One of the loveliest probability puzzles of all times. Yes I have probably blogged about it before but I shall do so again because it's just so cool. Yes it does have relevance to the Creationism probability argument post below this, so you might want to read that first.
It's a simple game. In fact, it's the American gameshow "Let's make a deal". There are three doors. Randomly arranged behind the doors are 2 goats and a car, each one assigned to one door. If you choose the door with a car you win the car, and if you choose one with a goat you lose and get nothing.
So first, you choose a door without opening it. Here is the twist : after you choose a door, the gameshow host opens one of the remaining two doors and reveals a goat, and asks you if you want to reconsider your decision.
What is the optimal solution to this problem? Either
A. you should always stick with your current decision
B. you should always switch to the last remaining door
C. it doesn't matter what you do at this point
Well you can post what solution you feel is best in the comment box, and I shall give a solution at a later date, and perhaps a more detailed history of this problem. This problem was catapulted into public consciousness in 1990 as a result of Marylin's vos Savant's column in Parade Magazine, where she posted a solution to this problem. (she had the highest IQ in the world according to the Guiness Book of Records). There were hundreds of letters of condemnation against her by mathematics PhD holders from around the world in reaction to her solution. And apparently now this problem is so instrumental in probability teaching that it is entering even High School probability syllabus now -> so I am not sure how many of you have already come across it. I know I've told some of my friends about the Monty Hall problem before, so hush =p
8 comments:
Always change your decision.
Hez
I've seen the problem before, but the solution was so complicated I never remembered it.
I'll go with Vimes' answer.
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CatR.
Stick with it. You were destined to choose that door even if the goat was shown to you or not.
Sanzo has been watching too much lovey dovey anime.
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CatR.
From my understanding let's see. You have 2/3 chances of picking a goat door and 1/3 of picking the right door. So most of the time you'll pick the one with the goat. The host shows gets rid of a goat door. Meaning most of the time you will be on the goat door and the switch will get you the car that is 2/3 times that will happen. So it would be better to switch due to this probability. The twist is pretty interesting...
mygo you have it exactly right.=p and you've just saved me the trouble of explaining.
wolf : destiny isn't everything =p
Shouldn't the probability be 1/2?
2:3 goat:door
1:3 car:door
a door and a goat both leave the equations
1:2 goat:door
1:2 car:door
Of course the probability of me landing on goat:door in the first place was more likely. But now that one goat:door is gone, I'm faced with a new equation, not an old one.
Unless we do it fractions-style which would make things wonky, but I fail to see why I would need to alter the 'door' numerical quantity to 6 in the first place. Because, uh, one door is gone.
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CatR.
ARGH bugger NOW I get it.
Most contestants would think like me - but the fact is that before you make your decision you are working on the 2:3/1:3 probability. Hence even if what lies ahead of you - the obvious - is a 50/50, your decision was made on your 2:3/1:3 knowledge. That decision and the basis of that decision remains, even if the numbers of the items change before you.
Aaaa now I get it!
YAAAAY
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CatR.
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