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In the Monty Halls problem, the players at the game show is given the choices of three
doors, where behind two doors there are donkeys and behind one door there is a car. The
strategy to win the game is that whenever the player is given a chance to switch, he should
take it. The strategy can be explained by the help of conditional probability.
Assuming that you have picked door No.1, there are 3 cases:
Case
Door 1
Door 2
Door 3
Action of the host
1
Car
Donkey
Donkey
Door 3 or door 2
2
Donkey
Car
Donkey
Door 3
3
Donkey
Donkey
Car
Door 2
Now if we assume to follow case 1 – the probability of opening door 3 is 1/3
P (case 1 opening door 3) = P (case1). P (case 1/ opening door 3) = 1/3 *1.2 = 1/6
Adding the probabilities of the both ways that the doors can be opened becomes
1/3 +1/6= 1/2
Now P (case 1 / open door 3) becomes = (1/6)/(1/2) = 1/3
If the player’s first choice was goat, then he can win 1 – 1/3 times by switching
That is 2/3 times.

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In the Monty Halls problem, the players at the game show is given the choices of three doors, where behind two doors there are donkeys and behind one door there is a car. The strategy to win the game is that whenever the player is given a chance to switch, he should take it. The strategy can be expl ...
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