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Riddle Me This Again


Tenacious_Peaches

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This is one of my favourites. I love the logic behind the solution. Perhaps you know it.

You are a contestant on the game show Let's Make a Deal, and emcee Monty Hall has a game for you to play. On stage are three huge doors, and he informs you that hidden behind one of the three doors is a brand new sports car, behind another is a donkey, and behind another is 400 pounds of bananas. (This may seem obvious, but your goal is to get the car, not the donkey or the bananas.) He asks you to choose one door, and you will win the prize behind that door. You choose, but before he reveals what you have won, Monty reveals what's behind one of the doors you did not pick, and it's not the car. Before revealing what lies behind the remaining two doors, he makes you one final offer: if you wish, you may switch from your choice to the other remaining door. The question is: Should you switch, should you not switch, or does it matter?

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Okay, say you picked the door stage right. Monty opens the center door to reveal the donkey. But the donkey is turned to stage right and chewing on bananas. You should switch your choice to the stage left door, because you now know that the donkey is in the center and the bananas are behind your original choice, the stage right door.

Sure, that's it. :)

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Das, you must be correct in some alternate dimension. Now since I love this riddle so much, I'm putting up the answer now:

You should switch. If contestants switch every time, one of

two things happens:

(a) With probability 2/3, the contestant picks a wrong door.

Monty Hall shows the prize behind the other wrong door, and the

contestant switches to the door hiding the car.

(B) With probability 1/3, the contestant picks the right

door, Monty Hall reveals either the donkey or the bananas, and

the contestant switches to the other "zonk" prize.

So, with probability 2/3, the "switch" strategy will elicit the car.

This is one of the most famous (and widely-debated) probability problems of

the past ten years. For more information, and a simulation to show that

the above solution is truly correct, check out the following website:

http://www.cut-the-knot.org/hall.shtml

The site gives a further explanation... I was utterly fascinated by the logic when I first learnt of this one :)

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The gist of the whole thing is that chances are in favour of the door you DIDN'T select having the car. So once the host has opened a door with the donkey/bananas... probability suggests that the remaining door is the one with the car.

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