In the Monty Hall problem, when the host picks a door and reveals an goat, does it make any difference if he did not know which door the real car was behind, and he just happened to pick a door with a goat?

5Short answer: Yes, this makes a difference (and it is the crux of the subtlety in this puzzle). – hardmath Nov 07 '14 at 11:59

2Why is this a duplicate? – quid Nov 07 '14 at 12:48

1yeah, this is not a duplicate but a question on a (very interesting) subtlety of the problem. – Nov 07 '14 at 13:02

1I nominate for reopening...its not about solving the usual monty hall problem. It's actually interesting. – Nov 07 '14 at 13:06

See this answer http://math.stackexchange.com/a/609302/ point 4. I still think this should be reopened. – quid Nov 07 '14 at 13:24
3 Answers
Ok, ill turn my comment into a full answer:
Let $C$ be the event where you choose the car and let $G$ be the event that Monty Hall shows the goat. Then the probability that the other door has a car given that monty shows you a goat is:
$P(\neg CG)=\frac{P(\neg C)(P(G\neg C)}{P(\neg C)(P(G\neg C) + P(C)P(GC)}=\frac{\frac{2}{3}\times\frac{1}{2}}{\frac{2}{3}\times\frac{1}{2} + \frac{1}{3}\times1} = \frac{1}{2}$, which is different from the original problem.
Intuitively, this is because Monty Hall is not really providing you with any additional information by showing you a goat (obviously he does if he reveals a car)
We can do this by computing conditional probability, and those numbers don't lie, but let me try to intuit more of the situation before I start doing numbers.
In standard Monty Hall, we can divide all contestants into two groups. Group A initially chose a door with a goat behind it, and Monty opened the other goat's door. Group B initially chose the car, and Monty randomly opened one of the other two doors and showed a goat. Up to this point in the game both groups had exactly the same experience; what's the likelihood I'm in Group A vs. Group B? There are twice as many contestants in Group A, so I'm twice as likely to be one of them, that is, $\frac23$ chance of Group A, $\frac13$ chance of Group B.
In modified Monty Hall, there are three groups of contestants. Group A initially chose a door with a goat behind it, and Monty then revealed the car. Group B initially chose a door with a goat behind it, and Monty then revealed the other goat. Group C initially chose the car, and Monty revealed one of the goats. I just saw a goat, so I'm not in Group A, but Group B and Group C had the same experience as I did up to this point; what is the chance I'm in Group B vs. Group C? Group C was onethird of all contestant to start with; Groups A and B were twothirds of all contestants, but Monty showed a car to half of them and a goat to the other half, so the two groups are equal, onethird of all contestants in each. So, same number of contestants in Group B and in Group C. I have the same chance to be in either group.
Now we can compute the conditional probabilities in the usual way and confirm that both answers are correct.
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This has been said in various ways in other answers/comments, but I think this is a clearer way to explain it that requires very little theory.
The most obvious difference is that sometimes when the host opened that door, the car would be there and the game would be over. This would happen 1/3 of the time. Also 1/3 of the time the car would be behind the door the player picked initially, and the other 1/3 of the time the car would be behind the door the player had the option of switching to.
If the question is "What is the probability of winning if you switch doors?" well it depends on whether you are including the instances where the contestant does not have that opportunity (on account of Monty opening the winning door by accident). If so the answer is 1/3. If, on the other hand, you are only looking at when Monty doesn't show the car prematurely (maybe the contestant gets to go again?) well then 1/3 out of 2/3 is 1/2.
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