Monty Hall Problem

Can someone coherently explain the Monty Hall problem to me? Its driving me insane i feel like a tard. Is this Jew trickery?

en.m.wikipedia.org/wiki/Monty_Hall_problem

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You are a tard.

Think about it like this:
When you change the door, you win if you pick the wrong one at the start.

>2/3 are wrong doors
>pick a door
>Monty opens the other wrong door
>if you picked a wrong door at the start and now switch you win

Can someone shoop that green beans to be challah instead?

Basically the trick is his choice of door is not independent of yours, nor is it independent of which door the goat is behind. Infact, most of the time he only has one option.

Since he always chooses a non-goat door, your odds don't improve despite an option being eliminated.

Think of the same problem but with 100 doors.
You choose one and then 98 wrong doors are closed. Obviously here you will change your pick to the other door, right?
With 3 doors the probability is only marginally improved as compared to 100 doors but it is improved.

If more often than not host doesn't have a choice then switching gives you bigger chance of wining.
Do it yourself no the paper. It works.

Imagine there are 1,000,000 doors and 999,999 goats.
You pick a door. The host opens 999,998 doors he knows have goats behind them, leaving the two unopened doors.
Should you switch? Of course.
There is only a 0.0001% chance of a car behind the door you picked.
There is a 50% chance the car is behind the other door.

No, it's obviously 50% chance because there's two doors.

Thank you. Im a confirmed tard.

I think this proof went with 1 goat behind 100 doors and hosts opens 98 doors that are empty.

It was used to show how we perceive problems.

The key is that information is added by the deliberate action of the host

Imagine there are 1,000,000,,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 doors and 999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999 goats.

You pick a door.
The host opens 999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,999,998 doors he knows have goats behind them, leaving the two unopened doors.

There is a less than 50% chance that the car is behind the door you picked.

There is a problem with the study of probability because it can either be approached a priori or a posteriori. In one the problem is that you can't define how big is big enough a sample to be reliable in the calculation of results, and in the other one the problem is that you don't actually know what's going to happen unless it happens. The Monty Hall is a scenario devised to illustrate how fucking stupid and illogical that is, but people who didn't actually understand the depth of it didn't realize that it was meant to be used ironically.

All you need to know about the Monty Hall is that it's not true that changing your choice improves your chance of winning, and that it's like when animes talk about the Schrodinger's cat to sound intelligent. Just stay away from it and the retards who believe it unironically.

but how can you be sure the goats turn into cars behind the doors???

Imagine there are 3 doors. Two of them are goats. You pick the right door so it's 50/50

>its like when animes talk about schroedingers cat to sound intelligent

K E K

god damn you are so fucking stupid while trying to sound smart, it's pathetic

33.3% of the time you'll choose the wrong door. Think of it as being given two shots at opening the right door, but obviously not quite.

Since you only have a 33.3% chance of picking the right door, the host is giving you a higher chance of winning (66.6% instead of 33.3%) because you probably would have picked the wrong door simply because the odds are against you.

>All you need to know about the Monty Hall is that it's not true that changing your choice improves your chance of winning
Every time. Trolls trolling trolls with this basic probability theory bullcrap a third grader could do on a piece of paper.

>there are 100 doors
>99 of them a goat another a car
>you pick one door
>host opened all doors except the one you chose and a random door and its all goat
>DO YOU WANT TO CHANGE DOORS? WHAT ARE THE CHANCES THAT ONE EXACT DOOR YOU DIDNT CHOOSE THAT THE HOST DIDNT OPEN CONTAINS A GOAT GEE WHIZ

this
is it really that hard to understand?

>mexican intellectuals

>All you need to know about the Monty Hall is that it's not true that changing your choice improves your chance of winning

>mexican intellectuals

Try it yourself pablo,
At first the chance that you picked the correct door was 1/3 after 1 is revealed the chance that there is a car behind the OTHER door is 50%

But who am i kidding you are not going to try it and will keep spouting this bullshit because you think that you are smarter than everyone...

You can predict that 2/3rds of the time, you'll pick a goat, and force him to reveal the other goat.

If you're implementing the switching strategy as recommended, switching will be better than staying 2/3rds of the time. And 1/3rd of the time, it will have a neutral effect.

