EDIT: I know that this is a trigger topic. I know there is a hell of posts out there. I don't need to be told, that I should switch, neither why. I want to know, what is wrong with those trees. If there is a post out there explaining it, I would happy accept that. But an answer like "The trees must be wrong because different calculation shows that odds are 1/3 and 2/3" does not answer why the trees are wrong.
Another question on Monty-Hall-Problem:
I totally understand all reasoning behind the classic problem and it pitfalls when it is not completely defined. I am also familiar with the Bayes' Theorem. I wanted to do a round up on the topic for a blog post and while I was tinkering with illustration I stumbled upon a thing I can solve:
I tried to illustrate the problem with a complete tree of possibilities. Without loss of generality I can assume that the car is behind door three and I have three start cases, where the player can choose door one, two or three:
Case 1 Case 2 Case 3
[G][G][C] [G][G][C] [G][G][C]
^ ^ ^
Case 1 expands to only one possible branch, Monty has to open door two. After that the player can stay or switch. Case 2 is similar:
[G][G][C] -> [G][X][C] -> [G][X][C] Stay: Goat
^ ^ | ^
-> [G][X][C] Switch: Car
[G][G][C] -> [X][G][C] -> [X][G][C] Stay: Goat
^ ^ | ^
-> [X][G][C] Switch: Car
Case 3 is the tricky one. Since the player picked the right door, Monty has two possible alternatives, the case is branching into two, then into four:
[G][G][C] -> [X][G][C] -> [X][G][C] Stay: Car
^ | ^ | ^
| -> [X][G][C] Switch: Goat
-> [G][X][C] -> [G][X][C] Stay: Car
^ | ^
-> [G][X][C] Switch: Goat
And here's my dilemma. I got 2 goats and 2 cars while staying and also 2 goats and 2 cars while switching. My first reasoning was to say, the both sub-trees in case 3 are equivalent, therefore I would have 2 goats and 1 car with Stay and 1 goat and 2 cars with Switch, as expected. But then case 1 and case 2 should also be equivalent and I would be down to 1 and 1.
I found another clue: I give Monty the possibility to choose. If I define the problem as: Monty must open the goat-door with the lowest number, everything would be fine again. But: How can the choice of the door by Monty affect the state-probability of being on a car-door?
Again: I know, this can't be right. I know, how to calculate all the stuff, but I can't figure out, what is wrong with making this tree and counting good events vs possible events.