Is the Monty Hall problem an idiot detector? Like, anyone who cares about it is just a try hard deep thinker right? Not even trolling. I was just thinking to myself, "I'm pretty sure if I explain the Monty Hall problem to someone and their first reaction is 'this is stupid,' then I'm probably not going to get along with them."
The whole argument was stupid. Probability would adjust itself. I wouldn't say that it's an "idiot detector", as it is a "pretentious deep thinker detector".
>implying a straightforward probability problem as if it's some sort of filter
>not the sleeping beauty problem, which has actually different nontrivial probabilistic interpretations
Why are OPs in /his/ always so fucking stupid?
The solution is.
It is in your solution not to switch because if you fail anyway your reaction is going to be like
>Ooooh better luck next time
But if you switch and fail you're going to be kicking yourself for the next while.
>Is the Monty Hall problem an idiot detector?
No, it’s a fun logic puzzle that demonstrates the counter intuitiveness of probability
>Like, anyone who cares about it is just a try hard deep thinker right?
No, nobody “cares” about it
>if I explain the Monty Hall problem to someone and their first reaction is 'this is stupid,' then I'm probably not going to get along with them."
So you have terrible teaching skills. How does that reflect on the MH problem?
And what do you mean by “stupid”? That the math is wrong because you fail to grasp it?
This. Also, people forget that at least one source of this "paradox" came from a TV game show. You know, where they would want to make the probability confusing so they did not have to give away the prize.
That's a bit mean.
History major here that stopped at high school math, it's not that hard.
You have three doors, one has a prize behind them. You are asked to pick one of them.
There is a 33.333 recurring percent chance that the door you picked has the prize behind it. There is a 66.666 recurring percent chance one of the other two doors has the prize behind it.
The "host" opens one of the doors you didn't pick. Note (and this is crucially fucking important) he could NEVER, ever, ever, ever have picked the door you originally chose or the door with the prize behind it.
There is still a 33.333 recurring percent chance you picked the right door initially. There is NOW a 66.666 percent probability the other, remaining, door is the correct one.
It's just as if you picked from two doors from the start
You're making a classic mistake and not understanding the basic premise of the problem. If you account for all probabilities, then getting the prize is a 50/50 chance. But the question is not "what chance is there you'll get the prize?", it's "is switching better than staying?".
No he is right.
>what stops the one you've allready picked being the right one
Nothing. That still has a 1/3 prob.
>It's just as if you picked from two doors from the start
No it is not. You are given information about one door. That is not the same as if the door was never there
>I'm probably not going to get along with them
If you bring this up the first time we meet I'm probably not going to get on with you either
> If you account for all probabilities, then getting the prize is a 50/50 chance.
What's this even supposed to mean? The whole point is that even though there are two doors remaining, one of which has the goat, your odds without switching are still the one you picked out of the three you picked from.
> Yea I'll have a big mac c/o thank you very much
> You see, I've hidden it under one of these 3 cups. One of them have it, one is empty, and the other one have fries.
> Ok I pick this one
> Before that, this is the ones with fries
Think of it this way: The number of possible situations where the door you picked is correct are less than the number of possible situations where a door you did not pick is correct (because there are more doors that you did not pick than you did pick). However, switching is normally bad because even if you knew for sure that the door you picked had no prize, you still wouldn't know which of the other two had the prize. So when the host eliminates one option, all you know is "the door i picked is probably wrong, and there is only one other door option"
My favorite explanation of this problems is the 100 door version where there are 100 doors and you pick 1. Then the host opens 98 of them to reveal all goats. It's pretty obvious at this point why you would want to switch doors everytime
Okay, so if I count sticking as just picking a new door wouldn't that just make the probability 1/2?
>So in 2/3 of the cases the game host can only open one of the two doors you didn't select because the other contains the car.
Obviously that's what game shows actually do but I'm not quite getting the real logic of the situation.
Please. They were just trying to mess with the contestants heads to add a little entertainment value. They didn't know the odds would change.
The rest of the show was about decisions that you don't have helpful information for. "Do you want to keep this prize, or take what's in the box?"
