Probability: A variant of the Monty Hall Problem Monty has a new game it's like his old one but now there are no goats just one car hiding behind one of three doors and this time he doesn't know where the car is.
To play you pick one of the three doors.  Monty now opens one of the other doors chosen at random.  If it's the car then you have lost and the game is over if not you get to choose if you want to change your mind or not and the door you choose is opened.  If the car is standing behind it, it's yours to drive away and keep.
The questions.


*

*What's the probability of the game getting to the final round.

*What's the probability of the game getting to the final round and you win the car if you switch.

*What's the probability of the game getting to the final round and you win the car if you don't switch.

My initial thoughts:
You have $\frac{1}{3}$ probability of picking the car in the first round so there is a $\frac{2}{3}$ probability that the host has a chance of picking the car meaning that $\frac{1}{3}$ of games finish on the second round.  So $\frac{2}{3}$ of games reach the final round and whether you swap or not you have a $\frac{2}{3} \cdot \frac{1}{2} = \frac{1}{3}$ probability of winning.
However, It's possible to list all possible games making all possible assumptions of winning door and both the contestants and the hosts choices.  When I do this I find there are only 30 possible games.
80% of which get to the final (third) round and I win 40% of the time if I swap and 40% of the time if I don't swap.
Which set of results is correct?  And what's wrong with the flawed argument?
 A: The point in the classical Monty-Hall Dilemma is that you can exploit the knowledge of the moderator (and therefore the information he gives you). So if the moderator chooses at random too, then switching should not give you any advantage.
To give a more detailed analysis, I draw a tree:

There is just one path in which the game is finished after two stages, and it has probability $2/3 \cdot 1/2 = 1/3$. So the probability of getting to the final round is $2/3$. For your other question we can read off:
ii) $1/3$
iii) $1/3$
So switching or staying with your first choice doesn't matter.
A: You could list the "possible games" by making a decision tree. 
At the first level of the tree someone puts the car behind one of the doors -- 
3 possible outcomes.
At the next level, you choose a door -- 9 possible outcomes altogether.
Then Monty opens one of the two doors you didn't choose, and now there are
a total of 18 possible outcomes. In six of those cases you lose immediately;
in the other twelve cases you can choose to "stay" with your
original choice of doors or "switch" to choose the other unopened door,
which brings us to 30 cases.
But those cases are not all of equal probability.
For example, take the two cases where the car was behind door 1, 
you chose door 2, Monty opened door 3, and you either "stayed" or "switched".
The sum total of the probabilities of those two cases is exactly the same
as the probability of the case where the car was behind door 1, 
you chose door 2, and Monty opened door 1, causing you to lose immediately.
So 80% of your cases get to the final round, but the probability that you
will reach one those cases is just 2 times the probability that you will reach
one of the cases where you lose before the final round, not 4 times that probability.
Your first analysis gives the correct odds:
lose $\frac{1}{3}$ of the games before the final round,
lose $\frac{1}{3}$ of the games during the final round,
and win $\frac{1}{3}$ of the games during the final round.
A: Your first analysis is correct, and the problem with the second argument is that the 30 games are not equally likely.  The 24 games ending in the final round each have a probability of 1/36, but the 6 games ending in the second round each have a probability of 1/18.
A: The flaw is in your 80%/40%/40%. You are correct in that if you write out all the possible paths, 20% of them will end after the second round. But that is not equal to the probability that path will occur. Assuming that each potential path of events is equally probable is a very common fallacy in calculating probabilities.
Here's a good example:
You flip a fair coin. If the first coin flip comes up heads, you flip again. If the first coin flip comes up tails, you stop. What's the probability of flipping 2 heads in a row?
The potential paths are:
T - game ends, you lose
HT - game ends, you lose
HH - game ends, you win
If all of these paths were equally likely, you'd have a $\frac 13$ chance of winning. However, they're not all equally likely. Tails coming up first (and thus ending the game) will happen $\frac 12$ of the time, whereas the other two paths will each happen $\frac 14$ of the time. So the probability of winning is $\frac 14$.
