Monty Hall variationThe question is a variation of the classic Monty Hall problem. Should one switch their first choice of door one, if there are four doors, with two cars and two goats, the second door containing a goat. How would one solve this using fractions?

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assumintdz · Accepted Answer

So the setup: You know there is some arrangement of two cars and two goats behind four doors. You pick a door, call it   A and then another door opens to reveal a goat.   You wish to know if you should stay with   A or switch to one of the other two doors, call it   Bedit: To be clearer, let   A be the event of a car being behind the first door you choose, and   B be the event of a car behind behind whichever other door you choose should you switch after one of the goats is revealed.There is equal chance that a car is behind the door you first picked.  P  (  A  )  =            1      2      If you were right first time, then there&#039;s one goat and one car behind the other two doors.      P    (  B  ∣  A  )  =            1      2      If you were wrong, then both cars hide behind the other doors.      P    (  B  ∣      A    ∁    )  =  1      P    (  B  )     =     P  (  A  )      P    (  B  ∣  A  )  +      P    (      A    ∁    )      P    (  B  ∣      A    ∁    )    =            2      4                  1      2        +            2      4                  1      1          =            3      4

Hailie Blevins · Accepted Answer

Suppose we switch. The only way to get a goat is to choose a car originally, that is       1    2   chance, and then after seeing one of the goats, choose the door with the other goat, that is       1    2   chance. That would be       1    4   chance of getting a goat. That leaves       3    4   chance of getting a car.Suppose we don&#039;t switch. Then there is a       1    2   chance of getting a car.

Monty Hall variation The question is a variation of the classic

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