people still aren't going to get it, but that's basically the heart of the issue. the host has access to knowledge that the contestant doesn't, and specifically alters the situation based on him being able to see/know what's behind the doors.
> Three doors > Choose a door > Wrong door > You are given a chance to choose from the remaining doors > 50/50 chance. Choose a door. > "Are you sure about that? You can still choose the other door." > Two possibilities. > A) This is an attempt to instill uncertainty in you, cause you to think and overthink. That rabbithole of psychological warfare extends infinitely, and no matter what angle it is approached from will not yield constructive conclusions. > B) This is mere showmanship, a ploy to generate tension and suspense in the audience. > No way to know for sure. No definite information gained. > Changing options effectively accomplishes nothing. > Either way, odds are still unchanged. Still two options. 50/50. > "~buuuh the host noe da ansssar so da chans go up~"
Shut the frick up moron.
You choose a door. You have a 1/3 chance of picking the correct door. This means there is a 2/3 chance the correct door is one of the other two doors. Monty opens one of the other doors to expose a goat. The odds have not changed, though; your door still has a 1/3 chance to `be correct while the other two doors have a 2/3 chance, but since we now know which one of those two doors is the goat, the other door itself has a 2/3 chance of being the winning door. Thus, by switching your choice, you double your chance to win.
Everyone who doesn’t understand the problem after it’s been explained a million times is a nitwit or troll and these threads should be banned.
people still aren't going to get it, but that's basically the heart of the issue. the host has access to knowledge that the contestant doesn't, and specifically alters the situation based on him being able to see/know what's behind the doors.
Glad to see that people on this board actually understand the problem.
A easy way to visualize the problem is to push it to the extreme.
There are a thousand doors. 1 Door has the perfect $100K crypto portfolio (99% Chainlink/0.5% ETH/0.5% BTC) and the 999 other doors each have a goat behind of them.
You choose 1 door, and then host will open 998 doors, all of them each revealing a goat behind the respective doors.
Do you stick with the door you chose at random, or do you switch to the other lone door (out of the 999 he could have left unopened) the host did not open up?
Also, this puzzle was used in Zero Escape Zero Time Dilemma, and people got screwed over by correctly choosing the right one the first time around lol
Still too ambiguous.
We can go further.
There are 10*e^999 doors. 1 Door hides a pristine turkey sanwich, but every other door contains and additional 10*e^999 doors, which themselves are characterized by an identical distribution of prizes.
How many doors would you need to open before you could fully reconstruct the turkey from the harvested sanwich meats?
Still not understanding that point people are making who are claiming that the problem boils down to a 50/50 chance are arguing that the statistical abstraction that you've created does not map on to reality. It doesn't matter how loudly you play the tune, you're still out of key.
That still doesn't explain things enough for most people.
Here's a better explanation:
Your first pick had a higher chance of being a goat. Since Monty removes the other goat door, the chances of the car being behind the remaining hidden door initially is higher than the door you picked first, so you should switch to have a higher chance of winning the car (since your initial pick was most likely wrong to begin with).
It's all because the initial odds were not in your favor (you had a 2 out of 3 chance to pick a goat).
The final hidden door has a higher chance of being the car to begin with.
Obviously switching doesn't guarantee anything, you could have gotten lucky on your initial pick, but you have no way of knowing that.
Because Monty can not reveal the door you picked, and always removes one of two losing doors.
Your initial pick was PROBABLY wrong, so the last remaining hidden door is PROBABLY the car.
Because your initial pick never stops being 33% even when there's just two doors. Monty opening a goat door allows you to capitalize on this and switch to a door with odds that are the reverse of yours.
Monty Hall knowing the answer has absolutely nothing to do with it. The only thing that matters is that the door he opens has a goat, which is always how the question is posed.
You're thinking about the choice to switch. You should be thinking about the odds of the individual doors themselves. Your first pick is always 1/3 odds no matter what happens after.
>pick is always 1/3 odds no matter what happens after.
Except it's not you moron
The moment you agree to switch, all doors gets odds redistributed
First one included
The problem is that your first choice is factored into the reveal. Anything you didn't choose, with the exception of one other, is removed.
It's not the odds of the apparent problem (2 doors or 1:2 odds) that you're playing, it's the odds that the reveal just gave you the correct answer (2:3 if you swap, 1:3 if you stay).
Long version:
A better way to think about it is that your inital choice always has a 2/3 chance of being wrong, this is pretty easy to grasp.
The host however knows where the prize is and will use that information to always open a door that has a goat regardless of your initial choice, thus the chance of your initial door being wrong has not changed, the location of the prize is always the same before and after he reveals the goat.
Since theres only one other option at this point switching gives you a 2/3 chance of getting the prize.
You have a 33% chance of having the correct door. There a 66% chance of one of the other doors being correct.
Monty opening or closing the doors does not ever change these odds, your choice is still "Do I stay with a 33% chance of being correct, or switch to a 66% chance that my original guess was incorrect"
>Make choice, Monty makes his >Go backstage and take out contestant who wasn't watching the show >Tell them to decide which door
Alright, so what are the odds now moron?
>oh yeah well if I completely change the question then the previous answer is no longer correct, CHECKMATE
If the contestant doesn't know which door was initially picked, it's 50/50
If they are aware of what you chose, and can see the opened goat door that Monty chose, then the best choice for them would be to choose the unopened door that you didn't pick. They'd be right 2/3 times
>Witnessing Monty pick a door changes the odds
So you admit the odds can/do change?
10 months ago
Anonymous
If Monty Hall shoots you in the dick and then rapes your bullet wound, do you still suck his dick?
10 months ago
Anonymous
oh whoops lol, I misread who you were replying to.
Yes the odds when switching do change
If they are aware of what you chose, and can see the opened goat door that Monty chose, then the best choice for them would be to choose the unopened door that you didn't pick. They'd be right 2/3 times
dude its bait, everyone actually gets its not 50/50 but they pretend it is so we can have these threads. literally the same points are parroted every time this is posted >just imagine 1000 doors! >its only 2 doors, so 50/50! >the host knows, thats why its not 50/50! >flowchart autism
at least the portal threads and coach/watch image baits are actually problems without objectively correct answers so its fun to argue in those, here you are just arguing with fishermen and their bait
This type of statistics shit is and will be always a meme because they don't take into account time.
If there were 100000 doors, you choose one and they open every other door except the one you chose and another one the odds are still 50/50 NOW
You have a 1 in 3 chance initially of picking right, but the parameters change when they remove one door that is explicitly the correct answer, now you have a 50/50 chance of getting it right instead of 1/3.
The second choice is not the same as the first.
The second choice is heavily informed by the first, anon. You can't just ignore the context of the situation.
If there are 1,000,000 doors and you pick one, then 999,998 doors are opened, leaving only your door and one out of 999,999 remaining, which one is more likely to have been the correct choice?
