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?

Solution 1:

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:

$P(\neg C|G)=\frac{P(\neg C)(P(G|\neg C)}{P(\neg C)(P(G|\neg C) + P(C)P(G|C)}=\frac{\frac{2}{3}\times\frac{1}{2}}{\frac{2}{3}\times\frac{1}{2} + \frac{1}{3}\times1} = \frac{1}{2}$, 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)

Solution 2:

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.