There are 3 doors and there's a car behind one of them. What's so difficult to understand about this. It doesn't make a difference if you change the door or if you don't. The car isn't moving.
it has nothing to do with pretension or deep thinking, or even the humanities really. it's just a quirk of statistics (probability) that a lot of people without study in the area intuitively get wrong
I just grasped it.
You should always change.
If you didn't choose the car in the first try then he'll always be on the door the host didn't get open.
But you may as well be. If the scenario was every so slightly different and you did get to pick a new door but picked the same one anyway this wouldn't mean that magically one door has a 1/3 chance but one has a 1/2 chance, that's fucking retarded.
Neither the doors nor the car have moved, and there isn't some bizarre 16.6% chance of something new randomly jumping out of either of them. It literally is a 50/50 bet.
If you pick a goat then there are two doors left, one has a car and another one has a goat. The host can only open the goat door in this case.
If you pick a car initially the host can open both of the two remaining doors since they both have goats.
The chance of initially picking a goat is 2/3, therefore the door that is left unopened contains the car in 2/3 of the cases, which is why you should switch.
Say the car is behind door 1. Then there are 3 possible scenarios:
You pick door 1. The host reveals a goat behind door 2. If you switch you lose.
You pick door 2. The host reveals a goat behind door 3. If you switch you win.
You pick door 3. Th host reveals a goat behind door 2. If you switch you win.
You win 2/3 of the time if you always switch
Try thinking about your choice in terms of your initial odds. They do not change once a goat is revealed. There is still a 2/3 chance your first choice of door was one with a goat behind it, so the odds of the other one hiding the car are substantially improved.
I get what you're saying. Originally the door you picked had a 1/3 chance, and the two you didn't collectively had a 2/3 chance. Then one of them was showed to have a goat in it.
I don't get why the math doesn't change. Now there are two doors so rightfully both of them should have a 1/2 chance (game-show logic aside) rather than the remaining unpicked door somehow still has a 2/3 chance.
You are not as bright as you want to believe. This problem is flawed and is used to determine who refuses to accept authority. The "popularity" of this problem is artificial, as so is used to test what I said across the internet. Translation: you are being data mined.
I might give arguments about this, but no, you are smart enough to understand it. Besides, the dataminer is probably going to deny it.
the math does not change because your choice does not exist in a vacuum
imagine there had been not three, but thirty or maybe a thousand doors
the chances that you picked the correct one out of all those is very slim
and now if the host opened all but two (your choice + one another), that does not make the probability _that you had picked the correct door in the first place_ different
the odds that you had picked the door with the prize stay the same no matter what you do - the entire switching or staying merely changes whether you exchange the worse probability of having picked the prize door from several doors, with the better probability of the prize being in one of the other doors (the remaining door retained by the show master)
IF, and this is a _different scenario_, as in probability of a different event, you were to walk into a room and saw the experiment being performed, and saw only two doors remaining, and only knew a car was behind one of those, then YOUR chances of correctly guessing where the car is would be 50/50
however, the chances of the participant having picked the correct door would be as in the first paragraph - remember, this again all boils down to the original chance of having picked the ONE correct door out of SEVERAL
Yes but all but two of those doors are totally irrelevant, from the information you're given only 2 of those doors have any mystery to them.
So for all uncertainty it only comes down to two. It can only be a 1/2 choice.
imagine this with more doors
like 123 doors
what are the chances you pick the ONE door with the prize out of 123? very low
so you pick a door
then the game show host, who knows where the prize is, opens ALL DOORS but one + the one you picked
meaning that either the prize is the door you picked OR the door he did not open
except when you were picking your door, you only had a 1 in 123 chance of picking the correct door, so the chances of the prize being in your door are exactly that low - and chances of the prize being in the other are the rest
so you switch and win (most of the time)
you are indeed asked to pick between two doors, but each door has a different probability of having the prize behind it
your originally picked door retains the 1 of 123 (to use the previous example) chance of having the prize behind it
the other door, by virtue of the knowing game host removing other doors, now has the (much) greater chance of having the prize
this is a very simple problem once you apply larger numbers of it - imagine two groups of 300 people, one switchers, the other non-switchers
each group of 100 picks either A/B/C and then behaves accordingly to their switching preferences, with the prize being behind door A
you will find that with the non-switchers, only those who had picked A will get the prize (100 people), but with the switchers, it is those who had picked B and C who win (200 people)
But it doesn't.