The monty hall problem is essentially betting on whether or not your first choice was correct. Since statistically your first choice only had a 33% chance of being correct, it is always a safer bet to bet "I was wrong the first time" and then switch
I used to think this way too, but it factually is. >pick an option, 1 in N chance to be correct >reveal: now majority of other options are removed except for your choice and one other option >one of those 2 is guaranteed correct
Think about it. First it's 1 in N because any of them could be correct, but then it's narrowed to 2 options and one of them is guaranteed correct.
If you choose not to swap to the other option after the reveal, you're still taking those 1 in N odds, BUT if you decide to swap, you actually have better odds since it's more likely that the other option is correct.
It's not readily apparent for just N=3 options, but for larger values of N, the likelihood of the non-chosen option being correct becomes much more favorable.
10 months ago
Anonymous
>>pick an option
it does not matter what you pick initially. you will always have a coin toss between a car and a goat. the only reason that you think that you're switching is because Monty said so. In reality is a coin toss. every goat is discarded except the goat necessary to manufacture a coin toss.
10 months ago
Anonymous
The part where your 2nd choice is informed by the reveal is that your odds have changed if you switch, but have not changed if you don't.
So using anon's 1,000,000 option example, you are taking 1 in 1,000,000 odds if you don't switch.
You are very much less likely to have chosen correctly if you stick to your first guess because the correct answer was *more likely to be isolated by the reveal*, and your first choice was really unlikely to be the correct choice.
imagine instead of 3 doors, there's 52. and instead of doors, they're playing cards. i, the host, tell you, the player, that if you draw 1 random card from a deck of playing cards, and it's the king of hearts, you get a prize. if it's any other card, you don't get anything.
so i shuffle the deck, and you draw a card and place it face-down infront of you. it's pretty low odds that it's the king of hearts, right? 1 in 52.
now, before anything else happens, i pick up the remaining 51 cards, and start looking through them, looking for the king of hearts. if i find it, i'll put it face-down infront of myself. if i don't find it (because you happened to draw it yourself) then i'll just put a random card face down infront of myself.
so i do that, and now there's a card face-down infront of me. i put aside the other 50 cards, and sure enough, you can confirm that the king of hearts is no longer among them. that means it's either face-down infront of me, or infront of you.
who is much MUCH more likely to have the king of hearts face-down infront of them? me, because as the host, i was allowed to look behind the scenes at the hidden information and specifically put the king of hearts infront of myself if it wasn't the one you selected originally.
now, say before we flip our cards face-up and see if you have the king of hearts, i offer that you can swap with mine. obviously you would, right? because clearly i almost certainly have it infront of me rather than you.
it's the same thing with the monty hall doors. even though you're more likely to have originally picked the prize in the 3 doors scenario than you were in the 52 cards scenario, it's still more likely that the host sets it up so that the last remaining choice is the prize, because he gets to see and manipulate the hidden information.
it's not really a problem, sure you have better odds with 1/3 as opposed to 1/52, but the host still has better odds than you do in either situation.
the scenario can just be redone with 3 playing cards rather than 52 (which is basically just the monty hall problem)
even though you have a higher chance of drawing the king of hearts from 3 blind cards rather than 52, the host still has higher odds of getting it himself by intentionally selecting it (if it's there) from the remaining 2 cards than you did drawing blind from 3, so it's still more likely to be infront of him when you have the chance to swap for it.
it's the same thing with the doors, the door he doesn't open and you didn't pick is like his 'choice', and his choice is more likely to be the car than it is that you originally picked the car
>look everyone I took statistics 1 in muh fancy pants state college and now I'm gonna explain this introductory Bayes theorem case study to the layman nonces of /vee/ aren't I such a special heckin smartypants
my statistics teacher was the worst professor i have ever had in my entire educational career
elementary, middle, high, college, he was the worst
he wouldnt even "teach" he would open up the statistics book and start reading
he wouldnt do problem, he wouldnt explain anything, he wouldnt take questions, and his homework was always like 50 questions every single night
he was old as frick too, it's obvious he had tenure and didnt give a shit anymore
and then that crusty old bastard died a few years after i graduated, good riddance
To simplify it, expand it to 100 doors, with 99 doors having a goat and 1 prize door. You pick a door. You are 99% certain to pick a goat door, 1% to pick a prize door. When he opens 98 of the goat doors, it is a 99/100 chance that the door you originally picked contains a goat, the other door a prize. So your chance if you switch is 99%. The same thing happens with 3 doors. There's a 2/3 chance you originally picked a goat door, so when he removes 1 goat door there's a 2/3 chance your original door contained a goat, so you should switch.
Now that that's solved, let's have a discussion about this baby
This is a trick question. Airplanes are propelled by their engines, not by their wheels. It's like asking if you could drive a hot wheels car across a conveyor that always matches the speed of its little wheels but in the opposite direction.
I'll leave proving that this means the conveyor must reach infinite speed immediately when the plane engines turn on as an excersize to the reader.
it's not a trick question. The treadmill is as big as a runway, the fact that it's moving is completely irrelevant to the plane. It would take off exactly the same as normal.
This is a trick question. Airplanes are propelled by their engines, not by their wheels. It's like asking if you could drive a hot wheels car across a conveyor that always matches the speed of its little wheels but in the opposite direction.
I'll leave proving that this means the conveyor must reach infinite speed immediately when the plane engines turn on as an excersize to the reader.
Because the acceleration of the jet is from the jets which are independent of the wheels, it accelerates forward regardless what happens with the wheels and treadmill. So it's now a question of which breaks first from overspeeding, the treadmill or the wheels. If it's the former, it takes off.
wheels on planes are free-rotating like on hotwheel cars, it's thrust is determined purely by the pushing of air by the engines and has nothing to do with the acceleration of wheels touching the ground.
if the engines are on it would start moving forward the same way a hotwheels car would move forward on the treadmill no matter how fast the tread is moving if you were pushing it forward with your finger.
The only reason staticians claim the odds aren't 50/50 is cause some women claimed it isnt and that the media stirred up accusations of "sexism" when all the other staticians called her dumb
(This is bait (I think))
This post might be bait but you do have better odds when you switch. You can find simulations online and test it yourself. I didn't believe it until I ran through some simulations myself.
> This computer program says it, therefore it must be true
Lol?
The reason it's not 50/50 is because the host KNOWS where the car is, and will NEVER pick the car.
Let's say the car is in Door C.
If you pick Door A, the host will ALWAYS open Door B. The correct answer is to switch.
If you pick Door B, the host will ALWAYS open Door A. The correct answer is to swtich.
If you pick Door C, then switching is a bad idea. The host doesn't need to pick a specific door here.
For two of the three scenarios here, switching is the correct choice. Picking any door and then switching grants you a 66% chance of getting the prize, because unless you picked the right door the first time (33%) the option that you're allowed to switch to will always contain the prize.
