>pick door >1/3 chance >one of remaining doors open >you picked your door with a 1/3 chance and it remains so, but now the odds are 2/3 if you switch to the last door
There aren't fresh odds in the Monty Hall Problem. It's a basic introduction to conditional probability.
3 months ago
Anonymous
All you are telling me is I should treat probability mathematicians with the same disdain as I do theoretical physicists. The starting situation ceases to matter when the first door is opened.
it's not that hard to understand
c- correct x-incorrect, picking door 1 first
door 1 door 2 door 3
c x x switching incorrect
x c x switching correct
x x c switching correct
so switching is correct 2/3 times
intuitively, if 2 has a goat, they HAVE to pick the other goat door (3), which means 1 will have the prize. so your odds are better than 50/50 if you switch
Suppose it was 100 doors, and when you chose one door they closed 98 doors and asked you if you want to switch. What's more likely, that you selected the right door first try, or that they opened every door except the one with the prize?
>having a goat and switching to a car is split into two outcomes for no reason
this is why nobody takes probability seriously
you have a goat and are switching to a car, or you have a car and are switching to a goat, there is no difference made by the order of events
the monty hall problem is a 50/50 chance disguised as a 2/3 chance for brainlets
This shit is so moronic. The moment 3 was revealed to have a goat behind it was the moment the odds of my door being the winning one became 50/50. There is quite literally no difference between picking a door at the start and picking a door now. >but muh 1/3 to 1/2 your odds are better!!
No they’re not. Yes I had 1/3 odds of picking the right door at the beginning and if I were to make my first pick after door 3 was revealed to have a goat behind it then my odds would be 1/2. HOWEVER picking a door that wasn’t 3 at the beginning changes my odds from 1/3 to 1/2 once door 3 is revealed to have a goat behind it. At that point there will be no difference whether I stick with my door or pick the other. My odds of it being the correct door are the same.
Imagine 100 doors
You pick one (1% chance it's the right one)
98 doors are revealed to be incorrect
So you think that remaining door is equally likely to be the right one as your first 1% chance?
Chances are relative, not static. So yes. Your 1% 1/100 door becomes a 50% 1/2 door when 98 doors are eliminated. the wrong doors are removed from the equation.
And yet, despite the "50% chance", 99% of people that stick with their door end up losing, while 99% of people who switch to the remaining unopened door end up winning.
Yes, moron. There’s a difference between statistics on paper and in practical function. There’s two doors left, I’ve either got the winning door or I don’t. To hell with the probability of me picking the right door at the beginning of the game, at this point in the game my odds are 50/50.
>50/50. There is quite literally no difference between picking a door at the start and picking a door now.
congratulations on being a brainlet
The correct answer to the Monty Hall problem is to switch
[...]
Why bloat Ganker with threads that can be settled with wikipedia?
>muh statistics
see above and then seethe more when you realize i’m right.
>I am smarter than what is agreed upon by wikipedia
Let's fact check wikipedia. >George floyd protests >ctrl+f peaceful >While the majority of protests were peaceful
A fish is smarter than someone who trusts wikipedia.
>There’s a difference between statistics on paper and in practical function
Sweet innocent moronic, Anon. This very problem is based on an actual gameshow. And yes, the real life data matches the on paper statistics.
>There’s a difference between statistics on paper and in practical function
No there isn't. This has been found experimentally to be the case. You Black folk always retreat to the same preset arguments when you should really just retake your remedial math class.
>50/50. There is quite literally no difference between picking a door at the start and picking a door now.
congratulations on being a brainlet
The correct answer to the Monty Hall problem is to switch
https://i.imgur.com/Zlvcpo6.jpg
Well adventurer, what's your choice?
Why bloat Ganker with threads that can be settled with wikipedia?
oh sorry I'll make a post that Ganker enjoys, ahem, oh no no no pokesissies palworld chads are eating good noooo there's trannies in the game have sex chud
This solution is arbitrary because it includes 2 goat doors into the calculation. one door does not exist, it is only there to show you the goat and thus has to impact on odds.