One door has car behind it. One door has a goat behind it.
The odds of either of the three doors having a goat behind it is 2/3 with the remaining 1/3 having a car behind it. If you elimate one of the goat duds you're left with 2 doors, each with a 1/2 chance of having either a goat or a car behind it.
I don't understand why if you remove the element of initially picking a door the chances are 1/2 for either. But if you've already picked a door it suddenly becomes 2/3 vs. 1/3.
but that is exactly the core of the problem - you are picking out of many doors, not the remaining two
remember, do not mix up the various events or probabilities - you are interested in the probability of having picked the prize door out of many doors in the first place, and that chance is low
again, imagine it with a much larger number of doors
>If you elimate one of the goat duds you're left with 2 doors, each with a 1/2 chance of having either a goat or a car behind it.
eliminating the goat duds does nothing to your (low) chance of having picked the correct door
you are still standing in front of one door out of many (again, imagine this with a hundred doors) which you have picked and hope for the best
But you are, the whole premise of the problem is that all but 2 doors you know for a fact are duds.
This is why choosing to stay or swap is exactly like just picking between two doors with a 50/50 chance. Whilst initially you were taking a pure 1/3 shot in the dark, now that you know one is a dud you're essentially being asked to pick between two options only knowing that one has a car and one has a goat but no way of knowing which so it's 50/50.
The opening premise of the situation just make it deliberately confusing, those two doors are all that matter.
So the original participant gets a 1/3 chance on his original or a 2/3 chance on the remaining, but an observer who also gets a choice has 1/2 for each door? Is this testable or just a thought experiment based on the way probability mathematics operates?
You can test it with online simulators or with just 3 cups and a penny.
When I tried it I got the exact opposite of the expected result, staying seemed to generally be the better option.
I’m going to give it one last try:
Think of it like this:
>When you pick a door you lay down the odds and *they never change during the game!*
You pick a door and that door has 1/3 chance of a car. And 2/3 of a goat.
The other 2/3 chance of a car now resides in the two other doors. Combined.
Because we have no further information it is fair to equally divide those 2/3 among the two doors. Giving each 1/3
Now one door is revealed to have a goat.
All this does is give you information about the distribution of the 2/3 chance of a car.
The odds did NOT change. The 2/3 chance of a car is still divided among the two doors. Only now you have information about HOW this is divided. You now know that one door has 0 probability and the other door has (all of the) 2/3. That’s the exact same 2/3 you first assumed was equally divided among the two doors.
So, if the scenario is pick a door; have a goat revealed; choose “stick” or “change”:
changing will double your chances of winning the car (from 1/3 to 2/3)
Think of it this way:
There are 100 doors. You pick one, and then the host opens 98 other doors with goats behind them. Now you are offered to switch or keep your current door after the other 98 were opened. Is the chance still 50/50?
no, i am convinced you are being deliberately obtuse or trolling, there is plenty of food for thought in this thread or on the internet in general, including simple calculations which will prove the truth to you
again, simply imagine the exercise with a hundred doors
if what you were saying was true, then your chances of picking the one single correct door out of a hundred would be 50%
which clearly is not the case
>you already know what is behind those doors, but you did not know when you picked your door
As I say, the premise of originally picking a door just makes it deliberately confusing. When it comes down to it you're picking between two doors.
>and since you still appear to claim that picking one door out of a hundred is a 50% chance of picking the correct door,
But you aren't picking between 100 doors. You're picking between two.
You are picking between two doors, but the chances of the Car being behind one door (the one you picked) are 1 in 100. While the chances of the Car being behind the other door are 99 in 100. Why? Because originally you had picked a single door out of a hundred (1 in 100 chance). Meaning that only one person out of a hundred competing would have had picked the Car, let's say the Car is hidden behind door number 49 every time and the hundred people competing in separate rooms each pick a different door from one to a hundred. Now 98 doors are revealed, only the participants' choices and number 49 remaining, showing Goats only. Out of those hundred people participating, only the person who had originally picked door number 49 would win. While the ninety-nine who had picked doors other than 49 would win by switching.
