I'm Frankie, I play TTRPGs too much. I will be reblogging and/or posting a lot of furry arte and some of that's going to be kink stuff so heads up
AKA Nerts but that's going back a while


GoopySpaceShark
@GoopySpaceShark

the monty hall problem is absolutely bullshit. the odds between the door you picked and the door you didn't are the same, they don't magically have different odds just because you picked one prior

you pick A. B is shown to be false, so you're given the choice between A and C. A and C are both still unknown to you and thus have equal odds. A no longer has a 1 in 3 chance, because there's only two doors to pick from now.


GoopySpaceShark
@GoopySpaceShark

sorry, but none of you can convince me that this isn't some sort of fucked up maths major in-joke or hazing thing


overnecked
@overnecked
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fwankie
@fwankie

Yeah, I feel like every explaination out there is for boomers/americans who know who Monty Hall is (not even a mathematician) and have watched a bunch of whatever gameshow the problem references, and already know the rules of the game being played, so they don't explain them very well.
It's like if I mathematically prove that running sideways around your target in a circle while shooting at it is mathematically optimal in terms of avoiding return fire and being able to hit the target, telling everyone they're wrong and going over the maths again if they say I can't run like that or that bullets are too fast to dodge, without mentioning I mean in Quake 3.

(also the problem also says that you know he knows the prizes and the effects that has on his choices, but not that he knows that you know that he knows, and intuitively you would think he might bluff, which is usually an unwritten rule of explanations of the problem that he's not playing against you, he's following a script. I assume that's because that's how the gameshow goes?)


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in reply to @GoopySpaceShark's post:

Consider this:
there are 1 million doors
you pick one at random
then all of them are opened except yours and another door
Do you switch?

Or, we can chart all the possibilities for the 3 door case
the prize is marked with "X", non-prizes with "N", and opened doors with "O", so our 3 door combo can be described as
|X N N| (behind door A) (1/3 chance)
|N X N| (behind door B) (1/3 chance)
|N N X| (behind door C) (1/3 chance)
let's say you pick door C, and the gameperson opens another empty door at random.
then our games would look like this
|X O N| (1/3)
|O X N| (1/3)
|O N X| (1/6) or |N O X| (1/6)
our odds of winning are still 1/3 at this point. (it's the 1/6+1/6 cases)

but what if we switch?
then we switch to the (1/3) and (1/3) cases, making it 2/3 in total.

i mean - i laid out every possibility, and the numbers add to 2/3.
you can take out a pencil and paper and write it out, and im sure you will come to the same conclusion.

Consider 10,000 doors.
if 9,999 are opened, except yours and one other, does that feel like 50/50? or does it feel like you gained new information when 9,999 doors were opened?

the opened door B is not chosen randomly, it depends on what's behind your choice A: if A is one of the junk prizes then B is always chosen to be the other junk prize. (it would be so cruel if B was the big prize and the host still asked you if you wanted to swap, wouldn't it?) so if A had junk in it (2/3 chance when you chose) then swapping to C will get you the big prize

you're right, A only has a 2/3 chance of being junk! in the remaining 1/3 chance, where A was the big prize, you'd prefer not to swap, because no matter what the host picks to be B, C is still junk.

but you don't know what kind of prize A holds, only the host knows! so from your perspective you can't tell if "A is junk, so i should swap after reveal" or "A is good, so i should keep after reveal" is true.

all you do know is there is a 2/3 probability that A is junk when you pick it, and a 1/3 chance it's the big prize. what's behind A doesn't change when the host picks and reveals B. so C is the big prize with 2/3 chance, and junk with 1/3 chance

I don't see how not switching increases the odds.
It was at 1/3 before, and you have done nothing. Why would the probability rise to 1/2?

Me listing out all the possibility leaves much to be desired, but did you see it? do you see how the math I have used makes the numbers equal to 2/3 if you switch?

How did you list the possibilities, and how do you think it adds to 1/2?

is that a sincerely held belief? it may be counterintuitive but there really is a 2/3 chance of winning monty hall if you swap. it's taking all my willpower not to launch into an explanation 😅 i am too easily baited into these things

I normally don't comment on your stuff but you prompted me to finally look up the problem and solution so thank you. My instinct was the same but it has to do with the host's choice of what to reveal not being random.

There's a 1/3 chance that both unchosen doors are false so it contributes 1/3*0=0 to the chance the unchosen, unrevealed door is true. Switching is wrong 1/3rd the time.

There's a 2/3 chance that one of the unchosen doors is true. The host always picks an unchosen door that is false. This removes the unchosen false door, meaning the unchosen, unrevealed door is true in both configurations (B true or C true given you picked A). This contributes 2/3*1=2/3 to the chance the unchosen, unrevealed door is true. Switching is right 2/3 the time.

The reason A doesn't become a 1/2 chance is because A can't be revealed by the host so not picking it doesn't tell you anything about it - it happens regardless of A being right.
B/C increases in chance because most of the time, it not being revealed means it has to be true. This is why the odds aren't equal despite both surviving doors being unknown.

EDIT: Sorry, I missed your follow-up post.

in reply to @GoopySpaceShark's post:

it might as well be because nobody ever mentions the the really crucial detail that monty knows where the car is and will always open an unopened door without the car in their explanations, and it's always just sort of assumed or glossed over in the description of the problem. if this isn't the case and monty just picks an unopened door at random, the intuitive answer is correct (and it's Deal or No Deal). if the behaviour of the host is unspecified either answer is just as valid. but everybody is just like "no it's unintuitive, look, imagine what happens if there's a million doors / play with these simulations" without realising that they're talking at cross purposes. sorry you probably don't want another explanation this thing just makes me upset. a lot of people come out of this problem thinking there's something wrong with their understanding of math because they've come up with a correct answer to a trick question.

wait what

I'm gonna be real, that crucial bit of context would've been infinitely more helpful than literally every single one of the answers on reddit, quora, etc combined, which could pretty much just be summed up as calling the OP an idiot in increasingly offensive ways. saw a few slurs, even.

yeah it feels like mathematics has a serious communication problem

in reply to @fwankie's post:

it doesn't really "make sense" it just becomes the optimal way to play when you realise that the problem operates on a script like a videogame rather than like a gameshow where the host can react to you

It's because time is linear. When you chose A, there was a 1/3 chance you were right. The revelation that B is false doesn't retroactively change that probability. There's still only a 1/3 chance that you picked correctly.

It's a trick, basically; a trap made of extraneous information. Nothing changes when one of the doors you didn't pick is revealed as fake. It doesn't matter. The probability that you picked right the first time doesn't change.

Yeah, it's a 1/3rd chance to pick the car the first time, but because the host isn't random and will never reveal the car, then switching every time means the 1/3 chance of picking the car first makes you lose, but your 2/3rds chance of not picking the car first means it'll be the one you switch to.