lunes, 17 de octubre de 2011

The Game Show Problem

Reading the Book "The man who loved only numbers: The Story of Paul Erdos and the Search for Mathematical Truth" I met "The Game Show Problem". This problem is very simple and doesn't need any background, only simple logic is needed.

I just copy the problem directly from

"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. You pick a door, say #1, and the host, who knows what's behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors?"

It is clear that if we don't change our choice, the probability to win is 1/3. But, what's the point here? We are not using an additional info we have: that the door #3 does not contain the car!! What if I change my choice when I have the opportunity and take door #2? Then, at the beginning we could win either if the car is in door #2 or door#3, then our chance to win is 2/3!!

The following tables clear all doubts.

Option 1 (mantaining the initial choice):

                   DOOR 1      DOOR 2         DOOR 3      RESULT
GAME 1     AUTO          GOAT             GOAT         Stay and you win.
GAME 2     GOAT          AUTO             GOAT         Stay and you lose.
GAME 3     GOAT          GOAT             AUTO         Stay and you lose.

   So the probability is 1/3

Option 2 (switching):

                    DOOR 1      DOOR 2         DOOR 3      RESULT
GAME 1     AUTO          GOAT              GOAT         Switch and you lose.
GAME 2     GOAT          AUTO              GOAT         Switch and you win.
GAME 3     GOAT          GOAT              AUTO         Switch and you win.

    So 2/3 !!!

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