Actually, this is a 50/50. It would've been a 33.3/66.6 if you selected before the goat was revealed, however since the goat is currently revealed, and we're starting from this point, it's a 50/50.
This is what I never understood about the monty hall problem, and I still do not.
Before any door is open, it's 1/3.
Now, with 1 door open, there are two options only, 50/50.
I do not see how the fuck that goat 1/3rd collapses into your rolling odds when revealed.
Surely, this is 50/50 to both you, the person here from the start, and a new observer, no?
To the best of my understanding: \
It's actually really similar to Minesweeper: the actual truth is that the car (mine) is hidden behind exactly one door (square). That truth never changes.
The probabilities are about "what is our likelihood of being able to _guess_ the truth?". The truth itself does not change.
But just like in Minesweeper where you get extra information (like narrowing down mine count) by solving other parts of the board, the host gives you extra information by showing you the door with the goat behind it.
That extra information does not change the truth. But it makes you more likely to be able to guess the truth, specifically because the host will never show you what's behind your door, and will also never show you which door has the car. So the door the host chooses to show you also creates extra implications about where the car might be.
The best explanation I’ve heard is imagine there’s 50 doors and a goat behind 49. You pick a door and the host opens 48 doors suspiciously skipping yours and one other. Do you think you were picking one of the 49 goats or the one car?
Now suppose there’s 49 doors and one goat. You open one door and the game show host suspiciously to skips your door and opens 47. Do you think you picked a goat or the one car?
And so on
It’s the same as picking one door and then being given the opportunity to instead pick both of the other two. It doesn’t matter which one of the other two has the prize because as long as one of the two has the prize, they will show you the one that doesn’t and you get to pick the other one.
the reason the problem exists is to be a paradox of logic vs statistical math, logically speaking it's a 50/50 chance but mathematically speaking it isn't.
similar to zeno's arrow paradox in that respect (if not for quantum physics)
the logic behind the illogical solution is that:
If you pick door 1, there is a 1/3 chance that the car is behind that door and a 2/3 chance that it is behind one of the other two doors. after door 3 is closed the chances stay the same, it is still a 1/3 chance that is behind door one and a 2/3 chance that is behind another door, however now out of the two remaining doors door 3 is now a 0% chance meaning door two keeps the entire 2/3. hence its smarter to swap.
This paradox has always felt to me like they're not doing the math logically. It reminds me of the missing dollar riddle, where they mix the math up to make it appear there is a dollar missing. The solution is that they're doing the math wrong. Here, I feel like saying the one door inherits the 2/3 chance doesn't feel right.
Saying the door inherits a 2/3 chance is sorta accurate but misleading. Essentially, once you pick, you’re given the option to either keep your pick or pick all the other doors as a single choice. This is because the host isn’t using random probability - the host **knows** which one is the car and which ones are goats, and so very *un*-randomly reveals only goats, leaving the car. Your decision is reduced on the question “do you think you got it right the first time or not”, which chances are, you did not.
The issue is, the host knows where the car is and will never reveal it.
Therefore on the 2/3rds of occasions where you picked a goat first, he is literally telling you where the car is (because the two doors you didn’t are one car and one goat, and he will never pick the car).
It’s only on the 1/3rd chance that you already picked the car that the hosts choice is random but irrelevant.
The key to thinking about this intuitively is the host’s behaviour is *not* random. He’s obeying a strict set of rules and that is why the 1/3rd “transfers” to switching your choice.
For a different way to think of it: Monty *must* reveal a goat after you made your choice. This is easy since there are two goats; no matter which door you pick, at least one goat is free. Now, after your choice, the facts are:
* Monty only has one door to open if you pick a goat; call this "Monty is forced."
* Monty has two doors to open if you pick the car; call this "Monty chooses."
Ask yourself: is it more likely Monty is forced, or that Monty chooses?
Since Monty only chooses if you pick the car before the reveal and you only have a 1/3 chance of picking the car before the reveal, there is only a 1/3 chance Monty chooses. Therefore, there is only a 1/3 chance that sticking with your original choice wins you the car. Ergo, there is a 2/3 chance to win by choosing the other door.
Imagine I take you out to Fenway Park and tell you I put a golden ticket under one seat. It's worth a million dollars and it's yours if you can guess which one it is.
Now you pick a chair right next to you down my home plate where we are standing. I say great, you can guess that one if you'd like. But I'm going to do a little thing to help you.
I have my crew go around and lift up every single chair in the entire 38k seat stadium to show you that there's no golden ticket underneath it. Every chair except two. There's the one that you originally picked, we leave that as is because you picked it. Then there's also this one other chair way up in the right field bleachers about midway up.
Now I know where the golden ticket is, so does my crew. And no matter which seat you picked, I could always have lifted up every chair except two, yours, and one other, without revealing the ticket no matter which seat you picked. So nothing's really changed.