Ah, now it makes sense.
No matter what your chances of picking a goat are 2/3 and no matter what Monty Hall HAS to reveal the other goat. So since it was so likely you picked a goat it makes sense to switch.
Why are all the posts ITT trying to explain it making it so complicated?
it is just a small fun puzzle, hardly takes up any time, if someone believes it makes them a super-genius they are trying too hard but it isn't a big deal
It is this thread that makes me feel like an idiot.
I don't understand what is going on, from OP to people explaining it, and people saying they don't get it, and the replies.
Why would people believe the problem is benchmark for either intelligence or tryhard?
It's a tricky counter-intuitive problem so simply at first sight that lots of intelligent people fall for it
Once you know there's a "twist" to it, the explanation is so fucking simply even stupid people will get it
Not giving two shits about the problem does not make you shallow in any sense
And the problem (and what leads to the common mistake) is in itself intersting to analyse and take apart, without amking you a tryhard
So how on Earth can you conclude anything from someone's attitude towards this problem?
I know the world is scary place full of uncertainty, but you are trying to find meaning in so little.
Also, there are seven different explanations, you have no excuse not to search the solution online.
Have another one:
>When you choose the first door, you are creating two sets of doors: one set with the single door you chose, and another with two doors. Probs are 1/3 and 2/3 of the car being in each.
>Given the chance, you'd choose the set with two doors, obviously, even after seeing which one of the two has a goat
Think of it with boxes instead of doors.
There's three boxes, two of which have small prizes and one of which has the keys to a new car. You pick one. Then, without you seeing, they dump the contents of one the two remaining boxes into the other.
So then, one box contains two of the three prizes, and your box contains one. The Monty hall problem is 100% exactly like this problem.
Or, even better, imagine there were only two boxes with three prizes distributed so that every box has at least one prize. And then, imagine every time you pick, you're told you picked a box with just one item, and you can switch to the box with two items.
That's the Monty Hall problem as simple as possible. Whether you "know" what one of the prizes was doesn't matter.
Look it's all in the grammar of the problem.
They only EVER flip a door with nothing behind it. If you picked one door, and they randomly flipped one of the other two, and that one HAPPENED to be empty, it would be a 50/50 shot. But in that case, sometimes they'll flip a a door to reveal the car. That chance that they would reveal the car, but don't, is why you have a better shot switching. But the way the problem is worded, instead of ever flipping the car door, they only flip one that's empty.
It's a grammar trick.
After the host has eliminated the other 98, there are 2 remaining.
The one you picked, or the one the host has been specifically avoiding. Why has he not picked this one? either it's random and you happened to guess right the first time (1% chance)
Or he has been avoiding it because it has a car behind it (99% chance)
I don't know how to make it any more simple than this. The host isn't opening doors at random, he's opening ones that you did not pick and that did not have a car behind them.
>people prefer a filthy common car to a Great Of All Time like pic related
An infinitesimal is not a number.
There's a well-known theorem that between any two distinct real numbers there's a rational number. Show me the rational number between 0.999... and 1, and I'll believe you.
Also, if you have a notion of a LUB, and inclusive and a non-inclusive interval on R will have the same LUB, which is essentially what it means to say 0.999...=1
If you take [0,1] and [0,1), you'll realize you can't find a LUB for either that is distinct, even though [0,1) is a proper subset of [0,1]
The whole problem boils down to what you pick at the beginning. You are more likely to pick the goat, so you are more likely to have the other goat when the goat is revealed. Because you probably have the goat you should switch.
You can do it on your own if you like. Its easy to do. Cups and coins. Work it out.
>You can do it on your own if you like. Its easy to do. Cups and coins. Work it out.
It can take hundreds or thousands or millions of tries to get close to the right odds. You need large numbers, which is why they usually do this stuff on computer simulations to be sure.
I tried 20-30 times and the switch was effective 90% of the times, when it should be closer to two thirds.