Is this numberphile math completely wrong or am I dumb?

[SirNumbers]

 

 

This is bullshit, she says if there 3 doors, and you choose one but dont open it, and someone else opens another door and there's nothing behind it, but gives you the option to switch to the third door and she says that there 2/3s chance the cars behind the third door. 

 

 

In my head that's just fucking stupid, If there are three cars and one has a car behind it. You choose 1 door you don't know what's there. That's one 1/3 chance there's something there. however if the game show host opens a door that's empty it makes only 2 options remain meaning there 50 percent chance the car could be under either door right?

[Nineshadow]

It is mathematically correct.

[iamdarkyoshi]

it makes sense to me. Perhaps try watching this: http://www.dailymotion.com/video/x2mykzd

[Ekpyrosis]

No, she's right. It's due to conditional propabilities.

Basically, if you use the strategy to always switch, you will always win if your pick before the switch is an empty door.

[Oshino Shinobu]

[minibois]

If there are 3 doors to choose from, you have a 33% chance of getting the right one and 66% wrong.

When they show you which one it is not, your chances for the 2 doors left is 50/50.. Now here's the part I don't understand too.. Apparently if you choose the third not opened, not chosen door.. That is suppose to make your chances higher.. But there's an equally high chance for the door you chose already

[Philosobyte]

This is the way it works. 

You have 2/3 chance of picking the wrong door.

You have 1/3 chance of picking the right door. 

 

If you decide to switch every time they ask you, then that means:

When you pick the wrong door, you're actually going to get the right door,

When you pick the right door, you're actually going to get the wrong door.

 

Therefore, if you decide to switch every time,

You have 2/3 chance of initially picking the wrong door and switching to the right door.

You have 1/3 chance of initially picking the right door and switching to the wrong door.

[SirNumbers]

If there are 3 doors to choose from, you have a 33% chance of getting the right one and 66% wrong.

When they show you which one it is not, your chances for the 2 doors left is 50/50.. Now here's the part I don't understand too.. Apparently if you choose the third not opened, not chosen door.. That is suppose to make your chances higher.. But there's an equally high chance for the door you chose already

Thats what im saying. He takes the door out of the equation so there should 50 percent chance

[manikyath]

--

this is a really good explanation actually...

[adhisakti]

If car at door 1, and you choose :

- door 1, door 2 or 3 opened, not switch = win, switch = lose

- door 2, door 3 opened, not switch = lose, switch = win

- door 3, door 2 opened, not switch = lose, switch = win

 

It is better to switch every time, because your probability to win is 2/3

[iamdarkyoshi]

Thats what im saying. He takes the door out of the equation so there should 50 percent chance

By switching, you essentially chose BOTH the remaining doors, because he shows you that one of the remaining doors has nothing behind it, letting you chose the other one.

[minibois]

Thats what im saying. He takes the door out of the equation so there should 50 percent chance

¯\_(ツ)_/¯

 

It's all the chance either way.

[SirNumbers]

By switching, you essentially chose BOTH the remaining doors, because he shows you that one of the remaining doors has nothing behind it, letting you chose the other one.

 

 

If car at door 1, and you choose :

- door 1, door 2 or 3 opened, not switch = win, switch = lose

- door 2, door 3 opened, not switch = lose, switch = win

- door 3, door 2 opened, not switch = lose, switch = win

 

It is better to switch every time, because your probability to win is 2/3

Okay I get it now

[Glenwing]

Thats what im saying. He takes the door out of the equation so there should 50 percent chance

Every individual door has a 33% chance. So when you pick one door, you had a 33% chance of picking the right one, and a 66% chance of picking the wrong one. Once the host reveals one of the doors, the chances don't reset, because the car hasn't been moved again now with a 50/50 split behind door 1 and 3. It's initially behind 1 (33% chance) or behind 2-or-3 (66% chance). Once door 2 is revealed those chances don't change. So if there's a 66% chance of it being behind 2-or-3, and you know it's not 2, then it's a 66% chance of being behind 3.

