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: Case 1 [G][G][C] -> [G][X][C] -> [G][X][C] Stay: Goat ^ ^ | ^ -> [G][X][C] Switch: Car ^ Case 2 [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: Case 3 [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.

  • $\begingroup$ This topic has been beaten to death at this and various other sites. See here for example $\endgroup$ – Shailesh Jan 19 '16 at 8:57
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    $\begingroup$ Also why are people always assuming the person actually wants a car? $\endgroup$ – mathreadler Jan 19 '16 at 9:18
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    $\begingroup$ Well thanks for be such a supportive community. After all I was able to figure out, what is wrong with those trees, but since everybody think, that it is already answered somewhere else without presenting how those answers are linked to counting paths in a tree, I am not able to answer my own question. $\endgroup$ – Oliver Jan 19 '16 at 9:52
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    $\begingroup$ Your tree is fine, but you should be weighting the branches by their probabilities. Contestant picks 1, 2, or 3 should each be given weight 1/3. In case 1, Monte opens door 2 should be given weight 1; in case 2, Monte opens door 1 should be given weight 1; in case 3, Monte opens door 1 should be given weight 1/2, as should Monte opens door 2. Now if we look at the switching branches, the contestant will win 1/3*1+1/3*1=2/3 of the time, and will lose 1/3 * 1/2+1/3*1/2 = 1/3 of the time. $\endgroup$ – Will Orrick Jan 19 '16 at 12:00
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    $\begingroup$ @WillOrrick Thank you very much. That's exactly what's missing there. After reformulate to colored balls and urns, I found out. I would have answered, but.. anyhow.. Thank! $\endgroup$ – Oliver Jan 19 '16 at 13:18