The interviewer is wrong in stating that Monty "clearly" picked door 3 and that therefore the chances are 50-50. Monty picked door number 3 in this particular case but he didn't say why.
The interviewer is correct that if we do not know why Monty showed door number 3 (does he always show door number 3 no matter what is behind it or what you pick-- does he always pick a door at random that you did not pick no matter what is behind it-- does he always show a door with a goat you did not pick?) then the question can not really be answered. (But he is wrong in assuming that it is 50-50). We have to make an assumption but... what assumptions are valid and which aren't?
Here are several possible rules Monty could be played by:
Classic: Monty always shows you a goat you do not pick. Strategy: switch. 2 out of 3 in your favor.
Random: Monty picks a door you did not pick and this time it just randomly happened to be a goat. Strategy: doesn't matter. 2 out of 4 whether you switch or stay.
Devious: If you pick the car Monty will show you a goat in the hopes that you will assume a classic game. If you pick a goat he won't give you a choice. Strategy: stay. 100% in your favor.
Tough luck: Monty will always show you the car if he can. He'll only show you a goat if you pick the car. Strategy: stay. 100% in your favor.
Warped: There is one goat with a spotted tail. Monty will always show you a door you did not pick that does not have the spotted tail goat. Strategy: stay. If you switch it is 2 in 3 that you will get the goat with the spotted tail. So it is 2 in 3 if you don't switch you get the car.
Hierarchical: If you pick the goat with spotted tail, Monty will show you the goat without the spotted tail. If you pick the goat without the spotted tail Monty will show you the car. If you pick the car Monty will show you the goat without the spotted tail. Strategy: 50-50.
etc.
Which is the more likely one he is playing? We can't tell. And obviously these are not the only strategies.
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Actually what would be fair is if it were worded like this:
You are an a game show and where you have a chance to pick a car or two goats. The hosts goal is to give you a goat and keep you from picking a car. You pick door 1 and he shows you door 3 has a goat and offers you a chance to switch to door 2. Should you?
Answer: it doesn't matter. Whichever door you pick he will put the goat behind it after you pick it.
Best Answer
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 $\frac14+\frac13=\frac7{12}$.