the monty hall problem kinda seems to be a big problem because it gets explained badly
The way I had it explained to me to finally get it was focusing on my odds of being wrong for each choice. As in the first three doors I have a high 2/3 chance of picking the wrong door. When one false door is taken away and I am asked if I want to switch to the only remaining door, I now have a 1/2 chance of being wrong for this pick, which are much better odds because by sticking with my original choice I am still in that 2/3 chance of wrongness.
yeah no this shit still seems fake to me
the final choice is still a 50/50 no matter how many extra steps it took to get there, which makes staying as valid as swapping
like it's not a matter of not getting the logic behind it, it's a matter of the logic being horseshit
I'm convinced that statisticians or whoever are fucking gaslighting us
And a 50/50 is much better than a 33/66.
that isn't relevant! the "do you swap" choice is entirely disconnected from the "pick one of three" choice!
hell, the promise of always revealing a bad door and then asking again makes the first choice pure theatrics, and in practical terms you may as well have started at the coin toss!
That's...kind of the point though. Instead of thinking of it as "do you want to swap" think of it as "do you really think you picked the right door, or now that we've given you another chance with fewer options do you want to pick again?"
If there are a hundred doors and you pick one at random, and then they open *every other door except one* to show you that the prize is behind none of them, the question becomes "do you really think you picked the right door, or do you think the prize might actually be behind this one other remaining door?"
when you first picked, you only had a 1/100 chance of picking the right door. now they've narrowed it down to two choices, and you have to decide if you think you actually picked the right door. you probably didn't, and now you have the chance to switch to the door that almost definitely has the prize behind it.
the exact same odds apply to 3 doors, it's just less of a guarantee.
the best odds you can get at that moment are only ever 50/50, yes, but that's only if you switch every time you have the chance. if you don't switch, your odds are locked in at whatever point they were when you made your initial choice.
this coin toss has a 50% chance of success if you pick tails, but a much smaller chance of success if you pick heads, because this coin used to be a d6, until very recently
this coin toss has a 50% chance of success if you pick tails, but a much smaller chance of success if you pick heads, because this coin used to be a d6, until very recently
this coin toss has a 50% chance of success if you pick tails, but a much smaller chance of success if you pick heads, because this coin used to be a d6, until very recently
this coin toss has a 50% chance of success if you pick tails, but a much smaller chance of success if you pick heads, because this coin used to be a d6, until very recently
What even is this metaphor
sorry, should I have said a d3
It's still wrong.
There's 100 doors. One of them has a prize behind it. You pick one. 1 in 100 chance of success.
The doors are reduced to 2. Your door and one other. The prize is guaranteed to be behind one of them. Ask yourself, out of 100 doors do you think you picked the winning door? Or is it more likely you picked the wrong door at the start and the other door has the prize?
this coin toss has a 50% chance of success if you pick tails, but a much smaller chance of success if you pick heads, because this coin used to be a d6, until very recently
What even is this metaphor
sorry, should I have said a d3
It's still wrong.
There's 100 doors. One of them has a prize behind it. You pick one. 1 in 100 chance of success.
The doors are reduced to 2. Your door and one other. The prize is guaranteed to be behind one of them. Ask yourself, out of 100 doors do you think you picked the winning door? Or is it more likely you picked the wrong door at the start and the other door has the prize?
like, I cannot stress this enough, there is no amount of reiterating the same points that yall could do that would convince me that anything before the final choice matters
this coin toss has a 50% chance of success if you pick tails, but a much smaller chance of success if you pick heads, because this coin used to be a d6, until very recently
What even is this metaphor
sorry, should I have said a d3
It's still wrong.
There's 100 doors. One of them has a prize behind it. You pick one. 1 in 100 chance of success.
The doors are reduced to 2. Your door and one other. The prize is guaranteed to be behind one of them. Ask yourself, out of 100 doors do you think you picked the winning door? Or is it more likely you picked the wrong door at the start and the other door has the prize?
that doesn't matter! there's only two doors now!
And you picked one of them out of 100 others.
So do you think you got it right the first time? That's the key point to understanding the whole thing.
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JedocIn the scupperswith the staggers and jagsRegistered Userregular
The only Monty Hall problem I recognize is how if I ever met the smug bastard he'd have the problem of catching these hands.
this coin toss has a 50% chance of success if you pick tails, but a much smaller chance of success if you pick heads, because this coin used to be a d6, until very recently
What even is this metaphor
sorry, should I have said a d3
It's still wrong.
