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Re: The Monty Hall Problem
by pramod srini - Tuesday, 22 June 2010, 04:36 PM
  This is a tricky problem. but a systematic analysis will leave no doubts in our minds.
Let Monty hall name the goats as goat1(G1) and goat 2(G2) and the car as "car" .well Monty can arrange the the three things  behind doors A,B,C in the following ways.
DoorA  DoorB  DoorC
  G1      G2    car
  G1      car    G2
  car      G1    G2
  car      G2    G1
  G2      G1    car
  G2      car    G1

Now when given the first choice,let us select door 'B'. According to the problem , Monty can only open the door which does not have a car(the logic is obvious). Hence if we select the door having goat G1 , he can only open a door containing goat'G2' but Not the door having the car.
Monty now offers us the choice of either switching from our choice of door'B' or staying with doorB.  Based on above logic, the different possible combinations are as follows.

let: original choice=C,  monty's choice of opening the door=M
    dont switch=DS      , switch=S

  O choice        Mont choice        DontS        Switch
  G2              only door A            G2        car
  car              door A or C            car        G1 or G2
  G1              only door C            G1        car
  G2              only door C            G2        car
  G1              only door A            G1        car
  car              door A or C            car        G1 or G2 

As we can see, the probability of finding a CAR behind those doors if we accept Monty's choice of switching is 4/6=0.6666.
The result is same if we choose door A or door C initially.We can  surely try and verify!
So,we can conclude that irrespective of what our sixth sense says, it is always more favorable to us to SWITCH if we want to find a Car behind those doors :)