why does swapping gives a 66% chance not sit right with me
there HAS to be more to this
what is missing
Try to imagine a slightly changed version:
1) There are 100 doors
2) The prize is behind 1 of them
3) You choose the door
4) The host removes 98 doors that surely don't have the prize behind them and asks you whether you stay with your choice or switch to the remaining door that's left
Now the choice seems pretty clear and it's easier to see why it pays to switch.
It becomes easy to understand once you realize that the showhost always know which the correct door is. First you have 1/3 chance, meaning you're more likely to pick the wrong door than the right one. The host then opens a door he knows does not contain the prize. Then, seeing as how its more likely that your first guess was wrong then correct, it's more likely that the prize is in the other door, so you should swap.
It doesn't matter whether it's minority or majority, the only thing that matters is that the host has removed the door(s) that surely don't have the prize. If your first choice was correct, you get the prize by staying at your choice; if your first choice was wrong, you win by switching. It's much more likely that your first choice was wrong. (99 times more likely in the 100 door scenario, but twice as likely in the 3 door scenario)
It's still 50/50.
Anyone who says it's 66% is falling for the gambler's fallacy. You simply can't look at an entire group of results to infer the probability of a single occurrence.
i think the same way but results dont lie
in the indepent event, is it REALLY 50/50 the results show 2/3 win if you switch every time
is it really just a huge coincidence? i dont think so
but i think something is missing
Nope, see >>26393875
Your first choice is more likely to be wrong, and if it's wrong then the prize is in the door you'll be swapping too, thus swapping is the best choice.
so that means all the people parroting the solution fail to take into account this?
in practice, it would be 66.6% because thats what experiments say
the accepted answer, not the correct explanation. science is about contesting accepted answers
Because a percentage chance of something is meaningless. Whether or not you have a 66% chance of winning the lottery or a 0.00001% change, there are still only two outcomes. It doesn't matter if one is more likely than the other, because there's still a chance you won't win.
The experiments themselves say switching doors improves the odds. The switching itself does nothing, especially since they usually only look at the second half of the problem directly and fail to take into account the first half of the problem. Usually the immediate wins.
To expand, I've seen several experiments with this. Most of them are always weirdly worded regarding the first half of the experiment and involve wording like, "Take it on faith that," or, "If we assume it's not the winning door..."
End of the day you calculate the probability of each step individually. The rest isn't a true representation of the chances you are taking.
Here's an explanation even a retard could understand:
Dont think of it as 3 cups/doors but as 2 groups: The one you picked and the one you didnt pick.
The probability of your group being the winner is 1 out of 3 = 33%.
The chance to win for the second group are 2/3 because it contains 2 doors/cups.
Now the host revelas one door from group B is a looser.
And gives you a choice to switch.
What the host in essence did is to allow you to switch from a low probability group to a high probability group.
In Group 1 you get 1 door , in group 2 you get 2 doors, one of the doors from group B is just being opened by the host for showmanship and to mislead you.
I'm saying that increasing your chance of something from 50% to 66% doesn't really mean much. Sure something is technically more likely, but there's no way of knowing if your 50-50 guess is objectively better or worse than a 33-66 guess.
SOMETHING IS MISSING
Let me break it down for you, again. Roll a di twice and try to get two sixes in a row. Most people would assume that your odds are 1:36 to roll a 6 a second time. It's not. It's still 1:6.
Your perception of the problem is clouding the issue. The same here. The odds for the second half of the problem is still 50/50.
It's 2/3rds because if you pick the car (1/3rd chance), switching means you get a goat. If you pick a goat (2/3rd chance), switching means you automatically get the car because the other goat is revealed and discounted from the equation.
67% chance of winning, 33% chance of losing. Depends on what's behind the door you pick.
two ways to think of this
then 1/6 again
meaning you're gated by two 1/6's
fucking retard normie now resorting to fallacies because i dont accept his parroting
you fail to understand that the door selected by the host IS NOT RANDOM
if it was random there'd be 1/3rd chance of car and the game could end right there but the host ALWAYS selects a goat and only AFTER you've chosen a door and NEVER selects your door
your first selection has a 1/3 chance of being right
once one of the goats has been removed the door unselected by either of you only has 1/3 chance of being WRONG because the car can never be selected and there's only 1 goat left and the host CANNOT open your door
thus changing doors gives you a 2/3rds chance at being right instead of your original 1/3rd selection
t. poster with actual ~130 range IQ
>everyone arguing about a theoreticall correct answer
It is more a case of how you approach the question as to which answer is right.
Statistically speaking, if an incorrect door is opened both doors have a 50% chance of being the car.
