For some reason the internet still has weird arguments about the Monty Hall "paradox". I think it might be because sometimes the way it is explained is a bit confusing to some people? Idk, but I do remember that when I first heard about it the explanation was kinda weird to me until I rephrased it in simpler terms.
Here is the thing. You have 3 doors, one of them has a car train behind, the others have goats. You pick a door at random, the host doesn't open it, but he eliminates one of the other doors by showing it has a goat behind it, and asks you if you want to change your choice. So perhaps you picked door A first, he eliminated door B, and now your choice is between door A and door C. If you change your initial choice, it turns out that you will be correct 2/3 times, whereas if you don't you will only be correct 1/3 times. This confuses some people because they expect that since it becomes a choice between 2 doors it should be 50-50.
The simple explanation is that if you picked the door with the train the first time, then by changing you will always lose. If you didn't pick the correct door the first time, then changing always wins. But you will only get it right the first time 1/3 times, whereas you will get it wrong the first time 2/3 times, so changing is advantageous. That's it. That's all there is to it. There doesn't need any more mystification.
If it is still not intuitive, imagine if you had a million doors, you picked one, and the host eliminated all the others except one. There is (almost) no way you picked the right one the first time, it is literally a one in a million chance. So if it is not that one, and it almost certainly isn't, it must be the other which the host practically hand picked for you.
EDIT: For the reasoning to work, the important assumptions are two. One, the host always eliminates a goat, never a train. Two, the host always reveals one of the other two doors, not the one you picked. Both of these are significant, without the first one you have only a 1/3 chance of winning regardless of whether you change or not, and without the second one it becomes a regular 50-50 choice.
Another easy way to lay it out that isn’t too tedious since there’s only 3 doors is to just make a table to see all the possible outcomes.
For example, if the train is in door A, the possible outcomes are:
- Choose door A and don’t switch, win
- Choose door A and switch, lose
- Choose door B and don’t switch, lose
- Choose door B and switch, win
- Choose door C and don’t switch, lose
- Choose door C and switch, win
So switching wins 2/3 of the time and staying only 1/3.
There's also the fact that the first person to famously solve the "paradox" was a woman. How is this possible? Am I truly being bested in logical reasoning by... a woman?
Edit: Marilyn vos Savant, her name sounds kinda badass, like a genius superhero or fantasy protagonist's name.
She wasn't the first person, it was an older problem which was restated in a few different forms, and it was solved basically the moment it was discovered, it's pretty trivial compared to what probability theorists and statisticians usually deal with. Vos Savant was a columnist who supposedly had a really high IQ and people sent her puzzles to solve. However, the puzzles usually already had known solutions. It is definitely true however that this is why a lot of people who read her got pissed, they were annoyed that she was a woman and were very intent on owning her.
Interestingly, she herself has gone against mathematical consensus for silly reasons a few times since.
to be fair, you have to have a very high IQ to understand decision-making trees applied to hypothetical game show conundrums
True, true. I meant that she did it first in a very public way. You know, it must really suck to be famous for being a genius. How are you supposed to deliver on these expectations?
You know, it must really suck to be famous for being a genius. How are you supposed to deliver on these expectations?
True.
It took me until the end of this post to realize I misread and you weren't talking about Monty Python
From what I remember reading in interviews with Monty Hall he said that they were rather inconsistent with offering the switch to begin with (sometimes opening right away, or offering a small money reward instead) and would normally do it whenever someone had picked the right door at the start so sorry, but you're actually wrong you shouldn't swap learn 2 real life
Yes, it wouldn't, which is why you can't use it to yourself. The reason is that if the host just randomly eliminated one, then he'd eliminate the train every now and then.
When I first heard of it I was in school and I tried to figure out how I could use it to do better at multiple choice questions, until I realized that's not how it works lol
The real potential of the Monty Hall problem is to assess the lengths people go to in order to avoid changing their minds.
The Monty Hall Problem is an example of language being mistranslated when modeled in the logic of computing.
Every other detail in the Monty Hall Problem is wordplay and mathematical sleight-of-hand distraction.
reddit twitter are worthless solipsists
The global math community has collectively gotten it completely wrong. The Monty Hall Problem should be a case study for at least the next hundred years of how an entire world of smart people managed to thoroughly fool themselves with bad logic and assumptions.
we already know :epstein:
These explanations are good, but you have to be a little careful because I think there's a risk of them also giving you the wrong intuition.
For example, imagine Monty has slightly different rules- instead of always eliminating a door with a goat, he randomly picks one of the doors the contestant didn't choose and eliminates it, even if it's the one with the train (in which case they're shit outta luck). Today you're on the show, you pick a door, he eliminates one at random and- good news, he eliminated a goat. Now do you switch or stay with your current door?
In this case, it actually doesn't matter whether you switch or not, it really is 1/2 either way. But it superficially sounds like your reasoning for the real Monty Hall problem should still work- if you initially picked a goat then the remaining door is a train, and vice versa.
it superficially sounds like your reasoning for the real Monty Hall problem should still work- if you initially picked a goat then the remaining door is a train, and vice versa.
