Monty Hall Problem

[Anonymous No.5911050]

Hey, so my girlfriend and I are having a disagreement with the Monty Hall Problem. I'm a chemist and she's microbio engineer so neither of us are mathematicians...

For those who don't know, the Monty Hall Problem is as such:
You're in a game show and you are presented with three doors. Behind 1 of the doors is a care. behind the other two are goats.
Doors:
| a | b | c |
You choose door | a |. The host knows what is behind each door and opens one that is a goat, door | c |. He/she presents you with a choice. Do you want to stick with your original door, door | a |, or do you want to change to door | b |?

Monty Hall Problem proves that you if you change your choice to door | b | you have a 33.34 % greater chance of winning the car.

That is all good, assuming the objects behind the doors are not switched. This is where the disagreement comes in. With round 2, she thinks that if you were to switch the objects, you would start from scratch with a 50/50 chance and disregard round 1, while I think it makes no difference to the odds. Neither of us can mathematically prove either way. Can anyone smarter than us help?

[Anonymous No.5911059]

>>5911050
Isn't this essentially superposition? You don't know until you open the door on round two if your odds are different because you have no idea how the odds have been affected? Switching answers instead of switching prizes should affect the percentages more concretely, but IDK really.

[Anonymous No.5911061]

>>5911050
the wikipedia entry on this is (surprisingly) informative

under the standard set-up:
a) you select (but don't open) one of the three doors: locked in 1/3 odds
b) Monty knowingly reveals a goat (behind a door other than the one chosen above)

the prize MUST be behind one of the remaining doors; if both are opened the prize would necessarily be revealed

you've locked in 1/3 odds on one door (your initial selection) so 1 - 1/3 = 2/3
switching to the other door (the one not selected initially) gives 2/3 odds

[Anonymous No.5911062]

>>5911050
tl;dr Monty Hall Problem, what if you switch door 1 and 2 in second round?

bumping with some tits i like

[Anonymous No.5911063]

You need to be clearer about what you mean by switching the objects. Which objects are switched? How are they switched? (Is it random?) Are they switched before or after the host eliminates a box and does the host know where the prize is after the switching?

[Anonymous No.5911064]

>>5911062
blue (sfw) board...

[Anonymous No.5911067]

>>5911059
>>5911061
Thanks for the replies...

>>5911061
It is quite informative, surprisingly! Yeah I get this... just wondering if someone can prove that it doesn't make a difference to the odds, whether or not THE ITEMS in the doors are switched... (with you also switching your choice).

The question I think is irrelevant, but I can't explain well enough why, to her... I'm not amazing at math and she's Danish so ESL so can't explain accordingly...

[Anonymous No.5911069]

>>5911064
oh yeah sorry... it's been a long time since i've been on /sci/

[Anonymous No.5911071]

>>5911064
fixed... sorry my years on /b/ have tainted me.

[Anonymous No.5911079]

>>5911067
On my mobile...
Also due to my mathematical ineptitude...

But I reckon that since the item behind the doors is completely unknown to the player, it makes no difference whether or not they are switched randomly or not...

[Anonymous No.5911384]

>>5911050
Watch this:

The guy is maths doctor. Also check the channel numberphile, it's amazing.
>http://www.youtube.com/watch?v=njqrSvGz8Ps

[Anonymous No.5911421]

>>5911050
I bump with this one
http://en.wikipedia.org/wiki/Sleeping_Beauty_problem

which is far more discussed (and not "solved" yet)

So you can discuss again with your grilfriend :)

[Anonymous No.5911428]

The reason it's not a 50/50 after the reveal is because the host has revealed information, the revealed door was not chosen at random.

[Anonymous No.5911453]

Here's a simulator that runs the problem.
http://www.grand-illusions.com/simulator/montysim.htm
Pic related, I ran 100 tests.

[Anonymous No.5911491]

>>5911453

I'm having a hard time trusting this simulator without knowing the mechanism behind it, since it's possible it's coded to cheat.