That does mean that if you change the door the odds of the door as you pick it have increased (as the door you pick will have been picked at 1/2 rather than 1/3), but the chance that the door is the car is not any higher.
The whole question is cleverly worded to have different Interpretations.
my iq is 127 im just gonna discredit what you say for doubting me
since you doubt me
ill just doubt you
Get fucking rekt by your own logic nigger
>The whole question is cleverly worded to have different Interpretations.
Exactly. The original problem poised to Marilyn vos Savant was much more clearly worded. Then idiots got ahold of it and turned into a, "What happens if a sun of fire and a sun of ice collide?" troll.
m8 you are retarded
it's 2/3 and this has nothing to do with science but it's high school mathematics
you have a 1/3 chance of being correct the first time so when he tells you that you can either keep your choice or open both the other doors it's obviously correct to open both (aka switch)
get it through your thick skull
It's still 50/50 because you either win or you lose
>I didn even read the thread bro
and you wondering why im calling you a fucking nigger?
its because you are the definition of nigger, ignorant person
by choice you chose to not read the thread, niggerlord
>just read your post and owned you with own logic
>durr he obviously cant read
>oh wait he read and replied to that post
>and did it again
>oh look he did it again
>damn he can read
>Under the "standard" version of the problem, the host always opens the door and offers a switch. In the standard version, Savant's answer is correct. However, the statement of the problem as posed in her column is ambiguous.
This shit has been a troll since before 4chan existed. Shush newfags.
no, this isn't true at all
it's not ambiguous, it's not poorly worded, you don't understand it
there is zero way the chance of the second choice is ever 50%, because guaranteeing a car and a goat are left as 2 choices in the second phase means the door removed in the first phase was not random
even if the choice in the first phase is random the probability in the second phase is not 50% because there was a 33% chance the right choice got removed in phase one
PROBABILITY ASSUMES RANDOMNESS EXISTS
RANDOMNESS DOESNT EXIST
>you have to read this thread to fully understand grade school maths
Yeah fuck off
If you cant grasp the difference between picking 2 door instad of 1, and that the rest is just a fucking show designed to fuck you up, you are beyond help.
she is correct that it's always correct to switch and the only "ambiguity" is if the host always offers the switch or not which you have to misunderstand on purpose not to get
and most of the retards not accepting this like you and OP are too stupid to even realize that
Nope that's not how it works
If you pick one of the goats, the host is forced to select the other goat door to open. Which means that switching has a 100% chance of causing you to win. There is a 67% chance of selecting a goat, so there is a 67% chance of winning. Like the other guy said, the door being opened by the host isn't random, it's determined using previous information.
If you select the car, which you have a 33% chance of doing, the host can open either of the other two doors, but both doors are goats so if you switch it doesn't matter which door you get, meaning that if you pick the car you have a 100% chance of losing if you switch. You have a 33% chance of selecting the car, so therefore a 33% chance of losing if you switch.
If you don't switch you're left with 100% - 67%, so that'd be 33%
The addition of new information is actually really consequential in this problem because it's based entirely off of previous information which is decided by different probabilities than just 50/50 or 33/33/33
>nothing I wrote
Good job, anonymous. Surely your towering intellect and grasp of the situation (such as posting anonymously) is ironclad. It's a good thing you read MY part of the post and not just my citation.
talking about newfags is not relevant to mathematics
I have a fine grasp at the situation, I'm finishing my BS in mathematics and have taken courses in probability theory in which this problem was discussed
WATCH THE FUGGING VIDEO
I think the explanation in the video is simple enough.
>I pick a goat, guy picks another goat, I change, I win
>I pick another goat, guy picks the previous goat, I change, I win
>I pick the car, guy picks a goat, I change, I lose
Now let's see if I don't change:
>goat, goat, don't change, lose
>goat, goat, don't change, lose
>car, goat, don't change, win
It's the choice you made at the beginning that changes the probability, because which door the host is forced to open is based on that choice
If you chose the car, switching is obviously bad
If you chose the goat, the other door isn't going to be a goat because the second goat has already been revealed, so you know you're going to get the car.
1/3rd chance of picking the car at onset
2/3rd chance of picking the goat
>1/3rd chance of picking the car at onset
Which, if you switch, leads to losing
>2/3rd chance of picking the goat
Which, if you switch, leads to winning, because the other door is always going to be a car
The Monty Hall Problem is actually a great example of how men constantly underestimate women.
After Marilyn vos Savant wrote about the solution to the Monty Hall Problem in her column, she received over 10,000 letters, many from noted scholars and Ph.Ds, informing her that she was a hare-brained woman.