There's 100 doors. One of them has a prize behind it. You pick one. 1 in 100 chance of success.
The doors are reduced to 2. Your door and one other. The prize is guaranteed to be behind one of them. Ask yourself, out of 100 doors do you think you picked the winning door? Or is it more likely you picked the wrong door at the start and the other door has the prize?
that doesn't matter! there's only two doors now!
But they weren’t opened randomly. The person opening them to show you they were empty knew which one had the car in it. So by opening 98 wrong options, the choices are now 99% if you switch, and 1% if you stay.
That said, the Monty hall problem is fascinating just for how much people cannot understand it. It’s like the plane on a treadmill thing.
the logic trick here is that the door that's kept because you originally picked it and the alternate door that's left for you to switch to have different conditions for being part of the new set of options, which give them different odds of being the correct door. The one you picked first is there because you picked it rather than because of its contents, so it retains the odds from the original choice. Since they always remove a losing door to offer the second choice, the other offered door is offered 2/3s of the time because it's the winning door (you picked a losing door initially), and 1/3 of the time because it's a losing door (you picked a winning door initially)
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webguy20I spend too much time on the InternetRegistered Userregular
like, I cannot stress this enough, there is no amount of reiterating the same points that yall could do that would convince me that anything before the final choice matters
Ok lets do a million. There is a prize behind one door in a million. You pick one of the million doors.
Now they keep the door you picked, and one other out of the million, and ask you to choose. Out of that first million do you think you got it right on the first pick?
Keep in mind the prize IS NOT randomized after you pick and they remove all the ones you didn't pick. It is still behind the same door as it was the first time. All they've done is removed the 999998 doors that are guaranteed losses.
Chances are you did not pick the one in a million door the first time, and in fact the prize is in the other one they are offering you, so you should swap.
like, I cannot stress this enough, there is no amount of reiterating the same points that yall could do that would convince me that anything before the final choice matters
But you can demonstrate that the answer is true, why are you dying on this hill?
I mean if you want to keep saying the same thing with a bigger number it's not like I can stop you
It really is a simple question you keep refusing to answer: Did you pick the right door the first time? The numbers are meaningless, scaling higher just reinforces that singular question.
I mean if you want to keep saying the same thing with a bigger number it's not like I can stop you
I was hoping with the bigger number it would be easier to see. Its almost paradoxical that way.
I assumed that most folks would believe they would not pick correctly the one in a million doors with a prize behind it, and it would help it make sense.
I think what people are forgetting about the math, and this is important, is that the Guardian of one door always speaks the truth and the Guardian of the other door always lies.
What broke the monty hall problem open for me was realizing the host will never open the door with the prize.
It's super basic, and people here definitely already know it, but that's what did it for me
the monty hall problem kinda seems to be a big problem because it gets explained badly
The way I had it explained to me to finally get it was focusing on my odds of being wrong for each choice. As in the first three doors I have a high 2/3 chance of picking the wrong door. When one false door is taken away and I am asked if I want to switch to the only remaining door, I now have a 1/2 chance of being wrong for this pick, which are much better odds because by sticking with my original choice I am still in that 2/3 chance of wrongness.
yeah no this shit still seems fake to me
the final choice is still a 50/50 no matter how many extra steps it took to get there, which makes staying as valid as swapping
like it's not a matter of not getting the logic behind it, it's a matter of the logic being horseshit
I'm convinced that statisticians or whoever are fucking gaslighting us
I've seen people use 1000 doors as an explanation but I think it's easier with 2.
There are 2 doors. Behind one of them is a prize.
What are the odds that you will pick the correct door? 50/50.
You pick door 1.
I am very helpful and this isn't a real game show, so I tell you look I'm gonna open the door that doesn't have a prize behind it, and you can pick again if you like.
I open door 1. There is no prize behind it.
What are the odds that you will win a prize if you stick with door 1?
like, I cannot stress this enough, there is no amount of reiterating the same points that yall could do that would convince me that anything before the final choice matters
you don't need it explained, run the simulation that was linked earlier. Run "keep the door" 100 times and "switch doors" 100 times.
My simulation had a 75% chance of winning when i switched, and a 30% chance of winning when I kept the door.
It's just how it works, it doesn't matter if you don't understand it, it is objectively true.
Basically even though the final decision is a choice between two options, it's not a coin-flip because the whole scenario is controlled by malevolent non-random forces who want to make you look dumb on TV.
they're separate simulations? The percentage is always going to be different because they're two different simulations.
