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I know this probably gets posted a lot, but why does the 2/3

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I know this probably gets posted a lot, but why does the 2/3 probability get shifted onto the one door, instead of becoming a 50/50 probability?
>>
Because your original guess had a 1/3 chance of being correct.
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>>6658253

but the fact there's 3 doors initially is a non-factor, because one of the empty ones gets removed.

the goat is still either in your door or the other door whether you switch or not, which is 50/50
>>
Two out of three doors get chosen. Why wouldn't the probability be 2/3?

Your focus shouldn't be on the door you initially choose; it's a red herring.
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>>6658249
yeah it does.

google monty hall problem, there are hundred excellent examples of why it works the way it does already on the internet.
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>>6658254
Extrapolate to 100 doors.

You choose 1 out of 100 doors and the prize is behind 1 door. The host then opens 98 losing doors. What's the probability that the door you did not choose has the prize?

Certainly you can't believe you had a 50/50 shot at getting it right at the start!
>>
The situation is simple enough that you can see why it works by just making a complete probability tree.

Let the doors be A, B, and C, with the prize behind A.

For your first choice, probabilities are

A (1/3)
B (1/3)
C (1/3)

If you pick A, he can remove either B or C, with equal probability. But if you pick B he has to remove C and vice versa, so there is only thing that can happen in those cases.

So, extending the tree into (first choice),(removed choice):

AB (1/6)
AC (1/6)
BA (0)
BC (1/3)
CA (0)
CB (1/3)

If it is AB or AC, you win by staying, but that totals only 1/6 + 1/6 = 1/3

If it is BC or CB, you win by switching, and that totals 1/3 + 1/3 = 2/3

So, there is a 2/3 you win by switching.
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>>6658249

Because the probability doesn't get shifted onto the door.

Probability exists in the mind, not in reality. In reality, two of the doors have 0% chance of having the car while one of the doors has 100% chance of having the car. A coin doesn't have 50/50 chance of coming up either side, it has a 100% chance of coming up the way it will come up and a 0% chance of coming up the other way.

Only your state of knowledge,* as influenced by the evidence presented to you, can have probabilities different from 0 and 100%, because you don't have all the evidence.

So, to start with, in your picture: Door 1 has .33 chance of having a car, and so does door 2, and so does door 3.

Picking any door does not change this, although it is helpful at this point to conflate doors 2+3 to collectively .66 chance of having a car, and 1.0 chance of having at least one goat.

Now, a goat is revealed in group 2+3. What is this evidence for?

Basically nothing. You already knew there would be at least one goat in group 2+3.

Since you have no new evidence, you have nothing to update probabilites on, so your probabilities are still .33 for your door and .66 for the other two doors.

*And quantum outcomes, if the collapse interpretation is correct.
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>>6658257

that's a completely different situation

for n doors the general rule will be that its better to switch, but for 3 doors it makes no difference
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>>6658266
>that's a completely different situation

No it isn't.
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>>6658257
Good explanation!
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>>6658266

Real quick here: You should have written

>For n>2 doors the general rule is that it is better to switch.

>For n=3 doors, this rule also holds but I do not intuitively see it.
>>
>>6658254
just think that your initial choice had a 2/3 chance of being the goat so its more likely the first door you chose had a goat behind it.
>>
>>6658266
Not sure if troll... Anyways...

The general rule is the probability of winning when switching is (n-1)/n for n > 2.

Not sure why you see that the rule applies for n = 100, but not for n = 3. Of course if n = 2 the host either reveals you've chosen the right door (you win) or that you've chosen the wrong door (so you switch), giving a 100% win rate.
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>>6658278
its actually a very poor explanation, because it just reframes the exact situation with a different number, hence the same question is asked with no new information.

in other words, it is just as easy to assume that the two remaining doors have an equal chance of being right. considering they both started out with 1/100 chance, now that they are the only ones remaining, why arent they both 1/2?

it takes a more careful approach to explain why this is not true, ie examining the probabilities as they evolve.
>>
>>6658257
see
>>6658344


try to think about things instead of just repeating them ad nauseum from /sci/, there are too many idiots reposting bullshit here.

this is the correct way to approach the problem
>>6658264
and in this case, increasing the number of doors to 100 just complicates things drastically and unnecessarily.
>>
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>>6658344
It's a good explanation because the issue here is false intuition.

