Can someone explain the fucking Monty Hall problem, I know it's

>>8438414
the prize (car) exists, total prize probability = 1

initially you choose one of the 3 doors, the probability that the prize is behind it is 1/3

Monty opens a door, revealing a goat, the probability that the prize was behind that door = 0.

1 = 1/3 + 0 + x
x is the probability that the prize is behind the unopened door which also was not the initially selected door
x=2/3

It would be 50/50 if there was no outsider influence (someone other than you knows which door has what. They open a bad door for you). It's simple, senpai.

You pick a door, there's a 2/3 chance you're on a losing door.

If the dude were to ask if you want to swap now to another door it would still be 2/3 wrong. Importantly if you were already on a goat you could still swap to a goat, it would basically like choosing for the first time.

The dude reveals which one of the other two doors has a goat behind it. Now if you swap you won't swap to the revealed door obviously so you know that if you swap doors now you'll also swap the thing behind the door. That is, after revealing a goat door you know that if you're on the car and you swap you'll get a goat, and if you're on a goat and you swap you'll get a car. It's no longer possible for you to go from a goat door to a goat door.

There's a 2/3 chance you're on a goat door still, when you swap you also swap the reward. So 2/3 of the time you swap from goat door to car door, 1/3 of the time you swap from car door to goat door. If you don't swap there's still a 2/3 chance your on a goat door.


Yes once he's revealed a door you now know there's a 50/50 chance for each door, but you have already chosen when it was 2/3 and that's what matters

100 doors. You pick a random one. 98 other doors are revealed showing 98 goats and one closed door. Do you switch to the closed door or bet that you picked right the first time?

Imagine that you don't have 3 doors, but 1000.

You choose one door, and Monty opens 998 of the remaining doors, which leaves you with the choice of your original door, or the one Monty left closed.
For me it would be intuitive to switch now, because the chance that you picked the correct door by choosing randomly is 1/1000, near zero.

That means the probability of winning the game by switching is now 999/1000.
You could generalize this for n doors, the probability of winning by switching is (n-1)/n , which leaves you at 2/3 for 3 doors.

>>8438414
The idea is picking the correct door at first pick without switching has a probability of 1/3. This doesn't change even if Monty later opened another door with a goat behind it.

Suppose you pick your door after, not before, Monty opened the dud door with a goat behind it. Of course if you only need to choose 1 out of 2 unopened door, it's 50/50. But remember, in the original problem, Monty can only open the door that YOU have not chosen. They are 2 different problems.

>>8438414
https://warosu.org/sci/?task=search2&ghost=&search_text=%22monty+hall%22&search_subject=&search_username=&search_tripcode=&search_email=&search_filename=&search_datefrom=&search_dateto=&search_op=op&search_del=dontcare&search_int=dontcare&search_ord=new&search_capcode=all&search_res=post

>>8438484
>>8438488
The thing is that 100 or 1000 doors its much more intuitive to believe that you're already on a goat door

But with exactly three doors, one picked, one disqualified and one mystery it's difficult to stomach that the mystery door is statistically better than the picked one.

>>8438414
The opening of the door is a red herring. Imagine instead that you first pick a door, and then you're allowed to switch to *both* of the other doors.

>>8438414

the main idea is that the host WILL ALWAYS REMOVE THE WRONG ONE.

>>8438540

Not really, since that exclude the possibility of choosing the correct one at first try.

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>>8438414

>>8438543
How so?

Think about it this way: You have cheated, and you know that door 1 is a goat. You pick door 1. Monty opens door 2, and reveals the second goat. You now know 100% that the car is behind door 3.

If you don't cheat, the probability that you pick a goat the first time is 2/3, not 100%. This is why you will win the car 2/3 of the time if you always switch.

Since we're talking about probabilities in this thread:
If I have to choose between
A. 5 rolls of a roulette, each one with a 10% chance of winning
B. 2 rolls of a roulette, each one with a 25% chance of winning
Which would give me more chance to win at least once, A or B?

>>8438449
you'd have a 67% chance of not killing anybody
weight is distributed over 5 people

>>8438414
It's NOT true in practice that you're "twice as likely to get the car if you switch doors" and you are right it makes no sense. This is just a myth that people tell to kids in middle school or elementary school, it's kinda funny you haven't heard or read any of the real theorems yet OP. haha.

