The goat either IS or ISN'T behind the door. If you take out the choosing door part, it's clearly 50% chance...

I'm a brainlet myself but you 50/50 fags take that shit to a whole new level.

It's like you are a child. You see 2 doors, it must be 50/50.

Of course it is a 50/50 chance if you take out all the stipulations and variables relating to the door. But that is missing the point on purpose

...

What are you even saying? I'm confused as to which side you're arguing for because you're not doing a good job at explaining youself

then you can't read, his post is completely clear.

i'm baffled at how Veeky Forums will still argue about a problem with a solution so easy to verify with a computer and any programming language. i get it that the problem confused even mathematicians when it was published because it's very non intuitive, but reading some of the replies here i can't believe i'm on the science board

your post is literally "im so smart look at me ooohhh"
even worse than the idiots here desu

if you think understanding this problem is being smart, that says more about you than about me

>hurrr hurrrr don't call me out you're dumb
you're a stuck up faggot, and your posts are useless trash which contribute to nothing.

I was baiting. I just wanted to see the retarded 50/50 man explanations. They are usually the quality of my original post

There are three doors. The game master will shoot you in the head until you are dead and then open a door at random. If there is no goat behind that door he will open another door at random to get that goat. He then proceeds to murder that goat and grill its meat for dinner.

What is the probability of your death?

50/50 either it happens or it doesn't

the door that you did choose could have been one of the 98 doors opened, there is no difference between the rest and the one you pick. Probability (1/100), but if you choose the other, the chance of it be the right is the the one you pick in a hundred less de 98 eliminated (1/100-98)=1/2

/\
/ \
/ \
/ \
/ \
/ \
1/100 99/100
/ \
/ \
/ \
R W
/\ /\
/ \ / \
/ \ / \
1/2 1/2 1/2 1/2
/ \ / \
/ \ / \
R W R W
1/200 1/200 99/200 99/200

.../\
.../ \
.../ \
../\ /\
../\ /\
./ \ / \
.FAG
./ \ / \
./\ /\ /\
./\ /\ /\
50/50
/ \ /\ / \
/ \ /\ / \
/ \ /\ / \
. ||
. ||

They can't perceive the problem in their head, even though it has been proven to be 2/3.

Then they use mental gymnastics to fool themselves that they are actually the ones who are right

The user you're replying to isn't agreeing with you, it has been proven to be 1/2.

The probability of winning regardless of choice by playing the game a single time is 50%, yes. You either win or you lose.

If you played multiple times though, you have a higher probability of getting more wins than losses if you always switch your choice.

You should always switch

Thanks for the spacer tip.
................../\
................./..\
................/....\
.............../......\
............../........\
............./..........\
.........1/100 99/100
.........../..............\
........../................\
........./..................\
........R..................W
......../\.................../\
......./..\................./..\
....../....\.............../....\
...1/2..1/2..........1/2..1/2
..../........\.........../........\
.../..........\........./..........\
..R.........W.......R.........W
1/200 1/200 99/200 99/200

Yeah I know, why are you telling me?

>i can't believe i'm on the science board
You aren't on a science board

By who?
You?

R gets 1/200 + 99/200 = 50%
(IF you choose at random at the end)

But that was never the question. The trick to this fucking question is, that it asks the probability of a fixed strategy (Always Switch) OR (Always Stay) which skews the probability.

(It's like: How many rectangles are there? --> Cue everyone taking it as "How many rectangular tiles are in this image?" and then complaining when people getting it wrong. Ask clearer questions.)

That's true, the random process is just the initial pick. Sorry.

................../\
................./..\
................/....\
.............../......\
............../........\
............./..........\
.........1/100 99/100
.........../..............\
........../................\
........./..................\
........R..................W

Afterwards, only two doors will be left, one W and one R, so that the operation of switching doors is effectively an inversion of the result (R->W and W->R), which yields the mentioned 2/3 of getting the right door in the 3 door case.

................../\
................./..\
................/....\
.............../......\
............../........\
............./..........\
.........1/100 99/100
.........../..............\
........../................\
........./..................\
........R..................W
........|....................|....IF FLIP
....... v...................v
.......W...................R
.....1/100...........99/100

Wow - sorry to hear about your low level of intelligence, OP. Maybe don't go around exposing yourself as a dummy.