Let's say you were playing a video game, and you discovered a treasure room that had 2000 chests in it, and only one has treasure (randomly assigned by the computer). You only have one key. You're allowed to pick one at random without opening it, say you pick chest number 1000, and when you do, all the other chests disappear except chest number 627, and you know it can't eliminate the one with the treasure, and it doesn't "re-assign" the treasure between the two remaining options; the treasure is still in the same number chest that the computer originally picked at random when there were still 2000 of them. Is it really a 50% chance between chest 1000 and chest 627? I don't think so.

[Glenwing]

In case anyone wanted to try it out a bit, I wrote a little program to simulate this You will need to download Python to run it if you don't have anything to run it with already though.

https://www.python.org/ftp/python/3.5.0/python-3.5.0-amd64.exe

Just paste this into a txt file and rename it to the .py extension:

#Python 3.3.3#Monty Hall Demonstration#Created by Glenwing: http://linustechtips.com/main/user/2466-glenwing/from random import *def initialize():    print('This is a demonstration of probabilities with the Monty Hall problem.')    print('')    print('First set up the program by entering the number of doors you want.',    'During each test trial, the computer will randomly assign "treasure" behind one',    'door. You will pick one door at random, and the computer will eliminate bad',    'options until there is only one other option left (in addition to the one you',    'guessed). You will then pick whether you want to open the door you originally',    'chose, or the door left over by the computer. The treasure must be behind one',    'of the two.', sep='\n')    print('')    print('To see the probabilities in action, use "manual" mode. To ensure fair results,',    'the computer will tell you upfront which door the treasure is behind before any',    'choosing or eliminating takes place, but it will be salted with a random number',    '(provided) and converted to binary so you cannot tell what number it is. If you',    'wish to verify that the computer has not changed the treasure door after',    'originally choosing it, you can of course convert the number back to decimal',    'with any online converter, and subtract the provided salt to see that it',    'is the same door as revealed in the end result.', sep='\n')    print('')    print('To see bulk statistics over many trials, use "auto" mode. The computer will',    'randomly pick a door for you. You will specify whether you want to always keep',    'your original door choice, or always switch to the remaining choice provided by',    'the computer.', sep='\n')    print('')    main()def main():    print('MAIN MENU ================================')    print('')    check = False    while check == False:        d = input('Enter the number of doors: ')        try:            d = int(d)        except TypeError:            print('Please enter an integer.')        except ValueError:            print('Please enter an integer.')        else:            if d < 3:                print('Please enter at least 3. ')            else:                check = True    check = False    while check == False:        m = str(input('Manual or Auto mode? (m/a): '))        m = m.upper()        if m == 'MANUAL' or m == 'M':            check = True            manual(d)        elif m == 'AUTO' or m == 'A':            check = True            auto(d)        else:            pass    print('')    main()def manual(d):    print('')    print('MANUAL MODE ==============================')    print('Enter "done" to go back. Shortcuts: You can enter "k" to keep the same choice or "s" to switch choices if you don\'t want to type in the specific numbers.')    