[Anonymous No.5911509]

>>5911491
So here is one in JS :
<script type="text/javascript">
var games = 1000000;
var winsWithSwitch = 0;
var winsWithoutSwitch = 0;

document.writeln("<pre>After " + games + " tries…");

for(var i = 0; i < games; i++) {
// Put the prize behinf a door and let the player choose
var prizeDoor = Math.floor(Math.random()*3);
var choice = Math.floor(Math.random()*3);
// Open a door without the prize behind it
var openedDoor;
if (choice == prizeDoor)
openedDoor = (prizeDoor + 1 + Math.floor(Math.random()*2)) % 3;
else
openedDoor = (0 + 1 + 2) - choice - prizeDoor;
// Allow the player to choose the door which haven't been opened.
var switchDoor = (0 + 1 + 2) - choice - openedDoor;
if (choice == prizeDoor)
winsWithoutSwitch++;
if (switchDoor == prizeDoor)
winsWithSwitch++;
}

document.write("Succes rate when player does not change");
document.writeln(winsWithoutSwitch / games);
document.write("Succes rate when player chages his mind");
document.writeln(winsWithSwitch / games);
document.write("</pre>");
</script>

[Anonymous No.5911565]

if you can't pinpoint the problem your disagreement is based on, you're both idiots and should either kill yourselves or at least gtfo

[Anonymous No.5911567]

>>5911565
What a useful intervention. Thnak you very much, Sir

inb4 u2

[Anonymous No.5911572]

>>5911567
If you are op I'm afraid I have to inform you that you can't inb4

[Anonymous No.5911576]

>>5911491
Enumerate all possibilities, you'll see why it's 2/3 and not 1/2.

Let's say the prize is behind door c.

1) In the following configurations you always change your initial choice to the remaining door.
You pick a. Host reveals b. You switch to c. You win.
You pick b. Host reveals a. You switch to c. You win.
You pick c. Host reveals a (or b). You switch to b (or a). You lose.
By always switching you win in 2/3 of cases.

2) In the following configurations you always stick with your initial choice.
You pick a. Host reveals b. You stick with a. You lose.
You pick b. Host reveals a. You stick with b. You lose.
You pick c. Host reveals a or b. You sitck with c. You win.
By always sticking with your initial choice you win in 1/3 of cases.

So if the prize is behind door c, you win twice more often by always switching than by always sticking to your initial choice, and the probability of winning by always switching is 2/3.
The reasoning is exactly the same if the prize is instead behind door a or door b, so the same conclusion always applies.

By always switching, you win when you initially picked a goat. Since there are goats behind 2 of the 3 doors, you win 2/3 of the time. By always sticking to your initial choice, you win when you initially picked the prize. Since the prize is behind 1 of the 3 doors, you win 1/3 of the time.

[Anonymous No.5911579]

The problem is invalid because I want a goat.

[Anonymous No.5911583]

>>5911567
I'm not.
I provided an intersting link
>>5911421
but nobody seem interested.

I also write this >>5911509
if it can help this guy>>5911491

(actually, poor translation of the part of the french wikipedia page I wrote)

[Anonymous No.5911588]

>>5911050
>That is all good, assuming the objects behind the doors are not switched. This is where the disagreement comes in. With round 2, she thinks that if you were to switch the objects, you would start from scratch with a 50/50 chance and disregard round 1, while I think it makes no difference to the odds. Neither of us can mathematically prove either way

You mean switch the objects after the host has revealed one of the goats?

Once again let's go with enumeration, there aren't many doors so it's not too much of a hassle.

Let's say prize is behind door c initially, then switched to the remaining door after host has revealed one of the goat.

1) In the following configurations you always change your initial choice to the remaining door.
You pick a. Host reveals b. Prize is switched to a. You switch to c. You lose.
You pick b. Host reveals a. Prize is switched to b. You switch to c. You lose.
You pick c. Host reveals a (respectively b). Prize is switched to b (respectively a). You switch to b (respectively a). You lwin.
By always switching you win in 1/3 of cases.

