We have discussed the Monty Hall problem already a couple of times. One reason why people make this mistake is that they forget the role of the host in reducing the uncertainty of the car’s location. In other words, when the host eliminates one wrong door, he doubles the chances for the participant.
100 doors
Imagine a modified game in which there are 100 doors. You pick one. There can not be two opinions here that the chances of guessing the right door (having a car behind it) is one in one hundred. The host then opens all but one opening and shows you 98 goats. Will you switch this time? Or do you still think your original choice has a 50% probability of finding the car?