If the host didn't know where the car was and truly picked a door at random, you would have a 1/3 chance of getting it right by switching, a 1/3 chance of getting it wrong by switching, and a 1/3 chance of the host accidentally revealing where the car is. It would be a true 50/50 then in scenarios where the car didn't appear.
if a goat is always removed then it always boils down to a 50/50 chance and the only reason in your head that it doesn't is because you're accepting the host's narrative that you're "switching" or not. the host is telling you this is to heighten the drama of game show and to throw a wrench into your feelz
No
No you fricking moron that's not how it works
I just LITERALLY explained every option to you and whether switching is the correct or incorrect choice. I'll go through it again, using all three doors to remove all shreds of doubt >>Car is in Door A >door A = don't switch >door B = switch >door C = switch >>Car is in Door B >door A = switch >door B = don't switch >door C = switch >>Car is in Door c >door A = switch >door B = switch >door C = don't switch
There are nine combinations here, nine different ways to arrange where the car is and what door you picked. In six of those nine scenarios you win by switching. You have 66/33 odds of winning if you switch/don't switch respectively. There are literally only nine possible results and we know what wins in each of them
The reason it's not 50/50 is because the host KNOWS where the car is, and will NEVER pick the car.
Let's say the car is in Door C.
If you pick Door A, the host will ALWAYS open Door B. The correct answer is to switch.
If you pick Door B, the host will ALWAYS open Door A. The correct answer is to swtich.
If you pick Door C, then switching is a bad idea. The host doesn't need to pick a specific door here.
For two of the three scenarios here, switching is the correct choice. Picking any door and then switching grants you a 66% chance of getting the prize, because unless you picked the right door the first time (33%) the option that you're allowed to switch to will always contain the prize.
If the host didn't know where the car was and truly picked a door at random, you would have a 1/3 chance of getting it right by switching, a 1/3 chance of getting it wrong by switching, and a 1/3 chance of the host accidentally revealing where the car is. It would be a true 50/50 then in scenarios where the car didn't appear.
(Not that guy but as an addendum, if the car is in Door A or Door B, you can swap the letters around and the math still checks out.)
>You can run it through a simulation
i did. the only way to get two-thirds is to run the simulation over and over again which is bullshit given that the contestant only plays the game once.
This post might be bait but you do have better odds when you switch. You can find simulations online and test it yourself. I didn't believe it until I ran through some simulations myself.
The revealed goat is actually a misdirection. The ACTUAL decision you're making is a choice between the following >Is your car behind the first door you picked? >Or is it behind one of the two doors you didn't pick
Your options are literally one door, or two doors. And wouldn't you know it, you're twice as likely to pick the door with the car if you pick two doors.
The correct answer is you dont switch because the door you picked has already got you so far don't betray it now. gays that switch are the same scum who abandon party members due to power creeps
PICTURE THE FOLLOWING SCENARIO: YOU PICK A DOOR. A RANDOM AUDIENCE MEMBER IS ASKED TO OPEN ONE OF THE OTHER TWO DOORS. THIS AUDIENCE MEMBER IS NOT TOLD THE CONTENTS OF THE DOOR BEFORE OPENING ONE, AND MAY OPEN THE PRIZE DOOR. THE AUDIENCE MEMBER OPENS A LOSING DOOR.
WHAT ARE YOUR SWITCHING:STAYING ODDS OF WINNING IN THIS SCENARIO? IF THEY CHANGE: WHY?
this is literally no different than the original problem statement. the only difference is that instead of monty removing the losing door, it's the person asking the riddle, aka (you)
>You pick between three doors >1/3 chance to pick car >2/3 chance not to >Audience member can pick between the two you don't pick >If the audience member doesn't get the car the game continues for you >You can now pick between your original door and the remaining door >Still a 1/3 chance you picked the car the first time >Still a 2/3 chance you didn't
The interesting thing is the audience member doesn't have a 1/2 chance of getting the car, either. There's that 1/3 chance that you picked the car first, in which case their probability is 0% because their options don't even include the door with the car. The remaining 2/3 of the time, they do have a 1/2 chance to pick the car. However, 2/3 * 1/2 is a 1/3 chance of picking the car, the same one you had at the beginning. There's a car behind only one of the three doors.
You have a $1000 promotional chip for a casino on its opening day. The chip can't be traded in for cash until it has been played in the casino that day.
You decide to play a roulette table. The screen above the table indicates that the past 24 rounds have all been landed on red. Should you bet on red or black?
You have a $1000 promotional chip for a casino on its opening day. The chip can't be traded in for cash until it has been played in the casino that day.
You decide to play a roulette table. The screen above the table indicates that the past 24 rounds have all been landed on red. Should you bet on red or black?
Theoretically, it doesn't matter since in this case the odds are not dependent on previous rolls
That said, in reality, if it's a casino on opening day and it keeps landing on red, there may be an issue with the table itself, so go ahead and vote red
Pic related sounds like the sickest videogame I've ever imagined.
I have no idea how it would work. It has to be some kind of third person beat 'em up to have the same energy and I have no idea how you would make an action game out of the trolley problem but it'd all be worth it for this climax.
You only perceive it as switching because Monty told you that it's switching. What if instead you tossed a coin to make you final decision between the two remaining doors?
There are 3 doors, 1 has the prize, so there are only 3 possible scenarios: >LOSE/LOSE/WIN >LOSE/WIN/LOSE >WIN/LOSE/LOSE
For the sake of simplicity let's say you choose the first door >L/L/W
If you swap doors, you win.
If you don't swap doors, you lose. >L/W/L
If you swap doors, you win.
If you don't swap doors, you lose. >W/L/L
If you swap doors, you lose.
If you don't swap doors, you win.
In 1 out of 3 scenarios, you win when you don't swap doors.
In 2 out of 3 scenarios, you win when you swap doors.
Doesn't matter, prize is placed at random at the beginning each time, chance divided equally between the doors. Situation where prize is behind door you didn't pick at the start still happens twice more often. Revealing door and asking for confirming choice are there just to confuse you.
>Revealing door and asking for confirming choice are there just to confuse you.
it's there to confuse, period. the final choice is 50/50. the only reason to believe otherwise is to play into monty's narrative. monty is not your friend.
10 months ago
Anonymous
You just like to play dumb, are you? It's been explained several times already and I just said it - revealing door does not remove it from the equation and in 2/3 cases prize is behind the one that you didn't pick. You act as if placement was rigged so one door never gets prize (and that's against the rules) or that prize is again randomly placed after one door removal (which also isn't true). Or keep acting like idiot to farm replies. Your choice.
10 months ago
Anonymous
Revealing a door absolutely removes it from the equation.
10 months ago
Anonymous
It does not because placement happens when there are three doors. There are still three doors after reveal, just one contents has been revealed. Use your brain for once.
10 months ago
Anonymous
You have a choice between two doors. Behind one is car and behind the other is a goat. Which do you choose?
10 months ago
Anonymous
You have the choice between three doors you mouth breathing moron. You got info about contents of one but placements and initial odds of placement that ALREADY happened stays the same.
10 months ago
Anonymous
The first "choice" isn't real. It is discarded. The entire notion that you have a choice to "switch" is a con.
10 months ago
Anonymous
you're absolutely right, the problem is, the equation is not about the doors but the car.
You pick a door, it has 1/3 chances to be the car. Therfore, the car has 2/3 chances to be in one of the other two doors.