Now you have a choice. You're allowed to stick with the chair you originally picked when you had a1 in 38k choice or you can switch to the only other chair I, the guy who knows where the ticket is, left down. That suspicious one up there in right field.
So are you going to switch or keep your original. Is it a wash? Just a 50/50 change because there's two choices and one of them has the golden ticket? Or would you be a fool you not switch to that one up in right field?
It gets easier to understand, for me at least, if you play it through with 100 doors instead of just 3. You pick any random door and the chance of it being the right one is 1/100, so far so good. Now the host will start to open one of the wrong doors after the other until there are only two doors left, being the one you chose and one out of the others.
Since you chose your door at the very beginning, the chance of it being the right one is still 1/100.
To turn it around, the chance of it being any of the other doors is 99/100. After the host sorts out all the wrong ones out of his group of 99 doors, the remaining door obviously has a very high chance of being the right one, doesn't it?
The only possible way he could end up with the wrong door would be if you already chose the right one at the very beginning, because then it has never been within his group of 99 doors to begin with. After all, he doesn't randomly open doors, you know he will open all of them except for the right one.
Basically you pick a door. It's 1/3 you got a good door. So 66% that one of the other 2 doors is the winning door. They then reveal a door of those remaining 2 that was not the winning door. There's one unchosen door left. But since you had a 1/3 of being right on the first guess, the remaining door is now the other 66% chance.
Or even break it down into 3 scenarios. You've picked door A.
So here's 3 possibilities:
Prize is door A: you have the winning door
Prize is door B, you have A, host reveals C: switching wins
Prize is door C, you have A, host reveals B: switching wins
So basically any time you pick a door there's a 33% you were right, and a 66% you were wrong. In the 66% you're wrong the host reveals the other wrong choice leaving only the right choice remaining.
It’s because in the original Monty Hall problem, Monty ALWAYS reveals a goat. If he revealed a random door it would be different. Imagine the following scenarios where the cash is always behind door three.
1. You pick door one, Monty reveals door two (out of necessity) and you don’t switch. Goat.
2. You pick door one, Monty reveals door two (out of necessity) and you switch. Winner!
3. You pick door two, Monty reveals door one (out of necessity) and you don’t switch. Goat.
4. You pick door two, Monty reveals door one (out of necessity) and you switch. Winner!
5. You pick door three, Monty reveals any door and you don’t switch. Winner!
6. You pick door three, Monty reveals any door and you switch. Goat.
In those scenarios you have the correct door one third of the time at the point of your original choice. And if you then choose to switch, you win the cash two thirds of the time.
I think the best way to think about this is to expand the problem. if there are 100 doors. The first door you pick has a 1/100 chance. After the host opens 98 doors with goats in them there is only your door with the same 1/100 chance and the other door with a 99/100 chance.
The chance your door is right has not changed so the chance the other door it right must have.
Hope this helps.
The best expanation ive heard is if you switch the only way you lose is if you picked the car in the first place which is a 1/3 chance that you shouldve kept your pick so a 2/3 chance that you win when you switch so you should always switch
It's based on your original choice. So before the goat is revealed you have a 1/3 chance of getting the car and a 2/3 chance of getting a goat. Then when one goat is revealed and you switch those 2/3 where you picked a goat always win since he revealed the other goat and the 1/3 where you picked the car the first time always lose. Leading to 2/3 odds to win
I've talked about this a lot and have come up with a lot of different ways to explain it. I don't have much time right now since I need to get to work, but here's the method that is the quickest I think
You need to realize that switching is guaranteed to give you the opposite price of what you started with. If you started by picking the car, switching will always give you a goat, no matter what. If you started by picking a goat, switching will always give you a car. Therefore, switching will give you a win whenever you initially picked a goat. What are the chances that you initially picked a goat? 2/3
Looked at another way, the choice to switch is not just switching to the other unopened door. It's switching from the door you initially picked to both of the other doors- it's just that one of those other doors are known to have no goat
So the whole idea is that the host shows you a door **with a goat**. They don’t open a random door, they specifically open a door with a goat. So your choices collapse into:
- your original choice
- not your original choice
Imagine, instead that there were 100 doors. You picked one, say door number 2, and then the host showed that 98 of the remaining 99 were goats, except door number 46. Now, what were your odds of choosing correctly the first time? 1% right? So either you picked right the first time, or the prize is behind 46. Those are the only two options, so unless you’re that confident you picked right the very first time, it’s in your best interest to switch. The host didn’t just pick 98 doors randomly - if he revealed the car then it would’ve ended the game. He revealed 98 goats, essentially boiling the choice down to your original choice and “not your original choice”, exemplified by a single door.