Perhaps it's really easier if you write it all down with a probability tree. You have your preliminary choice and the rest.
Yours has the probability n and the rest has the probability 1-n. This is true no matter how 1-n looks. The elimination of all doors but one is nothing more than the guy saying "Look, instead of opening that door, you can open AAAAALL of these, which have a 1-n chance of having the goat in it." The reason the chances don't change is because you already knew that at least all but one of them had to be losses. Opening them does not change this. It also does not eliminate the choice with n you made. What it does change is the way 1-n looks.


Apart from all of this:
It is demonstrably true. So whatever.
>>
>>6658347
what? the chances of you randomly picking correctly are much less than you randomly picking incorrectly on your initial choice. when he throws away 98 doors, you are left with the obvious difference of

a) i was insanely lucky in the beginning, and if i switch i will lose

OR

b) it was overwhelmingly likely that i was wrong and if i switch i WILL win.


its an obvious choice and the explanation is a good one for pointing it out.
>>
>>6658415
why does he open 98 doors insted of 1.
This example is useless
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>>6658427
Because under the false assumption that 2 doors mean that it is 50-50. It suggests that it is obviously not.
>>
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>>6658443
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>>6658446
He is just exaggerating to help you see the point.
>>
I just wrote a program to test it.
I get 1/3 chance of winning without a switch and 1/2 if the player does switch.

http://pastebin.com/mKMhfMJx
I don't think I have a bug, but I might have missed something. Feel free to take a look at the code.
>>
>>6658457
the exaggeration is wrong, there should be only on door opened.
>>
>>6658473
I think you don't know what exaggeration means.
>>
the odds dont change at all, its just a popular scam/troll. much like barnetts identity and tesla's tower
>>
host opens one goat door before letting you to choose.
Where is your god now?
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>>6658495
Then there wasn't a 2/3 chance that you stopped him to open a certain door.
>>
>>6658503
*forced
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>>6658466
>http://pastebin.com/mKMhfMJx
line 31. The problem is that the host does not randomly open a door. He KNOWS where the prize is, so he will always open a door that does not have the prize.
Great that you are taking the experimental approach!
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>>6658511
No it's only random if the player's pick is the correct door. In that case both of the other two doors have nothing/goats behind them, so in that case only a door is opened at random.

Though I did some debugging and found that after the switch I never end up with a 3rd door choice. So the problem is probably there.
>>
>>6658672
>>6658673
Why would you go trough all that work when you could just link the wikipedia page
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>>6658681
>I'm trying to refute that argument.

lol, but it's demonstrably true that switching results in a win 2/3 of the time. So whatever argument you're using (which I can barely read, so screw that) is necessarily flawed. What are you trying to accomplish?
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Well, the only time you lose if you switch is if you get the door with the prize at the start. Since there are 3 doors, that's a 1/3 probability. Since there is a 1/3 probability of losing, there must be a 2/3 probability of winning.
>>
What's the question for this anyways?
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>>6658701

see Wikipedia
>>
>implying you won't want a goat
>>
Guy deletes all of his posts too
At least hes not completely retarded or mods are based
>implying mods
>>
>>6658344
You're a fucking idiot if you actually believe this. The problem people have with Monty Hall is conceptual, not mathematical. When you show them conceptually why you have a greater chance at winning when wrong doors are revealed, they can grasp the problem better.

Showing people like this a probability tree is as useless and showing a young earth creationist a skeleton of Lucy.

Also the Monty Hall problem is identical with any n number of doors > 2 when n-2 doors are revealed. If you don't understand that you're just as fucking dumb as OP.
>>
>>6658721
yeah yeah yeah I checked my argument before you said anything

I forgot to take into account that 2/3 of the time you'll pick wrong initially.

Heil 2/3
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>>6658754
word, I was confused by the initial conditions, not math
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>>6658249
Because the host will never open the non-goat door.
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>>6658757
No you failed to take into account that you were retarded.
This is like 101 trolling post here, literally every day.
Like there is no argument in this but the fact that you thought there was is hilarious.
>>
>>6658473

No, just like in the original problem he opens all unchosen doors except one, leaving you with a choice of two doors.
>>
>>6658950
What do they say in the game:
>I'll now open all the doors that are not chosen leaving only one another closed
or
>I'll open one of the not chosen doors
?
>>
Let G1 be the first goat, G2 the second goat, and C be the car. Let's go throw each of the possible permutations of what can happen.