One of the other two doors is opened. This does not change the fact that the car still has 2/3 chances to be in the two doors, except now there's only one left.
Then it's just a matter of choosing to keep your 1/3 chances or going for 2/3.
I played that Money Case flash game once, and the first time I played I won without ever switching my choice, and it forever fricked up my reasoning towards this thought experiment.
So in a vacuum, if I was offered a choice between a door the first person picked, or the door he did not pick, after the third losing door had been eliminated, should I pick the door the first person did not pick, since his first choice could only have a 33 percent chance of being correct?
Why would it not be a 50/50 percent chance for me if I'm coming in once the doors have been narrowed down to two?
idiot. how do you not understand that the people arguing for 50/50 are de facto arguing that the abstraction created by the 33/66 mob is incorrect? you aren't even engaging with the question being asked. the only justification for believing that the first choice informs the second choice is that monty told you that the first choice informs the second choice and monty is the fricking conman running the game show.
There are 3 doors. I pick one. I have a 33% chance of being correct.
If I am then given the option to bet if I was incorrect with my first choice, I will always bet that my first choice was incorrect, because there was a 66% chance that I was.
The first choice only informs the second choice because you are told that it does. In reality they are completely distinct choices. For the second choice you will always have a choice between a car and a goat. The only reason that you think that you are "switching" is because monty tells you that you are "switching". You aren't actually locked into your first choice. Your first choice doesn't actually prune the probability tree.
>You are given 100 options >99 options give you nothing, 1 option gives you a billion dollars. >You pick an option. >You then have the option to pick either your original choice, or pick the entirety of the other 99 options.
If you still think this is "50/50" you are legitimately mentally moronic and I feel sorry for you.
10 months ago
Anonymous
>going from 3 to 2 choices is mathematically identical to going from 100 to 99 choices
the power of """math"""
10 months ago
Anonymous
I understand the 33/66 position, yet I'm the moronic because the 33/66 position can't understand the 50/50 position.
this is like assuming that W must be 2*V because hurr W looks like VV or that 8=2*0 because 8 is 2 0 on top of each other. the only answer to any of those problems is that there isn't enough information given.
Caterpillar is 7 when it has 5 segments + head + hat/
Caterpillar on final line has 6 segments plus head but no hat. To me that’s also 7 as there’s no other way to convert body segments of top line to 7 evenly - unless a decimal is the answer in which case fu
I'm a brainlet that doesn't understand why changing doors matters. I get that it goes from 33% to 50% but wouldn't picking the same door after showing a goat be the same odds?
All you know about your first pick is that it had a higher chance of being the goat.
Once the other goat is eliminated from your picking pool, you should probably switch since the final door had a higher chance of being the car.
>>pick an option
it does not matter what you pick initially. you will always have a coin toss between a car and a goat. the only reason that you think that you're switching is because Monty said so. In reality is a coin toss. every goat is discarded except the goat necessary to manufacture a coin toss.
>randomly pick 1 out of 500000000 >all except your pick, and one of the others is revealed to have nothing >"LOL I'LL STICK IT'S A 50/50"
you are moronic
Last time I mentioned Monty Hall at work lunch break, it spawned long and intense discussion with two opposite interpretations. What I found funny is how (almost) all guys were "1/3, 2/3 team" and all women were "50-50 team". I'm honestly curious why since it was proved numerous times, both theoretically and practically (via simulations) that odds are not 50-50. And I honestly don't know how can anyone come up with "explanation" of "it's always 50% because it either happens or it don't".
They are designed to match reality. You can't play 10000 games IRL in row to get statistics you need but can easily run simulation with even bigger numbers.
It becomes easier to understand if you increase the number of doors.
>1000 doors >999 goats >1 car >you pick a door (99.9% chance its a goat) >host opens another door (100% its a goat) >998 doors left to pick >stay with your 99.9% chance pick or change now that you have a 99.8% chance to pick correctly
Yes, it's in the rules. He also can't show you the door you picked, it's always the one you didn't - if you always chose A and prize is behind B, he HAVE to show C and so on (if prize is behind A, he can show either B or C). So he both try to confuse you and smuggle additional information.
Reminder that staying on the same door is functionally identical to making a second choice and re-selecting the same door, thereby making a 1/3 odds and 1/2 odds choice the same.
Monty Hall knows the answer. The odds change because he's forced to give you information you didn't have previously.
There, I saved you 500 posts of people not getting it.
it's either the goat door or it's not. it's 50/50, simple as.
people still aren't going to get it, but that's basically the heart of the issue. the host has access to knowledge that the contestant doesn't, and specifically alters the situation based on him being able to see/know what's behind the doors.
> Three doors
> Choose a door
> Wrong door
> You are given a chance to choose from the remaining doors
> 50/50 chance. Choose a door.
> "Are you sure about that? You can still choose the other door."
> Two possibilities.
> A) This is an attempt to instill uncertainty in you, cause you to think and overthink. That rabbithole of psychological warfare extends infinitely, and no matter what angle it is approached from will not yield constructive conclusions.
> B) This is mere showmanship, a ploy to generate tension and suspense in the audience.
> No way to know for sure. No definite information gained.
> Changing options effectively accomplishes nothing.
> Either way, odds are still unchanged. Still two options. 50/50.
> "~buuuh the host noe da ansssar so da chans go up~"
Shut the frick up moron.
You aren't describing the Monty Hall problem.
Nta but that is literally the Monty hall problem.
> Three doors
> Choose a door
> Wrong door
> You are given a chance to choose from the remaining doors
That's not the monty hall problem
You choose a door. You have a 1/3 chance of picking the correct door. This means there is a 2/3 chance the correct door is one of the other two doors. Monty opens one of the other doors to expose a goat. The odds have not changed, though; your door still has a 1/3 chance to `be correct while the other two doors have a 2/3 chance, but since we now know which one of those two doors is the goat, the other door itself has a 2/3 chance of being the winning door. Thus, by switching your choice, you double your chance to win.
Everyone who doesn’t understand the problem after it’s been explained a million times is a nitwit or troll and these threads should be banned.
This. Also the situation requires him to always show you the goat.
Glad to see that people on this board actually understand the problem.
A easy way to visualize the problem is to push it to the extreme.
There are a thousand doors. 1 Door has the perfect $100K crypto portfolio (99% Chainlink/0.5% ETH/0.5% BTC) and the 999 other doors each have a goat behind of them.
You choose 1 door, and then host will open 998 doors, all of them each revealing a goat behind the respective doors.
Do you stick with the door you chose at random, or do you switch to the other lone door (out of the 999 he could have left unopened) the host did not open up?
Also, this puzzle was used in Zero Escape Zero Time Dilemma, and people got screwed over by correctly choosing the right one the first time around lol
Still too ambiguous.
We can go further.
There are 10*e^999 doors. 1 Door hides a pristine turkey sanwich, but every other door contains and additional 10*e^999 doors, which themselves are characterized by an identical distribution of prizes.
How many doors would you need to open before you could fully reconstruct the turkey from the harvested sanwich meats?