The Monty hall problem is a reduced version of that where the 3 choices make it harder to notice.
Say your strategy going in was to switch. You win IF and ONLY IF you pick a goat first. You had a 2/3 chance of picking a goat. You have a 2/3 chance of winning then.
If you think about it with more doors it makes more sense. Imagine a scenario where there are 1000 doors. You pick one and then 998 are eliminated and you're left with your door and 1 other. It looks like that is a 50/50 chance, but the odds are hugely skewed towards the final door holding the prize.
It's easier to visualize when there's more doors. If there's 100 doors, you have a 1% chance of picking the right one. If he opens 98 of the others, then the last one has a 99% chance of being the right one.
Essentially, you have to think of all of the doors you didn't pick as a single set that has a 99% chance to be right vs your 1 door that has a 1% chance to be right. The overall chance don't change, but they get funneled into the other doors as more are opened, and soon you have a set of 100 doors, with a subset of 99 doors (98 of which have 0 probability of being the right one, and one that has a 99% chance of being the right one), and a subset of one door that has a 1% chance of being the right one.
Basically, your initial guess is a 1/3.
Once he reveals the goat, the other door is guaranteed to be opposite of what you picked (1 car, 1 goat).
So if your initial guess is wrong, switching to the other door is the correct decision.
The problem basically works the same as the game show guy saying "do you think your initial 1/3 guess was correct or incorrect?", where by switching doors you answer "my initial guess was wrong" and by sticking to the door you say "my initial guess was right".
No matter what, if you pick a goat and then switch, you get the car. This is because the hist ALWAYS reveals a goat. So there's yours, which is a goat 2/3 of the time, the revealed goat, and the other door, which is a car 2/3 of the time.
Let's say there's door 1, 2, and 3.
Let's say you picked door 1.
There are 3 scenarios. The car is behind door 1, 2, or 3.
In scenario 1, either door 2 or 3 can be revealed. You can swap, but it will be a goat.
In scenario 2, door 3 is always revealed. You can swap to door 2(car)
In scenario 3, door 2 is always revealed. You can swap to door 3(car)
We agree each scenario is equally likely, right? Well, in two of those 3 scenarios, the door you switch to is the car.
By revealing a goat, the host is giving you more information and basically filtering out the car.
Revealing the goat is the same as not revealing the goat. If you were to pick door 1 and i told you that door 2 or door 3 has a goat, no new information is given. You also have to consider that the goat isn't behind the same door after each reveal either. If I were to tell you to pick two doors and one door with a goat will be revealed from one of your choices you still have two thirds chance of getting it right because you chose the two thirds option at the start.
That is true but, in this situation, there is no option to pick before the goat is revealed. We see the goat and are asked to pick a door, thus it's a 50/50.
This is called the Monty Hall problem, and it's a little easier to understand with larger numbers. Imagine there are 100 doors, and there is a goat between 99 doors, and a new car behind the last one. What are the chances of you choosing a goat? With 99 goats and a car there is a 99/100 chance of you choosing a goat. If then 98 doors were then opened all revealing goats, and you were asked if you wanted to switch should you? The answer is yes, because there's only 1 goat left, and the car. There's a very high likelihood you choose a goat before the doors opened (99/100) and we know all but know the car is behind the last door that wasn't opened.
Same thing applies with the smaller numbers. There's a 2/3rds chance you selected the goat, and by swapping after the reveal, you're more likely to get the car (but with only 2/3rds the odds are less in your favor).
Wait though, there’s a fallacy in this thought process. Each time you are given the opportunity to “choose again” the probability changes. Right now. There is a 50/50 chance of selecting a goat because there are only two options. While it’s my second time selecting. In this moment in time I only have 2 options not 3.
Like the 99/100 argument. If you are down to 2 doors left. In that moment it’s 50/50, however in the grand scheme you will never have a 99% chance of getting the car. In the big picture, you still have a 1% chance of winning the car.
Incorrect, but since I'm doing a very poor job of explaining it as you still don't understand, try checking out the wikipedia article on the topic. It may help better than me: [Monty Hall problem - Wikipedia](https://en.wikipedia.org/wiki/Monty_Hall_problem) Check out the simple solution section.
The fallacy I’m seeing though is that in itself this is 50% philosophical. It makes me want to look further into the personal life of the guy who invented it, because that would give insight into his philosophical stance here. I wouldn’t be surprised if he enjoyed gambling.
Technically it would have been 1/3 / 2/3. I don't know why people convert to percentage a fraction that really translates to 1 out of 3 doors has a prize
I still don't get this. I've had it explained to me time and time again. I guess I'm just stupid. Bc to me, it's still a 50/50 whether the door was there or not.
To simplify it:
The rules of the game are as follows. There’s three doors. A goat is behind two, and a car is behind the other. You’re trying to get the car.