I don't think so, because the fact that if you change you win hinges on the fact that the train is still there to be found and hasn't already been revealed by the host. If the host revealed the train then obviously you've lost immediately. So it's not even 50-50 really, 1/3 times you will just lose immediately by having the train revealed.
You're right that before he gets involved it's 1/3. But I'm talking about the situation after Monty eliminates his door and reveals that it's a goat... which doesn't always happen- sometimes it's a train! But let's say you're actually on the show, and this has happened, and in this particular case he revealed a goat.
Yeah, what I am saying though is that if the host doesn't know which door has the train behind and he just reveals one of the doors, your total chance of winning either by changing or not changing is just 1/3, because you don't get any new information, if the host accidentally reveals the one with the train you just lose. You can just do away with the host entirely, you can just do it yourself.
Just being pedantic: There's three doors, the train has to be behind one of them, so the probabilities for each door have to sum to 1. He just revealed one door has a goat, so the probability of that door having the train is 0, so the other two can't both be 1/3, otherwise you're missing a bit of probability. You're right that the probabilities are equal for the two remaining doors, but that makes 1/2 each, not 1/3. You can use the million door analogy here- he opens 999,998 doors at random and there's goats behind all of them. You're now a lot more confident than when you started that you picked the right door, so your chances definitely feel better than 1 in a million now!
But my (hopefully less pedantic) point is that in this situation (three doors, you picked one, then Monty eliminated one at random and showed that it's a goat), you could still say: "The simple explanation is that if you picked the door with the train the first time, then by changing you will always lose. If you didn’t pick the correct door the first time, then changing always wins.". That's all still true: if you originally picked a goat and Monty eliminates a goat, you win by switching, if you originally picked a train and Monty eliminates a goat, you win by staying. But this time that would lead you to the wrong conclusion (that switching is better than staying).
He just revealed one door has a goat,
IF he revealed it has a goat, yes. I am talking about the full game though. It is equivalent to a problem without any host at all. Basically it is like picking one door at random, then looking at another one and deciding you're never gonna pick it, and then rethinking which of the other two you're gonna pick. You don't learn anything new, so it is just 1/3. You are right that it is 50-50 if he has just happened to reveal the goat, but 1/3 times he just reveals the train and you're fucked. So 1/3 times you're immediately fucked, and 2/3 times it becomes a 50-50 thing (although that needs a bit of explaining for why it is the case which I won't go into), so in total you have a 1/3 chance of winning overall.
Yep, I agree with most of this, I just think it may not be intuitively obvious that looking at a door picked randomly and seeing it's a goat gives you less information (or less useful information) than being told by somebody who already knew it was a goat.
Yes, it is slightly more confusing. The key to understanding why that is is realizing that if he happened to reveal a goat, then that lends slightly more credence to the hypothesis that you picked the right door in the first place. That's because of you picked the wrong door, the host revealing one of the other two at random has a 1/2 chance of revealing a goat. On the other hand, if you picked the right one, then the host will always reveal a goat. If you work it out, 1/3 times in total he reveals the train and you lose, 1/3 times he reveals a goat and your initial choice is incorrect, and 1/3 times he reveals a goat and your initial choice is correct. So if you see that he revealed a goat, whether you switch or you don't doesn't make a difference.
what helped me get it was splitting the choices into 2 groups:
your door = 1/3 chance of winning total
all the other doors = 2/3 chance of winning total
say somebody gets to pick 'all the other doors' as an option; we know they have a 2/3 chance of winning
that chance will stay the same no matter what, since the goats/trains aren't getting reshuffled
so if doors get revealed as duds in this group, you would now just know that 2/3 can all be placed on whatever door(s) remain
tired: believing that both doors in the Monty Hall problem have an equal chance of hiding the car
wired: believing that nothing happens when the guy in the blue-eyed islander problem points out that someone has blue eyes
inspired: rejecting Mochizuki's brilliant proof of the abc conjecture because you're a stupid Westerner who doesn't know what "and" means
Idk if it is a joke but from what I've read so far most people think Mochizuki is wrong and he is definitely very annoying about it and not doing a great job at explaining even if he is right. Unfortunately there is like 50 people in the world who understand enough about this to have a valid opinion and most of them won't bother to really engage very deeply with the subject. However it seems like most experts agree with Scholze and Stix, and none of the people who claim it is correct have really provided a decent explanation for why Scholze and Stix are wrong, so...
Have you not heard of the famous comedy troupe Monty Hall python? Not sure I trust your post now
I still don't understand the problem. Do you get to keep the goat? You have a 100% chance of success.
Only if you interact with game show hosts who do this particular thing I guess? Otherwise no, because for it to work you need that specific kind of set up. You need the host to come in and provide that extra bit of information in this way, otherwise it can be worked out that it won't make a difference. When I first learned of it I was in school and I tried to figure out how I could maybe use it to do better at multiple choice questions until I figured out you require a "host" who already knows exactly what the answer is for it to work, otherwise it doesn't matter. The important part that makes it work is that the host knows which door has it and always reveals a door which doesn't have it, and that the door they reveal is not the one you picked in the first place.
This problem only works assuming the host does not care if you win or lose. In real life they probably want you to lose, so no, don't trust them.