1.) You pick G1. The host reveals G2. You switch and get the car. (Switch, win)

2.) You pick G1. The host reveals G2. You stay and get the goat. (Stay, lose)

3.) You pick C. The host reveals G1. You switch and get the goat. (Switch, lose)

4.) You pick C. The host reveals G1. You stay and get the car. (Stay, win)

5.) You pick C. The host reveals G2. You switch and get the goat. (Switch, lose)

6.) You pick C. The host reveals G2. You stay and get the car. (Stay, win)

7.) You pick G2. The host reveals G1. You switch and get the car. (Switch, win)

8.) You pick G2. The host reveals G1. You stay and get the goat. (Stay, lose)

Odds of winning when keeping your choice: 2/4

Odds of winning when switching your choice: 2/4

What did I screw up?
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>>6659049
you don't pick car in half the cases.
3-6 are only half the chance of what you value them there
>>
>>6658754
Never judge a fish based on its ability to climb trees, otherwise it'll spend the rest of its life believing it's stupid. People like you that don't give other people a chance because you believe you're above everyone for as long as you know something they don't, need to die. I hope you have children and they watch you die in a fire. Or at least get your head caved in by someone you attempt to belittle.

>inb4umad
>inb4inb4
>>
>>6659056
Oh, also, I'm a fucking Navy Seal.
>>
>>6659056
Or maybe he expects that people lurk for at least a day before posting, as it stands this troll gets posted pretty much every day.
Oh the people these days, literally expecting posters to know the board instead of just starting instantly to spam their retarded opinions.
>>
>>6659068
>Or maybe he expects that people lurk for at least a day before posting,

If so, he is among the truly retarded. Newfags HAVE never lurked and they WILL never lurk. Expecting the world to change because it would be more convenient to you is just Sparta.
>>
Replace the 3 doors with the same set up except there are 1 million doors. The host removes 999,998 of them and asks if you want to switch. The odds you picked the right door in the first place are 1 in 1 million. You switch with absolute certainty. The probability that switching is successful is 999999/1000000.

Then you repeat it each time with 100000, 1000, 100, 10, etc. until you get to 3.
>>
>>6658249
because it's a door vs a group.
the door's P is 1/3,
the group's P is 2/3.
by opening one of the group's door
doesn't change it's P.
>>
>>6658249
because youre saying your original guess was wrong. i.e. your pick when you had a 1/3 chance was probably wrong.
>>
Let's assume you're playing Deal or No Deal, you've killed the 30 briefcase board except for 1 briefcase, #12 and you're holding # 24. Only the tree fiddy and 1millioonnnnen dollar brief case remain.

Howie asks you if you want to switch your briefcase for #12, do you?

Would #12 have a 29/30 chance of holding the million or 50/50?
>>
>>6659009
That's the point you stupid fuck. They say "one" because there is only "one" that is not chosen. The 100 doors explanation is good because it makes it very clear that "one" is just the stand-in for "All not chosen", when intuitively it feels like it's important that it's just one. It isn't; it's important that it's "all the doors not chosen", and the fact that it's "one" serves to obfuscate that.
>>
>>6658754
This
>>
>>6661752
It definitely isn't 50/50.
When you pick your case at the beginning, you are much more likely to not pick the million. If the million isn't in case #24, then it is most surely in case #12.
>>
>>6661752
fiddy fiddy
Not the same as the Monty Hall problem
>>
>>6661752
I don't think so, but I don't remember the rules of Deal/No Deal exactly enough.