The probability is obviously -1/12

Why don't you just stop posting shit and go one with fucking the goat?

I wonder if there are actually people out there whose intuition would lead them to believe this.

Are there still people who legit don't understand?

Cause you are retarded.

>you pick a goat or a car
>he reveals a goat
>if you picked a goat, switching will get you a car
>if you picked a car, switching will get you a goat
>you had a bigger chance to pick a goat at the beginning
therefore you switch, because you are more likely in the situation where the unpicked door is a car

literally as simple as this you fucking mongoloids

>2017
>still replying to b8 threads

sage

Well. The probability for getting a car is 50% IF you pick at random all the time.

The question does not really highlight obviously that it asks for the probability of a different strategy (Always switch) or (Always stay).

If you switch you will win in 2 of 3 cases (because you will always switch from a goat to a car) and loose in 1 of 3 cases (because you picked the car at the start and always switch to a goat).

what is the practical application of this?

none, also most retards argue without using the Bayes theorem

>what is the practical application of this?
>none

the entire question is a practical application of probability, what kind of stupid Q&A is this?

a tv show

the rules of the TV show didn't allow for the contestant to change doors after the host starts opening them.

bro...

the goat ain't real

heres a novel thought: the tv show is already fictional and you can therefore make up a similar one in which the rules do allow for the contestant to change doors after the host starts opening them

or, you could make a real tv show with these rules

>practical application
>fictional
these threads get worse every year

If you flip a weighted coin, it is indeed heads or tails.
If you pick a card from a deck that has 51 blue cards and one red, it is indeed red or blue.
But the odds are not equally distributed.

It is based on a real tv show, Let's Make a Deal, hosted by Monty Hall. Hence Monty Hall problem.

> If you take out the choosing door part, it's clearly 50% chance...

Trash

no they arn't everyone understands shit like this intuitively the people that don't are dumb or trolling and hence dumb.
You may be new but this is the not the first Monty Hall thread and it won't be the last

you pick goat = you win
probability of picking goat = probability of winning = 2/3
close this thread

>pick a door
>another door has a goat
>now you have two doors with a 50/50 goat or car
>for some fucking reason you're supposed to be more likely to win if you switch doors?
Mathfags are so fucking retarded.

>Demonstrably true probability phenomenon.
>"Hurr... stupid mathfags be retards."

For those of you who need pictures and scary looking hags to explain shit to prevent your head from deflating: youtube.com/watch?v=4Lb-6rxZxx0

>>for some fucking reason you're supposed to be more likely to win if you switch doors?
If you choose to ALWAYS switch doors.

Try imagining:
There are 99 doors. One single door has a car behind it.
If you choose a door the GM will open 97 doors that don't have a car behind them. You can then choose to switch to the remaining door.

What is the probability that you win the car, if you ALWAYS switch.

It's 1/99 * 0 + 98/99 * 1 = ~98,98%

Why? If you have chosen the 1 door out of 99 you will win 0% of the time if you switch. If you have chosen one of the 98 doors without a car you will win 100% of the time if you switch.

The chance of you getting a goat is actually 100% because he places the car after you've chosen if you'll switch or not.

this is good thank you

I wrote up a python(3.6) script for you user.
1/3 :

import random as rnd

def mghtnotexist(changeChoice):

D = [0,0,0,0,0]


w = rnd.randint(0,4) #randomly select door as winner

D[w] = 1 # set random element in D to 1 which is the winnig thing.


choice = rnd.randint(0,4)
# print (D)
# print(choice)
#remove all elements besides for the choice and one other.
DTwoChoices = []
DTwoChoices.append(D[choice])
del D[choice]

iterations = len(D)-1
for i in range(iterations):
s = rnd.randint(0,len(D)-1)
del D[s]

# print("We are going throught the iterations at ", i, "we randomly selected the index ", s, "and D is now: ", D)

DTwoChoices.append(D[0])
# print(DTwoChoices)
if changeChoice == 1:
return DTwoChoices[1]
else:
return DTwoChoices[0]

2/3

def tochooseornottochoose(changeChoice):

D = [0,0,0,0,0]


w = rnd.randint(0,4) #randomly select door as winner

D[w] = 1 # set random element in D to 1 which is the winnig thing.


