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?