clear statement of the problem you are actually interested in solving.
– lulu
Jan 7, 2022 at 18:08
• Perhaps your variant is: "in all cases, Monty has a preference for opening the door with the smallest number. Indeed, if both unselected doors are empty, Monty opens the one with the lower number with (known) probability $p\in [\frac 12, 1]$. Given this, and assuming optimal strategy, what is the probability a contestant will find the prize?" Just guessing here, but at least that variant makes sense.
– lulu
Jan 7, 2022 at 18:20
• No worries, good luck!
– lulu
Jan 7, 2022 at 18:29

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 $$\ 0, so you should switch to door $$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\$$.