Oh and those IQ tests that you robots love so much? Marilyn said such attempts to measure intelligence were useless, and she rejected IQ tests as unreliable. She who scored a staggering 228
You pick a door with a 1/3 probability of being correct. Host removes one and you now have two options total. If you swap, you're wagering 50-50. If you stay, you're wagering 1:3 despite one of the choices now having been eliminated
>a great example of how men constantly underestimate women
More like an example of how easily autists can be triggered, and how many scholars are autists. Particularly when an answer is not as clearly explained as it can be.
I didn't realise that Vos Savant was a woman until about four minutes ago though.
I thought that her name was a guys name (because of marilyn manson).
At least I can knowingly claim to be the dumbest person in the thread.
>there's no way of knowing if your 50-50 guess is objectively better or worse than a 33-66 guess.
It is easy to know why the second one is objectively better. That's because 0.66>0.50.
Let's try with another example.
What's the probability that a thunder will strike you in the next second? There are only 2 options (will strike/will not strike), so it's 50-50. Correct?
Let's say it's 100 doors. You play the game a few times. Are you saying if you stay with your door every time you'll win 50% of the games? No, you'll win 1% of the games. The initial guess you're only gonna hit the car 1/100 times.
nice false equivolancy rofl kid
comparing a thunder strike to a choosing between 2 thigns?
how do u think of this shit
no shit because you chose between 100 doors, not 2
Yep, I'm comparing a thunder strike to choosing between two things. There are two options in both cases, so if you think that 2 options -> always 50-50 chances, you should answer easily.
Here is a shitty explanation I drew up.
The key concept is:
IF YOU DIDN'T PICK THE CAR BEHIND 'YOUR DOOR' SWAPPING WILL ALWAYS WIN.
See the arrows.
you just asked me a loaded question and used false equivolancy and put in meme logic
the previous round has no implication on the current round
what if you didnt swap and you pick the door with the car?
The gambler's fallacy would only apply if he randomly picked the door.
The question clearly states that he will always knowingly pick a door with a donkey.
Look at the diagram.
I can't make it any clearer than this.
I failed first year maths at university and even I understand this.
You must be clinically retarded.
how am i the stupid one when you're using gamblers fallacy?
in this independent event you just told me theres 2 doors and one has a car and one doesnt
if im on the door with car and i switch 0
if im on the door with the carn and stay 1
if im on the door without car and switch 1
if im on the door without car and stay 0
when he removes the door, it boils down to only 2 doors, not 3, the third door is no longer relevant
The point of the problem is that the host knows where the car is. He will never open the door with the car in it. If he opened random doors then it would be 50/50, but he doesn't. Therefore it's not 50/50.
i have a computational mathematics degree and i dont understand this
because it is wrong
if he openened random doors yo ucould be in a scenario with 0% but thats not the case
BECAUSE you kNOW he knows where the car is and WONT get rid of the car
you are put in 2 scenarios
2 doors where you chose the car
2 doors where you chose the goat
both have 2 options
both have 2 answers
last explanation here because either you're trolling or simply too stupid to get it
there are 3 doors, A, B and C
the car is hidden behind door C
if you guessed A first the host reveals B and you should switch
if you guessed B first the host reveals A and you should switch
if you guessed C first the host reveals A or B (note that you don't count this twice) and you lose by switching
2/3 cases you win
Just think about it. Think about 100 doors. It's easier that way. You have a tiny probability of picking the car. You have a huge probability of not picking the car. Just because the host opened some doors doesn't change that probability.
Draw yourself a diagram.
When I first heard of the Monty Hall problem I basically sketched what I just did before.
It's how I understood it.
Also good to know they just give away degrees like that. I might try and get one.
why dont you understand that he removed a door so there is not 3 in the denominator
yes it does, because if he opened 98 doors then you're sutck with goat door and non goat door
why do i need a diagram to figure out that when you have 2 options and only one is correct that you have a 50% chance
>why do i need a diagram to figure out that when you have 2 options and only one is correct that you have a 50% chance
Because you ignore the first half of the problem.
If you have picked the car, swapping will always lose (because you will be guaranteed to be swapping to a donkey).
If you have picked a donkey, the other donkey will be removed (by the host) and you will always win by swapping.
You are statistically more likely to have picked a donkey than the car in the first part.
nevermind the number of doors idiot, what matters are the cases
since you pretend to have a meme degree just code it up in python
it will show you the correct result given a 1000+ trials
2/4 scenarios result in you getting the car
picking car !swapping = true
picking !car swapping = true
picking !car !swapping = false
picking car swapping = false
But the chances of you picking the car are one in three.