But I seem to remember that all briefcases are equally likely, the Host doesn't deliberately remove only unwanted cases.
>>
>>6661997

I just looked up the rules and you are, in fact, wrong. If there are only two briefcases remaining and the top prize hasn't been revealed yet, it is equally likely to be in either case. The probability distribution is very different from the MH problem.
>>
>>6658344
The point is, that it's more likely that you'll select the wrong door to begin with. If you're more likely to pick the wrong door and only wrong doors are eliminated, then it's more likely that the uneliminated door is the correct one.
>>
>>6659009
stupid fuck
>>
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>>6662299
I actually wrote this earlier tonight because I was bored.
you basically end up picking both of the other doors. you can look at it like this

door1 door2 door3
goat goat car

pick door 1
host opens door two
switch
you get car

pick door two
host opens door 1
switch
you get car

pick door 3
host opens door 1/2
you get goat
>>
>>6663606
I realize the variable naming is shit, by the way
>>
>>6663606

This program has been written so amazingly often. It might be the second-most common program after "Hello World."
>>
>>6658254
>the goat is still either in your door or the other door whether you switch or not, which is 50/50
I'm going for a walk.
I'll either be hit by lightning or not.
50/50
yolo, bitches
>>
>>6663640
doubt it, second probably belongs to Fibonacci
>>
>>6663651
or fizzbuzz
>>
>>6658249
>there are people who believe the second goat exists
>>
>>6658254
Because there is a 2/3 chance you were wrong the first time, knowing what was behind one of the 2/3 doesn't increase you probability to 1/2. That door is still there whether or not you acknowledge it, and it still factors in the probability. Another way to look at it, is if you could choose either two doors, or one, you'd probably pick the one. But if I told you that door one was had a goat, you still have both doors.
>>
>>6658446
It's to show that just because there is only two doors left, doesn't mean it's 50/50. Our minds have a hard time seeing that, but in this case it's obviously true, for the same reason it's not so obviously true in the three door example.
>>
>>6663670
>knowledge does not increase probability
Er... I don't think your argument works.

It's better to look at it as groups.
You have your chosen group (one door) and the unchosen group (two doors). The probability that the car is in the unchosen group is 2/3. Showing you which of the doors does not change this, since you already knew at least one of them contained a goat. This does not change the fact that the probability of the defined groups. But it does change the probability within the unchosen group.

I don't know why we are still discussing this. It's a proven fact. Both mathematically and demonstrably. It has been explained from every angle.
If you still don't get it, give up or read a book.
>>
just don't be stupid to think that you are picking the right door from a million doors, that's how you get the probability of putting greater chance on the other door.
>>
There is only one scenario where win the prize after switching: having guessed a goat in the first place. Guessing either door with a goat and then switching results in a win. The choices are independent, so the additive rule says that the probability of making either choice is the sum of their individual probablityes: 1/3 + 1/3. You have a 2/3 probability of winning if you make the switch.
>>
The host flips a coin. You are asked to select the result. The host reveals that the coin has NOT landed on its side and asks whether or not you'd like to switch your answer. What is the probability you would win the toss if you switch your answer?
>>
Three ( or N ) eggs are in the bowl.
Only one of them is hard boiled.
You select one egg, put it on the table.
The chance that the table egg is hard boiled is 1/3 ( or 1/N ).
The chance that the bowl has the hard boiled egg is 2/3 ( or (N-1)/N ).
The show host breaks one egg ( or N-2 eggs ) from the bowl.
The chance that the bowl has the hard boiled egg is STILL 2/3 ( or (N-1)/N ).

note:

1)
The host does NOT open a random door, the opened door always reveals a goat.

2)
If the problem had 10 doors instead of 3,
after you pick a door, the host would open 8 doors instead of only 1 door.

3)
Question:
How would the problem work (still with 3 doors)
if the host randomly opened 1 of the 2 remaining doors,
and it HAPPENED to be the one of the empty doors
(or the one with the goat, etc.)?

Answer:
The state of the host's mind doesn't matter when the goat is revealed.
So for that one time, nothing new - it still pays off to switch.

For the times (50/50 chance) that the host randomly reveals the car,
the show doesn't make sense.
>>
>>6664849
Why doesn't removing an option affect probability?
>>
The big reason why people find this so hard to understand people are under the impression scenario that you should pick 1, view 2, and change to 3.

It only refers if you pick 1, view 2 -finding it's a goat- and then swapping to number 3.

You probably picked the wrong door with a 1/3 choice. If you know one of the two reminding doors is a goat it's probably behide the other door.

It's not really a hard or clever puzzle at all, it's just that it's poorly explained and worded.
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