2) In the following configurations you always stick with your initial choice.
You pick a. Host reveals b. Prize is switched to a. You stick with a. You win.
You pick b. Host reveals a. Prize is switched to b. You stick with b. You win.
You pick c. Host reveals a (respectively) b. Prize is switched to b (respectively a). You stick with c. You lose.
By always sticking with your initial choice you win in 1/3 of cases.

So when the objects are switched are round 2, the odds are reversed. I guess you could prove it with a formula, but exhausting all the possibilities has the same proof value.

[Anonymous No.5911593]

>>5911588
Shouldn't have copy/pasted too much, in 2) the result is you win in 2/3 of cases of course, not 1/3

[Anonymous No.5911600]

imagine if there are 1000 doors, and just one with the goat. you choose one, and then 998 are eliminated, letting you with only 2 doors. can you understand now which door is probably the one with the goat?

[Anonymous No.5911689]

>>5911600
i said it like the goat was the winning prize

[Anonymous No.5911945]

>>5911689
As far as I'm concerned, a goat is a much more beautiful guest than a car. So I totally agree with your post.

More seriously :

It's still amusing to see how much it's difficult for us to really "feel" probabilistic problems.
Monty Hall is a famous example
(http://en.wikipedia.org/wiki/Marilyn_vos_Savant))

But also :
http://en.wikipedia.org/wiki/Bertrand_paradox_%28probability%29

And as mentionned before
http://en.wikipedia.org/wiki/Sleeping_Beauty_problem
(>>5911421)

Why are we so bad ? I don't troll, it's a serious question. What make us having so wrong intuitions most of the time ?

We = Most of the population, maybe not you, please don't spread your ibtelligence, that's not the point.

[Anonymous No.5912104]

>>5911945
Bump for interest :

>http://en.wikipedia.org/wiki/Bertrand_paradox_%28probability%29
I don't undestand what the fuck is happening.

>http://en.wikipedia.org/wiki/Sleeping_Beauty_problem
This one, I don't even understand the terms of the problem

Any help ?

[Anonymous No.5912428]

op here... sorry got super drunk last night hence mia.

>>5911565
Thank you for such a constructive comment, fine sir!

>>5911384
wicked, thanks mate

>>5911421
>>5912104
>>5911945
>>5911588
>>5911576

awesome!!! I've got some reading to do...
thanks heaps

[Anonymous No.5912436]

>>5911050
With three doors, it means that the host opened a door that he knows that is a wrong one, so, your chances rise, BUT, your chances are even higher if you choose the other door instead of the one you choose in the first place, since the host opened a wrong door, so it's most likely that the door he ignored is the right one.

[Anonymous No.5912438]

>>5911384
>math
>doctor
Pick one.

[Anonymous No.5912441]

You have more probabilities to pick a goat than a car. If you pick a goat and the guy opens the goat door, you may swap to get the car.

[Anonymous No.5912484]

This goes by the principle that:

(1) There are three doors. Two of which have no special prizes; one has a special prize.

(2) You pick one door.

(3) Host then later opens of the two unselected doors.

Now if he was given to open one wrong door of the three doors instead of the two in the previous case, it wouldn't matter whether you keep or change. There is a 50% chance between both doors in the end.

[Anonymous No.5912538]

>>5911050
So you're more or less suggesting the following scenario:

1) The player picks a door
2)The host reveals a bogus door that is not the door you've chosen
3)The items behind the doors may or may not be switched without your knowing

Your question: Does this change the probably of switching winning 2/3 of the time, or is the probability unaffected?

[Anonymous No.5912548]

>>5911050

If the objects behind the doors are switched after the first round, I think the problem is pretty much meaningless and yes it's 50/50.

If the objects AREN'T switched, you gain better odds by changing your initial choice.

[Anonymous No.5912549]

>>5911050
I disagree with you and your gf

I'd say the odds if you SWITCH after they switch equals 1/3

The odds if you STAY after they switch is 2/3.

you choosing to switch+them switching= staying.

Round 1 still effects the odds, however

[Anonymous No.5912616]

>>5912549
same guy here, to clarify, the new issue you bring up is the MAY vs MAY NOT do they switch it after round 1. This makes the whole thing moot, I think. So it's 50/50

[Anonymous No.5912625]

>>5912616
switch is the most advantageous because you know for sure that the host won't open the door with the main prize. Your chances gets higher, but of course that doesn't mean that you gonna get a hit. If he opened the door with the gold, then you could imagine that the chances are 50/50, but he will NEVER do that.

