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Post by abacus9900 on Sept 21, 2010 15:48:37 GMT 1
The Monty Hall problem is a problem about probability which I find perplexing. Imagine being on a game show and are given the chance to win a car by being presented with three doors and choosing one of them. The car is behind one door and goats are behind the other two. Let us say you pick door number one. Then, the host picks a door which has a goat behind it. The host knows what is behind the doors, BTW. The host then gives you the opportunity to change your choice of doors from number one to number two. The crucial question is: is it advantageous to switch doors or not?  en.wikipedia.org/wiki/Monty_Hall_problem |
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Post by pumblechook on Sept 21, 2010 23:52:12 GMT 1
Door one has a 1/3 chance. This does not change.
The chance that it is 2 or 3 is therefore 2/3 (must add up to one).
If 3 is revealed to be a goat then 2 takes all the 2/3 chance. Two doors left.
Other words... If one (1) has a 1/3 chance the other (2) must have a 2/3 chance. (3) Now known to have zero chance.
Might be easier to think of 100 doors and the host reveals 98 to be goats. The one you have picked always has a 1/100 chance. The 99 have a 99/100 chance. The host avoids the one with the car and is 'pointing' you towards the one he doesn't pick which then has a 99/100 chance. Only two doors left 1/100 + 99/100 = one.
Always switch.
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Post by StuartG on Sept 22, 2010 0:14:38 GMT 1
If the host is always going to open a door to reveal a goat, then effectively it becomes a 'two horse race' and then to be sure whether to change or not, the amount of times that the doors have revealed 'winners' in the past should be known, otherwise it's a 'two door' race odds even on either outcome.
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Post by pumblechook on Sept 22, 2010 0:20:15 GMT 1
You might have lots of cars and prefer a goat.
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Post by abacus9900 on Sept 22, 2010 8:40:06 GMT 1
pumblechook, you're a hero sir! Thank you very much.  |
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Post by abacus9900 on Sept 22, 2010 9:13:06 GMT 1
The only way I can get my head around it is to say whatever choice of door you make has 2/3 probability of being a goat, since there are two goats and one car.
When the host reveals a goat behind one of the doors then the door you chose, with a 2/3 probability of being a goat, means the other unopened door now has a 1/3 probability of being a goat (there is no reason to suppose that your probability has changed from the starting position). Now, looked at another way, my door, which has a 2/3 chance of being a goat, has only 1/3 chance of being a car. The other door, which only has a 1/3 probability of being a goat has a 2/3 probability of being a car because there are only two doors to consider now and all the odds must add up to one. The odds of either being a goat or a car are simply mirror images of one another.
It all hinges on the initial position where the chances of you picking a door with a goat behind it is 2/3 (two goats and one car) and after the host has opened one of the doors revealing a goat, remains 2/3, from a probabilistic point of view.
So, from a purely logical perspective, it makes sense to switch doors, although of course, there is no guarantee that you will pick the correct door and win the car in any particular instance.
The reason many people have trouble with understanding this puzzle is because it is so anti-intuitive. Many people assume that the odds are even that the car will be behind either remaining door and so see no value in switching doors.
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