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The Monty Hall problem (wiki) is described as follows:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

In the analysis, it reads,

Most people come to the conclusion that switching does not matter because there are two unopened doors and one car and that it is a 50/50 choice. This would be true if the host opens a door randomly, but that is not the case; the door opened depends on the player's initial choice, so the assumption of independence doesn't hold.

I am interested in the variant in which the host opens a door randomly and try to formalize it.

First, it is crucial to explicitly identify the assumptions for this variant:

  • $A_1:$ The host must always open a door that was not picked by the contestant.
  • $A_2:$ The host must always offer the chance to switch between the originally chosen door and the remaining closed door.
  • $A_3:$ The host does not know what lies behind the doors and opens one at random that happens not to reveal the car.

By Bayes' theorem, we can obtain that

\begin{align*} \Pr\{C_2 \mid H_3, Y_1\} = \frac{\Pr\{H_3 \mid C_2, Y_1\}} {\Pr\{H_3 \mid C_1, Y_1\} + \Pr\{H_3 \mid C_2, Y_1\} + \Pr\{H_3 \mid C_3, Y_1\}}, \end{align*}

where, the event $C_i$ denotes that the car is behind the door $i$, $Y_1$ denotes that you pick door 1, and $H_3$ denotes that the host opens door 3 and happens to reveal a goat.

Therefore, the task is to assign probabilities to $\Pr\{H_3 \mid C_1, Y_1\}$, $\Pr\{H_3 \mid C_2, Y_1\}$, and $\Pr\{H_3 \mid C_3, Y_1\}$.

Questions:

  1. Are the three assumptions correct and sufficient to characterize the situation where the host opens a door randomly?
  2. Should the assumption $A_3$ enforce the requirement that "... happens not to reveal the car"? Similarly, should the event $H_3$ specify that "... happens to reveal a goat"?
  3. How to assign probabilities to $\Pr\{H_3 \mid C_1, Y_1\}$, $\Pr\{H_3 \mid C_2, Y_1\}$, and $\Pr\{H_3 \mid C_3, Y_1\}$?
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  • $\begingroup$ Are your assumptions for the standard game or your proposed variant? They don't seem much different. If you retain $A_3$ then it would seem to be the same as usual. There will be only be a choice of doors for the host to open if the contestants choice is the car. How this choice is made does not seem to matter. If you drop $A_3$ then the host will sometimes reveal the car. We would need to know whether the contestant is allowed to switch to it. $\endgroup$
    – badjohn
    Commented May 8, 2018 at 8:54
  • $\begingroup$ @badjohn The assumptions are for the variant. IMO, the key of $A_3$ is that the host does not know what lies behind the doors; this is not the case in the standard version. $\endgroup$
    – hengxin
    Commented May 8, 2018 at 8:57
  • $\begingroup$ Then how is the "happens not to reveal the car" enforced? Or are you saying that it is not enforced but you are only considering the case when it does not occur. I would guess that this case is the same as the original. I don't see that it will make a difference whether it opens a door because he knows that it is a goat or just happens to by luck. The information provided to the contestant is the same. $\endgroup$
    – badjohn
    Commented May 8, 2018 at 9:00
  • $\begingroup$ @badjohn I am also confused with this part "happens not to reveal the car". Now I think the explanation of "it is not enforced but you are only considering the case when it does not occur" is reasonable. In this case, I think $\Pr\{H_3 \mid C_2, Y_1\}$ is $1/2$, instead of 1 in the standard version. $\endgroup$
    – hengxin
    Commented May 8, 2018 at 9:05
  • $\begingroup$ But you was you that wrote: "happens not to reveal the car". I am asking you to clarify what you mean. $\endgroup$
    – badjohn
    Commented May 8, 2018 at 9:06

5 Answers 5

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Your formulation is correct.

Now $\Pr\{H_3\mid C_1,Y_1\}$ is simply the probability that the host opens door $3$ given $C_1$ and $Y_1$ (because you are given $C_1$, the host can't open door $3$ to reveal a car), and this is $1/2$ because he randomly chooses between the two doors you didn't open. Similarly $\Pr\{H_3\mid C_2,Y_1\}=1/2$. However, $\Pr\{H_3\mid C_3,Y_1\}=0$ - if the car is behind door $3$, he can't open that door and reveal a goat. This gives $\Pr\{C_2\mid H_3,Y_1\}=1/2$ as expected.