[Anonymous No.5914175]

By switching you reduce the probability of winning a goat.

[Anon No.5914187]

Solved by simple diagram. 2/3 time switching is the better choice.

[Anonymous No.5914271]

>>5914187
best explanation

[Anonymous No.5914664]

>>5911050
The issue is where you draw the lines, also look at that OR statement.
We have door car, door goat A, door goat B

1. pick door car, do not switch and win
2. pick door car, do switch and lose
3. pick door goat A, do not switch and lose
4. pick door goat A, do switch and win
5. pick door goat B, do not switch and lose
6. pick door goat B, do switch and win
If you look at it this way switching is good as switching wins 2/3 of the time.

But what if I give you the options 1. 2. 3. 4. 5. and 6.
1. win
2. lose
3. lose
4. win
5. lose
6. win
Now it looks like a 50/50 chance.
Now I tell you that both sets are the same thing, the first one just has more words that confuse you. Now math says I am wrong, but math never has explained how this is as my argument seems very valid. One thing that comes up is whether there are two rounds or one round, I do not see how this changes things in a practical sense.

Example: If I have a person in a strength lifting contest who wins two round to be the grand winner or the same person wins one round to be the grand winner, that person is still the grand winner which is the same outcome because their strength was greater then any of the competitors (simple highest number wins). {yes i know this ignores muscle fatigue and other variables, but they do not matter in this argument} It is the number of contestants that changes the odds not the number of rounds, as if we grab more people we increase the likelihood of adding a new person who has greater strength(a even higher number).

Do you see any connection to the gamblers fallacy here? Because that one makes sense to me, this doesn't as I see it as a 50/50 with switching being irrelevant.

[Anonymous No.5914687]

>>5914664
>But what if I give you the options 1. 2. 3. 4. 5. and 6.
>1. win
>2. lose
>3. lose
>4. win
>5. lose
>6. win
>Now it looks like a 50/50 chance.
it doesn't look like you're not accounting for the fact that monty can reveal either goat door a or goat door b if you choose the car, but he will be forced to reveal goat door a if goat door b is chosen and vice versa. I'm really tired right now though so I don't know if that's the issue

for those who are having trouble, draw out a probability tree

[Anonymous No.5914711]

>>5911050

when the contestant makes his/her first choice, his/her odds of getting the car are 1/3

when the host reveals one of the goat doors, it gives the contestant new information that allows him/her to eliminate one of the three doors

when the contestant chooses to pick a new door, the odds of choosing a door are now 1/2 because of the knowledge that allowed him/her to eliminate one of the goat doors

without switching your choice the contestant is guessing between three doors; with switching the contestant is choosing between two doors

[Anonymous No.5914718]

>>5911050
You're not describing the Monty Hall problem in full detail. The host always selects a door that doesn't have a car behind it, so therefore he changes the odds.

[SciDropout No.5914774]

According to Monty Hall, the Monty Hall problem is based on the false assumption that he knew what was behind the cards, doors, or whatever. He said he did not. If you watched the show, you would see that he did not by the fact that he often picked the prize himself.

[Anonymous No.5915986]

>>5914774
That's retarded. Why would he open the door with the prize?

[Anonymous No.5916551]

this one cleared it up for me, it took lots of other ones before I found it
http://www.youtube.com/watch?v=koPBkK_Ra-k

[Anonymous No.5916566]

if there are 1000 doors, 999 goats and only one car.
you'd pick one and he'd reveal every other door but one.
what do you think your chances are now

[Anonymous No.5917249]

>>5916566
50/50, because the other reviled doors are meaning less to the outcome. If you don't use your option to switch and it goes away, did that option change anything?

[Anonymous No.5917944]

>>5911061
>switching to the other door (the one not selected initially) gives 2/3 odds

Don't you mean 1/2 odds?

Why is the goat that has been revealed, still calculated as a potential choice?