(1/3 chance) picking car !swapping = true
(2/3 chance) picking !car swapping = true
(2/3 chance) picking !car !swapping = false
(1/3 chance) picking car swapping = false
If you didn't pick the car (which you are statistically more likely to do) then swapping is a better option.
the previous round has no implication on the current round
It has nothing with to do with gamblers falacy.
Imagine you pick the door with the car which happens 1/3 of the time. In that case switching looses.
If you didn't pick the one with the car which happens 2/3 of the time, switching wins.
It is very counter intuitive because if you switch winning doors become loosing doors and loosing doors become winning ones.
and it changes to 1/2 when you remove one of the choices
3 is the denominator
1 is correct
1:3 ratio or expressed in fractons 1/3
>guy removes door
the choice CHANGES
It's not hard at all, guys.
Let's say that I'm a game show host and you're a contestant, but this time there are 100 doors. You pick door number one, and you have a 1% chance of being right. I know which door has the car, and open the other 98 doors. So, there's a one percent chance that your initial guess was right, and a 99% chance that the door I left closed was the right one.
Let's say that you decide to stay with your initial door - what are the chances that you're right? If door one is right, and I open any other door, then you win. If it was door two, I win. Door three, I win. Door four, I win. Door five, I win.... Door one hundred, I win. See? You're basically selecting between taking your chances with your one initial door, or every other door minus the first one.
But with just three doors, you have a one in three chance that your first guess was right, and a two in three that switching is the best thing to do.
Numberphile has a good video on this.
You play 2 different games.
1. If you decide to switch you have 2/3 to be landed on a door with a goat. So you win 2/3 of the time.
2. If you don't decide to switch the probability is 1/3.
Does pretending to be mentally retarded get your rocks off, or what?
> 100 doors
> Car is behind door 50, but you don't know that.
> you choose door 38
> doors 1-37, 39-49, and 51-100 are opened
> you decide not to switch.
> You lose
Now replace 38 with every other number. You will only win if you choose 50 on your first pick. You will lose if you pick any other number if you decide not to switch.
I'll try to explain it slowly for all the normies in here.
What this problem does is turn 3 choices into 2.
On the surface, it looks like your choices are:
When in reality, you're choosing:
>Door 2 OR Door 3
Now, I know what you may be thinking:
>But wait! The wrong door is revealed either way, so isn't it 50/50?
And here's the trick: The information you get from the incorrect door changes depending on whether you switch.
If you decide to stay, what the door really tells you is:
>One of the other two choices was wrong
Which you already knew when you made your 1/3 guess. Essentially, this information means nothing.
But if you switch, you learn the contents of BOTH of the other doors. And the problem guarantees that if one of them is correct, you automatically select that one. And that's why switching increases your chance of winning to 66%.
Nice trolling there, got me to reply.
Are you sure that your IQ is 127 and not 27?
it's because in the beginning you only have a 1 in 3 chance of guessing the right one
so it's only in that scenario (when you guess right from the first time) that you lose when you decide to switch
Imagine the same thing, but with 10000 doors. You pick one at random and 9998 doors are opened, leaving the door you picked and one other door.
Now would you rather swap, or keep the same door?
The best way to get it is this:
choose a door. The host offers you to choose the other two together and discard one goat from the prize. Opening a door and revealing a goat is the exact same thing as discarding one goat from the prizes.
Just draw 2 tree diagrams of all possibilities for choosing once and changing your choice.
if you only choose once the choice is pretty obviously 1/3. We're happy with that so focus on the tree diagram when you change your choice.
Start from a dot and at the first stage split into 3 'branches' for the car being put behind door A, B,C and draw 1/3 on each of them.
Then for each branch draw another 3 branches each with the player's first choice of A, B , C
Then for each of the 9 new branches where the player's choice does not match the car door, the game master has to reveal the only remaining goat door. There are 6 such branches in total, all with equal probability.
In these branches, if the player automatically changes his choice he will win the car because apart from his initial choice there is only one door left, which has the car behind it.
Then for the 3 remaining branches where the player originally chose the door which had the car behind it, the game master can reveal the gist behind either of the two remaining doors, and in these circumstances, if the player decides to change his choice he is guaranteed to lose.
All of the 9 branches have equal probability 1/3 times 1/3, so there are 6 charm he's where automatically changing to the other door will win you the car, and 3 where it will lose.
Math doesn't change though. Addition will always be addition, geometry will always he geometry, calculus will always be calculus. The only mathematical innovations than can come about are solving things that have yet to be solved. Science on the other hand is always changing based on new information.