You are probably familiar with the Monty Hall Problem. There are 3 doors before the player, 2 with nothing behind them (goat doors), and one with a car behind it (car door). The player does not know which door has the car behind it, but his objective is to win the car by choosing the correct door.
The player is asked to first choose a door, an assistant will then open one of the doors that: 1. The player did not choose. 2. Is a goat door (nothing behind it).
The player is then asked if they would like to switch doors, to instead choose the other unopened door or if they would like to stand by the door they first chose.
The assisstant will then open the door that the player chose in the end revealing if they won the car or not.
Mathmaticiens today will insist that the player has a 2/3s chance of winning the car if they switch, but only a 1/3s chance if they stand by their initial choice.
Obviously the player has a 1/3s chance whether they switch or not. What do you think?
You learn nothing further about your original choice, so it's probability remains 1/3, but you do learn that one of the other doors was ruled out, so the probability of the final door is 1 - 1/3 = 2/3, so you should switch.
Dominic Lewis
How come it isn't:
1. Pick whatever door. 2. Assistant reveals a goat door. 3. Now there are only 2 doors. One is a car, the other is a goat. Your chances are now 50-50. So by switching, or by NOT switching, you end up with a 50-50 chance of winning the car. Switching doors has no effect on your chances.
Zachary Powell
You know that Monty is never going to show you the car. So if you picked a goat first, let Monty show you the other goat, you know for sure that the car is in the door that you didn't pick, nor Monty opened. So if you picked a goat first, you will win a car everytime you switch. And since the probability that you picked a goat first is 2/3, switching results into winning the car with 2/3 probability.
Henry Smith
So if you pick door 1, they're gonna open door 3. If you pick door 3, they're gonna open door 1. If you pick door 2, they're either gonna open door 1 or door 3.
But in any case you'll still have two closed doors in front of you. You'll have learned nothing about your original choice. The car could still be behind your door or the other door. This shit makes no fucking sense.
Brandon Clark
lets assume you know that door 3 is a goat. picking door 3 first will definetely win you the car. why? because monty is going to show you the other goat. so lets say he opens door 1 and reveals the other goat. you know your own door 3 is a goat, and you know now that door 1 is a goat, too. so where can the car only be now? behind door 2. so you switch to door 2 and win the car.
now the probability to pick the goat first, without actually knowing which door is which, is 2/3. this is why switching will win you the car 2 out of 3 times. if you actually knew which door is a goat, you would win the car 100% of the times, if you switch.
Chase Thompson
kys retard
Nathaniel Brown
Because you were more likely to choose a goat initially, so it can't be 50/50.
If you chose a goat door and you switch, you get a car. This is more likely since there are two goats initially
If you chose a car door and you switch, you get a goat. This is less likely since there is only 1 car.
Thomas Nelson
This is a really nice way of putting it, user. Well done.
Parker James
Let's take an extension to describe why you're a fucking moron, shall we? There's a hundred thousand doors. 99 999 goats, 1 car. You pick a door. The gameshow host opens 99 998 goat doors. This leaves one door somewhere in there closed, and the one you picked blindly. Of these 99 998, they're nothing but goats, as per the problem's statement. The gameshow host will only reveal goats, not the car.
Consider that there's 100 000 doors and 1 car. There is a 99.999% chance you picked a goat initially. Conversely this implies a 0.001% chance you picked the car initially. That leaves two options: - a blind guess out of a hundred thousand - of the other 99 999 doors, the one that wasn't opened Thus, we can state that the door left closed has a 99.999% chance of being the car. We therefore can conclude user is a brainlet.
Nathaniel Kelly
What if monty opens the door randomly and it just so happens a goat is behind it?
Joshua Cox
Then you should still switch.
There are three available choices.
Each choice has a 1/3 chance of being correct.
That means that you have a 2/3 chance of picking the WRONG choice.
After you pick a choice, a wrong choice is revealed (randomly or not is irrelevant).