Like I just ran both again out of 100 and I won 71% on switch and 41% on keep.
The point isn't that it's a fixed percent, it's that it is overwhelmingly better to switch. Odds are still odds. You could still lose, but you have a way better chance if you switch.
Hm, maybe that doesn't make sense. This one works for more folks I think:
There are 10,000,000 doors infinitely arrayed before you, behind one of them is a very cool goat you want.
You pick a door. Your odds of correctly guessing which door has a cool goat are 1/10,000,000
I am very nice so I say wow that was a ridiculous thing to do, I'll open all but two doors: the one you picked, and the one with a goat behind it.
This leaves 2 doors. You now have a 9,999,999/10,000,000 chance of getting an awesome goat only if you switch which door you picked, because this new decision that you're making between two doors is between one that's really unlikely to have it (1/10,000,000) and one that's almost guaranteed to have it (every other door opened was opened because it either did not have a cool prize or was the door you picked. What are the odds that the one you picked also was the one with a cool prize? 1/10,000,000. So the other door will have the cool prize 9,999,999/10,000,000 times).
Edit: Also I still like my two-door extremely silly version. It's a bad game show, but the point is that if you remove the ability to stick with a closed door you've picked you're forced to confront the fact that if you do not switch you are actively deciding to have no chance of winning. In the 1/10,000,000 version you're deciding to have a super-small chance of winning.
Edit again: I explained it wrong at first! Just goes to show why it's a classic brain-teaser.
Basically even though the final decision is a choice between two options, it's not a coin-flip because the whole scenario is controlled by malevolent non-random forces who want to make you look dumb on TV.
okay but if I'm being conspired against then why is this being presented as a math problem
Librarian's ghostLibrarian, Ghostbuster, and TimSporkRegistered Userregular
edited July 2021
Okay so I legit don’t get it. I get the initial guess is 1/3 because there are three doors, but once they open a door that door is no longer a choice so wouldn’t it then be a 50/50 and therefor a toss up which of the two remaining doors your choose?
Okay so I legit don’t get it. I get the initial guess is 1/3 because there are three doors, but once they open a door that door is no longer a choice so wouldn’t it then be a 50/50 and therefor a toss up which of the two remaining doors your choose?
the hurdle is not thinking of the percent as a fixed state through both circumstances. it just absolutely does not work that way, even though it feels like it should.
edit: when inantp says that humans do not understand statistics, they're not joking. Our brains are not wired for understanding statistics at a fundamental level. They go against any rational instinct. It's what makes trusting a computer so hard because it feels wrong.
Okay so I legit don’t get it. I get the initial guess is 1/3 because there are three doors, but once they open a door that door is no longer a choice so wouldn’t it then be a 50/50 and therefor a toss up which of the two remaining doors your choose?
the logic trick here is that the door that's kept because you originally picked it and the alternate door that's left for you to switch to have different conditions for being part of the new set of options, which give them different odds of being the correct door. The one you picked first is there because you picked it rather than because of its contents, so it retains the odds from the original choice. Since they always remove a losing door to offer the second choice, the other offered door is offered 2/3s of the time because it's the winning door (you picked a losing door initially), and 1/3 of the time because it's a losing door (you picked a winning door initially)
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Librarian's ghostLibrarian, Ghostbuster, and TimSporkRegistered Userregular
Okay so I legit don’t get it. I get the initial guess is 1/3 because there are three doors, but once they open a door that door is no longer a choice so wouldn’t it then be a 50/50 and therefor a toss up which of the two remaining doors your choose?
the hurdle is not thinking of the percent as a fixed state through both circumstances. it just absolutely does not work that way, even though it feels like it should.
But if offered to change my selection, I step back and in my head consider the two doors as a new problem, I got two doors, one has a prize that is 50/50.
Okay so I legit don’t get it. I get the initial guess is 1/3 because there are three doors, but once they open a door that door is no longer a choice so wouldn’t it then be a 50/50 and therefor a toss up which of the two remaining doors your choose?
My thought process:
At first each door has a 1/3 chance of having the prize.
Then you choose one, it locks that door in at a 1/3 chance of having the prize.
The other two doors have a combined 2/3 chance of having the prize.
Then the host reveals one of the doors. The secret to this problem is: The door he opens is never the one with the prize. If the door he opened was random, then the monty hall problem wouldn't exist.
Now you have one door which inherited a 2/3 chance of having the prize, and the one you picked, which has a 1/3 chance of having the prize.
