Marilyn's Six Games

Over the past two weeks I have argued that, at precisely the moment at which a contestant in a Reverse Monty Hall game is required to assess the likelihood of benefitting from switching from the one remaining closed door of her selected two doors to the door she did not originally select, the likelihood is 1/2.  A key part of my argument was what I present in Monty Does Play Dice and another key part was this, my argument that there are as many as six mini-games that a contestant going into the game will face and that when it is time to make an assessment of likelihood, there are only two remaining.  This is still true.  However, it does not change the likelihood of benefitting from a switch, which remains 2/3.

Other than adding this introduction and a note at the end, I have not edited this article from what I intended to write when I was still fully convinced that the answer was 1/2.

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In Marilyn Gets My Goat, which was an attempt to crystallise the scenario introduced in The Reverse Monty Hall Problem, I tried as much as possible to show that when a contestant is playing the game, it’s a real game.  You can do it in real life, with a real contestant, a real host, real doors, real live goats and a real car.  This was because there were a few people who keep saying that, hypothetically, even though a particular door was opened, the host could have opened another door, and revealed a different goat.  Well, yes, this is true but not after the door was opened.

Anyway, another major objection relates to the fact that I agree that, over multiple instances of the Reverse Monty Hall Problem (and the Monty Hall Problem), the contestant will win 2/3 of the time by adopting a policy of staying (with the Reverse Monty Hall Problem, which is equivalent to switching with the classic Monty Hall Problem) while at the same time maintaining that, in a single iteration, one shot instance of the game, there is no benefit in switching (because the likelihood of winning as a consequence of switching or staying is equal at 1/2).

This is a reasonable objection, because it appears to fly in the face of the Law of Large Numbers, a fact which /u/ChalkboardCowboy and /u/OmarDiamond particularly object to.  Note I said “appears”.  Neither Chalkboard nor Omar make that distinction, so when I say something like “my solution doesn’t fly in the face of the Law of Large Numbers” we end up talking past each other.

What I’d like to do here is explain what is going on and how you can go from a likelihood of winning of 1/2 in a single instance to a likelihood of 2/3 over many instances as a consequence.

Note that I’ve already addressed it partially in various comments.

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The thing is that a contestant in the Reverse Monty Hall Problem could be playing one of six different games, they are the game in which she tries to win the car when the distribution of goats (called Ava and Mary, as per Marilyn Gets My Goat) and car are:

Ava-Car-Mary (ACM)

Ava-Mary-Car (AMC)

Car-Ava-Mary (CAM)

Car-Mary-Ava (CMA)

Mary-Ava-Car (MAC)

Mary-Car-Ava (MCA)

If you think that the goats are indistinguishable and/or interchangeable and/or represent only otherwise undefined losing options, then there are still three potential mini-games.

Car-goat-goat (Cgg)

goat-Car-goat (gCg)

goat-goat-Car (ggC)

When you think of multiple instances of The Reverse Monty Hall Problem, you are considering a uniform distribution of these mini-games without the player having any idea which one they will be playing.  You know that if the goats and car are randomly (and uniformly) distributed, then each time you select two doors you have a 1/3 likelihood of getting the car with each door.  As a combination, having two doors gives you a 2/3 likelihood of having the car, so you should (as a policy) stay with your two doors.

However, when you are in the middle of a single iteration, one shot instance of the game, after the door has opened, you have more information.  We’ll use the situation that the contestant Marilyn found herself in in Marilyn Gets My Goat.  Marilyn chose the Red Door and the Green Door and the host (Holly Mant, played by Angelina Jolie) opened the Red Door to reveal Mary the Goat.

When this happens, there are no longer six possible games that Marilyn could be playing.  She can only be playing the two games that have Mary behind the Red Door, namely:

Mary-Ava-Car (MAC)

Mary-Car-Ava (MCA)

Or, if you are not in the habit of naming your goats:

goat-Car-goat (gCg)

goat-goat-Car (ggC)

My argument is that because the goats and car were placed at random and the doors were selected by Marilyn at random, then we have no reason to think that, prior to the game commencing, there was any more likelihood associated with the MAC (ggC) distribution that there should be with the MCA (gCg) distribution.

There is certainly a greater likelihood that the Red Door was opened as a consequence of the car being behind the Green Door – because if there is a car behind that door the host would have been obliged to open the Red Door while he has a choice between the Red Door and the Green Door if Ava was behind the Green Door.  However, prior events do not always affect the present.  If there was a goat behind both the Red Door and the Green Door then the host would open a door at random (strictly speaking on instruction from her producer, who selected a door at random).  However, the likelihood of having opened the Red Door to reveal Mary is 1/1 once that door is opened.

This is a similar situation to tossing a fair coin and getting seven heads in a row.  The a priori likelihood of getting eight heads in a row is 1/256.  However, the likelihood of getting a head after seven heads in a row is 1/2.  It might be less likely to see Mary behind the Red Door in the instances where Ava is behind the Green Door, but once the door is opened and we see Mary behind the Red Door, it’s now no longer a likelihood, it’s an absolute certainty.

This might seem either quite sensible or totally counter-intuitive.  If you feel the former, you should read up on the Monty Hall Problem, because this answer here is not the standard solution.  If you feel the latter, you might want to take a look at Monty Does Play Dice, if you have not already done so.

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In conclusion, the reason why we get 2/3 over multiple iterations is because when we have a Red Mary scenario (being a game in which Mary is behind the Red Door) we only see it half the time.  The rest of the time we will see Ava behind either the White Door or the Green Door.  The same sort of spread happens with White Mary and Green Mary scenarios and the cumulative effect of this over multiple iterations is that there is an increased likelihood to win if you have a policy of staying in multiple iterations of the Reverse Monty Hall Problem (and a policy of switching in multiple iterations of the classic Monty Hall Problem).

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Please remember that while I found this argument quite convincing only a few days ago, it is nevertheless wrong in so much as the likelihoods are 2/3-1/3 and not 1/2-1/2.  I still think that it is true that the contestant is playing one of two specific mini-games, but the ramifications of that appear to be negligible.