applecinnabun

in the monty hall problem, if you remove monty hall and replace him with an actual random number generator, that actually has a chance to open a door with a car, and it happens not to, now it's actually 50/50 between the two remaining doors right? it's specifically monty hall, his knowledge of where the car is, and his methodology for opening a door that creates the unintuitive effect? or is that still there in this case...

applecinnabun @applecinnabun4/5/2024, 3:15 AM

ok i did not explain this very well, but if you're curious, i wrote a little python script showing that the thing i am poorly trying to explain was correct, i think?:

as @belarius helpfully put it in the comments, the situation i'm explaining is that rng monty hall could have opened the door with the car, and could have opened your own door, but didn't. the situation where he didn't has already occurred. it's like how the odds of flipping two heads is 1/4 but the odds of flipping a second heads given that the first has already occurred is 1/2.

here's a python script that rerolls car locations and opened door locations until the "keep or switch" scenario from the original monty hall problem occurs (a door has been opened, it's not your door, the door did not have the car) 3000 times. if i've coded that correctly, it shows that keeping your door #1 under these conditions results in winning the car ~1500 times out of every 3000, or 1/2.

import random

# the number of times our door, door 1, contained the car, given that the scenario has occurred
successes = 0

# the number of times the specific scenario has occurred, where a door that isn't yours has been opened, and that door did not contain the car
montyhallscenarios = 0

while(montyhallscenarios < 3000):
    #car and opened door can be in doors 1 through 3
    car = random.randrange(1,4)
    door = random.randrange(1,4)
    
    #if the opened door has the car, then this isn't the "stay or switch" scenario
    if door == car:
        continue
    
    #if the opened door was our door #1, then this isn't the "stay or switch" scenario
    if door == 1:
        continue

    #if this *is* the "stay or switch" scenario, then we get a point if the car is in door 1!
    else:
        montyhallscenarios += 1
        if car == 1:
            successes += 1

print(successes)

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in reply to @applecinnabun's post:

@wooby4/5/2024, 2:33 AM

i'd imagine it's still 66/33? you could model monty hall as a random number generator that just so happens to open a non-car door every time, which would still fit your description. the 2/3rds chance comes from the fact that at the beginning you have a 1/3 chance to choose wrong, not from any hidden knowledge.
imagine a hundred door problem; you pick one with a 1/100 chance of having the car, monty/rng opens 98 doors with goats behind them, there are 2 doors left. your door still only has a 1/100 chance! that didn't change! but the other door stayed closed through 98 door opening events; the chances that it has a goat are pretty astronomical!

Robin @RobinProblem4/5/2024, 2:36 AM

Yeah, basically this. It's the simple fact that your first choice has a 1/3 chance of being correct and the car being behind one of the other doors has a 2/3 chance of being correct (and a goat door being opened for you effectively means you get to open both of those doors). Mr. Hall has nothing to do with it.

Signed by someone with a math degree.

applecinnabun @applecinnabun4/5/2024, 2:38 AM

but in the world where rng chooses the doors, and there are 100, the odds that rng hasn't already revealed the car are also extremely small either way i would think?

the thing i'm trying to get at is that if it's monty hall, there's a 0% chance he opens the door with the car, and if it's rng, there's a 33% chance it opens the door with the car, and i'm wondering if that changes things...

Robin @RobinProblem4/5/2024, 2:41 AM

For the situation where the rng opens a goat door it does not change things whatsoever. It's the exact same scenario as Monty Hall opening a goat door. It doesn't change where the car is at all. The odds still remain the same as the original scenario.

invertigo hayfever @the-doomed-posts-of-muteKi4/5/2024, 3:21 AM

The trick is that if you're throwing out any trials in which the RNG opens up the door with the car, outside of the total number of possible accepted outcomes you should see that the improved odds of switching will hold.

If you don't remove the cases where this happens then you'll include all the cases where switching doesn't improve your chances, because your only other choice is another empty door.

(EDIT: also the case mentioned by Belarius in the example below me)

Ultimately: if the RNG can't choose your door there is either an advantage to switching or the game is over already.

applecinnabun @applecinnabun4/5/2024, 3:27 AM

this is true! the reason i throw out the trials in which the RNG has already opened up the car door is related to the phrasing of the original problem.

in the original phrasing, you are presented with a scenario where a door has already been revealed to not have the car, and you're asked if you should switch or not. people's intuition about this is that it's 50/50, which is false.

but if it were rng, as people's intuition tells them, and if the trials in which the RNG has already opened up the car door have already been thrown out, because we're already at the scenario where a door was revealed to not have the car, as i think people's intuition might tell them, then they would be correct in saying it's 50/50. i think. and my hope was that that was helpful to people but it's so hard to communicate that it clearly isn't lol

Belarius @belarius4/5/2024, 2:39 AM

If Monty Hall's behavior is truly random, and he has already opened a door you didn't choose to reveal the dud option, then you are correct: It means the remaining uncertainty is distributed among the two doors, one being your original pick, and the other being the third untouched door. Put another way: You've imposed a condition (that an event had already unfolded a certain way), so you're asking about a different conditional probability than that in the original scenario.

What's tricky about RNG Monty Hall is that any way you slice it, he's sometimes going to open a door that has a car behind it, and your description leaves ambiguous whether he can open the very door you chose or not. Presumably, we constrain him so he must choose randomly between the two as-yet unchosen doors (so your eventual choice is to stay or to switch), in which case he'll reveal a car by accident 1/3 of the time (in which case you should always switch to the open door, obviously). Your even having an ambiguous option to stay or switch (which assumes no car has been revealed) is being divvied up between the other 2/3 of the possible timelines.

applecinnabun @applecinnabun4/5/2024, 3:07 AM

thank you! yes, you got to the bottom of what i was trying to say. the problem i had was in accurately describing the scenario without decades worth of statistics discourse baggage coloring what i'm saying lol. gotta word statistics scenarios like wishes to genies i swear...

the important part is that yes, RNG monty hall can open the door we chose, and yes he can open the door with the car, and we're now discussing the situation where he already hasn't done either of those things. an event has already unfolded a certain way, as you said.

the reason i was thinking about this is that i think the reason monty hall is so unintuitive to people might have to do with a belief that they are living in the world i'm describing here! but seeing as i've failed to communicate that scenario to anyone but you this might not be the helpful tool towards helping people understand it that i hoped it would be :p

Belarius @belarius4/5/2024, 3:21 AM

These problems are good as brain teasers but really treacherous as teaching tools, because intuitions about conditional probabilities routinely fail to take into account all the counterfactuals. What's nice about this situation is that the number of possible timelines is small enough that one can just chart all of them and then see what's left. In your case, RNG Monty failing to reveal a car does indeed give you information, but it's different information than Omniscient Monty.

Extending this to the "Monty opens N doors" scenario is instructive. If there are 100 doors and one car, and RNG Monty opens 98 of them, he's very likely going to reveal the car, and he's also very likely going to pick your door. Only a tiny number of contestants will be left with two ambiguous doors. In that very small percentage of cases that a genuine choice remains, the two remaining doors (yours and the other unopened one) end up once again at 50/50 odds. However, if Omniscient Monty opens 98 doors, your chances of getting the car are 99% if you switch because Omniscient Monty will never pick your door and never reveal a car, causing all the uncertainty from the original 99 unselected doors to pile up in one place.

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