I hate statistic morons. A pattern emerges when you view hundreds of examples of any given scenario as a group, however, that pattern becomes useless when broken down to a singular event.
For example, you have flipped a coin 99 times and gotten heads as a result each time. What are the odds you will get heads on your next flip? Statistic tards will throw down some impossible small percentage, but the real world odds are 50% because every coin flip is a unique circumstance no matter how hard statistic homosexuals argue otherwise.
Same with the monty hall problem.
If you chose door A but could switch to doors B and C at the same time, would you switch? That's what happens when the game show host confirms there's a goat behind door C.
People who understand statistics have absolutely zero issue understanding that the odds of 1 coin flip being heads are 50/50, and the odds of 99 consecutive coin flips being heads is something infinitesimally small. Those are two very different scenarios.
In a discrete, individual game of the Monty Hall problem, switching gives you better odds.
because what is really being asked is if you picked the right door originally; the opening of doors is irrelevant because it doesn't change the odds of your initial pick. if the question was worded >the host doesn't open any doors and instead asks if you'd like to wager that the car is behind the other two doors
everyone would understand you have a better chance switching, but they get bamboozled by the door opening as if that affects anything.
3 months ago
Anonymous
But the game state changes based on additional information. If you walked up to a set of 3 doors with a written explanation of possible prizes and one was already open to show a goat, you wouldn't humor the idea of picking that door in the first place.
Guys the problem isn't a neutral coin flip
The host knows which door has the car
Think about it this way
1000 doors
You pick one
He opens 998, leaving 1
What's more likely, your door is a car or a goat?
No, anon
You had a 1/1000 chance of getting it right
He opens 998 fricking doors, leaving #455 SPECIFICALLY
This is something that actually holds up in mathematics, I'm literally taking a statistics course in university right now
>I'm literally taking a statistics course in university right now
then you have probably realised that 99% of math is literally made up with specific rules for specific cases to explain stuff that couldn't be explained otherwise
especially statistics are just a scam
I think that what the 100 door gays don’t realise is that 3 door gays look at the problem like this:
1) pick one of three doors
2) 1/3 chance to select the right option
3) one door is eliminated
100 door gays view:
4) switching gives you a 1/2 chance of winning
5) staying gives you a 1/3 chance of winning
2 door chads view:
4) you choice is wiped clean
5) a new game has commenced with only two options
6) each door results in 50% chance of winning
you're the fool who didn't ask what was behind the other doors, instead you focused on your silly mathematics in an attempt to "ouwit" me without even considering what kind of game you were even playing in the first place
>2 mystery boxes >Or a goat
I'm not actually moronic enough to think this must automatically be related to the statistics problem, simply due to the shared commonalities.
That goat is at least a known quantity and those 2 guards are armed and armored, but clearly they aren't trying to stop me from accessing those doors.
Moreover, door 1 has the fricking knob on the other side and is slanted differently when compared to the safe goat door and the door that I "chose", that may as well be a neon sign indicating that it could possibly be a mimic.
To add insult to injury, the trend of armor for the guards seems to indicate that whatever they're guarding is more likely to sexually assault a male, or at least has a preference for anal.
The context is it was part of a 1960's game show. The point was adding spectacle and fake tension for the board housewives watching it without having any production complexity. They couldn't just play a loud recorded music sting and flash some lighting effects like modern game shows do.
I rape the guard on the left with the other guard, we open all the doors, spitroast the goat, spitroast the woman again, and drive away with her in the trunk
>switching is only bad if you initially picked the car (1 in 3) >therefore switching gives 2 in 3 to win
how is it any harder than this, holy shit. please tell me none of you vote
The lesson to be extracted, never second guess yourself. An alpha gets it right the first time, if he didn't get it right the first time, then he didn't deserve the car anyway.
The Gold Ball question, I kind of get it. It isn't the simplest question asked, and even though the recourse of "b-but the w-wording is different!1!!" is still incorrect; but how are you brainlets still confused by the fricking Monty Hall Problem?
I prefered the yanderedev code thread a few days ago for this sort of thing. You had anons coming up with impressively intelligent ways to be wrong as a joke. This one's just full of people making the same argument over and over and refusing to budge, there's no skill to it.