The myth is as follows (it's pretty stupid so bare with me):

You make a guess and there's a 1/3 chance it's right.

They eliminate an option and ask if you want to pick again.

The "1st grade level" paradox is that they want you to think that you picked from a 1/3 chance initially and if you were to switch doors it would be a 1/2 chance now because there's only two doors and your first choice was out of three.

Anyone understands that by sticking with your initial choice it's the same thing is being presented with the 1/2 chance initially and just picking any option.

Switching doors DOES NOT change anything in fact check out the simulators, probability analyzers as well as the papers debunking this elementary myth.

It's quite funny actually.

They figured out that elementary schoolers aren't intelligent enough to see through the myth so they use it a lot in movies and shit.

>>8438414

Stick with your choice because he is trying to trick you into giving up the goat, which is better.

>>8439129
This is correct. If anyone didn't agree with this then I'm never coming to /sci/ again.

Think of it this way: You have a 1/3 chance at the beginning of picking the door with the car, and a 2/3 chance that you didn't. When the host opens the door, it does not change the fact that the car has a 2/3 chance of being in a door you didn't pick; what it does is "collapse" the probability down to a single door, the one you should now switch to.

Alternatively, imagine taking the problem to a ridiculous extreme. Instead of 3 doors, suppose there were a million doors, 999,999 of which have goats.You pick one, and the host opens 999,998 of the unselected doors to reveals the goats, so that only two doors remain; the one you chose, and the one you can switch to. For the car to be in the door you first chose, you needed to make a 1 in a million guess. Clearly, it's best to swap.

Why is there more than 5 replies to this thread? This puzzle is very basic.

>>8438540
>The opening of the door is a red herring
NO
The opening of the door is information.

Relabel them as 1 terrorist and 2 hostages, one of which you have to shoot. You don't think him opening the door makes you feel better?

>>8439152
I didn't agree with it

guess you have to leave

see you tomorrow.

>>8439290
Maybe my wording wasn't clear. What I meant was that the original game is equivalent to the following game: You pick a door, and I let you switch to both of the other doors (i.e. you win if the car is behind either of the other two doors). Obviously this increases your chances from 1/3 to 2/3, and having one the goat doors open doesn't change anything, because you already knew that at least one of those doors had to have a goat behind it before it was opened.

Cant u just fucking apply bayesian theory to solve this

https://betterexplained.com/articles/understanding-the-monty-hall-problem/
There is also a simulator with as many doors as you want, so the effect is bigger.

>>8438414
Dude moved the car and goats around behind the scenes to make it less likely that the person would get the car.

Not really related to the math problem, but, hey....

>>8439460
you can even solve it with some basic tree diagrams and counting then adding up the outcomes.

Contrary to popular belief, I will disagree with the answer that switching gives car 2/3 times.

Switching door has the same chance of getting car, as not switching. For any number of doors, the answer is the same – 1/2 chance of correct answer.
The final problem is known before the first pick even happens.

Given the definition of problem:
There are N initial picks.
N-1 picks are wrong.
2 picks remain for the final problem.
N-2 picks are irrelevant.
Irrelevant picks are always removed from the final problem. I.e. the correct pick is always left in the game.
In the standard version of the problem, N is 3.

Given these facts, can decide which 1 out N picks will be carried to the final problem.
However, there will also be 2nd pick carried to the final problem. If initially our pick was wrong, the other pick will be good. Otherwise it will always be random.
This gives us 2 out of N picks with 2/2 – 100% – probability of the good pick being carried to the final problem.
In detail, the pick is either {1/N, 1/1} or {1/1, 1/N} which are exactly the same set, the only difference being who decided on which pick.

The final problem is bet which of the 2 picks is good. Here we always have 1/2 chance of making good decision – be it either switching or not.

This answer applies to the standard Monty Hall problem, where all wrong picks – but one – are always rejected, and the option to switch is always present.

>>8438422
>implying the probabilistic space didnt change

>>8439264
It's a meme thread. Like P=NP threads.

Lrn2wiki fgt pls
https://en.wikipedia.org/wiki/Monty_Hall_problem

>>8440233

nah, it's >>8438548

>>8440233
You can visualize it yourself if you dont believe it.

Lets say you pick door 1 and switch after the moderator opens a door. Lets also say there are two possible ways the Monty ones to play, first he always wants to open door 2, second he always wants to open door 3.

