Game:

Zero Time Dilemma (2016)

Duration: 40:08

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1

I mean, this is the Monty Hall Problem, after all. For those of you that never saw Let's Make a Deal, the standard set up is a player is offered three doors. Two of them contain, say, a goat, while one contains a car. Whatever is behind the door the player picks is what the player wins. Very simple so far.
After the player picks, the host will open up one of the two remaining doors to show the goat behind it, and then offer the player the chance to change their pick to the other unopened door.
After the second pick, all the doors are opened and the player gets their prize. You'd think that the player's odds just improved to a 1:2 choice, but mathematically, their odds have stayed exactly the same for the first door at 1:3. However, the odds for the other unopened door, mathematically, have increased to 2:3. Why?
Well, I'm not the greatest at explaining it, but essentially, by revealing all but the two doors, all the probabilities for the doors the player did not pick get combined. After all, originally, if the car had a 1:3 chance to be in the door the player picked, than it had a 2:3 chance to be in a door the player did not pick. That doesn't change because a door gets revealed...there is still a 2:3 chance the player did not pick a winning door, and now, that chance is reduced down to being in one door having a 2:3 chance to have the car behind it.

The same principle applies here, only with more doors. When we originally pick a door, that means there was a 9:10 chance it was behind one of the other doors. Revealing eight of the nine doors doesn't change those odds, it just condenses them down to one door, so the original door picked still has a 10% chance to have a gas mask in it, but the other closed door has a whopping 90% chance.

In short, whenever you get a Monty Hall Problem, you always change to the other door.