trials = 0    success = 0    while True:        print('')        t = randint(1, d)        salt = randint(4096, 16777216)        print('Trial ', trials + 1, ': [The treasure is behind door ', '{:b}'.format(t+salt), ' (salt: ', '{:,d}'.format(salt), ')]', sep='')        check = False        while check == False:            print('Pick a door between 1 and ', d, ': ', sep='', end='')            try:                choice = input()                if (str(choice).upper() == 'DONE') or (str(choice).upper() == 'BACK'):                    return                choice = int(choice)                check = True            except ValueError:                print('\nPlease enter an integer.')            except TypeError:                print('\nPlease enter an integer.')        if (str(choice).upper() == 'DONE'):            print('')            return        if (choice != t):            randomdoor = t        elif (choice == t):            randomdoor = choice            while (randomdoor == choice):                randomdoor = randint(1, d)        print('The treasure is behind door ', '{:,d}'.format(choice), ' or ', '{:,d}'.format(randomdoor), '! Pick a door: ', sep='', end='')        check = False        while check == False:            finalanswer = input()            try:                if finalanswer == 'k':                    finalanswer = choice                elif finalanswer == 's':                    finalanswer = randomdoor                else:                    finalanswer = int(finalanswer)                if (finalanswer == randomdoor) or (finalanswer == choice):                    check = True                else:                    print('Please pick door ', '{:,d}'.format(choice), ' or ', '{:,d}'.format(randomdoor), ': ', sep='', end='')            except ValueError:                print('Please enter an integer.')            except TypeError:                print('Please enter an integer.')        print('')        check = False        if (finalanswer == t):            success = success + 1            print('Correct! ', end='')        elif (finalanswer != t):            print('Incorrect! ', end='')        trials = trials + 1        print('The treasure was behind door ', '{:,d}'.format(t), '!', sep='')        print('So far you\'ve gotten ', success, ' out of ', trials, ' correct (', '{:.1%}'.format(success/trials), ').', sep='')def auto(d):    print('')    print('AUTO MODE ================================')    print('Enter "done" to go back.')    while True:        trials = 0        success = 0        print('')        print('Keep choice every time or switch doors every time? (k/s): ', end='')        check = False        while check == False:            mode = str(input()).lower()            if (mode.upper() == 'DONE') or (mode.upper() == 'BACK'):                return            elif (mode == 'k') or (mode == 's'):                check = True            else:                print('Please enter "k" (keep choices) or "s" (switch choices): ', end='')        check = False        while check == False:            try:                trials = int(input('How many trials?: '))                check = True            except ValueError:                print('Please enter an integer.')            except TypeError:                print('Please enter an integer.')        count = 0        while count < trials:            count = count + 1            t = randint(1, d)            choice = randint(1, d)            if (choice == t) and (mode == 'k'):                success = success + 1            elif (choice != t) and (mode == 's'):                success = success + 1            else:                pass        print('')        print(success, ' out of ', trials, ' successful cases (', '{:.1%}'.format(success/trials), ') by ', sep='', end='')        if (mode == 'k'):            print('keeping the original door choice.')        elif (mode == 's'):            print('switching to the remaining option.')initialize()