Sad part is you can tell it's the same guy over and over, and he's going out of his way to be as aggressive as possible about it just to secure the (You).
I dunno, she's a midwit so she might know the monty hall problem and instantly choose it instead of thinking over if this really presents the monty hall problem or not
a cheeky scenario where an adventurer is presented with 3 doors and 2 guards talking to him, seemingly resembling a famous math problem but not actually declaring that it's the famous math problem thus fooling people
>The problem was originally posed (and solved) in a letter by Steve Selvin to the American Statistician in 1975. >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?
the game show stuff was just flavor text, it was made by a mathematician
i get shitposting over the portal scenario, as that can't actually be proven one way or another, but this one is experimentally and mathematically demonstrated. switching is always advantageous. that's not an opinion. you can simulate it yourself at montyhall.io; if you always switch your win rate converges to 66%.
This thread is stupid because you don't know what's behind the doors other than a goat on the 3rd door and that a guard is asking if you wanna change tour choice. My first question is this situation what is behind each door. Would one door have a price and another nothing? Would one have said price and the other a punishment. Are both door punishments. Point is that without said information the logical choice is the goat because at least you can frick the goat.
Switch, because I'm not a brainlet that had to take remedial math.
Switch because math will teleport the goat to the new door.
I choose the goat
Attack both guards.
If I die, I die.
If I win, I open all the doors.
bros i want to do unspeakeable things to the guard at the left
switch because odds are better but im not smart enough to understand why this works
>pick door
>1/3 chance
>one of remaining doors open
>you picked your door with a 1/3 chance and it remains so, but now the odds are 2/3 if you switch to the last door
Yes but in not switching after the first door opens you're picking your door again. You get fresh odds.
>You get fresh odds
It's a new choice. The situation changed when the first empty door was revealed.
There aren't fresh odds in the Monty Hall Problem. It's a basic introduction to conditional probability.
All you are telling me is I should treat probability mathematicians with the same disdain as I do theoretical physicists. The starting situation ceases to matter when the first door is opened.
>the host is gaslighting you into picking the wrong door
Yeah? no
it's not that hard to understand
c- correct x-incorrect, picking door 1 first
door 1 door 2 door 3
c x x switching incorrect
x c x switching correct
x x c switching correct
so switching is correct 2/3 times
intuitively, if 2 has a goat, they HAVE to pick the other goat door (3), which means 1 will have the prize. so your odds are better than 50/50 if you switch
Suppose it was 100 doors, and when you chose one door they closed 98 doors and asked you if you want to switch. What's more likely, that you selected the right door first try, or that they opened every door except the one with the prize?
suppose it was 2 doors, you choose door 1 and the host says that door 2 has a goat behind it, do you switch?
I would take a goat over one of those smartcars tbh
>take car
>sell
>buy multiple goats
>???
>profit!
>having a goat and switching to a car is split into two outcomes for no reason
this is why nobody takes probability seriously
you have a goat and are switching to a car, or you have a car and are switching to a goat, there is no difference made by the order of events
the monty hall problem is a 50/50 chance disguised as a 2/3 chance for brainlets
moronbro..
No, he does have you there. The option would be better presented as two negative outcomes not 3 positive outcomes. Goat pussy is ridiculously tight.
okay Ahmed..
Exactly. Like I keep telling 'em, Russian Roulette? 50/50. You either have a nice day in the head or you don't.
t. brainlet
if it helps you, you can present it as "a goat, a donkey, and a car" and they reveal to you there's an animal in one of the doors you didn't choose.
Second with left
Why is door 1 facing a different way?
open door 3 have sex with goat
I don't recall posting in this thread earlier, yet I posted something so profoundly based so it must be me.
Opinions on the aesthetics of goat pussy, lads?
Pick door number 3
Door number 3, the other two could be empty or contain something less valuable than a goat
Take the goat so I have a companion that will probably ram someone off a cliff
This shit is so moronic. The moment 3 was revealed to have a goat behind it was the moment the odds of my door being the winning one became 50/50. There is quite literally no difference between picking a door at the start and picking a door now.