What if there's a car worth 30k, a goat and an anti-car that's worth -30k

You know that the presenter will reveal the least valuable object that you haven't picked as your first choice. Should you switch?

>the chance that there is a car behind the OTHER door is 50%

Exactly, thus the chance that the car is behind THIS door is also 50%. If your initial choice had a chance of success of 33.333%, then if you keep the same choice your chance of success grow to 50%, and if you change your choice your chance of success grows to 50% as well.

Believing in the Monty Hall thing is like believing that if you throw a coin 3 times and get 3 heads, your chance of getting a tail on the fourth try is 87.5%. Stop being retarded.

Just have a muslim be the one to pick the door.
They can smell their wife from 60 cubits away.

It doesn't work the jews want you to think. Every question is an independent state only in theory.

>"gives you bigger"
nope, it is determinated from the beggining since that one which you choose first cannot be removed.

Just play every scenario out on paper.
Then you'll see.

Lets say the prize is behind door 3 and you pick door 1. Door 2 is opened.

>Switch to 3. Win
>Stick with 1. Lose

Okay, so now rewind and pick door 2

>Switch to 3. Win
>Stick with 2. Lose

So far we're even
But what if you pick door 3?
>Switch - Lose
>Stick - Win

NOW look at all the above scenarios and tell me, which option has the better outcome? :^)

People in this thread already highlighted how the problem can be more easily understood by increasing the number of doors. There's been computer simulations done to prove it as well.

At this point you deserve nothing less than being spammed with

>mexican intellectuals

kek

>Believing in the Monty Hall thing is like believing that if you throw a coin 3 times and get 3 heads, your chance of getting a tail on the fourth try is 87.5%. Stop being retarded.


No, you are retarded

Chance doesn't mean probability, the chance that you will end up with heads is always 50% the probability of that happening decreases with ever consecutive heads you get.

Like many other mentioned, look at it in bigger scale, 100 doors, 1 car 99 goats, you pick door number 20
Monty reveal all the doors except for door number 50, do you really think that it doesn't matter if you will swap or not because "It's 50% after all, its either behind 50 or 20"
It would be the case if you had 2 doors to begin with
You had 3 doors and he only REVEALED 1 door it didn't change your chance to pick the right one because you chose a door before that when you had 1/3 chance, now if you swap it you have 1/2 chance that you will get it right, if you stay its 1/3 that you get it right.


Again, try it yourself, it was proven you can prove it yourself yet you keep trying to sound smart.
You are the reason we laugh at mexicans on this board.
>intellectuals

Assuming the host is not retarded, he will not offer for you to swap your door unless you picked the correct one first.
So, pick a door and stick to it.

These are the possible outcomes:

Pick car > goat 1 or 2 revealed > stay > win
Pick car > goat 1 or 2 revealed > switch > lose
Pick goat1 > goat2 revealed > stay > lose
Pick goat1 > goat2 revealed > switch > win
Pick goat2 > goat1 revealed > stay > lose
Pick goat2 > goat1 revealed > switch > win

You can never tell which goat he's revealed, so you have to treat it as if it were pre-determined.

what the fuck
I can see why Trump wants to build a wall

The number of doors doesn't matter because the domain of the probabilistic space will always be only two doors. When you pick a door for the first time you're just picking one of the two elements that will be in the domain of the experiment (the other one will be determined by the game master or whatever), you're not actually making a decision that's relevant to your final success rate.

I could flip a million coins behind a curtain each and then ask you to predict the result of the last coin, the chance is the same as with a singe flip because all the other elements were there only to confuse you, they aren't relevant at all.

>the probability of that happening decreases with ever consecutive heads you get.

No. What changes is the probability of getting a strike of n anything, the probability of any given flip is the same. Get back to high school. You both clearly need to relearn the basics of probability.

What's this big-nosed motherfucker called.

Yeah, this is also the point I gave up on trying to understand probability and dropped out.

The probability of you picking the right door on the first try is 1/3.
When one wrong door is removed there's a 50% chance the other door is the right one, while the one you picked first is still at 1/3. It's crazy but it's true. I actually did a recursive simulation of this in a beginners computer science course. With 10 000 simulations there was a clear advantage of switching the door if given a chance.

You can test this yourself
>Get 3 (or to better test it 10) opaque cups
>Turn around and have someone put an object under one of the cups
>Choose which cup you think has the object
>Have them show you all but two of the cups, none of which have the object
>Switch or don't switch your choice
If you switch you will find the object 50% of the time, if you don't you will find it 10% of the time.

