6 ms·

I absolutely cannot psychologically deal with the Monty Haul paradox. https://en.wikipedia.org/wiki/Monty_Hall_problem https://en.wikipedia.org/wiki/Monty_Hall_

I absolutely cannot psychologically deal with the Monty Haul paradox. https://en.wikipedia.org/wiki/Monty_Hall_problem https://en.wikipedia.org/wiki/Monty_Hall_problem I've coded it multiple times in multiple programming languages as a cathartic exercise. Despite seeing the results obviously printed out in a console in front of me, I just can't handle it. I lose SAN just thinking about it.

spurgu 6y ago
A bit late to the party (came from Hacker Newsletter) but it seems obvious to me: - If you happen to choose a door with a goat (⅔ of the time), the host has to choose a goat door, leaving "switch" a guaranteed win - If you happen choose the door with the car (⅓ of the time), the host opens a random goat door and you have 50/50 between staying and switching So always switching is the correct move - if you do that you'll win the ⅔ of the time (every time you open a goat door).
cbruns 6y ago
The explanation that makes the most sense to me, from wiki: Yet another insight is that your chance of winning by switching doors is directly related to your chance of choosing the winning door in the first place: if you choose the correct door on your first try, then switching loses; if you choose a wrong door on your first try, then switching wins; your chance of choosing the correct door on your first try is 1/3, and the chance of choosing a wrong door is 2/3.
- JohnDeHope 6y ago
  The bit that gets me... If he opens one door, which is a goat, so there is 1 goat and 1 prize remaining... then the chance should be 50% staying put or switching, right? That's the part that gets me. The whole bit about 3 doors and opening one of them and "do you want to switch" is all a ruse, it seems to me. At the end of the day there are two doors, one with a goat and one with a prize, and however you choose between 2 doors, you will win or lose 50% of the time. Right? Aargh the SAN loss...
  - emiliobool 6y ago
    It would only be 50-50 if you could only pick from 2 options, but your initial choice is between 3 options.
mercer 6y ago
The only explanation that 'felt' acceptable to me is to imagine that you're really playing two games instead of one. In the first game, you're right 1/3rd of the time. Then you get to play a new game where you're right half the time, but you can also choose to not play that game and stick with the results of the first game. Playing the second game (picking another door) is more obviously the right thing to do in this case.
bjourne 6y ago
If it makes you feel better, famous Hungarian mathematician Paul Erdös didn't understand it either. One of his colleagues had to write a computer program simulating it before he accepted it to be true.
- BrandoElFollito 6y ago
  Not only him, the publication of the paradox in a column in Science (if memory serves me right) dragged thousands of condescending answers from top mathematicians (some of them apologized afterwards, including the one who called the journalist the goat of the paradox)
shaftway 6y ago
The explanation that makes it more visceral for me is to increase the number of doors: - You pick a door (let's say 17) - I open 98 doors (let's say 1 - 16, 18 - 78, and 80 - 100) - Only doors 17 and 79 are still closed Does that help make it more intuitive?

spurgu 6y ago

A bit late to the party (came from Hacker Newsletter) but it seems obvious to me: - If you happen to choose a door with a goat (⅔ of the time), the host has to choose a goat door, leaving "switch" a guaranteed win - If you happen choose the door with the car (⅓ of the time), the host opens a random goat door and you have 50/50 between staying and switching So always switching is the correct move - if you do that you'll win the ⅔ of the time (every time you open a goat door).

cbruns 6y ago

The explanation that makes the most sense to me, from wiki: Yet another insight is that your chance of winning by switching doors is directly related to your chance of choosing the winning door in the first place: if you choose the correct door on your first try, then switching loses; if you choose a wrong door on your first try, then switching wins; your chance of choosing the correct door on your first try is 1/3, and the chance of choosing a wrong door is 2/3.

JohnDeHope 6y ago
The bit that gets me... If he opens one door, which is a goat, so there is 1 goat and 1 prize remaining... then the chance should be 50% staying put or switching, right? That's the part that gets me. The whole bit about 3 doors and opening one of them and "do you want to switch" is all a ruse, it seems to me. At the end of the day there are two doors, one with a goat and one with a prize, and however you choose between 2 doors, you will win or lose 50% of the time. Right? Aargh the SAN loss...
- emiliobool 6y ago
  It would only be 50-50 if you could only pick from 2 options, but your initial choice is between 3 options.

mercer 6y ago

The only explanation that 'felt' acceptable to me is to imagine that you're really playing two games instead of one. In the first game, you're right 1/3rd of the time. Then you get to play a new game where you're right half the time, but you can also choose to not play that game and stick with the results of the first game. Playing the second game (picking another door) is more obviously the right thing to do in this case.

bjourne 6y ago

If it makes you feel better, famous Hungarian mathematician Paul Erdös didn't understand it either. One of his colleagues had to write a computer program simulating it before he accepted it to be true.

BrandoElFollito 6y ago
Not only him, the publication of the paradox in a column in Science (if memory serves me right) dragged thousands of condescending answers from top mathematicians (some of them apologized afterwards, including the one who called the journalist the goat of the paradox)

shaftway 6y ago

The explanation that makes it more visceral for me is to increase the number of doors: - You pick a door (let's say 17) - I open 98 doors (let's say 1 - 16, 18 - 78, and 80 - 100) - Only doors 17 and 79 are still closed Does that help make it more intuitive?