>but muh 1/3 to 1/2 your odds are better!!
No they’re not. Yes I had 1/3 odds of picking the right door at the beginning and if I were to make my first pick after door 3 was revealed to have a goat behind it then my odds would be 1/2. HOWEVER picking a door that wasn’t 3 at the beginning changes my odds from 1/3 to 1/2 once door 3 is revealed to have a goat behind it. At that point there will be no difference whether I stick with my door or pick the other. My odds of it being the correct door are the same.
>t. Terrence Howard
Imagine 100 doors
You pick one (1% chance it's the right one)
98 doors are revealed to be incorrect
So you think that remaining door is equally likely to be the right one as your first 1% chance?
Chances are relative, not static. So yes. Your 1% 1/100 door becomes a 50% 1/2 door when 98 doors are eliminated. the wrong doors are removed from the equation.
And yet, despite the "50% chance", 99% of people that stick with their door end up losing, while 99% of people who switch to the remaining unopened door end up winning.
It's a conspiracy by mathematicians to gaslight people.
Yes, moron. There’s a difference between statistics on paper and in practical function. There’s two doors left, I’ve either got the winning door or I don’t. To hell with the probability of me picking the right door at the beginning of the game, at this point in the game my odds are 50/50.
>muh statistics
see above and then seethe more when you realize i’m right.
>see above and then seethe more when you realize i’m right.
"I am smarter than what is agreed upon by wikipedia and the rest of humanity"
k.
Yes.
the majority of humanity believes men can be women so i don’t really see what point you’re trying to make about their collective intelligence
>I am smarter than what is agreed upon by wikipedia
Let's fact check wikipedia.
>George floyd protests
>ctrl+f peaceful
>While the majority of protests were peaceful
A fish is smarter than someone who trusts wikipedia.
To be fair, there were a shitton of protests around the world back then, most of which were not chimpouts.
>There’s a difference between statistics on paper and in practical function
Sweet innocent moronic, Anon. This very problem is based on an actual gameshow. And yes, the real life data matches the on paper statistics.
>This very problem is based on an actual gameshow
uhmmm no it isn't, a mathematician made it up with the scenario of "what if there was a game show"
>There’s a difference between statistics on paper and in practical function
No there isn't. This has been found experimentally to be the case. You Black folk always retreat to the same preset arguments when you should really just retake your remedial math class.
>50/50. There is quite literally no difference between picking a door at the start and picking a door now.
congratulations on being a brainlet
The correct answer to the Monty Hall problem is to switch
Why bloat Ganker with threads that can be settled with wikipedia?
oh sorry I'll make a post that Ganker enjoys, ahem, oh no no no pokesissies palworld chads are eating good noooo there's trannies in the game have sex chud
This solution is arbitrary because it includes 2 goat doors into the calculation. one door does not exist, it is only there to show you the goat and thus has to impact on odds.
I hate statistic morons. A pattern emerges when you view hundreds of examples of any given scenario as a group, however, that pattern becomes useless when broken down to a singular event.
For example, you have flipped a coin 99 times and gotten heads as a result each time. What are the odds you will get heads on your next flip? Statistic tards will throw down some impossible small percentage, but the real world odds are 50% because every coin flip is a unique circumstance no matter how hard statistic homosexuals argue otherwise.
Same with the monty hall problem.
>Statistic tards will throw down some impossible small percentage
What statistic tards? The ones you made up in your brain?
That’s literally not how it works but okay.
If you chose door A but could switch to doors B and C at the same time, would you switch? That's what happens when the game show host confirms there's a goat behind door C.
That's why casinos glefully display past roulette rolls because midwits and their juju come in and think they can crack the code
roulette is deliberately completely random though, that's why no juju works on it, this is just a straightforward likelihood calculation
Congratulations on failing statistics class.
That's a failure of elementary school, not just statistics class.
Based
>moron still can't into conditional probability
People who understand statistics have absolutely zero issue understanding that the odds of 1 coin flip being heads are 50/50, and the odds of 99 consecutive coin flips being heads is something infinitesimally small. Those are two very different scenarios.
