Modified monty hall problem

Question

Hello how to show the following

You are given the choice of 3 doors. Behind one is a car and the other two are goats. You pick a door uniformly at random say 1, and Monty opens another door, say 3 which has a goat. Monty asks "do you want to switch your choice to door 2?" If we have the following modified assumptions:

*The car was put behind a uniformly chosen random door.

*Monty knows where the car is and always opens a door with a goat behind it.

*When Monty has a choice(i.e. the contestant picked the door with the car behind it), he chooses the door with the larger number with probability 3/4.

Show that if the contestant switches, she wins with probability 4/7 not 2/3.

Thanks a lot!

Answer 1

If she always switches, she wins exactly when her original choice was wrong, which means that she wins with probability $\frac23$.

Suppose that her strategy is to switch if and only if Monty opens the lower-numbered of the two doors that she did not choose initially. She wins if she chose right initially and Monty opened the higher-numbered of the other two doors, or if she chose wrong initially and Monty opened the lower-numbered of the other two doors.

The probability of the first of these alternatives is $\frac13\cdot\frac34=\frac14$.

Now suppose that she guessed wrong, which of course occurs with probability $\frac23$. Consider the $3$ equally likely arrangements, CGG, GCG, and GGC. If she guessed wrong in the first one, Monty didn’t open the lower-numbered of the other two doors, since the car is behind it. If she guessed wrong in the second one, Monty opened the lower-numbered of the other two doors with probability $\frac12$. And if she guessed wrong in the third one, Monty had to open the lower-numbered of the other two doors. In short, if she guessed wrong, the probability that Monty opened the lower-numbered of the other two doors is $\frac12$. The probability of this alternative is therefore $\frac23\cdot\frac12=\frac13$.

Her probability of winning if she does switch according to this strategy is then $\frac14+\frac13=\frac7{12}$.

Answer 2

For your question to make sense, I assume you want to show that $\frac{4}{7}$ is the probability of winning if the contestant switches given that the contestant chooses door 1 and Monty opens door 3. Let $A, C, M$ be the random variables representing the number of the door which the car is behind, which the contestant chooses, and which Monty opens respectively. \begin{align} & \text{P(car is in door 2|contestant chooses door 1 and Monty opens door 3)}\\ =&P(A=2|C=1, M=3) \\ =& \frac{P(A=2,C=1, M=3)}{P(C=1, M=3)} \\ =&\frac{P(A=2,C=1, M=3)}{P(A=1,C=1, M=3)+P(A=2, C=1, M=3)} \\ =&\frac{\frac{1}{9}}{(\frac{1}{9})(\frac{3}{4})+\frac{1}{9}} \\ =&\frac{4}{7} \end{align}

Because which door Monty opens makes a difference, the probability of winning if the contestant switches depends on the doors chosen by the contestant and opened by Monty.

Edit: I suppose you can write it as follows:

$P(A \not =C|M \text{is the larger of the 2 numbers other than } C)=\frac{4}{7}$

Similarly we would have

$P(A \not =C|M \text{is the smaller of the 2 numbers other than } C)=\frac{4}{5}$