Still not understanding that point people are making who are claiming that the problem boils down to a 50/50 chance are arguing that the statistical abstraction that you've created does not map on to reality. It doesn't matter how loudly you play the tune, you're still out of key.
Back to your containment board stinkie.
The 10 door version in ZTD unironically made the problem click in my head, expanding it like this really does help you understand the 3 door version.
This also helps.
That still doesn't explain things enough for most people.
Here's a better explanation:
Your first pick had a higher chance of being a goat. Since Monty removes the other goat door, the chances of the car being behind the remaining hidden door initially is higher than the door you picked first, so you should switch to have a higher chance of winning the car (since your initial pick was most likely wrong to begin with).
Why do you consider it switching, rather than tossing a coin?
It's all because the initial odds were not in your favor (you had a 2 out of 3 chance to pick a goat).
The final hidden door has a higher chance of being the car to begin with.
Obviously switching doesn't guarantee anything, you could have gotten lucky on your initial pick, but you have no way of knowing that.
>the initial odds were not in your favor
Correct, yet what difference does that make to your odds when you've got two doors to choose from?
Because Monty can not reveal the door you picked, and always removes one of two losing doors.
Your initial pick was PROBABLY wrong, so the last remaining hidden door is PROBABLY the car.
Because your initial pick never stops being 33% even when there's just two doors. Monty opening a goat door allows you to capitalize on this and switch to a door with odds that are the reverse of yours.
Monty Hall knowing the answer has absolutely nothing to do with it. The only thing that matters is that the door he opens has a goat, which is always how the question is posed.
>choose door
>le 1/3 chance
>get offered to rechoose door after goat is revealed
>agree
>choose same door
>1/3 = 1/2
You're thinking about the choice to switch. You should be thinking about the odds of the individual doors themselves. Your first pick is always 1/3 odds no matter what happens after.
>pick is always 1/3 odds no matter what happens after.
Except it's not you moron
The moment you agree to switch, all doors gets odds redistributed
First one included
The problem is that your first choice is factored into the reveal. Anything you didn't choose, with the exception of one other, is removed.
It's not the odds of the apparent problem (2 doors or 1:2 odds) that you're playing, it's the odds that the reveal just gave you the correct answer (2:3 if you swap, 1:3 if you stay).
the host won't ever reveal your initial choice
The Monty Hall problem makes absolutely no sense because what if he's just lying to get you to pick the wrong door?
lying about what? That there is a car?
he opens the door so you can see a goat
Anon that's-
>check Monty Hall's early life
actually a very valid concern.
Long version:
A better way to think about it is that your inital choice always has a 2/3 chance of being wrong, this is pretty easy to grasp.
The host however knows where the prize is and will use that information to always open a door that has a goat regardless of your initial choice, thus the chance of your initial door being wrong has not changed, the location of the prize is always the same before and after he reveals the goat.
Since theres only one other option at this point switching gives you a 2/3 chance of getting the prize.
Yeah this. No "what if le 100 doors?" shit, no excel spreadsheet shit either. The odds change. That's that.
The odds do not change though.
You have a 33% chance of having the correct door. There a 66% chance of one of the other doors being correct.
Monty opening or closing the doors does not ever change these odds, your choice is still "Do I stay with a 33% chance of being correct, or switch to a 66% chance that my original guess was incorrect"
>Do I stay with
is a monty mind frick. no information is actually carried over unless you buy into monty's congame
the host opening a wrong door changes the odds of you being correct if you switch
that is what people refer to when they mean the odds change
>Make choice, Monty makes his
>Go backstage and take out contestant who wasn't watching the show
>Tell them to decide which door
Alright, so what are the odds now moron?
>oh yeah well if I completely change the question then the previous answer is no longer correct, CHECKMATE
If the contestant doesn't know which door was initially picked, it's 50/50
>Witnessing Monty pick a door changes the odds
So you admit the odds can/do change?
If Monty Hall shoots you in the dick and then rapes your bullet wound, do you still suck his dick?
oh whoops lol, I misread who you were replying to.
Yes the odds when switching do change
If they are aware of what you chose, and can see the opened goat door that Monty chose, then the best choice for them would be to choose the unopened door that you didn't pick. They'd be right 2/3 times
dude its bait, everyone actually gets its not 50/50 but they pretend it is so we can have these threads. literally the same points are parroted every time this is posted
>just imagine 1000 doors!
>its only 2 doors, so 50/50!
>the host knows, thats why its not 50/50!
>flowchart autism
at least the portal threads and coach/watch image baits are actually problems without objectively correct answers so its fun to argue in those, here you are just arguing with fishermen and their bait
>objectively correct answers
monty hall problem lacks one too
This type of statistics shit is and will be always a meme because they don't take into account time.
If there were 100000 doors, you choose one and they open every other door except the one you chose and another one the odds are still 50/50 NOW
>The way to win is to completely switch your original choice
Deep
except that's not true. switching changes nothing.
It literally changes your answer.
You have a 1 in 3 chance initially of picking right, but the parameters change when they remove one door that is explicitly the correct answer, now you have a 50/50 chance of getting it right instead of 1/3.
The second choice is not the same as the first.
Meant to say explicitly the incorrect answer.
The second choice is heavily informed by the first, anon. You can't just ignore the context of the situation.
If there are 1,000,000 doors and you pick one, then 999,998 doors are opened, leaving only your door and one out of 999,999 remaining, which one is more likely to have been the correct choice?
The monty hall problem is essentially betting on whether or not your first choice was correct. Since statistically your first choice only had a 33% chance of being correct, it is always a safer bet to bet "I was wrong the first time" and then switch
>The second choice is heavily informed by the first
Great rebuttal
Feel free to test it yourself: https://www.mathwarehouse.com/monty-hall-simulation-online/
Why can't I run the simulation just once?
I... I can't believe what I'm seeing... No... It's 50%... It has to be... That's right! It's rigged! You're playing me for a fool!
Hmmmm
>sample size of 4
Ruh roh
Wow! If you swap doors, you have a 100% chance of winning every time!
>2nd choice is not informed by the reveal
I used to think this way too, but it factually is.
>pick an option, 1 in N chance to be correct
>reveal: now majority of other options are removed except for your choice and one other option
>one of those 2 is guaranteed correct
Think about it. First it's 1 in N because any of them could be correct, but then it's narrowed to 2 options and one of them is guaranteed correct.
If you choose not to swap to the other option after the reveal, you're still taking those 1 in N odds, BUT if you decide to swap, you actually have better odds since it's more likely that the other option is correct.
It's not readily apparent for just N=3 options, but for larger values of N, the likelihood of the non-chosen option being correct becomes much more favorable.
>>pick an option
it does not matter what you pick initially. you will always have a coin toss between a car and a goat. the only reason that you think that you're switching is because Monty said so. In reality is a coin toss. every goat is discarded except the goat necessary to manufacture a coin toss.
The part where your 2nd choice is informed by the reveal is that your odds have changed if you switch, but have not changed if you don't.
So using anon's 1,000,000 option example, you are taking 1 in 1,000,000 odds if you don't switch.