[SirNumbers]

In case anyone wanted to try it out a bit, I wrote a little program to simulate this You will need to download Python to run it if you don't have anything to run it with already though.

https://www.python.org/ftp/python/3.5.0/python-3.5.0-amd64.exe

Just paste this into a txt file and rename it to the .py extension:

Well thank you for that thats great

#Python 3.3.3#Monty Hall Demonstration#Created by Glenwing: http://linustechtips.com/main/user/2466-glenwing/from random import *def initialize():    print('This is a demonstration of probabilities with the Monty Hall problem.')    print('')    print('First set up the program by entering the number of doors you want.',    'During each test trial, the computer will randomly assign "treasure" behind one',    'door. You will pick one door at random, and the computer will eliminate bad',    'options until there is only one other option left (in addition to the one you',    'guessed). You will then pick whether you want to open the door you originally',    'chose, or the door left over by the computer. The treasure must be behind one',    'of the two.', sep='\n')    print('')    print('To see the probabilities in action, use "manual" mode. To ensure fair results,',    'the computer will tell you upfront which door the treasure is behind before any',    'choosing or eliminating takes place, but it will be salted with a random number',    '(provided) and converted to binary so you cannot tell what number it is. If you',    'wish to verify that the computer has not changed the treasure door after',    'originally choosing it, you can of course convert the number back to decimal',    'with any online converter, and subtract the provided salt to see that it',    'is the same door as revealed in the end result.', sep='\n')    print('')    print('To see bulk statistics over many trials, use "auto" mode. The computer will',    'randomly pick a door for you. You will specify whether you want to always keep',    'your original door choice, or always switch to the remaining choice provided by',    'the computer.', sep='\n')    print('')    main()def main():    print('MAIN MENU ================================')    print('')    check = False    while check == False:        d = input('Enter the number of doors: ')        try:            d = int(d)        except TypeError:            print('Please enter an integer.')        except ValueError:            print('Please enter an integer.')        else:            if d < 3:                print('Please enter at least 3. ')            else:                check = True    check = False    while check == False:        m = str(input('Manual or Auto mode? (m/a): '))        m = m.upper()        if m == 'MANUAL' or m == 'M':            check = True            manual(d)        elif m == 'AUTO' or m == 'A':            check = True            auto(d)        else:            pass    print('')    main()def manual(d):    print('')    print('MANUAL MODE ==============================')    print('Enter "done" to go back. Shortcuts: You can enter "k" to keep the same choice or "s" to switch choices if you don\'t want to type in the specific numbers.')    trials = 0    success = 0    while True:        print('')        t = randint(1, d)        salt = randint(4096, 16777216)        print('Trial ', trials + 1, ': [The treasure is behind door ', '{:b}'.format(t+salt), ' (salt: ', '{:,d}'.format(salt), ')]', sep='')        check = False        while check == False:            print('Pick a door between 1 and ', d, ': ', sep='', end='')            try:                choice = input()                if (str(choice).upper() == 'DONE') or (str(choice).upper() == 'BACK'):                    return                choice = int(choice)                check = True            except ValueError:                print('\nPlease enter an integer.')            except TypeError:                print('\nPlease enter an integer.')        if (str(choice).upper() == 'DONE'):            print('')            return        if (choice != t):            randomdoor = t        elif (choice == t):            randomdoor = choice            while (randomdoor == choice):                randomdoor = randint(1, d)        print('The treasure is behind door ', '{:,d}'.format(choice), ' or ', '{:,d}'.format(randomdoor), '! Pick a door: ', sep='', end='')        check = False        while check == False:            finalanswer = input()            try:                if finalanswer == 'k':                    finalanswer = choice                elif finalanswer == 's':                    finalanswer = randomdoor                else:                    finalanswer = int(finalanswer)                if (finalanswer == randomdoor) or (finalanswer == choice):                    check = True                else:                    print('Please pick door ', '{:,d}'.format(choice), ' or ', '{:,d}'.format(randomdoor), ': ', sep='', end='')            except ValueError:                print('Please enter an integer.')            except TypeError:                print('Please enter an integer.')        print('')        check = False        if (finalanswer == t):            success = success + 1            print('Correct! ', end='')        elif (finalanswer != t):            print('Incorrect! ', end='')        trials = trials + 1        print('The treasure was behind door ', '{:,d}'.format(t), '!', sep='')        print('So far you\'ve gotten ', success, ' out of ', trials, ' correct (', '{:.1%}'.format(success/trials), ').', sep='')def auto(d):    print('')    print('AUTO MODE ================================')    print('Enter "done" to go back.')    while True:        trials = 0        success = 0        print('')        print('Keep choice every time or switch doors every time? (k/s): ', end='')        check = False        while check == False:            mode = str(input()).lower()            if (mode.upper() == 'DONE') or (mode.upper() == 'BACK'):                return            elif (mode == 'k') or (mode == 's'):                check = True            else:                print('Please enter "k" (keep choices) or "s" (switch choices): ', end='')        check = False        while check == False:            try:                trials = int(input('How many trials?: '))                check = True            except ValueError:                print('Please enter an integer.')            except TypeError:                print('Please enter an integer.')        count = 0        while count < trials:            count = count + 1            t = randint(1, d)            choice = randint(1, d)            if (choice == t) and (mode == 'k'):                success = success + 1            elif (choice != t) and (mode == 's'):                success = success + 1            else:                pass        print('')        print(success, ' out of ', trials, ' successful cases (', '{:.1%}'.format(success/trials), ') by ', sep='', end='')        if (mode == 'k'):            print('keeping the original door choice.')        elif (mode == 's'):            print('switching to the remaining option.')initialize()

[Admiral Naismith]

On a side note, I had a class discussion where we were basically discussing the Monty Hall problem, except my math 'teacher' never had heard of it and gave me an F when I thought I had brilliantly explained the situation. She said the correct answer was 1/3 1/3 1/3 then 1/2 1/2....