choice = rnd.randint(0,4)

#remove all elements besides for the choice and one other.
DTwoChoices = []
DTwoChoices.append(D[choice])
if w == choice:
DTwoChoices.append(D[choice-1])

else:
DTwoChoices.append(D[w])

if changeChoice == 1:
return DTwoChoices[1]
else:
return DTwoChoices[0]

3/3

winningrecords_dontchange = []
winningrecords_change = []
winningrecords_notgaranteed_dontchange = []
winningrecords_notgaranteed_change = []
for i in range(1000000):
winningrecords_notgaranteed_dontchange.append(mghtnotexist(0))
winningrecords_notgaranteed_change.append(mghtnotexist(1))
winningrecords_dontchange.append(tochooseornottochoose(0))
winningrecords_change.append(tochooseornottochoose(1))

winningrecords_notgaranteed_dontchange_Average = sum(winningrecords_notgaranteed_dontchange)/1000000
winningrecords_notgaranteed_change_Average = sum(winningrecords_notgaranteed_change)/1000000


print("Chances of winning without changing: ", winningrecords_notgaranteed_dontchange_Average)
print("Chances of winning with changing: ", winningrecords_notgaranteed_change_Average)


dontchangeAverage = sum(winningrecords_dontchange)/1000000
changeAverage = sum(winningrecords_change)/1000000
print("Chances of winning without changing: ", dontchangeAverage)
print("Chances of winning with changing: ", changeAverage)

# Not Garanteed one will win:
# Chances of winning without changing: 0.200017
# Chances of winning with changing: 0.199422

# Garanteed one will win:
# Chances of winning without changing: 0.200583
# Chances of winning with changing: 0.799636

These results illustrate two things.

First, that when we are guaranteed that one of the choices is a 'winner', then what is essentially happening is that the rest of the choices are "squished" result into just the correct one. It is unlikely you randomly picked the correct one. This explains why changing your choice is much better since your first choice was out of the whole set but the second choice is a "squished" set.

Second is that when we are not guaranteed one of the choices is correct it resembles our intuitive reasoning. That is, if we take out our choice from the set of possibilities, and randomly select another one, they are both equally as likely. Thus changing one's choice is pointless.

Hope this helps user!

Son. If you need whitespace: code-BB-Tags exist.

(Look into the "normal" posting forms on subboards of Veeky Forums to look if there are special rules or tags for that board or ability to upload more or different files.)

>having to code simulations to understand a simple problem than any 5 year old can solve
The absolute state of Veeky Forums

Five year old would get tripped up by shoddily written questions that don't really highlight that the problem is asking "Imagine if contestant has decided that he will ALWAYS switch.".

The problem with this question lies mainly with shitty question authors.

I mean look at this shit!
>Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

>Is it to your advantage to switch your choice?
If you decide randomly on the spot instead of choosing a strategy like "always switch", the probability of this is actually 50/50.

THIS is the bullshit this gotcha question hinges upon.

Similar:
>How many rectangular tiles do you see in this image?
AND
>How many different rectangles could you draw on these rectangular grid lines?
GETS CONFLATED TO
>How many rectangles do you see?

Those questions hinge on different "normal" interpretations of what was actually asked.

You have a 2/3 chance to pick a losing door.

One losing door is always removed after picking, leaving only the winning door if you picked a losing one initially.

Picking a losing door and switching thus always yields the winning door since it is the only other door remaining.

Thus, switching carries the 2/3 chance of having picked the wrong door into a 2/3 chance of switching to the correct one.

does this question depend upon the knowledge the person has of what had happened?

Say there's a 100 doors, 1 has been picked by the player and the other 98 shown to have goats by the host. But now a third person who has no idea of what went on comes into the scene. Would this person be playing by the first person's possibilities or would they reset making it a 50/50?

Also, if the guy chose to get rid of a door but in completely random way, would this change the probabilities or would it still be a 2/3 chance by switching?

Even if the first person decides to choose between the two doors randomly the probability will be 50%. If he decides to ALWAYS switch it will be ~66,6*%.

If the third person learns or knows that the guy is in such a situation and gets to choose doors and decides to ALWAYS switch it will be again ~66,6*%. If he chooses randomly 50%, wheter he knows about the situation or just chooses between two doors (even though the probability calculation is different).