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If the host opens a door without knowledge and behind that door a goat is found then the new situation is that you can choose from two doors that have equal probability (so both $0.5$) to hide a car.

To come to this conclusion plain thinking is enough and the rule of Bayes is not needed.

There is no profit in changing your original choice and no profit in keeping to it.

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    $\begingroup$ Clear explanation! $\endgroup$
    – Macrophage
    Commented May 8, 2018 at 9:11
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What are problems like the Monty Hall Problem?

Here’s one: Originally urn 1 contains 1 black ball and 9 white balls. A ball is picked at random from urn 1 and placed in urn 2. Balls are then drawn from urn 1 (now having one less ball) without replacement: This illustrates/demonstrates the probabilities of Monty revealing the prize,and/or you getting it depending on your choice, as doors are shown not to have the prize behind them. Probabilities of what color the ball is on each particular pick are shown below depending on what happened when the first ball was taken from urn 1 and on the previous pick(s). Note N is the number of the “pick” after you’ve chosen which ball to put in urn 2 and the probabilities in a particular row are the apriori probabilities of getting a particular color ball on the pick you are making. The last column represents(9/10 * the probabilities in the columns for when urn 2 contains a white ball) Note that as you go down the yellow shaded column those probabilities are essentially conditional probabilities, in that they reflect having the event(s) in the cells immediately above not occur on the previous picks,i.e. not getting a black ball=not revealing the prize and, correspondingly, the event(s) to the immediate left occur. urn 2 having the black ball corresponds to you having made the correct choice for the prize. we are interested in the probability of getting the prize(black ball) as Monty opens doors equivalent now to picking from urn1. Monty is not allowed to touch urn 2. Of course if he were, the probability he would reveal the prize would be 1/10.

This is exactly what happens in the Monty Hall problem. Say instead of 3 doors, there were 10 doors, with the prize behind only one. A door is first chosen, but then eliminated from further participation in the selection process. The last row in the table above, highlighted in green, represents the probabilities associated with you choosing the last ball from urn 1 or, equivalently, choosing the final remaining unopened door in the Monty Hall problem, instead of your original choice. Just for fun let’s make the same table for urn 1 originally containing 3 balls, 2 White, 1 Black. We have:

So, the black ball (the prize) has a 2/3 probability of being behind the remaining unopened door, and if there were 10 doors to begin with, a 9/10 chance of the same. And not a word has been spoken about whether Monty knows where the prize is or not. Only tabulating the probabilities for getting the prize (or not) as doors besides the one you originally chose are opened…

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Without loss of generality, you may simply label door 1 as whatever door you choose, door 2 as whatever door the host opens and door 3 as whatever door is left.  The only random variable is: behind which of these is the car?

So: Given that the host has not revealed the car (but choose without bias), what is the probability that the car is behind door 2?

Then by your formulation:$$\mathsf P(C_2\mid Y_1,H_3,(C_1\cup C_2))=\dfrac{\mathsf P(C_2\mid Y_1,H_3)}{\mathsf P(C_1\cup C_2\mid Y_1,H_3)}= \dfrac{1/3}{2/3}$$

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It does not matter if Monty knows where the prize is or not.

Trying to work what Monty knows or doesn’t know into the problem is a tempting (apparently) but needless distraction.

This is all there is to the Monty Hall problem: What’s the probability the prize is behind the doors you didn’t pick? Does this probability depend upon whether those doors are opened simultaneously or in sequence? I think you might agree that it doesn’t. So as doors (say there are N doors to start with, not necessarily limited to 3) are opened, the probability that the prize is in the group of doors you didn’t pick remains the same. Which means that the probability it’s behind the door you did pick also remains the same. That’s all you need to know.

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    $\begingroup$ Over-emphasizing various words in your answer does not make it more convincing... $\endgroup$
    – user856
    Commented May 13, 2019 at 21:06
  • $\begingroup$ True. It's convincing enough without the emphasis, I would hope. There is a fair amount of controversy about Monty knowing or not knowing where the car is, however. $\endgroup$
    – Marshall
    Commented May 14, 2019 at 1:33

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