Another way of thinking of it is that after they show you the door that doesn't have the reward you can keep playing the same game or play a new game with just two doors.

First choice has a 1/3 chance of being right.

After one of the wrong doors is opened you have new information about the door you didn't choose (it could have been opened, but wasn't, therefore it could be the winning door), but no new information about the door you originally chose.

If you switch doors you have a 50% chance of winning. The odds of the original door being the winning door don't change and are still only 1/3.

Always switch doors.

>Chance doesn't mean probability, the chance that you will end up with heads is always 50% the probability of that happening decreases with ever consecutive heads you get.

Actually chance and probability are synonyms and the probability of a coin flip being heads or tails is entirely independent of how it has landed in previous flips .

YOU are retarded.

...

It's not making more sense. You're not explaining why your odds of having picked the right door first stay the same after the amount of doors possibly having a car are reduced to 2.

No, there is a 99.9999% chance that the car is behind the other door. The chances have to add up to 100% because the car only can be behind one of those two doors.

Probiscus monkey

It's pure bullshit math. The confusion stems from the fact that probability is not properly defined. It depends on subjective knowledge.

Consider this:
You go into a room and there are two doors. Behind one door is a goat, behind the other is a car. Does it make sense to assume anything other than 50-50 probabilities? Of course not. In that situation it doesn't matter whether there were three doors at some point in the past and a host opened one of them.

It's a thought experiment that plays with the probabilities of being right, and changing the problem halfway through. Pick door, all of the other ones are removed except one. Now, switching will always be right.

A fun thing to bring up whenever someone starts talking about this is that they - HAVE TO - belive that we live in a simulated universe. They have to believe it because theoretical probabilities make it so.

>So, pick a door and stick to it.
But-but- the reward door always switches like in the examples, remember, to illustrate that whoever made the example is right? It's never random or the same door twice in a row or anything.

When you're told what the one door contains, the door doesn't stop existing. You simply now know what 1/3 of the options are and don't pick it accordingly. When you first picked, you had a 1/3 chance of being correct, when he told you one you went up to 2/3 chance of being correct instead.

>Britbongs can't into math

obviously, but the point is that the probability of getting multiple heads in a row.

Probability of getting 4 heads in a row.
First flip: 50%
Second flip: 50%
Third flip: 50%
Fourth flip: 50%

.5*.5*.5*.5 = 0.0625

Sure the last flip is always a 50% chance, but the probability of you even having got that far is slim to begin with and can't be discounted.

>The missing link between apes and jews, undeniable proof of evolution

>It's pure bullshit math
are you implying that when you test it empirically, the results don't show exactly what the calculations show? Because you most certainly are better off in real life switching your guess.

Probability is the measure of the likelihood that an event will occur.

Chance is a possibility of something happening.

Probability that you will get 5th heads in a row is low but the chance that you will get heads on 5th try is just 50%

>empirically
Math is not an empirical science.

The door doesn't stop existing, but you can write it off as a choice because you know it's wrong. There are 2 possible correct options remaining, so each has a 50% chance of being the correct door. Switching won't change that, unless you're trying to argue that some people (who, presumably, want to win) will knowingly pick a door that was revealed to have a goat behind it, in which case the premise isn't rooted in reality.

stayorswitch.com/
Is this game rigged ?

Group A:
1 door

Group B:
999 doors

The guy opens 998 doors from group B and says
"you have a chance to swap from door A to door B"

Which do you do? Swap to B, because it's probably in the biggest group.

It works like this;

The prize is ALMOST CERTAINLY in group B, and for each door that opens, the probability increases.

Once there are only 2 doors left, it isn't 50/50, because you can't dismiss the probability of it being in the larger group (B).

It's not the probability of being behind the only door that's left in the largest group vs the smallest group (50/50), it's the probability that it was PROBABLY in the biggest group from the beginning, i.e. there were more chances of it being there.

>if you picked a wrong door
See that's why it's bullshit, you don't know this.

Okay I made a simulation to confirm that you're all retarded and it turned out that you were right. I swear last time I checked this didn't work like this, this must me the Mandela effect. You're all still retarded for being unable to properly explain it to me. Sage.

