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#1
okay, at the beginning of the movie, when kevin spacy is teaching the class and does the example about the game show with the 3 doors, did that make any sense to anyone?

Like if there are 3 doors and 1 has a prize, and you choose door number 1. The host reveals that door #3 did not contain the prize, and asks if you want to change your guess to door #2. Then it goes on to explain how you SHOULD change your choice to door #2 because some how the 1/3 chance that the prize was in door3 was transfered to door 2, so door 1 has only 1/3 chance of being the correct door, and door2 has 2/3. Did that scene make any sense to you guys?

Its pretty pointless, but it just doesn't make any sense to me.
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#4
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exactly
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#7
Mathematically... what he's saying is true. I don't quite understand it, but there is an extensive explanation offered by the mathematician who first suggested it (and backed it up with a diagram and some words on the subject).
#8
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#9
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#10
spacy's wrong. it just means that door 1 and door 2 have 1/2 chance each
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#11
That doesn't make sense.

Eg. (x) ( ) ( )

It's not number 3, but there's still an equal chance that its 1 or 2.
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#12
Quote by PitLurker
Mathematically... what he's saying is true. I don't quite understand it, but there is an extensive explanation offered by the mathematician who first suggested it (and backed it up with a diagram and some words on the subject).

iv never seen the movie, but that can't be right
i think all they did was apply the wrong concepts to a problem so you would get a different answer then the correct
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#13
i havent seen the movie, but with what you described i would think that the chances are 1/2 for door #1 or #2. it cant be 2/3 for door #2 because there is an equal chance for either door to have the prize.
#14
Quote by spiderjerusalem
spacy's wrong. it just means that door 1 and door 2 have 1/2 chance each


na, in the movie, somehow it turned into 33% 67 % not 50 50

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i havent seen the movie, but with what you described i would think that the chances are 1/2 for door #1 or #2. it cant be 2/3 for door #2 because there is an equal chance for either door to have the prize.


yea, but somehow through some wierd math is said otherwise, which is why i look like this
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#15
How can it be a 3rd when there's only two outcomes?
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#16
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na, in the movie, somehow it turned into 33% 67 % not 50 50

the movie's obviously wrong
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#18
Havent seen the movie, but seen the problem before. It was on the show Numb3rs, (and talked yto a few friends about it who are math majors). Spacy is right (or at least his character is). Dont ask me to explain it, but googleing might reveal an explanation
#20
Quote by sacamano79
yea, kevin spacy is a lying sack of ****

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i don't know who kevin spacy is, so i really can't tell if your sarcastic or not
so im gonna say, yea, he is
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#21
dude it obviously makes sense,

the probability of door three containing the preize was 33.33333...%, when it is revealed that it does not contain the prize the percentage of contaiment is transferred to door two beca-EPIC MALFUNCTION


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#23
Quote by aaron6890
yes what he is saying is that by eliminating one of the 3 doors it shifts from a 1 in 3 chance to a 1 in 2 chance. but i think in the movie they used %s instead of ratios.


no, thats not what it was saying, it turned from an equal 1/3 to 33 67 not 50 50
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#24
I just got home from watching that.

Each door has a 33.3% chance of being the right one.

(1) (2) (3)


If you KNOW one isn't right, you now have and increased chance of getting the right door. 66.7% to be exact. If your first choice had 33.3% chance also the second door has an increased chance. The first door could have yield the car though, just the middle had a greater chance. They also did this exaple in and episode of Numb3rs.
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#25
Quote by sacamano79
no, thats not what it was saying, it turned from an equal 1/3 to 33 67 not 50 50

but thats what happens, by taking out one choice of the three, it makes it a 50-50 chance that you'll have the right one. instead of the original 33-33-33 chance.


EDITh i get it, the guy above explained it perfectly.
#26
google: Monty hall problem.

think about this way: if you switch the door and win, then that means your original guess was wrong. so the only way to win by switching is if your original guess was wrong, and your original guess has a 2/3 = 66.66667% chance of being wrong.
#29
i just read the three lnks of proof that other guy posted and it makes somewhat sense, but not total.
because if you look at it like this: if you had chosen the 2nd door in the first place, then the first door would have a 2/3 chance
this can't be correct, because by doing the same thing in two possible scenerios, you get do different answers.
probability does not change with what one chooses, but by what is, and the probability changes with the two doors when you choose a different door, it must be wrong.
therefore, each door has a 50% chance
keep in mind, none of what i said is founded by proof, just based on my own logical thinking, so it could be completely wrong
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#30
it's because there are three different outcomes when one door is shown as being empty. in 2 out of 3 of these outcomes, the player wins by switching whereas theres only a 1 in 3 chance of winning by not switching.
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#31
just think of it this way, there are 3 possible situations:

1. you chose goat #1, the host opens the other goat, SWITCHING WINS
2. you chose goat #2, the host opens the other goat, SWITCHING WINS
3. you chose the car, the host opens one of the goats, SWITCHING LOSES

so you have a 2/3 chance to win if you switch. thats how i made sense of it anyway
#32
Hehe thanks to the guy who posted the Wiki for Monty Hall problem, I had forgotten the name/mathematician that proved it.
#33
My maths teacher explained it to my class, but it's pretty hard to explain over the internet.