This would result in Monty opening 1 out of 3 times Monty opening the door with the car for you. He will not do this, and therefore giving you a hint that these 1 out of 3 times that the car is definetely behind the other door. Of course you dont know when exactly Monty actually wanted to open door 2, but didnt do so because door 2 is the car. This is why basically you pretend he ALWAYS gives you the hint. This will end up with you being right 1 out of 3 times that Monty has given you the hint. So you have 1/3 chance of getting right yourself, and 1/3 chance that Monty has given you the hint if you always switch, which ends up with 2/3 probability to win it if you always switch.

>>8440233
This is good reasoning, but you overlooked one important fact.
{1/N, 1/1} and {1/1, 1/N} are not exactly as simple, and the correct values make the two sets unequal.
You even mentioned yourself that they differ by who made the pick, and it's where the probability difference lies.
Incorrect player choice has N-1/N probability.
Correct player choice has 1/N probability.
The 2nd choice added by host depends on player pick and is always the same, so it's not taken into account.
This gives us N-1/N vs. 1/N chance of switching being good vs. bad option.

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Here's how to calculate the odds, the numbers in the blue cells match the standard tv game,
you can fiddle with the contestant selecting several doors, with the total number of doors, and with the number of goats Monty reveals

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>>8438488
this

>>8439129
this, use the fucking Bayes theorem. don't say something asinine like 1000 opened doors.

>>8440319
What fucking use is that. Bayes works, but gives no intuition. Read the OP's question.

Let's reformulate the problem:
This time monty opens at random one of the two remaining doors, it's a goat.
What is the prob of the two closed doors?

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>>8440323

>>8440357
Reminds me of the old quip that a camel is a horse designed by a committee.

>>8440357
fail

>>8440378
thanks for pointing out the mistake

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https://www.youtube.com/watch?v=5d_3IEofXfY

>>8440380
Sorry. Allow me to point it out.
*points to you*
Glad I could clear that up for you.

Probability that you initially pick a losing door: 2/3.

Probability that the host has only one possible losing door to open (meaning the unselected door is a winner): 2/3

Probability that you're initial guess was correct: 1/3

Creating this thread on /sci/ for the 1000000th time? Priceless.

>>8438414
look up bayes theorem

>>8440384
seems like we need /pseudosci/ for people like you

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>>8438414

>>8438414
>Can someone explain the fucking Monty Hall problem

Bullshit math-pop made to make pure mathematics look relevant.

>>8438484
YOU BET 50/50 YOU FAGGOTS

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Practically the door was always behind either 1 or 2. If you choose 1 or 2, then the probability of you winning the car wouldn't change if you switched your pick between them.

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>>8438464
This is a nicely worded explanation.
Well done, anon.

1 person goes to take a medical test that is 95% accurate for a disease that affects 1 every 5000 people, what is the chance of the person being sick if the test came back positive?

>>8441325
hint: it is neither 1/5000 nor 19/20

well like this

choose 1 door or choose 2 doors

All of you fags act like you "get" it.

Well riddle me this: What if Monty reveals the door with the goat randomly?

>>8441325
19/5018=0.378%

>>8441325
>>8441370
19/20 + 1/10000

>>8441420
19/5018=0.379%

>>8441420
>>8441426
right

>>8441423
wrong

now try to get 2/3 of switching doors using the Bayes theorem

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now tell that popsci people didn't fall for this meme.

>>8438414
simply because your chance of selecting a goat to begin with is 2/3

>>8441835
this image makes me want to barf

>>8441835
You showed that if the prize is not behind any of the first n-2 doors, then it's equally likely to be behind either of the two remaining doors.

>>8441845
j =/= z,y

>>8441850
Yeah, j indexes the first n-2 doors and Z and Y are the last two. Your point?

>>8441852
the problem does not imply a natural order of the doors.

>>8441862
>Door i

This is an ordering on the doors

>>8439085
B is about 3% better

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>>8438449
>door opens to reveal 5 people
>drive trolley through open door
>guaranteed high score

>>8438414
To me this only intuitively made sense when it was explained in extreme.

Imagine a million doors with just one car behind a door. You pick a door at random, and all the other doors that aren't the car go away leaving only one left. Intuitively, would you stick with your initial choice or switch?