Christ, are you people this daft?

You don't know this (picked a wrong door), but what you DO know is that it's more likely that you did ( 2/3 chance vs 1/3).

It's basic fucking probability. Jesus.

>Heads or tails
>1/3
nope

Don't be silly, it's still yes or no.

> I have the combined intelligence of a small yeast population fermenting on a month old box of cereal
> You are the retarded ones for not being able to explain a concept beyond my intellect to me

No.

>java
Opinion discarded.

It's 50/50. It's a test to see how easily gulled you are by appeals to authority.

>Heads or tails isn't 50/50 because coins have been tossed in the past.

Or to put this the other way:

You pick 1 door from 1000. There's a 1/1000 chance that you have picked the door with the prize.

998 doors are eliminated, which are guaranteed NOT to have the prize in.

So there are two doors left - the one you picked that has the 1/1000 chance, and the one that now has the 999/1000 chance (because 998 non-prize doors were removed).

There's 2 doors, it's 50/50. It's just an appeal to authority.

Why complicate things?

Group A:
1 door (1 chance in 1000 to contain the prize)

Group B:
999 doors (999 chance in 1000 to contain the prize)

>which group is the prize probably in?

>open door #1 B
>not there
>it's still probably in group B
>open door #2 B
>not there
>it's still probably in group B
>...
>open door #998 B
>not there
>it's probably where?

>B

I'm sorry that you are unable to understand.

It's just mathfags getting extremely autistic over a very simple problem. The scenario with 3 doors, it never changes your probability, its always 50/50 with wrong and right door. If you add more doors, then yes it makes sense to switch.

But with 3 doors, its 50% and doesn't matter if you switch or stay, despite you have a 1/3 chance to pick the right door to start off with.

See, now it just sounds like you're making shit up.

No, now there are 2 doors. It's 50/50, you don't know anything else.

>its always 50/50 with wrong and right door
Except it's not.

>i know better than mathematician professors, statistical analysis and proof
t. Sup Forums pro

You're just appealing to authority, there are 2 doors, it's 50/50. You don't know anything else.

>You're all still retarded for being unable to properly explain it to me.

O_________O

you probably picked a goat in your first pick cant you understand that? you had 66% of having a goat behind the first door you picked

the host removed the other goat so now you have 66% chance of winning if you switch

No. Say there are 5 doors. You pick one. 3 doors are removed. The initial door you picked had a 1/5 probability of being correct. The probability does not change just based on the fact you are now left with two options. It is more likely that you picked a wrong door at the start. Not guaranteed, but more likely.

Just physically draw it out and you'll understand it.

see
And maybe you'll understand how you are mistakenly applying that rule to what are actually two completely separate probabilities.

It is precisely because the 1/3 chance of the first door being correct is INDEPENDENT of the 1/2 chance of the remaining unopened door being correct, that the odds of the first door you picked being the right one (choosing to stay) do NOT change.

Meaning they do not fucking become 50/50. They stay 1/3. They remain a lower chance than that of the remaining door.

The first time I was presented with this problem I was making the exact same "safe bet" mistake that you are, in using the "past events don't influence future probabilities" rule wrong. It applies here as always, but you're applying it wrong.

Think. You really do have more information about the remaining door, should you choose to make a new choice now (clearly a 50/50 situation) than you did when you chose the first door (clearly a 1/3 situation, and precisely because of the rule, this 1/3 probability DID NOT change!). That has to be reflected in the probability.

It's a matter of basic sense, you should be able to reason that this is true and if you actually play the game with someone you will see in very few matches that it's true.

Meanwhile, claiming that it must just be an 'authority gullibility" thing is just plain rationalization and delusion..

C'mon, seriously, surely you must see it already

here you go autists.

save this for the next 10000 times this thread comes up

Basically when you pick a door, you have a 2/3 chance of being wrong, yes?

Monty always has to get rid of a goat door. That means either you picked one of the goat doors (he gets rid of the other goat door) which is again a 2/3 odd, and a 1/3 chance of your being on the car first place (at which point which goat he shows doesn't matter.)

The switch flips the odds to 2/3 win (switching from goat to car) and 1/3 lose (car to goat.)

It's easier to imagine the more you stretch the number of doors out. If there were 99 goats and 1 car, you pick a door, and he got rid of 98 of the goat doors, it's way easier to visualize why switching works.

