Comment on What are the most mindblowing fact in mathematics?
Artisian@lemmy.world 11 months ago
For the uninitiated, the monty Hall problem is a good one.
Start with 3 closed doors, and an announcer who knows what’s behind each. The announcer says that behind 2 of the doors is a goat, and behind the third door is a car student debt relief, but doesn’t tell you which door leads to which. They then let’s you pick a door, you get what’s behind the door. Before you open it, they open a different door than your choice and reveals a goat. Then the announcer says you are allowed to change your choice.
So should you switch?
The answer turns out to be yes. 2/3rds of the time you are better off switching. But even famous mathematicians didn’t believe it at first.
But even famous mathematicians didn’t believe it at first.
They emphatically did not believe it at first. Marilyn vos Savant was flooded with about 10,000 letters after publishing the famous 1990 article, and had to write two followup articles to clarify the logic involved.
Oh that’s cool - I had heard one or two examples only. Is there some popular writeup of the story from Savant’s view?
I couldn’t tell you - I used the Wikipedia article to reference the specifics and I’m not sure where I first heard about the story. I just remember that the mathematics community dogpiled on her hard for some time and has since completely turned around to accept her answer as correct.
I know it to be true, I’ve heard it dozens of times, but my dumb brain still refuses to accept the solution everytime. It’s kind of crazy really
To me, it makes sense because there was initially 2 chances out of 3 for the prize to be in the doors you did not pick. Revealing a door does not reset the odds of the whole problem, it is still more likely that the prize is in one of the door you did not pick, and a door was removed from that pool. Imo, the key element here is that your own door cannot be revealed early, so it is never “tested”, and this ultimately make the other door more “vouched” for, statistically, and since you know that the door was more likely to be in the other set to begin with, well, might as well switch!
like on paper the odds on your original door was 1/3 and the option door is 1/2, but in reality with the original information both doors were 1/3 and now with the new information both doors are 1/2.
Yes, you don’t actually have to switch. You could also throw a coin to decide to stay at the current door or to switch. By throwing a coin, you actually improved your chances of winning the price.
This is incorrect. The way the Monty Hall problem is formulated means staying at the current door has 1/3 chance of winning, and switching gives you 2/3 chance. Flipping a coin doesn’t change anything. I’m not going to give a long explanation on why this is true since there are plenty other explanations in other comments already.
This is a common misconception that switching is better because it improves your chances from 1/3 to 1/2, whereas it actually increases to 2/3.
This explanation really helped me make sense of it: Monty Hall Problem (best explanation) - Numberphile
It took me a while to wrap my head around this, but here’s how I finally got it:
There are three doors and one prize, so the odds of the prize being behind any particular door are 1/3. So let’s say you choose door #1. There’s a 1/3 chance that the prize is behind door #1 and, therefore, a 2/3 chance that the prize is behind either door #2 OR door #3.
Now here’s the catch. Monty opens door #2 and reveals that it does not contain the prize. The odds are the same as before – a 1/3 chance that the prize is behind door #1, and a 2/3 chance that the prize is behind either door #2 or door #3 – but now you know definitively that the prize isn’t behind door #2, so you can rule it out. Therefore, there’s a 1/3 chance that the prize is behind door #1, and a 2/3 chance that the prize is behind door #3. So you’ll be twice as likely to win the prize if you switch your choice from door #1 to door #3.
It’s a good one.
Evirisu@kbin.social 11 months ago
I know the problem is easier to visualize if you increase the number of doors. Let's say you start with 1000 doors, you choose one and the announcer opens 998 other doors with goats. In this way is evident you should switch because unless you were incredibly lucky to pick up the initial door with the prize between 1000, the other door will have it.
Artisian@lemmy.world 11 months ago
I now recall there was a numberphile with exactly that visualisation! It’s a clever visual
TheSaneWriter@lemmy.thesanewriter.com 11 months ago
It really is, it’s how my probability class finally got me to understand why this solution is true.
dandroid@dandroid.app 11 months ago
This is so mind blowing to me, because I get what you’re saying logically, but my gut still tells me it’s a 50/50 chance.
But I think the reason it is true is because the other person didn’t choose the other 998 doors randomly. So if you chose any of the other 998 doors, it would still be between the door you chose and the winner, other than the 1/1000 chance that you chose right at the beginning.
Kissaki@feddit.de 11 months ago
I don’t find this more intuitive. It’s still one or the other door.
crate_of_mice@lemmy.world 11 months ago
The point is, the odds don’t get recomputed after the other doors are opened. In effect you were offered two choices at the start: choose one door, or choose all of the other 999 doors.
Not_Alec_Baldwin@lemmy.world 11 months ago
This is the way to think about it.
moreeni@lemm.ee 11 months ago
The thing is, you pick the door totally randomly and since there are more goats, the chance to pick a goat is higher. That means there’s a 2/3 chance that the door you initially picked is a goat. The announcer picks the other goat with a 100% chance, which means the last remaining door most likely has the prize behind it
SgtAStrawberry@lemmy.world 11 months ago
Don’t be sorry, your comment was the first time I actually understood how it works. Like I understand the numbers, but I still didn’t get the problem, even when increasing the amount of doors. It was your explanation that made it actually click.
Elderos@lemmings.world 11 months ago
I think the problem is worded specifically to hide the fact that you’re creating two set of doors by picking a door, and that shrinking a set actually make each individual door in that set more likely to have the prize.
Think of it this way : You have 4 doors, 2 blue doors and 2 red doors. I tell you that there is 50% chance of the prize to be in either a blue or a red door. Now I get to remove a red door that is confirmed to not have the prize. If you had to chose, would you pick a blue door or a red door? Seems obvious now that the remaining red door is somehow a safer pick. This is kind of what is happening in the initial problem, but since the second ensemble is bigger to begin with (the two doors you did not pick), it sort of trick you into ignoring the fact that the ensemble shrank and that it made the remaining door more “valuable”.
SnowmenMelt@lemmy.world 11 months ago
The odds you picked the correct door at the start is 1/1000, that means there’s a 999/1000 chance it’s in one of the other 999 doors. If the man opens 998 doors and leaves one left then that door has 999/1000 chance of having the prize.
UntouchedWagons@lemmy.ca 11 months ago
Same here, even after reading other explanations I don’t see how the odds are anything other than 50/50.
eran_morad@lemmy.world 10 months ago
read up on the law of total probability. prob(car is behind door #1) = 1/3. monty opens door #3, shows you a goat. prob(car behind door #1) = 1/3, unchanged from before. prob(car is behind door #2) + prob(car behind door #1) = 1. therefore, prob(car is behind door #2) = 2/3.
clumsyninza@lemmy.world 10 months ago
How do we even come up with such amazing problems right ? It’s fascinating.
Sharkwellington@lemmy.one 11 months ago
This is fantastic, thank you.