Okay so I legit don’t get it. I get the initial guess is 1/3 because there are three doors, but once they open a door that door is no longer a choice so wouldn’t it then be a 50/50 and therefor a toss up which of the two remaining doors your choose?
the hurdle is not thinking of the percent as a fixed state through both circumstances. it just absolutely does not work that way, even though it feels like it should.
But if offered to change my selection, I step back and in my head consider the two doors as a new problem, I got two doors, one has a prize that is 50/50.
You're correct, if the host opened one of the other two doors at random. He doesn't though, he wants to draw it out, and so he only ever reveals an empty door.
Psykoma on
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Librarian's ghostLibrarian, Ghostbuster, and TimSporkRegistered Userregular
Okay so I legit don’t get it. I get the initial guess is 1/3 because there are three doors, but once they open a door that door is no longer a choice so wouldn’t it then be a 50/50 and therefor a toss up which of the two remaining doors your choose?
My thought process:
At first each door has a 1/3 chance of having the prize.
Then you choose one, it locks that door in at a 1/3 chance of having the prize.
The other two doors have a combined 2/3 chance of having the prize.
Then the host reveals one of the doors. The secret to this problem is: The door he opens is never the one with the prize. If the door he opened was random, then the monty hall problem wouldn't exist.
Now you have one door which inherited a 2/3 chance of having the prize, and the one you picked, which has a 1/3 chance of having the prize.
That’s the part I absolutely can’t wrap my head around.
Oh gosh I actually totally forgot it's better than 50/50 odds when swapping because I just had it in my head "the odds get better". It's 2/3, because the start is:
1/3 chance it's the door you picked (door 1, say)
1/3 chance it's door 2
1/3 chance it's door 3
So 1/3 chance it's the door you picked, 2/3 chance it's either of the other two doors.
The host opens door 3, it's not door 3. Now it's a 2/3 chance that it's door 2 because we know it isn't door 3 and we know it's a 2/3 chance that it's either door 3 or door 2.
I mean it's def a problem designed to spin you around on probability, which is incredibly unintuitive in the first place.
This actually makes me miss talking stats, it's fun playing around with these problems.
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That's...kind of the point though. Instead of thinking of it as "do you want to swap" think of it as "do you really think you picked the right door, or now that we've given you another chance with fewer options do you want to pick again?"
If there are a hundred doors and you pick one at random, and then they open *every other door except one* to show you that the prize is behind none of them, the question becomes "do you really think you picked the right door, or do you think the prize might actually be behind this one other remaining door?"
when you first picked, you only had a 1/100 chance of picking the right door. now they've narrowed it down to two choices, and you have to decide if you think you actually picked the right door. you probably didn't, and now you have the chance to switch to the door that almost definitely has the prize behind it.
the exact same odds apply to 3 doors, it's just less of a guarantee.
the best odds you can get at that moment are only ever 50/50, yes, but that's only if you switch every time you have the chance. if you don't switch, your odds are locked in at whatever point they were when you made your initial choice.
this coin toss has a 50% chance of success if you pick tails, but a much smaller chance of success if you pick heads, because this coin used to be a d6, until very recently
What even is this metaphor
what I hear when reading houk's post
It's still wrong.
There's 100 doors. One of them has a prize behind it. You pick one. 1 in 100 chance of success.
The doors are reduced to 2. Your door and one other. The prize is guaranteed to be behind one of them. Ask yourself, out of 100 doors do you think you picked the winning door? Or is it more likely you picked the wrong door at the start and the other door has the prize?
that doesn't matter! there's only two doors now!
And you picked one of them out of 100 others.
So do you think you got it right the first time? That's the key point to understanding the whole thing.
But they weren’t opened randomly. The person opening them to show you they were empty knew which one had the car in it. So by opening 98 wrong options, the choices are now 99% if you switch, and 1% if you stay.
That said, the Monty hall problem is fascinating just for how much people cannot understand it. It’s like the plane on a treadmill thing.
Ok lets do a million. There is a prize behind one door in a million. You pick one of the million doors.
Now they keep the door you picked, and one other out of the million, and ask you to choose. Out of that first million do you think you got it right on the first pick?
Keep in mind the prize IS NOT randomized after you pick and they remove all the ones you didn't pick. It is still behind the same door as it was the first time. All they've done is removed the 999998 doors that are guaranteed losses.
Chances are you did not pick the one in a million door the first time, and in fact the prize is in the other one they are offering you, so you should swap.