In a discrete, individual game of the Monty Hall problem, switching gives you better odds.
>more reddit pseud "math"
Why is it considered together rather than starting a new "game" once one door is ruled out?
because what is really being asked is if you picked the right door originally; the opening of doors is irrelevant because it doesn't change the odds of your initial pick. if the question was worded
>the host doesn't open any doors and instead asks if you'd like to wager that the car is behind the other two doors
everyone would understand you have a better chance switching, but they get bamboozled by the door opening as if that affects anything.
But the game state changes based on additional information. If you walked up to a set of 3 doors with a written explanation of possible prizes and one was already open to show a goat, you wouldn't humor the idea of picking that door in the first place.
It has been both mathematically and programatically proven to be the correct choice to switch.
not vidya
what the frick are you all talking about
euhh
I choose door 3 because a goat is way easier to deal with than whatever might be behind doors 1 or 2 like a fricking basilisk or a hydra or something.
It's a 50/50, so it doesn't matter if I switch or not. Trick question
I choose something different. I choose the impossible. I choose... the goat.
i don't switch because i trust my gut feeling
i don't give a shit that it's the "incorrect" answer
You Black folk haven't told me what my supposed price is
for all i know there are 2 empty doors with a goat in which case i'd take the goat
Door 3, they already told me where the goat is.
Sumata with female guard's big meaty thighs.
I don't read the instructions and open a random door
Guys the problem isn't a neutral coin flip
The host knows which door has the car
Think about it this way
1000 doors
You pick one
He opens 998, leaving 1
What's more likely, your door is a car or a goat?
50/50
either you do or you don't
No, anon
You had a 1/1000 chance of getting it right
He opens 998 fricking doors, leaving #455 SPECIFICALLY
This is something that actually holds up in mathematics, I'm literally taking a statistics course in university right now
>I'm literally taking a statistics course in university right now
then you have probably realised that 99% of math is literally made up with specific rules for specific cases to explain stuff that couldn't be explained otherwise
especially statistics are just a scam
this is the real answer
anything else is midwitt cope
my odds of winning the lottery are 50/50
either i do or i don't
so true
Door 3, I want the goat
I switch to door #3.
can't afford taxes or insurance for the car, i just take the goat
>not taking the car but opting for the hidden lump sum amount equal to the car as is required to offer
Have you considered becoming a plumber or electrician? Or a refrigeratian repairman?
>refrigeratian
Pretty sure Let's Make A Deal had a lump sum clause.
I turn 360° and walk away
Turns around, leaves
I think that what the 100 door gays don’t realise is that 3 door gays look at the problem like this:
1) pick one of three doors
2) 1/3 chance to select the right option
3) one door is eliminated
100 door gays view:
4) switching gives you a 1/2 chance of winning
5) staying gives you a 1/3 chance of winning
2 door chads view:
4) you choice is wiped clean
5) a new game has commenced with only two options
6) each door results in 50% chance of winning
Wuh oh!
No fair, the OP pic presented a false dichotomy
Such is life
you're the fool who didn't ask what was behind the other doors, instead you focused on your silly mathematics in an attempt to "ouwit" me without even considering what kind of game you were even playing in the first place
>Not an actual goat
Can I change my answer to door 2?
>2 mystery boxes
>Or a goat
I'm not actually moronic enough to think this must automatically be related to the statistics problem, simply due to the shared commonalities.
That goat is at least a known quantity and those 2 guards are armed and armored, but clearly they aren't trying to stop me from accessing those doors.
Moreover, door 1 has the fricking knob on the other side and is slanted differently when compared to the safe goat door and the door that I "chose", that may as well be a neon sign indicating that it could possibly be a mimic.
To add insult to injury, the trend of armor for the guards seems to indicate that whatever they're guarding is more likely to sexually assault a male, or at least has a preference for anal.
So I'm going to pick door 1.
You should be able to solve this.
She is not white unlike the furry with the watch
Is the watch's time compared to the clock time
I'm too lazy to look up what it would in indicate but that's the answer right
the watch is showing a different time than the clock on the wall
>33
Oy vey
Kill the dude, do unspeakable things to the woman until she reveals the correct answer, then do unspeakable things to the goat.
