Problem no. 186 at mathproblem.info concerns a realistic dilemma in game show probability that inveterate geeks, such as myself, may find interesting. Here is the way host Michael Shackleford puts it:

On a game show there are three doors. Behind one door is a new car and behind the other two are goats. Every time the game is played the contestant picks a door. Then the host will open one of the other two doors and always reveals a goat. Then the host gives the player the option to switch to the other unopened door. Should the player switch?

As in so many of these problems, I suspect the wrong choice is psychologically alluring. The player's chance of winning is doubled by switching, though adopting this strategy means opening oneself to the emotional travail of abandoning the right answer. Think of it like this. Let the car be behind door one. A third of the time, the player will have chosen door one, and in those cases she trades away from the right answer and loses. If, however, she has chosen door two or three--a two-thirds probability--the detail about the host always revealing a goat means she can only switch to the winning door.

In other words: player chooses door two (or three), host reveals that a goat is behind door three (or two). If at this point the player switches to door one, she'll lose all the times she otherwise would have won (one-third) but will win all the times she otherwise would have lost (two-thirds).

The same logic applies to a multitude of "shell game" problems.

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