Origin ID: Discgolfer27
Untappd ID: Discgolfer1981
But you can demonstrate that the answer is true, why are you dying on this hill?
https://youtu.be/yTHQOwbzDNc
It really is a simple question you keep refusing to answer: Did you pick the right door the first time? The numbers are meaningless, scaling higher just reinforces that singular question.
Okay, let's do a billion.
There's a billion goats behind those doors. They're out for blood. What do you do?
die, probably
I was hoping with the bigger number it would be easier to see. Its almost paradoxical that way.
I assumed that most folks would believe they would not pick correctly the one in a million doors with a prize behind it, and it would help it make sense.
Origin ID: Discgolfer27
Untappd ID: Discgolfer1981
No, you obviously get in the car. Sheesh!
Let's try it again:
There's 10 billion goats...
It's super basic, and people here definitely already know it, but that's what did it for me
I've seen people use 1000 doors as an explanation but I think it's easier with 2.
There are 2 doors. Behind one of them is a prize.
What are the odds that you will pick the correct door? 50/50.
You pick door 1.
I am very helpful and this isn't a real game show, so I tell you look I'm gonna open the door that doesn't have a prize behind it, and you can pick again if you like.
I open door 1. There is no prize behind it.
What are the odds that you will win a prize if you stick with door 1?
you don't need it explained, run the simulation that was linked earlier. Run "keep the door" 100 times and "switch doors" 100 times.
My simulation had a 75% chance of winning when i switched, and a 30% chance of winning when I kept the door.
It's just how it works, it doesn't matter if you don't understand it, it is objectively true.
https://www.mathwarehouse.com/monty-hall-simulation-online/
they're separate simulations? The percentage is always going to be different because they're two different simulations.
Like I just ran both again out of 100 and I won 71% on switch and 41% on keep.
The point isn't that it's a fixed percent, it's that it is overwhelmingly better to switch. Odds are still odds. You could still lose, but you have a way better chance if you switch.
There are 10,000,000 doors infinitely arrayed before you, behind one of them is a very cool goat you want.
You pick a door. Your odds of correctly guessing which door has a cool goat are 1/10,000,000
I am very nice so I say wow that was a ridiculous thing to do, I'll open all but two doors: the one you picked, and the one with a goat behind it.
This leaves 2 doors. You now have a 9,999,999/10,000,000 chance of getting an awesome goat only if you switch which door you picked, because this new decision that you're making between two doors is between one that's really unlikely to have it (1/10,000,000) and one that's almost guaranteed to have it (every other door opened was opened because it either did not have a cool prize or was the door you picked. What are the odds that the one you picked also was the one with a cool prize? 1/10,000,000. So the other door will have the cool prize 9,999,999/10,000,000 times).
Edit: Also I still like my two-door extremely silly version. It's a bad game show, but the point is that if you remove the ability to stick with a closed door you've picked you're forced to confront the fact that if you do not switch you are actively deciding to have no chance of winning. In the 1/10,000,000 version you're deciding to have a super-small chance of winning.
Edit again: I explained it wrong at first! Just goes to show why it's a classic brain-teaser.
okay but if I'm being conspired against then why is this being presented as a math problem
the hurdle is not thinking of the percent as a fixed state through both circumstances. it just absolutely does not work that way, even though it feels like it should.
edit: when inantp says that humans do not understand statistics, they're not joking. Our brains are not wired for understanding statistics at a fundamental level. They go against any rational instinct. It's what makes trusting a computer so hard because it feels wrong.
But if offered to change my selection, I step back and in my head consider the two doors as a new problem, I got two doors, one has a prize that is 50/50.
My thought process:
Then you choose one, it locks that door in at a 1/3 chance of having the prize.
The other two doors have a combined 2/3 chance of having the prize.
Then the host reveals one of the doors. The secret to this problem is: The door he opens is never the one with the prize. If the door he opened was random, then the monty hall problem wouldn't exist.
Now you have one door which inherited a 2/3 chance of having the prize, and the one you picked, which has a 1/3 chance of having the prize.
That’s the part I absolutely can’t wrap my head around.
1/3 chance it's the door you picked (door 1, say)
1/3 chance it's door 2
1/3 chance it's door 3
So 1/3 chance it's the door you picked, 2/3 chance it's either of the other two doors.
The host opens door 3, it's not door 3. Now it's a 2/3 chance that it's door 2 because we know it isn't door 3 and we know it's a 2/3 chance that it's either door 3 or door 2.
I mean it's def a problem designed to spin you around on probability, which is incredibly unintuitive in the first place.
This actually makes me miss talking stats, it's fun playing around with these problems.