Pretty troll to make this thread and not post the full problem statement to the Monty Hall problem
The what problem
pseud "math"
frfr
I can hear it.
I switch to door 3. I've always wanted a goat.
Always switch, you have 2/3 chance that way.
There's no point in having three doors if one of them is always revealed to be the wrong choice, it's a 50/50 shot from the beginning.
The context is it was part of a 1960's game show. The point was adding spectacle and fake tension for the board housewives watching it without having any production complexity. They couldn't just play a loud recorded music sting and flash some lighting effects like modern game shows do.
>One of the guards always tells lies
>The other always tells the truth
Our chances went from 1 in 3 to 50:50?! That's awesome!
I rape the guard on the left with the other guard, we open all the doors, spitroast the goat, spitroast the woman again, and drive away with her in the trunk
>monty hall problem
always switch
>switching is only bad if you initially picked the car (1 in 3)
>therefore switching gives 2 in 3 to win
how is it any harder than this, holy shit. please tell me none of you vote
I voted trump
>switchgay votes
of course
what car?
So you discard reality (you chose the car) for the hypothetical (you might not have) and second guess your decision and lose? Wonderful.
The lesson to be extracted, never second guess yourself. An alpha gets it right the first time, if he didn't get it right the first time, then he didn't deserve the car anyway.
The Gold Ball question, I kind of get it. It isn't the simplest question asked, and even though the recourse of "b-but the w-wording is different!1!!" is still incorrect; but how are you brainlets still confused by the fricking Monty Hall Problem?
I prefered the yanderedev code thread a few days ago for this sort of thing. You had anons coming up with impressively intelligent ways to be wrong as a joke. This one's just full of people making the same argument over and over and refusing to budge, there's no skill to it.
Gold Ball threads are generally superior
Sad part is you can tell it's the same guy over and over, and he's going out of his way to be as aggressive as possible about it just to secure the (You).
You should be able to solve this
2/3
Will the chance be the same, if the box of silver balls is gone?
It's a guarantee that you picked from one that has a gold ball.
If you were your sister, which door would you say had the prize?
I dunno, she's a midwit so she might know the monty hall problem and instantly choose it instead of thinking over if this really presents the monty hall problem or not
But I'm not my sister.
Pick Door 3. There's a 1/3 chance I picked wrong the first time, and a 50/50 chance the guard always lies.
I will switch to Door 1.
OR
You give me both goats right now. The choice is yours.
is the "no racism outside of /b/" part of global rule 3 enforced on Ganker?
Only if you report and the jannies aren't asleep.
what if I report and I see jannies deleting normal fun threads but don't touch the posts with racist rants?
Naw, homie.
oh okay, there's no point reading the rules then
It's not enforced on any boards unless janny feels like it. Protip: He doesn't.
Can someone explain to me what's going on? Like what are the rules of this game.
a cheeky scenario where an adventurer is presented with 3 doors and 2 guards talking to him, seemingly resembling a famous math problem but not actually declaring that it's the famous math problem thus fooling people
>a famous math problem
No, it was a real-world game show bonus round.
>The problem was originally posed (and solved) in a letter by Steve Selvin to the American Statistician in 1975.
>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?
the game show stuff was just flavor text, it was made by a mathematician
Where the frick do you think the name "Monty Hall" comes from, idiot?
https://en.wikipedia.org/wiki/Let%27s_Make_a_Deal
i get shitposting over the portal scenario, as that can't actually be proven one way or another, but this one is experimentally and mathematically demonstrated. switching is always advantageous. that's not an opinion. you can simulate it yourself at montyhall.io; if you always switch your win rate converges to 66%.
That is run by mathematician with a vested interest in programming it to lie to you.
This thread is stupid because you don't know what's behind the doors other than a goat on the 3rd door and that a guard is asking if you wanna change tour choice. My first question is this situation what is behind each door. Would one door have a price and another nothing? Would one have said price and the other a punishment. Are both door punishments. Point is that without said information the logical choice is the goat because at least you can frick the goat.
Math not real nerd