>>8441393
Then he would reveal the car in 1 out of 3 cases. Then obviously you would switch, because he just showed you the car. This actually makes sense to explain switching is always better. You have a second, now way better shot at getting the right door.

>>8438414
>You have a 66% chance of picking a goat door.
>66% chance of being on a goat door means 66% chance of not being the prize door
>66% chance of one of the 2 other doors being prize door
>Other goat door is shown.
>probability is still the same since you made your choice with 3 unknowns, 66% chance shifts to that one door alone.

There is a 1/3 chance the initial door you chose has the prize, that never changes since you made your choice with 3 options. However when they open a door, the remaining 2/3 chance is shifted to the closed door.

if you pick a door and stick to it, did your chance of winning suddenly rise because Monty has opened a door? No, it is still 1/3. Now logically speaking, if not swapping leaves you with your initial 1/3 chance, then what is the chance of swapping? All probabilites need to add up to 1, so its 2/3.

The most important thing to keep in mind, which is not stated, is that Monty knows where the prizes are.
In the situation (1/3 of the time) where the player selects the correct door initially, he can simply open one of the remaining doors at random.
In the situation (2/3 of the time) that they initially pick a goat door, he will pick the door, he will pick the other goat door 100% of the time, to ensure that a choice has to be made.
So you have a 1/3 chance you guessed right, and a 2/3 chance you guessed wrong, and Monty just limited your other options to the correct answer.

>>8438548
If you're given situation 2 is the probability of choosing the right door 1/2?

>>8440233

with the effort it took you to write up this shite you could have just programmed a quick simulator to see if your asinine ideas were correct or not

>>8438414
What changes is that a guy that knows the answer gives you a clue. The first door you pick was a 33% chance before, and you haven't gained any information to change that. The other door is one out of two possible, so it's a 50% chance.

>>8439129
as no one in this thread that says 2/3, as posted a proof using(or tried to), defined the events, set theory and Bayes theorem, it is clear they say it out of a dogmatic position.

I didn't believe it at first, but I mapped it out and sure enough, it's correct

possibility 1: you pick the door of goat 1, he opens the door of goat 2, the car is in the other door
possibility 2: you pick the door of goat 2, he opens the door of goat 1, the car is in the other door
possibility 3: you pick the door of the car, he opens the door of goat 1 OR goat 2, the car is in the door you chose

the probability of choosing a goat door initially instead of the car door is 2/3, and he will always show you one of the goats regardless of whether you pick the car door or not
the trick here is it makes you think that the choice is between whether your intuition is right or not, which has the illusion of being a 1/2 chance, however when you consider the problem from the start, the ONLY way the door you picked is the car is if the goat door he opens to you is one of two remaining goat doors rather than the one remaining goat door that you didn't pick

>>8439129
>>8439152
>>8442880
>Choose A:
P(A) = 1/3
P(¬A) = 2/3
>B is opened (it's a goat):
You've gained no new information on what's behind A. P(A) remains 1/3. P(¬A), therefore, remains 2/3. Since you're down to two choices, probability of C is 2/3.

Please point out where I went wrong.

>>8442896
>You've gained no new information on what's behind A. P(A) remains 1/3. P(¬A), therefore, remains 2/3. Since you're down to two choices, probability of C is 2/3.

exclusive events =/= independent events

>>8442927
Sorry, I don't see what you're implying. What's independent about this scenario?

>>8442953
>B is opened (it's a goat):
>You've gained no new information on what's behind A.

this 2 lines imply independence of events A,B and C, that is not true. following you same posting structure.
>B and C are opened (it's a goat on both doors):
You've gained no new information on what's behind A. P(A) remains 1/3.

and this is clearly not true.

>>8442973
If you choose A, whichever door is opened is going to be a goat. That's part of the initial constraints of the problem. The host of the show knows what's behind each door and opens one that is neither your initial choice nor the car. If you picked a goat door, he'll open a goat door. If you picked the car door, he'll still open a goat door. You've learned something, but you haven't learned anything about what's behind the door you chose.

>>8442814
Those aren't independent situations 1. and 2. ,
2 comes after 1

at 1. the contestant has chosen door #3
at 2. Monty reveals the goat behind door #1

first choice=goat1 -> shown=goat2 -> unknown=final choice=car
first choice=goat2 -> shown=goat1 -> unknown=final choice=car
first choice=car -> shown=goat1or2 -> unknown=final choice=goat1or2