[Anonymous No.5917996]

>>5917944
I mean.... seriously.... after one of the doors has been opened, and revealed to be a goat....

There is STILL only one car, and now there are only TWO doors that it may be behind.

Hence 1 out of 2, or one half, or 50%

IT isn't out of three, because there aren't three doors to choose from any-more....

and it isn't 2, because you still only get ONE fucking choice.

it's 1/2, not 2/3... seriously... stop being this stupid.

[Anonymous No.5918014]

>>5912104
These are very interesting problems

>http://en.wikipedia.org/wiki/Bertrand_paradox_%28probability%29
This is saying random numbers don't mean anything unless you say how the random numbers are generated. For example, suppose I picked a random number x between 0 and 1, and I asked you to state the probability that x is between 1/2 and 1. You would probably immediately say it's 1/2, since "half" the numbers in [0,1] are between 1/2 and 1. But I never told you how I was generating the numbers. What if I picked x by first picking y uniformly between 0 and 1, then computing x = y^2. Technically x is a random number between 0 and 1, but the probability that x is between [1/2, 1] is only 7/24, not 1/2 as you might expect.

This probably seems stupid, like my method for making random numbers is only there to trick you. In this case that's reasonable, since it seems natural to pick the numbers uniformly. Bertrand asked a similar question, but instead of numbers between 0 and 1, he was picking a random point inside of a circle. There's lots of ways to pick a random point in a circle, but this time all of them are arguably the "natural" way. He's saying probabilities are only well-defined if you specify how the random number was generated in the first place.

[Anonymous No.5918061]

>>5917944
>>5917996

Not sure if you're trolling or retarded, but please just read the wiki.

[Anonymous No.5918083]

If you certainly switch the objects it is better to stay with the first choice.

If you have a 50% chance of the objects being switched both are equally likely.

The changing of choice is only an improvement if you assume that the set remains static otherwise you are faced with something new.

Take one million doors as an extreme example:
1 000 000 doors
1 prize
You choose one (Probability of being correct 1/1000000)
They remove 999 998 doors
You now have the one you picked and one more, it is "almost certain" that changing will win.

But if there is a 50% chance of them switching the prize to your first choice changing is not an improvement, it is equally likely both ways.

[Anonymous No.5918096]

L2Bayes

G represents "Hall reveals a goat behind door B or C"
A represents "The car is behind door A"

If Hall looks behind the other doors and always opens one with a goat:
P(A|G) = P(G|A)*P(A)/P(G) = (1)*(1/3)/(1) = 1/3

If Hall doesn't look behind the other doors and just happens to have opened one with a goat:
P(A|G) = P(G|A)*P(A)/P(G) = (1)*(1/3)/P(G|A)*P(A) + P(G|!A)*P(!A)
= (1/3)/(1/3) + (1/2)*(2/3) = (1/3)/(1/3) + (1/3) = 1/2

If the items are randomly switched or not switched before the second round then switching your door choice doesn't matter because half the time the prior odds will be reversed by the item switch

[Anonymous No.5918099]

>>5918061
>Not sure if you're trolling or retarded, but please just read the wiki.

You have got to be fucking kidding me.

At what point in your illicit, fever fuelled incoherent madness did you glean that two possible doors are really three?

One car, two doors = 1/2

Period.

[Anonymous No.5918114]

>>5911067

If the items are switched, behind the doors, then you SHOULD NOT switch, because then staying will give you a 66.6% better chance than switching. Done. Anyone who says otherwise is fucking idiot.

Think about it. The only reason that it's a 2/3's chance of getting the car when switching is because if you were wrong initially, switching makes you right, so there was 2/3s chance of being wrong initially. If you switch the doors, that means that the probabilities are reversed. Because if you were wrong initially, switching will now make you still wrong, and if you were right initially, you'll still be right if you switch. So You have a 1/3 chance of being right initially. In this scenario, you should always STAY.

Both you and your girlfriend are retards.

[Anonymous No.5918128]

>>5918099

So trolling, then. Ok.

[Anonymous No.5918172]

>>5918128
>So trolling, then. Ok.