You are very much less likely to have chosen correctly if you stick to your first guess because the correct answer was *more likely to be isolated by the reveal*, and your first choice was really unlikely to be the correct choice.
oh my gosh its like transitioning can make you happier who'd have thought?
>what if there were 100 doors!!!!!
Doesn't mean shit because there's only ever 2 after the goat.
It's 50/50 for all intensive porpoises, midwits.
good bait
you're not thinking about it correctly
imagine instead of 3 doors, there's 52. and instead of doors, they're playing cards. i, the host, tell you, the player, that if you draw 1 random card from a deck of playing cards, and it's the king of hearts, you get a prize. if it's any other card, you don't get anything.
so i shuffle the deck, and you draw a card and place it face-down infront of you. it's pretty low odds that it's the king of hearts, right? 1 in 52.
now, before anything else happens, i pick up the remaining 51 cards, and start looking through them, looking for the king of hearts. if i find it, i'll put it face-down infront of myself. if i don't find it (because you happened to draw it yourself) then i'll just put a random card face down infront of myself.
so i do that, and now there's a card face-down infront of me. i put aside the other 50 cards, and sure enough, you can confirm that the king of hearts is no longer among them. that means it's either face-down infront of me, or infront of you.
who is much MUCH more likely to have the king of hearts face-down infront of them? me, because as the host, i was allowed to look behind the scenes at the hidden information and specifically put the king of hearts infront of myself if it wasn't the one you selected originally.
now, say before we flip our cards face-up and see if you have the king of hearts, i offer that you can swap with mine. obviously you would, right? because clearly i almost certainly have it infront of me rather than you.
it's the same thing with the monty hall doors. even though you're more likely to have originally picked the prize in the 3 doors scenario than you were in the 52 cards scenario, it's still more likely that the host sets it up so that the last remaining choice is the prize, because he gets to see and manipulate the hidden information.
the problem is that you're more unlikely to guess 1/52 than 1/3
it's not really a problem, sure you have better odds with 1/3 as opposed to 1/52, but the host still has better odds than you do in either situation.
the scenario can just be redone with 3 playing cards rather than 52 (which is basically just the monty hall problem)
even though you have a higher chance of drawing the king of hearts from 3 blind cards rather than 52, the host still has higher odds of getting it himself by intentionally selecting it (if it's there) from the remaining 2 cards than you did drawing blind from 3, so it's still more likely to be infront of him when you have the chance to swap for it.
it's the same thing with the doors, the door he doesn't open and you didn't pick is like his 'choice', and his choice is more likely to be the car than it is that you originally picked the car
>3
Goat upper body
>2
Goat Lower body
>1
me fricking it from behind
What of there was a trolly problem behind the door?
Is it really that fun to pretend you are stupid?
>look everyone I took statistics 1 in muh fancy pants state college and now I'm gonna explain this introductory Bayes theorem case study to the layman nonces of /vee/ aren't I such a special heckin smartypants
my statistics teacher was the worst professor i have ever had in my entire educational career
elementary, middle, high, college, he was the worst
he wouldnt even "teach" he would open up the statistics book and start reading
he wouldnt do problem, he wouldnt explain anything, he wouldnt take questions, and his homework was always like 50 questions every single night
he was old as frick too, it's obvious he had tenure and didnt give a shit anymore
and then that crusty old bastard died a few years after i graduated, good riddance
>nonces of /vee/
While this is a correct description of Ganker, that word doesn't mean what you think it means.
bait thread, but i'll still explain it
To simplify it, expand it to 100 doors, with 99 doors having a goat and 1 prize door. You pick a door. You are 99% certain to pick a goat door, 1% to pick a prize door. When he opens 98 of the goat doors, it is a 99/100 chance that the door you originally picked contains a goat, the other door a prize. So your chance if you switch is 99%. The same thing happens with 3 doors. There's a 2/3 chance you originally picked a goat door, so when he removes 1 goat door there's a 2/3 chance your original door contained a goat, so you should switch.
Now that that's solved, let's have a discussion about this baby
I dunno. But I think it should be since the jet engines are the things generating the lift.
The wings and flaps provide lift, moron
Oh okay. Thanks for educating me.
Wings aren't catching air.
No air, no lift.
No lift, no takeoff.
it's not a trick question. The treadmill is as big as a runway, the fact that it's moving is completely irrelevant to the plane. It would take off exactly the same as normal.
Read the second part of my post to find out why it's a trick 😉
It's not a propeller plane that needs to wind up and pull air, a large jumbo jet needs lift on the wings.
This is a trick question. Airplanes are propelled by their engines, not by their wheels. It's like asking if you could drive a hot wheels car across a conveyor that always matches the speed of its little wheels but in the opposite direction.
I'll leave proving that this means the conveyor must reach infinite speed immediately when the plane engines turn on as an excersize to the reader.
Yes, but only because the convey would be rolling at the same speed on the ground and therefore moving it fast enough backwards to lift it
>plane goes forward
>plane goes up
>plane goes backward
>plane goes ???
The conveyor is held in place, the same way your grocery store conveyor doesn't walk out the door with your food.
Because the acceleration of the jet is from the jets which are independent of the wheels, it accelerates forward regardless what happens with the wheels and treadmill. So it's now a question of which breaks first from overspeeding, the treadmill or the wheels. If it's the former, it takes off.
wheels on planes are free-rotating like on hotwheel cars, it's thrust is determined purely by the pushing of air by the engines and has nothing to do with the acceleration of wheels touching the ground.
if the engines are on it would start moving forward the same way a hotwheels car would move forward on the treadmill no matter how fast the tread is moving if you were pushing it forward with your finger.
Better way to understand the problem
Ah!
I see, suddenly I understand everything!
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The only reason staticians claim the odds aren't 50/50 is cause some women claimed it isnt and that the media stirred up accusations of "sexism" when all the other staticians called her dumb
(This is bait (I think))
You can run it through a simulation if you don't believe it. In fact, it's already been done. But you can do it yourself if you don't trust it.
> This computer program says it, therefore it must be true
Lol?
Try it in real life then. It's easy to replicate.
Probably will. It sounds like bullshit, but there does seem to be concensus on the answer so I must be just not getting it
The reason it's not 50/50 is because the host KNOWS where the car is, and will NEVER pick the car.
Let's say the car is in Door C.
If you pick Door A, the host will ALWAYS open Door B. The correct answer is to switch.
If you pick Door B, the host will ALWAYS open Door A. The correct answer is to swtich.
If you pick Door C, then switching is a bad idea. The host doesn't need to pick a specific door here.
For two of the three scenarios here, switching is the correct choice. Picking any door and then switching grants you a 66% chance of getting the prize, because unless you picked the right door the first time (33%) the option that you're allowed to switch to will always contain the prize.
If the host didn't know where the car was and truly picked a door at random, you would have a 1/3 chance of getting it right by switching, a 1/3 chance of getting it wrong by switching, and a 1/3 chance of the host accidentally revealing where the car is. It would be a true 50/50 then in scenarios where the car didn't appear.
if a goat is always removed then it always boils down to a 50/50 chance and the only reason in your head that it doesn't is because you're accepting the host's narrative that you're "switching" or not. the host is telling you this is to heighten the drama of game show and to throw a wrench into your feelz
nice bait.
