The Reverse Monty Hall Problem and Conditional Probability

Please note that since I wrote this article, I have been persuaded that the argument it contains is wrong.  The correct answer for the scenario as it is worded is not 1/2 but rather 1/3 (meaning that the likelihood of winning as a consequence of staying is 2/3).

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Originally, this was to be my last article on The Reverse Monty Hall Problem – or at least the last on why the solution is “it doesn’t matter whether you switch or stay, the likelihood of benefitting from either choice is 1/2".  It didn't work out that way ...

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As the title suggests, we need to consider conditional probability.  The relevant equation is:

Pr(A | B) = Pr(A ∧ B) / Pr(B)

This means “The probability of A happening, given that B has happened, is equal to the (a priori) probability of both A and B happening, divided by the (a priori) probability of B happened”.  It may be phrased more briefly as “the probability that A given that B is equal to the probability that both A and B divided by the probability that B”.

In this case (please follow the links above if you don’t yet know the context), what we have is:

Pr(A) = the (a priori) likelihood that a car is hidden behind the door that you chose, but the philosopher didn’t open = 1/3

Pr(B) = the (a priori) likelihood that a goat is hidden behind a specific selected door = 2/3

Pr(A ∧ B) = the (a priori) likelihood that the car is hidden behind the unselected door AND a goat is hidden behind a specific selected door – noting that this is one of only three possible, equally likely permutations = 1/3

I have to stress, and I cannot stress this enough, that in the scenario I described, the philosopher had already opened the door before offering you the opportunity to switch.

I might need to ram this home, so please forgive me if you have already had a blinding flash of light and you now agree with me.  Let’s get more specific.  The doors don’t have any indications on them when you get there, but you are a paranoid sort and are convinced that there is some Derren Brown/Dynamo-type shenanigans going on, so you tell the philosopher you want to mark the doors just after he tells you which one he is going to open, but before he opens it.

• You walk to the door that you selected that will remain closed, and you mark it as Door 1,
• You walk to the door that the philosopher has told you will be opened, and you mark it as Door 2,
• You walk to the unselected door, and you mark it as Door 3, and
• Then you return to the philosopher to continue the show.   

The questions you must ask yourself are now made a little more simple:

Before I got here, what was the likelihood that the car would be behind Door 1?

There is only one car and three doors, therefore, Pr(A) must be 1/3

Before I got here, what was the likelihood that Door 2 would have a goat behind it?

There are two goats and three doors, therefore, Pr(B) must be 2/3.

Before I got here, what was the likelihood that the car would be behind Door 1 AND a goat would be behind Door 2?

This is precisely the same likelihood that there would be a car behind Door 2, because if there is a car behind Door 2, then there must be a goat behind Door 1, therefore Pr(A ∧ B) must be 1/3

Then it’s just a simple process of plugging in the numbers:

Pr(A | B) = Pr(A ∧ B) / Pr(B) = (1/3) / (2/3) = 1/2

It’s not rocket science, but it seems to have confused the heck out of a lot of people (including, to be totally honest, me).

Now that I have gone through this process, I am really hoping that the cognitive dissonance stays away!  However, there remains a feature that may still confuse people.

If you were asked, prior to entering the arena with the philosopher, whether it would be better, on average across multiple runs, to stay with your door (the one of two that remains to you) or to switch, the correct answer is it is better to stay – because, exactly as many have argued, this would be equivalent to the Monty Hall Problem.

This seems to be hugely counter-intuitive.

However, you have to remember that at this point in time, the philosopher has not yet opened any doors.  Therefore, at this point in time, the door that the philosopher opens could be any of the three (depending on which doors you select and which doors have goats behind them).

This seems to be the default position taken by all armchair statisticians, despite being encouraged repeatedly to consider the scenario from just after the door has been opened.

Someone did ask to explain how my two images could be reconciled (from The Reverse Monty Hall Problem – The Solution):

And 

Someone also complained about my graphical representation of the solution to the classic Monty Hall Problem: 

This might be difficult to explain, but please bear with me. 

First I need to re-jig the graphical representation of the classic Monty Hall Problem, not because what I did was wrong so much as it was not adequately explained.  I am assuming that, when given the opportunity to select between two goats, the philosopher chooses at random.  Therefore, the possibilities with the classic Monty Hall Problem are:

Now, imagine that we run the Reverse Monty Hall scenario over and over again, still assuming that when there is an option to select between two goats, the philosopher makes that selection at random.  This means that the possibilities are:

Then, removing invalid scenarios, we have: 

This can be compared with the results from the classic Monty Hall Problem.  So, yes, if you run the scenario over and over again, you will get the 2/3 - 1/3 split.  But ... there was never any intention that Reverse Monty Hall Problem be performed repeatedly.

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Once again, please note that the argument above is incorrect.  The correct answer is that the 2/3 - 1/3 split is as true (in this scenario) as it is for multiple iterations of it.