It's not 50/50. It's been proven.

You are not able to comprehend what is happening.

You are literally separating the end scenario and placing it on it's own. If I have 2 doors, and give you a choice of 1, it's 50/50. Yes, congratulations you understand this basic principle.

But that is not the issue. This problem involves 2 groups, one larger than the other, it is not 50/50.

Do this thought experiment, or better yet if you have some sweets like M&Ms, use them.

Take a huge handful of M&Ms, put them on the desk in front of you. Imagine they are all red.

Take a single M&M out of the bag, and put it on the other side of the desk. Imagine it's blue.

Now, you are told that 1 single M&M on the table is actually a gold nugget that's been painted to look like an M&M.

I pick one of the red M&Ms from pile A on your desk and eat it because I know which one is the gold nugget but I'm not telling you.

Which pile is the gold nugget likely to be in? The red pile (29 M&Ms)?
Or the blue pile (1 M&M)?

If you had to gamble, which pile to pick from, which would you choose?

If I ate 28 more red ones, and said choose which one probably contains the gold nugget, which pile would you pick?

If you say blue, or 50/50 you are a shitposter or troll.

Fuck this shitty bait thread.
Fuck all you retards.

If you pick a door initially with a goat behind it (2/3 chance) switching gets the car.

So explain then. If there are two choices, how is it possible to list probability in terms of threes? We know for a fact that one of the three doors does NOT contain it, so we can remove it from the equation.

>Pick door 1
>Door 3 is opened to reveal goat
>Have the choice of switching options.
>Assume that we want to win the car, and therefore will not pick door 3, knowing it's wrong.

If we switch our answer to door 2, there are TWO possible outcomes:
>The car was behind door 1, and we get a goat
>The car was behind door 2, and we win a car.

Therefore, there is a one out of two chance of picking the car by switching. Explain to me how my reasoning is faulty.

Finally someone explains it in a non-retarded way. Thanks, leaf bro.

The goat is behind one of 2 doors.

>>Pick door 1

YOU PROBABLY PICKED A GOAT BECAUSE THERE ARE TWO FUCKING GOATS SO SWITCHING HAS A HIGHER CHANCE OF WINNING NOW

>>Door 3 is opened to reveal goat
>>Have the choice of switching options.
>>Assume that we want to win the car, and therefore will not pick door 3, knowing it's wrong.
>If we switch our answer to door 2, there are TWO possible outcomes:
>>The car was behind door 1, and we get a goat
>>The car was behind door 2, and we win a car.

thanks Germany-bro

making a tree of all the possibilities is the easiest way to end any probability related argument

2 doors on the left is group A
1 door on the right is group B

Currently, the prize has a 66% chance of being in group A.

I open a door on the left, it's empty.

There's a 66% chance, and always has been, that the prize is behind a door in group A.

If there are 3 groups, A B C, and you have to pick a door, it's 33% chance that you will be correct.

If there are 3 doors, all in the same group, you have 100% chance to pick the prize.

The problem lies with people being unable to see the distinction between "2 doors is 50/50" with "Group A has a larger probability of containing the prize".

>Sup Forums struggles with basic pop-sci probability
embarassing

I guess we should email some maths professors and tell them they are retarded, then. Since obviously some guy on Sup Forums just proved maths wrong.

Its funny how nearly every paradox has something do about conditional probabilities.

Now I know you're just a shitposter.
What the fuck are you doing when you describe the first situation?

You pick door 1, and then door 3 is opened to reveal a goat. So? It was revealed AFTER you picked door 1, so your chance of being correct did not fucking increase to 50/50. Get it?

When you chose you had 2 goats and a car, your odds were 2 to 1 against you, they didn't magically raise to 50/50 just because a door was opened.

Monty ALWAYS has a door with a goat to open, that changes literally NOTHING about the probabilities involved in a choice you had ALREADY made.

What the fuck man, how do you not get this?

Spoiler: it's not everyone else in the thread that was being retarded, it was you.

To everyone else in the thread too retarded to understand it, imagine that the host doesn't flip open a door after you've picked, he just offers you the option to switch to the other two doors. If even one of the doors has the car behind it, you win. Certainly in this case it is obvious to switch, because you are giving yourself a 2/3rds chance as opposed to a 1/3rds chance.