There is a 66% chance that the door you can choose to switch to contains the prize.
Robert Williams
Why does door 2's chance change when door 3 is revealed randomly while door 1's chance doesn't change? Answer carefully.
Oliver Roberts
It doesn't matter if it's a million doors, only one has a car and you have one choice and no possibility to switch. The probability is 50/50 because you either win or you don't.
Sebastian Williams
You're basically right, but if I'm not mistaken, the probabilities should not be assigned to the individual doors, but rather to each particular act of choosing.
The first choice has a 1/3 probability of being correct.
After we eliminate one option, each door individually now has an equal probability of being a correct (namely 0.50).
However, if you don't make another choice, your original choice still has a 0.3333333 probability, regardless of the probability we to each door.
However, if you make a new choice, even if you end up picking the same door, it now has a 1/2 probability of being correct.
E.g. if you have a deck of cards, and pick 2, there is a 4/52 probability that the first card is a 9, and given that the first was a 9, there is a 3/52 probability that the second is a 9. However, if you reinsert the second card back into the deck, shuffle it, and once again pick a random card (even the same one) the probability is now 3/51 that the card is a 9 (a slightly higher probability than the first case).
Henry Hernandez
Assume you always switch. When will you win?
John Kelly
>You are probably familiar with the Monty Hall Problem. Yes, Veeky Forums used to have this thread daily.
>Obviously the player has a 1/3s chance whether they switch or not. Am I missing something or misread? Is this the standard version of the MH problem? It’s 1/3 if stay and 2/3 if switch.
Noah Gonzalez
>Obviously the player has a 1/3s chance whether they switch or not. You mean 1/2
Anyway, to answer your question it depends on whether the "assistant" is omniscient and opens a door that he knows conceals a goat, or is not omniscient and opens a door without knowing what it conceals and it happens to conceal a goat.
Also, I'd much rather a goat door because goats are fucking awesome.
Jose Bell
Don't think of each door as having its own 1/3 probability. There is only the probability that you were right the first time you chose, ~33%, and the probability you were wrong the first time you chose (~66%).
If forced to make a decision when there were only two doors available from the start, there would be a 50% chance of picking right or wrong. But there isn't, we have more information than that.
We know that there was a 66% percent chance that the prize was behind one of the two doors that you didnt pick, we know that only one of those doors is available to be chosen after the reveal.
Imagine the game in reverse: you can pick two doors. You have a 66% chance of having the prize in one of those two doors. One of them is revealed to have a goat. What is the probability that the prize is behind the door you didnt pick?
Jaxon Nelson
note that by switching you win 2/3 of the time
Jackson Ramirez
>Anyway, to answer your question it depends on whether the "assistant" is omniscient and opens a door that he knows conceals a goat, or is not omniscient and opens a door without knowing what it conceals and it happens to conceal a goat. >Also, I'd much rather a goat door because goats are fucking awesome. These are the correct answers.
Mason Phillips
Only drones and mental slaves reply "right" by trusting the host, MHP is designed to detect who has not a dissident mind.
THIS IS WHY PSYCHOLOGY IS A SCIENCE YOU FUCKING DRONES
Dylan Robinson
This is wrong. In the case where monty chooses randomly, the odds are 50:50.
Christian Wright
You have a higher pobability of choosing a goat on your first turn.
Matthew Reyes
That's not quite how it works. Refer back to the ten thousand doors. It's not a new scenario, which is why switching provides P>0.5. If the information was scrambled after the revealing of the doors, THEN you have a new choice between two equally likely options, instead of one equal chance, and one that by definition can't be the car.
Given n doors, there's a 1/n chance of you picking the right one, and a n-1/n-chance of you picking the wrong one. It is most likely you picked the wrong
Benjamin Miller
fuck sent early It is most likely you picked wrong one in the first go-around, instead of the right one. (pRight=1/n, pWrong=n-1/n) Furthermore, the door opened cannot be the car, it can only be a goat.