No
No you fricking moron that's not how it works
I just LITERALLY explained every option to you and whether switching is the correct or incorrect choice. I'll go through it again, using all three doors to remove all shreds of doubt
>>Car is in Door A
>door A = don't switch
>door B = switch
>door C = switch
>>Car is in Door B
>door A = switch
>door B = don't switch
>door C = switch
>>Car is in Door c
>door A = switch
>door B = switch
>door C = don't switch
There are nine combinations here, nine different ways to arrange where the car is and what door you picked. In six of those nine scenarios you win by switching. You have 66/33 odds of winning if you switch/don't switch respectively. There are literally only nine possible results and we know what wins in each of them
Nice bait moron
I finally understand it now.
Now explain what monads are.
(Not that guy but as an addendum, if the car is in Door A or Door B, you can swap the letters around and the math still checks out.)
>You can run it through a simulation
i did. the only way to get two-thirds is to run the simulation over and over again which is bullshit given that the contestant only plays the game once.
just apply multiple times moron
You're a special kind of stupid
explain
This post might be bait but you do have better odds when you switch. You can find simulations online and test it yourself. I didn't believe it until I ran through some simulations myself.
Maybe it'll make more sense if it's worded like this
>What are the odds your initial guess is wrong?
Irrelevant redditry.
Bet you'd put some more serious thought into it if the game was choosing between 3 girls but 2 had massive wieners
>pick a girl
>host shows one of the other two girls has a massive wiener
>pick that girl instead
>win every time
gg ez no re
it's 1/3 chance for double crit
The revealed goat is actually a misdirection. The ACTUAL decision you're making is a choice between the following
>Is your car behind the first door you picked?
>Or is it behind one of the two doors you didn't pick
Your options are literally one door, or two doors. And wouldn't you know it, you're twice as likely to pick the door with the car if you pick two doors.
The correct answer is you dont switch because the door you picked has already got you so far don't betray it now. gays that switch are the same scum who abandon party members due to power creeps
THIS POST IS FOR ALL MONTY HALL UNDERSTANDERS
PICTURE THE FOLLOWING SCENARIO: YOU PICK A DOOR. A RANDOM AUDIENCE MEMBER IS ASKED TO OPEN ONE OF THE OTHER TWO DOORS. THIS AUDIENCE MEMBER IS NOT TOLD THE CONTENTS OF THE DOOR BEFORE OPENING ONE, AND MAY OPEN THE PRIZE DOOR. THE AUDIENCE MEMBER OPENS A LOSING DOOR.
WHAT ARE YOUR SWITCHING:STAYING ODDS OF WINNING IN THIS SCENARIO? IF THEY CHANGE: WHY?
"Random" audience member? Nice try, Monty (if that even is your real name). She's got plant written all over her.
this is literally no different than the original problem statement. the only difference is that instead of monty removing the losing door, it's the person asking the riddle, aka (you)
>You pick between three doors
>1/3 chance to pick car
>2/3 chance not to
>Audience member can pick between the two you don't pick
>If the audience member doesn't get the car the game continues for you
>You can now pick between your original door and the remaining door
>Still a 1/3 chance you picked the car the first time
>Still a 2/3 chance you didn't
The interesting thing is the audience member doesn't have a 1/2 chance of getting the car, either. There's that 1/3 chance that you picked the car first, in which case their probability is 0% because their options don't even include the door with the car. The remaining 2/3 of the time, they do have a 1/2 chance to pick the car. However, 2/3 * 1/2 is a 1/3 chance of picking the car, the same one you had at the beginning. There's a car behind only one of the three doors.
You have a $1000 promotional chip for a casino on its opening day. The chip can't be traded in for cash until it has been played in the casino that day.
You decide to play a roulette table. The screen above the table indicates that the past 24 rounds have all been landed on red. Should you bet on red or black?
Red, of course.
did you know most boomers say black?
Theoretically, it doesn't matter since in this case the odds are not dependent on previous rolls
That said, in reality, if it's a casino on opening day and it keeps landing on red, there may be an issue with the table itself, so go ahead and vote red
The virgin Monty Hall
The Chad Trolley
This is now a trolley thread.
Post trolleys.
only a straight up sociopath would ever consoder the trolley problem a “problem”
Pic related sounds like the sickest videogame I've ever imagined.
I have no idea how it would work. It has to be some kind of third person beat 'em up to have the same energy and I have no idea how you would make an action game out of the trolley problem but it'd all be worth it for this climax.
Where’s Saddam Hussein?
it's easier to understand with a larger group like say 100 doors
once you explain it that way people get it
You only perceive it as switching because Monty told you that it's switching. What if instead you tossed a coin to make you final decision between the two remaining doors?
it still comes down to two doors. 50/50. not sure how you brainlets don't get this.
There are 3 doors, 1 has the prize, so there are only 3 possible scenarios:
>LOSE/LOSE/WIN
>LOSE/WIN/LOSE
>WIN/LOSE/LOSE
For the sake of simplicity let's say you choose the first door
>L/L/W
If you swap doors, you win.
If you don't swap doors, you lose.
>L/W/L
If you swap doors, you win.
If you don't swap doors, you lose.
>W/L/L
If you swap doors, you lose.
If you don't swap doors, you win.
In 1 out of 3 scenarios, you win when you don't swap doors.
In 2 out of 3 scenarios, you win when you swap doors.
Therefore, the optimal strategy is to swap doors.
There aren't three doors. There are two doors.
And one have prize twice more often than the other. Happy?
You're assuming that a contestant plays the game more than once.
Doesn't matter, prize is placed at random at the beginning each time, chance divided equally between the doors. Situation where prize is behind door you didn't pick at the start still happens twice more often. Revealing door and asking for confirming choice are there just to confuse you.
>Revealing door and asking for confirming choice are there just to confuse you.
it's there to confuse, period. the final choice is 50/50. the only reason to believe otherwise is to play into monty's narrative. monty is not your friend.
You just like to play dumb, are you? It's been explained several times already and I just said it - revealing door does not remove it from the equation and in 2/3 cases prize is behind the one that you didn't pick. You act as if placement was rigged so one door never gets prize (and that's against the rules) or that prize is again randomly placed after one door removal (which also isn't true). Or keep acting like idiot to farm replies. Your choice.
Revealing a door absolutely removes it from the equation.
It does not because placement happens when there are three doors. There are still three doors after reveal, just one contents has been revealed. Use your brain for once.
You have a choice between two doors. Behind one is car and behind the other is a goat. Which do you choose?
You have the choice between three doors you mouth breathing moron. You got info about contents of one but placements and initial odds of placement that ALREADY happened stays the same.
The first "choice" isn't real. It is discarded. The entire notion that you have a choice to "switch" is a con.
you're absolutely right, the problem is, the equation is not about the doors but the car.
