The Monty Hall problem

[nizar]

This problem is loosely based on a game show in the US back in the 1970s.
Its a play on probabilities, i spent a fair bit of time trying to figure this out.
Thought it would be interesting what sort of responses we get here.

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. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you "Do you want to switch to Door Number 2?" Is it to your advantage to change your choice? (Krauss and Wang 2003:10)

 

[Stan 101]

Yes.

The first choice was made with a 1 in 3 chance of being correct.
The second choice gives you a 1 in 2 chance of being correct.

Personally, I'll just listen for the goat. They can't be kept quiet for long

cheers,

 

[MichaelD]

I'll have a go at this without looking up the reference.

Success = car

Pre-test probability of incorrect choice = 66% (2 goats, 1 car)

Following the opening of one door with a goat, that leaves 2 doors. New probabilities of incorrect choice = 50% (1 goat, 1 car).


I conclude that you would be better off changing your choice as the probability of an incorrect choice was higher before the elimination of one incorrect door.

 

[bigt]

..though as the goat was behind the Monty door, the odds are the same i.e you still have a 1 in 2 chance for the car if you stick to your original choice..aren't they

Too early in the morning for this

Wont it also depend on probability being measured as at the time the choice was to be made?

 

[nizar]

50/50 is incorrect.

It WOULD however, be correct, if the host chose a door at random, and it just happened to be a goat.

BUT this is not what he does.
He knows where the goats are and must reveal a goat.
He never reveals the car, and he never reveals your original choice (ie. he always asks you to switch.

 

[nizar]

I conclude that you would be better off changing your choice as the probability of an incorrect choice was higher before the elimination of one incorrect door.

Sort of on the right track with this statement.

 

[bigt]

50/50 is incorrect.

It WOULD however, be correct, if the host chose a door at random, and it just happened to be a goat.

BUT this is not what he does.
He knows where the goats are and must reveal a goat.
He never reveals the car, and he never reveals your original choice (ie. he always asks you to switch.

Ahh yes, didnt digest that part nizar..need coffee