Will you change to the other door??
There are three doors, A, B & C. I've place a billion US dollar behind one of them.
Now, I let you choose one of the door (e.g. door A). After making your choice. I open another (e.g. door B) and show you that there is nothing behind.
Now, I ask you "Will you change to the other door (e.g. door C)?"
You will
a) stay unchanged; or
b) change to door C; or
c) nevermind, as both are the same.
????
Dennis Chow (from Hong Kong)
Monday, January 31, 2005
Of course, if you choose the right door, you can take and keep the billion dollars.......
^o^
Dennis Chow (from Hong Kong)
Monday, January 31, 2005
You choose the door you haven't opened, choice B.
You have a 1in3 chance on your initial guess. Just because someone opens another door, doesn't change your odds on that guess. When you get to change your guess though, you instantly get upgraded to a 1in2 chance.
Andy Visser
Monday, January 31, 2005
Someone's been reading the Curious Incident, then?
Iago
Monday, January 31, 2005
Interesting problem... this is the basis of a TV game show in India. My answer is it doesn't matter, at least mathematically.
If we have N doors, and I choose one of them. Then you open one of the other doors and ask me to choose one of the remaining N2 if I like. This keeps going on until only 2 doors are left.
This means irrespective of the value of N, the probability of my getting the $$ is the same at the end (= 0.5), even though the probability at the start of the problem depended (inversely) on the value of N. This is because you eliminate the wrong choices at each step, ending with 2 doors, one of which is 'correct' and one which isn't. It's a half chance either way.
thelearner
Tuesday, February 1, 2005
If you stay with your current door you maintain a 1/3 probablity. If you switch doors you upgrade you probability to 2/3.
I am not sure if I can prove this over email but her is a similar problem.
Imagine 1,000,000 doors with exactly one prize.
you pick a door x. Door x has a 1/1,000,000 probability of being correct.
The game keeper now opens 999,998 doors to show that they do not contain the prize.
A few things become clear.
1. The probability that your current door is correct is certainly not 1/2
2. the probability that the remaining door is correct is very high (999,998/1,000,000)
The crux of the problem is that your original door (x) is not subject to being open. Opening other doors gives you *no* information on your current door (x). Opening other doors gives you lots of information on the remaining doors.
maxell
Tuesday, February 1, 2005
This is already on the site, under "Monty Hall".
http://www.techinterview.org/Puzzles/fog0000000045.html
David Clayworth
Tuesday, February 1, 2005
Actually the odds are 50/50.
Either you open the right door or you don't.
:)
Mike Drips
Monday, February 7, 2005
Same with the lottery. That's 50:50. Either you pick the right numbers or your don't. :)
David Clayworth
Tuesday, February 8, 2005
1/3 if you stay 2/3 if you change.
Anonymous
Wednesday, February 9, 2005
"You can slightly improve your chances to win a lottery if you buy a ticket" :)
Dmitri Papichev
Saturday, February 12, 2005
maxell explains it very well. An other explanation goes like this: you might as well decide about changing or not *before* being shown "the empty door".
Then if you chose:
a) stay unchanged.
You know you are not changing, so its not worth being shown "an empty door", just go and see if you won a million – clearly you have one chance out of three.
b) change to door C
Well, door B and C will end up being opened; it doesn't matter who opened which one. By deciding before hand to switch, in effect you decide to open two doors => you have two chances in three.
AlphaBeta
Wednesday, February 16, 2005
the million doors analogy explains it very well.
there are only two doors out of the million that the host cannot open
(1) The on you chose
(2) The on with the money.
Now how confident are you that it doesn't matter?
mark jones
Sunday, February 27, 2005
on=one.
I *am* litrate <sigh>
mark jones
Sunday, February 27, 2005
To the folks talking about 2/3 or 999,998/1,000,000. Sorry folks, but the denominator is 2. That's right, TWO. You are making a NEW choice, which is 1 out of 2. The helpful moderator opening all but one false door after your initial choice is irrelevant to the current 1/2 odds, because he was constrained to not open your door, nor the prize door should you have picked false.
argh
Wednesday, March 9, 2005
for good info on why 1/2 is wrong and why 2/3 relies on the unstated assumption that monty MUST open a goatdoor:
http://www25.brinkster.com/ranmath/marlright/montynyt.htm
WanFactory
Wednesday, March 9, 2005
WanFactory, the problem as stated does not include the goat door (see top of thread).
argh
Thursday, March 10, 2005
goat door = empty door
the question is: why was the empty door is opened?
is the person running the game just opening a door at random and it happens to be empty? in this case 1/2 and switch or stay does not matter.
is the person running the game opening an empty door because they have to (promised before hand to) open an empty door? in this case switching is 2/3 likely to succeed.
is the person running the game opening an empty door and offering you a switch because you picked the right door to begin with, in which case switching is 0% likely to succeed.
That is, do you have some sort of guarantee before even picking a door that the person running the game will always open an empty door? the problem cannot be answered without first making an assumption about the answer to the previous question.
WanFactory
Friday, March 11, 2005
Enough talk, let's play the game for $1 repeatedly. We'll take turns guessing while the one who does not guess has to show an empty spot.
I'll always switch, you can switch or not randomly. Or heck if you think it wont matter, just stick with your initial choice. Dont wanna argue no more, will just take yer money.
WanFactory
Friday, March 11, 2005
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