You pick a door, it has 1/3 chances to be the car. Therfore, the car has 2/3 chances to be in one of the other two doors.
One of the other two doors is opened. This does not change the fact that the car still has 2/3 chances to be in the two doors, except now there's only one left.
Then it's just a matter of choosing to keep your 1/3 chances or going for 2/3.
That's a complicated way of saying "the odds were 1/3 that you picked right the first time and that hasn't changed"
just because monty got you emotionally invested in your first choice, doesn't mean that it informs your second choice
I played that Money Case flash game once, and the first time I played I won without ever switching my choice, and it forever fricked up my reasoning towards this thought experiment.
take the red pill - realise that the first choice is irrelevant to the second choice
All I remember from gambling is that if you keep guessing wrong your odds increase exponentially that you'll guess right.
I hate gachas.
Society is collapsing. I'd rather have a goat than a car.
So in a vacuum, if I was offered a choice between a door the first person picked, or the door he did not pick, after the third losing door had been eliminated, should I pick the door the first person did not pick, since his first choice could only have a 33 percent chance of being correct?
Why would it not be a 50/50 percent chance for me if I'm coming in once the doors have been narrowed down to two?
Because his first pick is a 33% chance of being correct. The switch is 66.7% chance of being correct.
idiot. how do you not understand that the people arguing for 50/50 are de facto arguing that the abstraction created by the 33/66 mob is incorrect? you aren't even engaging with the question being asked. the only justification for believing that the first choice informs the second choice is that monty told you that the first choice informs the second choice and monty is the fricking conman running the game show.
Whether Monty knows or not is irrelevant.
There are 3 doors. I pick one. I have a 33% chance of being correct.
If I am then given the option to bet if I was incorrect with my first choice, I will always bet that my first choice was incorrect, because there was a 66% chance that I was.
The first choice only informs the second choice because you are told that it does. In reality they are completely distinct choices. For the second choice you will always have a choice between a car and a goat. The only reason that you think that you are "switching" is because monty tells you that you are "switching". You aren't actually locked into your first choice. Your first choice doesn't actually prune the probability tree.
You are overthinking it, anon.
>You are given 100 options
>99 options give you nothing, 1 option gives you a billion dollars.
>You pick an option.
>You then have the option to pick either your original choice, or pick the entirety of the other 99 options.
If you still think this is "50/50" you are legitimately mentally moronic and I feel sorry for you.
>going from 3 to 2 choices is mathematically identical to going from 100 to 99 choices
the power of """math"""
I understand the 33/66 position, yet I'm the moronic because the 33/66 position can't understand the 50/50 position.
>monty hall
LOL, lmao even. Try a real puzzle for once
14?
the clock in the last line is a new variable
all 3 are new variables. these aren’t actually math problems, they’re childrens’ “spot the difference” puzzles masquerading as math problems.
No one said it's a math puzzle, have you ever played a Layton game?
caterpie = 5
flower = 2
last line
caterpie + 2 flowers + 5 o clock
5+4+5 = 14
CF= caterpillar with flower
CL = clock at 6
FL = flower
CF*3 = 21
CF = 7
CL + CL + 7 = 19
CL = 6
FL + 6 + 7 = 15
FL + 13 = 15
FL = 2
Caterpillar alone = CF - 2
Final line:
2 + (2*2) * (Clock at 5) = ?
>we assume since "Clock at 6" = 6, then "Clock at 5" = 5
2 + 4 * 5 = ?
2 + 20 = 22
why the frick did I do this
The flower is indeed 2 points but the caterpillars value is equal to the number of segments of his body
this is like assuming that W must be 2*V because hurr W looks like VV or that 8=2*0 because 8 is 2 0 on top of each other. the only answer to any of those problems is that there isn't enough information given.
No amount of information is enough for autistic morons who can't use logic and reason
Caterpillar is 7 when it has 5 segments + head + hat/
Caterpillar on final line has 6 segments plus head but no hat. To me that’s also 7 as there’s no other way to convert body segments of top line to 7 evenly - unless a decimal is the answer in which case fu
Thus, my answer is 7+2*2*5 = 27
Absolute fricking moron, also there are no hats anywhere in that picture
Oh it’s a flower. 26 then. Take it easy, sperglord, image is shit quality
55
>a problem discovered by a woman bother Ganker this much...
This is the most concise explanation I have seen of this, you would have to be REALLY low IQ not to understand it with this
I'm a brainlet that doesn't understand why changing doors matters. I get that it goes from 33% to 50% but wouldn't picking the same door after showing a goat be the same odds?
>I get that it goes from 33% to 50%
All you know about your first pick is that it had a higher chance of being the goat.
Once the other goat is eliminated from your picking pool, you should probably switch since the final door had a higher chance of being the car.
This is the topic that's guaranteed to reach bump limit every time
Monty hall threads are obviously bait. We need more portal threads where there are morons who actually think "A" is correct.
The last mine in Minesweeper.
feels bad man
>randomly pick 1 out of 500000000
>all except your pick, and one of the others is revealed to have nothing
>"LOL I'LL STICK IT'S A 50/50"
you are moronic
>you are moronic
how many times do you need to play the game to win the car?
What the frick are you talking about?
If it's a 3 door scenario and you switch, you have a 66% chance of winning the car
If it's a 100 door scenario and you switch, you have a 99% chance of winning the car
If it's a 100,000 door scenario and you switch, you have a 99.999% chance of winning the car
There is no guarantee you will ever win the car, but you improve your odds dramatically in every scenario by switching
Someone actually made a version of Minesweeper that doesn't have luckshittery like this. I guess the first click could still be lethal.
>would you let monty frick your girl?
33/66 say yes
50/50 say no
Last time I mentioned Monty Hall at work lunch break, it spawned long and intense discussion with two opposite interpretations. What I found funny is how (almost) all guys were "1/3, 2/3 team" and all women were "50-50 team". I'm honestly curious why since it was proved numerous times, both theoretically and practically (via simulations) that odds are not 50-50. And I honestly don't know how can anyone come up with "explanation" of "it's always 50% because it either happens or it don't".
simulations aren't reality
They are designed to match reality. You can't play 10000 games IRL in row to get statistics you need but can easily run simulation with even bigger numbers.
It becomes easier to understand if you increase the number of doors.
>1000 doors
>999 goats
>1 car
>you pick a door (99.9% chance its a goat)
>host opens another door (100% its a goat)
>998 doors left to pick
>stay with your 99.9% chance pick or change now that you have a 99.8% chance to pick correctly
in*correctly
Does monty hall ALWAYS show the contestant one of the goat doors?
Yes, it's in the rules. He also can't show you the door you picked, it's always the one you didn't - if you always chose A and prize is behind B, he HAVE to show C and so on (if prize is behind A, he can show either B or C). So he both try to confuse you and smuggle additional information.
Reminder that staying on the same door is functionally identical to making a second choice and re-selecting the same door, thereby making a 1/3 odds and 1/2 odds choice the same.
the host removes one door. the host cannot remove the door that has no prize. this fact raises your chances when choosing the door the host didnt pick