The act of selectively opening a door/doors reveals information, which is more oblique when n=3, but becomes more obvious when n increases to bigger numbers.That is why changing gives you a greater probability, because there's a reason that one of the doors not picked are left shut: Case 1) You picked a goat, and that means it has to be a car behind it Case 2) You picked a car, and that means it has to be a goat behind it. Because there's a 1/n chance you were right to begin with, that means that there's only a 1/n chance of case 2, meaning case 1 is the most likely and therefore the one to act in accordance to.
Julian Parker
imagine there's 100 doors instead. just because the all knowing host opens 98 empty doors doesn't mean your one suddenly has a 50% chance of having a car.
Jace Diaz
>0.333333 >wrong about everything Kys brainlet
Zachary Price
What if you already have a car but not a goat? You might as well pick the goat that is revealed to you and walk away a winner 100% of the time.
Christian Gomez
You just need to understand that Monty has left you with a new choice: goat and car. So unless you originally picked the car, which you did with a probability of 1/3, switching will win you the car. this means switching will win you the car with a 2/3 probability. Why can't brainlets understand this?
Logan Stewart
One of the choosing a goat paths will cease to exist once the door is opened meaning changing and sticking are equally likely to win the car
Angel Williams
I hate this problem because the answer is 2/3. It's so counterintuitive.
Andrew Reyes
>will cease to exist
ooh *puff* magic!
Camden Baker
having the ability to change or stick on one of the goat doors is impossible since it will always be revealed
Ryan Rogers
This.
Imagine at the start of the game Monty offered you the choice to either open 1 door or 2 doors. Also, since opening 2 doors guarantees you'll get at least 1 Goat, he says he'll keep one for himself.
Obviously you choose to open 2 doors.
It's the exact same scenario.
Christian Scott
2 doors will be opened whether or not you change doors
Michael Davis
fun fact: paul "brainlet" erdös didnt believe in this until he saw a computer simulation lol.
Imagine one goat is red and one is blue, and you decide to switch. Theres a 1/3 chance you choose the red goat right off the bat, the blue gets revealed, and you switch to get the car. There's a seperate 1/3 chance you pick the blue goat instead, where the red goat gets revealed, and switching gets the car. Then finally, the 1/3 chance you pick the car first, where one of the 2 goats gets revealed, you switch, and get the other goat
Gavin Robinson
WRONG! You are a reddit-tier psuedointellectual RETARD!
Experimental results:
>Test 1 D D D C G G
>Test 2 D D D G C G
>Test 3 D D D G G C
>Test 1 Pick door 1 Monty opens door 2 Switch is unsuccessful
>Test 2 Pick door 1 Monty opens door 2 Choice is never presented
>Test 3 Pick door 1 Monty opens door 2 Switch is successful
CONCLUSION: Switch is successful 50% of the time when Monty is not picking intelligently (always picking 2). Basically, you're a fucking RETARD!
The player has a 50 50 shot just like with everything else. Either you get the car or you don't
Luis White
The best explanation even the biggest brainlet will understand is the following:
Imagine the same situation but instead of 3 doors there are 100 doors. Still only one price. You pick one door, the guy in charge opens 98 of the remaining doors and now you have the choice to switch doors or stay with your original choice.
You would obviously choose the other door if you're not a dumbass.
Daniel Jones
You actually wasted those digits with something that retarded
Brandon Baker
How wonder how brainlet someone has to be to not understand that isn't 50:50..
Brandon Adams
>How wonder how brainlet yes.
Ryder Ward
*I wonder
When I can't sleep I start writing shits in English since I'm not English, sorry man
Better write "how" instead of "I" because I'm fucking insomniac instead of being a brainlet who cannot understand maths problem even after reading the solutions
Isaac Martinez
Suppose you start dating a woman. It's too early in the relationship to determine if she is a piece of shit. You learn that your buddy breaks up with his girl because she is a piece of shit. Should you find a new girl?