It's pretty much saying you have three doors. Lets just say you pick 3 AND 2, which you pretty much have done, you have a 2/3 chance of getting it right.

However, if you look at it another way you now have a 1/2 chance of getting it right.

It helps if you scale up the numbers. Lets just say there are 30 doors. You pick 1, then 2..then 29. Now you know its either in 29 or 30. The chances of it being in 29 are either 29/30 or 1/2 depending on what way you look at it.

Does that help at all?
#34
Look at it this way.

3 doors: 1 has a brand new car, the other have goats.

You pick door 1. There is a 1/3 chance that door 1 has a car. There is a 2/3 chance that 2 OR 3 has a car.

it is revealed that 3 has a goat. There is still a 2/3 chance that one of them has a car, and a 1/3 chance that door one has a car because probability cannot retroactively change. It is also clear that among those two, door 2 is the only one that could have a car.

The 2/3 chance that it doors 2 OR 3 have the car is not replaced, but rather complimented by the 100% probability that if they do, the car is behind 2. But the original probability cannot retroactively change. it is still a 1/3 chance that the car is behind door 1, and a 2/3 chance that it is not.

IF the prizes were to be rearranged randomly after one door was removed, it would be a case of 50-50. Because they are not, the case of 1/3 - 2/3 remains.
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#35
Quote by iforgot539
Look at it this way.

3 doors: 1 has a brand new car, the other have goats.

You pick door 1. There is a 1/3 chance that door 1 has a car. There is a 2/3 chance that 2 OR 3 has a car.

it is revealed that 3 has a goat. There is still a 2/3 chance that one of them has a car, and a 1/3 chance that door one has a car because probability cannot retroactively change. It is also clear that among those two, door 2 is the only one that could have a car.

The 2/3 chance that it doors 2 OR 3 have the car is not replaced, but rather complimented by the 100% probability that if they do, the car is behind 2. But the original probability cannot retroactively change. it is still a 1/3 chance that the car is behind door 1, and a 2/3 chance that it is not.

IF the prizes were to be rearranged randomly after one door was removed, it would be a case of 50-50. Because they are not, the case of 1/3 - 2/3 remains.

so the probability of door 1 can't change, but door 2 can? that doesn't make much sense to me
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#36
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so the probability of door 1 can't change, but door 2 can? that doesn't make much sense to me


The probability of either doesn't change, just the probability of the contestant being right or wrong if you get what I mean. It does make sense, it's just complicated.
#38
Think I've got it now

The key is to assume you were wrong first time

Your first choice you've got a 2/3 of getting it WRONG.

Assuming the host knows which one the prize is in the host will eliminate the other wrong door.

So if you chose wrong in the first place (odds are more likely) the other WRONG door will be eliminated.

This leaves the odds of winning more likely behind the other door as one wrong has been taken out and you think your first choice was wrong.

Again, assume you were wrong, this means you know which doors are wrong after the host takes one out leaving one shiny door left.

(N) (n) (X)

N-Your choice (assumed wrong)
n-door host picked
X-Prize
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#39
The original probability of 2 does not change. But, a new set of circumstances is introduced, you have more information, and the new probability of 2 being correct is significantly better than the original. Because 1 was picked under the original circumstances, the original probability applies.

When you pick door one originally, there is a 1/3 possibility that you are right.
When door 3 is removed from the equation, that original probability cannot magically retroactively change to 50%. There is still a 1/3 chance that you were originally right. There is still a 2/3 chance that you were originally wrong.
The fact that there are only two doors now is irrelevant. When you picked there were three and whether or not you were right they would have revealed one of the goats. The original probability cannot change because the information it was made with doesn't change.

Originally, if you had switched, you would have divided your chances among the other doors. In the 2/3 chance door one is wrong, there is a 50-50 split among the other two; they each would end up as a 1/3 chance of originally being right.

After door 3 is opened, you know something you didn't know when you made the original decision: which of the other two not to pick.
There is still the 2/3 chance that your original guess (door 1) was wrong. That cannot change. And if that is the case (which it is more likely than not to be), instead of a 50 percent chance that the right door is actually 2 (as opposed to 3), there is a 100% chance.
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#40
I've seen it, and it makes sense, but not really.
he says it is a 66.6% chance now, because in theory you are choosing 2 doors (3, since it has been opened, and another, each door is 33.3%)
So yeah, you have increased your odds in picking either door, but over all it does next to nothing, because it's presenting you a brand new question, one that is 50/50 not 1/3

(sorry i didn't look at page 2, so if someone else said this, sorry)
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