Subtle Twist on the Monty Hall Problem---Does It Make a Difference? 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?

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 one-third of all contestant to start with; Groups A and B were two-thirds 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, one-third 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.