Jonathan Hernandez
The question is literally "do you want to select a group containing two doors or one door". If you pick incorrectly initially- as you have a 33% chance to do- you are absolutely guaranteed to pick the correct door by switching. Meaning, you always have a 66% chance to guess correctly by switching IF AND ONLY IF you ignore the possibility they only let you change because you picked correctly initially. Yeah, I rounded, sue me.
Luis Nelson
True, but if you actually knew if one of the doors had a goat behind it, which in your example was 3, then it doesn't truly matter whether you pick either 2 or 3. You could pick two and not switch, or pick 3 and switch. If you actually know one of the goat doors you have a 100% chance of selecting the correct door whether you switch or not.
Owen Gomez
...
Wyatt Brooks
>If you pick incorrectly initially- as you have a 33% chance
67%
Angel Peterson
That one chick van Savant or whatever with the 210 IQ says you should change doors. I'm going with that. I mean, look at the size of that fucking cranium.
Leo Nelson
Here's the thing that bothers me.
If the rules are explained beforehand you of course have 2/3 chance to get it right by switching, HOWEVER.
In most cases when people present the problem it looks like the show host makes the player aware of this rule after he has already chosen a door. If the rules are not told beforehand, you don't know whether he only would've given you this opportunity had you chosen correctly in the beginning, trying to trick you. I think people catch this subconsciously and that is why so many don't switch.
Christian Roberts
Can i screw the goat?
Mason White
>with the 210 IQ 228. I still don't understand how Guinness made that category into "get a psychologist who doesn't understand how to score the test to give you a value that's outside the range for the test".
Jack Kelly
Duh
Easton Cook
Only if you win it.
Jayden Brown
Quit it with this brainlet bait. When you get the option to switch monty has imparted some of his knowledge about the doors. Your first choice is a blind guess between 3 doors. Monty eliminates one door that he knows is the goat. Now either your door or the remaining door are the correct door. Im not sure if I believe the 2/3rds thing(yes ive seen the math proofs for it but it doesnt seem like they are accounting for all the variables), but your chances are in fact greatly increased by switching. More than a 50/50 chance.
Andrew Myers
What if I want the goat, tho?
John Morgan
>Savant sees IQ tests as measurements of a variety of mental abilities and thinks intelligence entails so many factors that "attempts to measure it are useless" >only qualification is iq score
Grayson Adams
holy shit this problem is ridiculously easy
>What is the probability you pick a door with a goat
2/3
>take out a losing door >switch doors >1/3 probability
in fact
>suppose you have n doors and n-1 losing doors >the probability of picking a losing door is going to be (n-1)/(n) >now wait until there are two doors left >switch doors, thus maximizing your probability of winning >>if you switch more than once you decrease your chances of winning
Take a course on Probability Theory ffs, this shit is easy pickings.
Logan Butler
no gas, no repairs, free lawn care, and cheese. user you might be on to something
Dominic Parker
ITT: people who still don't know what Bayesian probability is
Brandon Long
New Game:
There are 4 doors. Behind 1 door is a car, behind the other 3 are goats.
After you pick a door, the host gives you a choice. He can either: a) Open your door, if it's the car YOU WIN, if it's not you get to open one of the THREE remaining doors.
b) Open a different door that he knows contains a goat, and give you the option to switch to one of the TWO remaining doors.
What's your best strategy?
William Ross
Can't we just fucking test this using an experiment and find out what the actual no bullshit solution is?
James Martinez
No, in fact you don't. If you know 3 is a goat, and pick 2, not knowing if it is a goat or not, Monty might open door 3 (which you didn't pick), and you would still not know if door 2 is the car or not. However, in this scenario switching and not switching are 50/50.
Ethan Murphy
Don't forget the warm pussy
Carson Thomas
Pretend there are 100 doors. No matter which door you pick, the assistant opens 98 doors. Would you switch or no?
