Four-door Monty Hall problem, but two doors are opened, one by one. Suppose a Monty Hall problem with 4 doors, and that after picking a door, another door--with one of the donkeys--is opened. The player now can choose to switch (between two other doors) or stay. After making the decision, another door--with one of the donkeys--is opened. The player once again can choose to switch or stay. (The player may switch back to his original door if he/she switched the first time.)
I understand that staying then switching in addition to switching then switching to not the original door yields the highest probability ($\frac{3}{4}$) of winning, and that staying then staying in addition to switching then switching back to the original door yields a $\frac{1}{4}$ chance of winning. 
Is it possible to calculate the individual probabilities for each individual case? 
 A: Case 1 (Switch, switch): You have a $1/4$ probability of being in front of the right door at the outset. In that case, switching the first time puts you in front of the wrong door. In the case you're in front of the wrong door at first, switching gives you a $1/2$ probability of landing at the right one. So the overall probability of landing at the right door after you switch the first time is $3/4*1/2=3/8.$ Now of course we know from the normal monty hall problem that if we're in front of the right door and we switch we lose and if we're in front of the wrong door and we switch we win, so the probablity of winning is $5/8.$
Case 2 (Switch, Stay): Same analysis as above, but staying the last step gives you a $3/8$ probability.
Case 3 (Stay, Switch): If you opt to stay the first time, you have a 1/4 probability of being at the right door. Now if you switch the second time you have a $3/4$ probability of winning since you win provided you were in front of the wrong door.
Case 4 (Stay, Stay): Worst of all. $1/4$ probability.
