Monty Hall Variation: Choosing 2 Doors I'm learning about the Monty Hall problem, and found a question which has confused me:

Consider the following four-door Monty Hall problem:
  Step 1: you
  choose one door such as door 1.
  Step 2: Monty Hall opens one door
  with no car behind, such as door 2.
  Step 3: Monty Hall asks you to
  decide whether you like to “stick with your current door or choose
  another door” such as door 3 or “switch to the other two doors” such
  as doors 3 and 4.    
You will win the car if it is behind either of the two chosen doors.    
Now apply Bayes’ theorem to calculate the probabilities of winning the
  car by the following two choices: 1) stick and choose another door;
  2) switch to the two other doors.

Original question image:
http://i.imgur.com/xjjtmcF.png
I know to set my initial probabilities of the car being behind each door to:
{1/4, 1/4, 1/4, 1/4} 
How does my working of the problem change, now that I am choosing 2 doors, rather than one?
I have tried reducing the problem to it being between:
"Do we stick and choose, or switch and switch?"
This is quite confusing. Any pointers would be really appreciated!
 A: To be clear about the assumptions: let us suppose I picked door $A$, and Monty Hall revealed door $B$ because he knew there was no car behind door $B$. Then I have three options: I can open doors $C$ and $D$ ("switch"), or I can open doors $A$ and $C$, or I can open doors $A$ and $D$ (both these options being "stick").
With probability $1/4$, we originally picked the door with the car. Then we get a car if we stick, and we lose the car if we switch.
With probability $3/4$, we did not pick the door with the car. Then: if we switch, we get the car. If we stick, we have a $1/2$ chance of getting the car.
So, if we switch, the probability is $\frac{3}{4}$ of getting the car. If we stick, the probability is $\frac{1}{4} + \frac{3}{4} \times \frac{1}{2} = \frac{5}{8}$.
So it is better to switch. (Thanks, Mathmo123, for pointing out that $\frac{3}{4} > \frac{5}{8}$.)
A: For convenience, I'll call the door you first chose $A,$ the empty door shown as $B,$
and the other two doors as $C$ and $D$
P(win by not switching) $= \dfrac14$
Scenario $1$, you switch to $C$
P(win) = P(not behind $A$) * P(behind $C$ | not behind $A) = \dfrac34\cdot\dfrac12 = \dfrac38$
Scenario $2$, you switch to cover both $C$ and $D$
P(win) = P(not behind $A$) * P(behind $C\cup D$| not behind $A) = \dfrac34\cdot\dfrac22 = \dfrac34$ 

If scenario $1$ means you are in effect choosing $A$ and one from $C$ or $D$, say $C$,
P(win) = P(behind $A$) + P(not behind $A$) * P(behind $C$ | not behind $A) = \dfrac14 + \dfrac34\cdot\dfrac12 = \dfrac58$
A: After the revelation, the probabilities are
$$(1/4,0,3/8,3/8) $$
So switching to a single other door will increase the winning probability from $\frac14$ to $\frac38$. Switching to both other doors will increase it to $\frac 34$.
A: Before he opens the door, the chance your original choice is correct is $\frac{1}{4}$. After he opens the door, that remains true, so the chance that the other two doors are correct is $\frac{3}{8}$ for each. If you take the option of picking one to add to your existing choice, your chance of winning is $\frac{5}{8}$. If you switch to the other two your chance of winning is $\frac{6}{8}$.