So, trying to undermine logic with appeals to authority, and argument from verbosity... eh?

[Anonymous No.5918190]

>>5918172

Perhaps he doesn't want to bother explaining something to you when there are already a million clear explanations already easily available to you.

[Anonymous No.5918195]

By opening one of the doors, and revealing a goat, you have removed an ENTIRE POSSIBLE CHOICE from the equation.

it's 1/2, and anyone who disagrees, is a gullible fool.

[Anonymous No.5918218]

>>5918195
3 doors
2 goats
1 car

There are three Doors to choose from, and behind them are a goat, a goat, and a car.

By opening one of the doors, and revealing a goat, one of the doors has been removed.

3 doors - 1 door = 2 doors.

And by revealing the goat, one of the goats has been removed

2 goats - 1 goat = 1 goat

Therefore, you are left with 2 doors

Therefore, you are left with a car, and a goat.

You only have one choice to pick between the two remaining doors....

Hence, a 1/2 chance of picking the car.

[Anonymous No.5918216]

>>5918195

Stop.

[Anonymous No.5918225]

>>5918216
>Stop.

[Anonymous No.5918230]

>>5918195
But you made the choice before that possibility was removed.

If you add more doors to remove the principle becomes trivially clear see: >>5918083 .

[Anonymous No.5918253]

To all skeptical people
I gave you one way to SEE the result
>>5911509

So, 2 possibilities :
-> you disagree with the way I programmed the thing, and explain why. Cause I simulate the very definition of the problem.

-> you admit the program is well written. You run it and you see you're wrong. It's maybe not satisfying, cause it's a simulation and doesn't explain why "it works how it works", but srsly, you have tons of explanatios everywhere.

You just don't want to admit that your intuition is wrong

[Anonymous No.5918258]

>>5918253
Shut the fuck up, you imbecile

[Anonymous No.5918260]

>>5911050
This isn't real math, just stats.
It relies on the assumption that the game show host knows which door contains the goat, and deliberately didn't open the door with the car behind it first to build suspense.
There's a 2/3rds chance your initial guess was wrong, and a 1/3rd chance it was right. If your initial guess was wrong, and the judge reveals the other wrong answer, you should switch. If your inital guess was right, and the judge reveals one of the wrong answers, you shouldn't switch. Because it is more likely that your initial guess was wrong than right, it is rational to switch.
Please note I have not read any of the thread, so if we've already moved on to other subjects, I apologize.
Also, didn't you dudes learn this shit in high school?

[Anonymous No.5918268]

>>5918260
By removing one of the selections, you have changed the odds.

on initial selection, you have a 1/3 (one in three) chance of being correct, because you have one choice, there are three possible selections, and only one of them is correct.

When you open one of the doors to reveal the goat, there are no longer three possible selections... there are only two.

Therefore, stating that you have a 2/3 chance if you switch, is retarded, because there are only two doors remaining, and you STILL can only choose 1 door, and there is STILL only one fucking car.

[Anonymous No.5918270]

>>5918260
Fuck, I didn't even read the OP, disregard me, I'm a retard

[Anonymous No.5918284]

>>5918258
What an impressive argumentation !

>>5918268
But do you even read the thread before doing the same errors that everybody ?

I don't know, we're supposed to be on /sci/, right ?

All the scientific community agree on the fact that the prob is 2/3. How do u explain that ? They're all wrong ? Conspirancy ?

Yes, it's an "argument from authority", but you don't read the scientific proofs or programs we provide to you, so...

[Anonymous No.5918292]

>>5918284
>All the scientific community agree

http://www.youtube.com/watch?v=3oRy46XQ0es

Oh, and all of the arguments that you believe validate your "2/3" bullshit, are just that... bullshit.

2/3 is wrong because you don't have 2 choices, and you don't have three doors.

You have ONE choice, and TWO doors.

it's 1/2.

Period.

[Anonymous No.5918318]

>>5918292
>You have ONE choice, and TWO doors.

So what? You are aware that events can have different probabilities of occurring, right? That just because there are 2 possibilities doesn't make each one have a probability of 1/2?

