November 9, 2009

Monty Hall Paradox

I have spent the better part of the day trying to figure out and explain the Monty Hall Paradox:

Suppose you're on a game show and you're given the choice of three doors. Behind one door is a car; behind the others, goats. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you "Do you want to switch to Door Number 2?" Is it to your advantage to change your choice?

The answer is "yes", it is to your advantage to change your choice. And here is why (best I can do to try to make sense out of this):

The goal is to win the car.

My door has a 1 in 3 chance (33%) of having a car.

The "other two doors" (think of them like one door) have a 2 in 3 chance (67%) of having a car.

Monty tells me that one of the "other two doors" (the 67% f having a car) doesn't have a car.

The 2 in 3 chance (67%) of the "other two doors" now only applies to one of the "other two doors". Monty has essentially transferred the 67% split between "other two doors" into one door.

So, here is what we have as it relates to the math:

My door (not been opened yet): 33%
"Other two doors" #1: 67% (not been opened yet)
"Other two doors" #1: 0% (already been opened so I know it doesn't have the car)

So, probability wise, it is much better to have a 67% chance of winning a car than have a 33% chance of winning a car.

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