>python
>posting raw code
openprocessing.org/sketch/480963

Fuck are you on about m8

In a given instance of the game being played, switching is twice as likely to get you the car than staying. That's all there is to it. What's this nonsense about choosing strategies? You don't have to play more than once, multiple trials just demonstrate the odds.

I mean you're on the game show once. You don't decide to "always" switch. You switch once. Which, I guess, is always, for as many times as you get to.

Twice as likely to win but still barely better than 50/50 is not an argument.

It's as good as it's going to get for this problem, which is the point. It also gets better with more doors. What do you actually want?

It matters.

Choosing randomly at the end will get you odds of 50%.

1/3 * 1/2 + 2/3 * 1/2 = 1/2 = 50%

If he decides whether or not he switches, the probability will be 66.6% for the times he switches and 33.3%for the times he does not switch. Assuming he survives being beaten by his wife's purse for being a dumbass for not switching.

you change = 2/3 car

you stick = 2/3 goat

>You either DO or DON'T get hit by lightning. It's a 50% chance.

...

I'll just leave this here.

math.ucsd.edu/~crypto/Monty/montydoesnotknow.html

>math.ucsd.edu/~crypto/Monty/montydoesnotknow.html

I just played this for a lil bit and got 6 cars out of 11 attempts without switching... so..

>I was only pretending to be retarded.

>intuitively
lol

This is an altered version of the game where you can immediately lose just by picking a door since the option exists for the car door to be revealed.

I'm well aware but that's not the point at all. Please explain to me the functional difference between "choosing the strategy to always switch" and just switching the one time it comes up because after all you're only on the game show once. Tell me why that is a relevant distinction. I don't understand the whole point of that massive post which seems to be very careful and deliberate in its wording but makes no sense. I don't see the problem with the original question.

"Switching" implies an awareness of the original choice, or at least an awareness that a choice was previously made. It's entirely different from just having to choose between two doors.

Choosing to switch is not a random choice. You don't flip a coin. Think of it like ... he learns of the game and chooses to switch before the game has even started. Doesn't even really matter if he does it on the spot.

You take a decision and stick to it, no matter what. You can calculate the odds of winning based on THAT choice. Choosing between the two options randomly has different odds.

That you only appear on that game show once does not matter for the odds.

Mind that the value of the odds will approached exactly in infinitely many games, but you could still suffer an improbable loosing streak of n loosing games where you decided to switch. It's unlikely but that's the way it is.

Switching randomly (with or without knowledge of the previous step) or choosing randomly between two doors have different probability calculations but the same odds of 50%.

You told me literally nothing I didn't already know and I'm still confused as hell as to what the fuck is actually getting at. As far as I'm concerned the original question covers the problem in its entirety and the answer is a straightforward "yes".

Also, with regards to switching, the point is that switching means you're aware that you're switching from something to something else. Even if you don't know which door was picked, switching will work just as well for you as if you had picked it yourself.

>
Basically the way the question is most of the time asked turns it into a trick question, probably by conflating some views.

It does not help that the probability is 50% if you choose randomly.

>switching means you're aware
Well choosing randomly would probably be the choice choose wehter you switch randomly.

So we have with different odds each.
* Switch
* Stay
* Choose randomly whether to stay or switch

If you play the game once you will either win or lose.

If you play multiple times, you will win more often by switching, but still have chances to lose by switching. There is no 100% win method and there is little to console a contestant who switched yet lost.

>christmas tree ascii on Veeky Forums
reported for religion

no. one of the three blue goat boxes disappears after the GM opening the first door

MERLIN'S FUCKING BEARD Veeky Forums IF THIS IS REALLY TOO DIFFICULT FOR YOU BRAINDEAD AUTISTIC FUCKING RETARDS TO FIGURE OUT THEN WHY DON'T YOU RUN A GODDAMN SIMULATION TO PROVE THE ODDS IN PRACTICE

If you don't understand why something is happening then the simulation is probably wrong.

...

Isn't there a proof that switching gives a higher probability of winning the prize? If so, why the fuck are these threads still a thing?

It's an extremely simplified form of superposition, some people can only think in terms of absolute states.