Thursday, 5 March 2015

Monty de Sade

The puzzle below was written when I firmly believed that the answer to the question at The Reverse Monty Hall Problem was 1/2 (as argued in my solution as well as in later articles and in many comments, both here and at reddit).

Now I have finally been convinced that the answer is, as many said right from the beginning, 2/3.  That said, I still find the exercise below vaguely interesting.

(For those of you who have suffered through my journey, I am pretty sure that this is an example of a “Monty Faffs About” game.)


Monty de Sade decides that the Monty Hall Problem is too easy, so he devises a new variant.

There are still three doors (Red to the left, White in the middle and Green to the right).

There are still two goats (Mary and Ava, but we no longer care about their names).

There is still a car.

The goats and the car are still arranged at random behind the three doors.

However, the contestant is required to pick between two options, A and B, rather than selecting a door.

A and B may be as follows:

A – G … B – CG

A – GG … B – C

A – CG … B – G

A – C … B – GG

In other words, A may be a door with a goat behind it (the contestant knows not which) while B would be a grouping of two doors, one of which hides the other goat and the other hides the car, and so on.

The contestant makes her choice between A and B (we can assume a random selection is made).

Monty opens a door revealing the goat without telling the contestant whether the door was from A or B, only that it was from the grouping of two doors so that A and B are still viable options for selection (both contain either a car or a goat, not both).

Monty asks the contestant if she wants to stay with her original choice or switch.

Should the contestant switch, or stay?  Why?


  1. It's fifty/fifty, but *not* for the same reason as it is in Monty Faff.

    In Monty Faff what you get is a 50/50 choice because 2 of 6 possibilities (if you distinguish the goats) are eliminated. Because the host is never forced the remaining choices are all equally likely.

    In this scenario the contestant is either in a normal monty hall problem (if he selected the lone door), or in the reverse monty hall problem (if he selected two doors). So what are his chances of winning if he switches?

    P(switch wins | single door selected) * P(single door selected) + P(switch wins | double door selected) * P(double door selected) =
    2/3 * 1/2 + 1/3*1/2 = 2/6 + 1/6 = 1/2

    One way to see that this scenario is different from monty faff is that if you switch the reveal and the contestants choice in Monty Faff you get a normal Monty Hall scenario. If you switch the reveal and the contestants choice in your new scenario the chances don't change at all.

    Also: you forgot 2 possible A/B distributions:
    A – G … B – GC
    A – GC … B – G

    1. I deliberately left out the distributions that you identify. I admit that I should have used the word "combination" to make this more clear (as opposed to "permutation", a word that I didn't use either), but in the paragraph after where I laid out what A and B might represent, I did actually describe combinations rather than permutations.

      I agree that, strictly speaking, this is not a "Monty Faffs About" game because the host is constrained. And I also agree that it is a mix of potential Monty Hall and Reverse Monty Hall games, in which the contestant does not know which form of game she is playing. From the contestant's point of view, therefore, I think it is functionallyequivalent to a "Monty Faffs About" game (while still accepting that it isn't actually one). All the contestant knows is that Monty de Sade will open a door to reveal a goat. She has no basis on which to establish which door is more or less likely to be opened (barring the original single constraint of the "Monty Faffs About", namely that the car is not revealed, but she has not idea where the car is). Once the door is opened, she has no basis on which to establish which of the remaining closed doors hides the car. Remember that in the Monty Hall and the Reverse Monty Hall, it's this knowing the consequences of the host's constraints that leads to the 2/3-1/2 result, in this case the contestant is denied the necessary information.

      It is possible to remove the constraint on the host in four slightly more diabolical variants of the game, all based on the host being able to open a door with a goat behind it completely at random:

      Variant 1: The host reveals a goat at random. If this goat was in a grouping by itself, the game show's producer will select one of the doors in the other grouping at random and reassign it to the grouping that the host just emptied. For example, the grouping would have been either CG-G or G-CG (where CG is a combination, not a permutation) and if Monty de Sade removed the lone goat, his producer would toss a fair coin with a goat on one side and car on the other and make a reassignment on that basis.

      Variant 2: The host reveals a goat at random. If this goat was in a grouping by itself, the game show's producer will move the goat from the other grouping to fill the gap.

      Variant 3: The host reveals a goat at random. If this goat was in a grouping by itself, the game show's producer will move the car from the other grouping to fill the gap.

      Variant 4: The host reveals a goat at random. If this goat was in a grouping by itself, the gap will remain and the contestant will will nothing if she selects that group or both the car and the goat is she selects the other.

      My gut feeling is that in each of these, the likelihood of winning is 1/2 irrespective if the contestant stays or switches, but admit to being a little uncertain with respect to the Variant 1 and Variants 2/3. What I think might be the case, is that each of them ends up being 1/2, but the probabilistic route to that 1/2 figure varies slightly. If this is actually case, then I feel justified in thinking of them all as being functionally equivalent variants of "Monty Faffs About".

    2. I used a probability tree approach and am now more certain that my gut feeling in this case is correct. I've posted an image here. I was going to do it in this comment, but there is too much faffing about required :)


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