The Monty Hall Decision

It turns out that if you switch doors you will win the car 2/3 of the time! Why?

What is tricky is that the information which door is opened by the host (door B or door C?) does not reveal any information at all about whether or not the car is behind door A (that is still 1/3). But it does give valuable information about the remaining unopened door since opening the door with the goat has reduced the possibility that the unopened door has a goat.

To understand this problem more intuitively, think of the related exaggerated problem where there are 100 doors, 99 of which have goats while 1 has a car. You pick a door (call it "Door A"). The probability that is has the car is 1/100. The game show host opens 98 of the doors, all of which have goats. Should you stick with Door A or switch to the other door (call it "Door B")

Opening the 98 doors gives you no additional information on if your Door A has a car since you knew that 98 of the other doors had goats. However, in order for Door B to survive the drastic pruning (comparing it with each of the other 98 doors), its chance of having the car increased to 1/2. So it is smart to switch.

Door A wins when it has the car, while, because of the pruning, B wins when any of the other doors have the car.

The following table explicitly shows all of the 9 possibilities with 3 doors.

Door
You Pick
Door
Has Car
You
Don't Switch
You
Switch
11 Win Lose
12 Lose Win
13 Lose Win
21 Lose Win
22 Win Lose
23 Lose Win
31 Lose Win
32 Lose Win
33 Win Lose

Question: With the 100 door version, what is the analysis if he only opens 50 doors (all having goats)?