Monty Hall Problem-Probability Paradox

The big difference between the two scenarios is that Monty deliberately avoids choosing the door the contestant chose, but he isn't even aware of what the man in the audience chose so that doesn't factor into his decision.

Let's do the math explicitly. Suppose (as you say) the contestant chooses $1$, the audience member chooses $2$, and Monty opens $3$. The audience member knows that Monty was choosing between doors $2$ and $3$, and could not open door $1$. Thus the audience member knows, as does the contestant, that Monty was forced to avoid door $1$ but elected to avoid $2$ as well. That deliberate selection is evidence that the prize is behind door $2$, so the audience member reasons in exactly the same way the contestant does, and concludes that he should stick with his original mental vote.

As another, similar, variant of the problem, suppose that Monty himself has no idea where the prize is, so just chooses between the unselected doors uniformly at random (possibly exposing the prize, in which case the game ends). In that case, convince yourself that there is no value to switching.

lulu's answer is perfectly correct, but I wanted to show an alternative (frequentist) way of arriving at the same result. For very small problems, like Monty Hall, it is often possible to just list out the entire sample space:

1. Car, Goat, Goat.
2. Goat, Car, Goat.
3. Goat, Goat, Car.

Each of these is equally probable, so they all have 1/3 probability. So far, so simple.

Suppose as before that the contestant chooses door #1, the audience member chooses door #2, and Monty opens door #3. Then, translating our cases into outcomes:

1. Monty opens either door #2 or door #3 (1/3 probability). Assuming he picks one at random:
   1. Monty opens door #2 (1/6 probability).
   2. Monty opens door #3 (1/6 probability).
2. Monty opens door #3 (1/3 probability).
3. Monty opens door #2 (1/3 probability).

I have struck out the cases where Monty opens door #2, since we assumed that those cases don't happen. That eliminates cases (3) and (1.1), leaving us with cases (1.2) and (2). Based on the probabilities shown above, case (2) is twice as probable as case (1.2), meaning that the car is twice as likely to be behind door #2 as door #1. These probabilities are "objective" (frequentist), so both the contestant and the audience member arrive at the same numbers by the same reasoning process.

However, we added an assumption. We assumed that Monty chooses a door to open at random in case (1). If (for example) Monty always picks the door on the right (with the greater number), then it might appear that this logic no longer goes through. That is not the case. If Monty always picks the door on the right, then the doors are no longer symmetrical, and we now have to account for the cases where the contestant and audience member select different doors. When this is done, we recover the same result.