Monty Hall Problem - But with 4 doors and two opportunities to switch

[x^2]

I had a question regarding a variation of the infamous Monty Hall problem. Assume four doors instead of the standard three. There is still one car and three goats, and of course the objective is to select the door with the highest probability that it is hiding a car.

You select a door. The game host reveals a joke door. You are given the opportunity to switch your selection to one of the remaining two doors. The game host then reveals a second joke door. You are given another opportunity to switch doors. Should you do it?

I think yes - regardless of the decision of your first selection, by the time of the second selection two doors have been revealed and only your selection and a final door remain. If I understand the probability correctly the door you have selected has a 1/4 probability of hiding a car. The remaining door has a 3/4 probability of hiding the car. But does the intermediate selection somehow effect this problem in a manner I have yet to recognize?

 

[Mensanator]

Finding out is a doddle.

import random
hist = {'won':0, 'lost':0}
for i in range(10000):
  game = ['c','g','g','g']
  p = game.pop(random.choice(range(4)))
  game.remove('g')
  game.remove('g')
  # always switch
  p = random.choice(game)
  if p == 'c':
    hist['won'] += 1
  else:
    hist['lost'] += 1
print(hist)

## trial run
##  {'won': 7527, 'lost': 2473}