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#1
10-26-2004, 08:27 PM
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Ok, another math problem (harder)
At a TV show, there are "n" number of doors. Behind 1 of those doors there is a prize, and behind all other doors there is no prize.
The Show host asks for a guest to chose a door. So the guest choses a door but does not open it. The host (wich knows where the prize is) opens one of the doors that haven't been picked yet and opens it, revieling that there is no prize. The guest then choses another door, and the same procedure repeats itself, until only two doors are left, the one the guest choses and the one that is still remaining. So, still following the rule, he swiches to the remaining door and sees if he won or not. 1- What is the probability that the guest wins the prize? (present the formula the simplest way possible) 2- What is the probability when "n" (number of doors) is infinite? To write down the answer, use only the four basic operations and the number e (Euler = 2,71828...) 
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#2
10-26-2004, 08:31 PM
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Re: Ok, another math problem (harder)
ok we get it you know the birthday problem and the monty hall problem. no one is impressed
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#3
10-26-2004, 08:56 PM
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Re: Ok, another math problem (harder)
Isn't the answer (n-1)/n --> 1 as n --> infinity?
Where does e come in? John
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#4
10-26-2004, 09:48 PM
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Re: Ok, another math problem (harder)
[ QUOTE ]
ok we get it you know the birthday problem and the monty hall problem. no one is impressed [/ QUOTE ] I read this question on a website and figure you people know how to answer it. I DON'T know the answer. Are these problems supposed to be famous or something like that? Please, I just want to know how to calculate these problems. Please teach me
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#5
10-26-2004, 10:22 PM
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Re: Ok, another math problem (harder)
These are both classic problems and appear about once a month on these forums (one more reason we need stickies)
Very simply, your first guess is right 1/3 of the time. More precisely it's one of the other doors 2/3 of the time (this should be easy to follow). Because he shows you precisely which one of the other doors would be the right one that 2/3 of the time, your odds, if you switch, go up to right 2 in 3 (from 1 in 3). So if you switch, you're right 2 in 3. That's the english language proof. The mathematical proof should follow pretty easily if you cared to. And remember, the search function will probably give you an answer to many of the questions you're likely to ask.
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