[Anonymous No.5918320]

>>5918292
>http://www.youtube.com/watch?v=3oRy46XQ0es
Are u serious... ? youtube ?

We're speaking of scientists...
http://www.math.leidenuniv.nl/~gill/mhp-statprob.pdf
http://econpapers.repec.org/paper/wpawuwpex/9906001.htm
http://www3.nd.edu/~rwilliam/stats1/appendices/xappxd.pdf
http://www.cs.dartmouth.edu/~afra/goodies/monty.pdf
http://scimath.unl.edu/MIM/files/MATExamFiles/JohnsonB_FinalPaper_FINAL.pdf
...
and so on

What's happening there ? Mondial consiprancy ? Are we on /x/ ? Illuminati ? Aliens ?

Actually, I'm sad for u, it means that u can be easyly cheated in your daily-life with simple "paradox".

>Ending an argument with "period".
Seems legit.

[Anonymous No.5918321]

>>5918318
>You are aware that events can have different probabilities of occurring, right?

You have two doors.

Behind one door, there is a car, behind the other, there is a goat.

If you pick one at random, what is the probability that you pick the car?

[Anonymous No.5918329]

>>5911050
Probability is fantamath

[Anonymous No.5918332]

>>5918284
Ignore him, it's a trick. I'm the dude he was replying to. Don't be foolish.

[Anonymous No.5918333]

>>5918321

What else do you know about them?

[Anonymous No.5918330]

>>5918320
>What's happening there ?

[Anonymous No.5918339]

>>5918333
>What else do you know about them?

Nothing.

[Anonymous No.5918346]

>>5918339

Then it's 1/2. But what does that have to do with the Monty Hall Problem?

[Anonymous No.5918348]

>choose B
>SURPRISE! door A actually had a car

gameshow logic

[Anonymous No.5918352]

>>5918346
>Then it's 1/2.

That is correct.

>But what does that have to do with the Monty Hall Problem?

IT *IS* THE MONTY HALL PROBLEM.

[Anonymous No.5918395]

>>5918352
>IT *IS* THE MONTY HALL PROBLEM.
Ok, there is the bug.
You didn't understand the problem...

What about redaing the wiki page ?

[Anonymous No.5918391]

[Anonymous No.5918399]

>>5918391
Damn'it !
Where is my goat ?

[Anonymous No.5918400]

>>5918395
>You didn't understand the problem...

ARE YOU GOING TO PICK THIS FUCKING DOOR?

[Anonymous No.5918402]

>>5918395
You're being tricked. Let the thread die, it's over.

[Anonymous No.5918408]

>>5918400
easy boy.
Using capital doesn't make you beeing right.

>>5918402
Actually, I'm doing serious stuff on other tabs.
That's my coffe break to come here.

/sci/ is perfect for that

[Anonymous No.5918414]

>>5918408
>easy boy.
>Using capital doesn't make you beeing right.

http://www.youtube.com/watch?v=4I-efBSAUiY

[Anonymous No.5918452]

>>5918402
>it's over

[Anonymous No.5918457]

[Anonymous No.5918560]

I bet.

[Anonymous No.5918659]

you both are saying the same thing.

[Talimar No.5918681]

I've always hated the explanation for this, from a Set standpoint. Mathematically, your "probabilities" end up being "absolute value" probabilities, to borrow a maths term.

When you proposition from N =n,n1,n2, each door has a specific whole value of a whose n-prime value for existence dictates that the other 2 n values cannot be apart of the Set n1-prime (when n-prime properties are not derived from n-element properties.)

What happens when you provide all values for a Whole (big N), you end up with a proposition for a n-prime which is actually n-prime+nx for both doors, but unfortunately ends up with a seed value from our choice selection.

This then calculates that you have more elements n for the Set N within a particular n-prime value. Ie. the door you didn't choose.

(never bothered working this out for more then 3 doors.)

[Anonymous No.5918909]

>>5918681
Medula Oblongata, Scwartzchild radius, Quantum Indeterminancy, Varicose, Holman, Schrodinger...

Checkmate.