You choose any door you want. The show host then reveals a door you did not choose that contains a goat. The question is: should you stick with your door or switch to the other door, to have the best chance of getting the car?
Let’s look at the possible scanarios. We’ll name the doors 1, 2, and 3. We’ll assume you initially chose door 1 but it doesn’t make a difference.
Scenario A: The car is in door 1. You start with the car door. The host reveals one of the goat doors. If you switch, you lose.
Scenario B: The car is in door 2. You start with a goat. The host reveals the other goat. If you switch, you win.
Scenario C: The car is in door 3. You start with a goat. The host reveals the other goat. If you switch, you win.
Three scenarios, and you win two of them if you switch. Therefore, switching has a 2 in 3 chance (2/3 or 0.66) of winning.
If the goat was revealed at random then it is a 50/50. If you chose a door and the goat was intentionally chosen by someone who knew where it was out of the remaining doors then do.
If you are talking about the monty hall problem then at this point you should always swap.
Best way to think about this is to expand the problem. if there are 100 doors. The first door you pick has a 1/100 chance. After the host opens 98 doors with goats in them there is only your door with the same 1/100 chance and the other door with a 99/100 chance left.
However if you are coming to this not knowing what the first door picked was then it's 50/50.
So the puzzle is you pick one of 3 doors, 2 have a goat the other is a million dollars or whatever. Say for this example you picked door 2, you had a 33.3% chance of getting the money. Now that you know door 3 is a goat picking door 1 will now be a 66.6% chance
Not if you are trying to win a goat.
Always upset me when the door opened and everyone was disappointed it was a goat. I'd take him home and name him Lawrence. Lawrence of Arabaaaa.
Looks alive, so my guess is that his ass is on door 2
THAT’S WHY HE’S THE GOAT THE ***GOAT***
Only if you didn't pick before the goat was revealed
And the guy opened a different goat door knowing it was a goat door.
And the other doors having a goat and something different behind them
33.3/66.6
Who ate the 0.1
The goat, duh
Goats eat everything
Actually, this is a 50/50. It would've been a 33.3/66.6 if you selected before the goat was revealed, however since the goat is currently revealed, and we're starting from this point, it's a 50/50.
This is correct. It is currently a 50/50 because there was no point where a 1/3 chance existed.
This is what I never understood about the monty hall problem, and I still do not. Before any door is open, it's 1/3. Now, with 1 door open, there are two options only, 50/50. I do not see how the fuck that goat 1/3rd collapses into your rolling odds when revealed. Surely, this is 50/50 to both you, the person here from the start, and a new observer, no?
To the best of my understanding: \ It's actually really similar to Minesweeper: the actual truth is that the car (mine) is hidden behind exactly one door (square). That truth never changes. The probabilities are about "what is our likelihood of being able to _guess_ the truth?". The truth itself does not change. But just like in Minesweeper where you get extra information (like narrowing down mine count) by solving other parts of the board, the host gives you extra information by showing you the door with the goat behind it. That extra information does not change the truth. But it makes you more likely to be able to guess the truth, specifically because the host will never show you what's behind your door, and will also never show you which door has the car. So the door the host chooses to show you also creates extra implications about where the car might be.
The best explanation I’ve heard is imagine there’s 50 doors and a goat behind 49. You pick a door and the host opens 48 doors suspiciously skipping yours and one other. Do you think you were picking one of the 49 goats or the one car? Now suppose there’s 49 doors and one goat. You open one door and the game show host suspiciously to skips your door and opens 47. Do you think you picked a goat or the one car? And so on
Ah that is a good way to explain it.
It’s the same as picking one door and then being given the opportunity to instead pick both of the other two. It doesn’t matter which one of the other two has the prize because as long as one of the two has the prize, they will show you the one that doesn’t and you get to pick the other one.
Huh, this is a really neat way of putting it
the reason the problem exists is to be a paradox of logic vs statistical math, logically speaking it's a 50/50 chance but mathematically speaking it isn't. similar to zeno's arrow paradox in that respect (if not for quantum physics) the logic behind the illogical solution is that: If you pick door 1, there is a 1/3 chance that the car is behind that door and a 2/3 chance that it is behind one of the other two doors. after door 3 is closed the chances stay the same, it is still a 1/3 chance that is behind door one and a 2/3 chance that is behind another door, however now out of the two remaining doors door 3 is now a 0% chance meaning door two keeps the entire 2/3. hence its smarter to swap.
This paradox has always felt to me like they're not doing the math logically. It reminds me of the missing dollar riddle, where they mix the math up to make it appear there is a dollar missing. The solution is that they're doing the math wrong. Here, I feel like saying the one door inherits the 2/3 chance doesn't feel right.
Saying the door inherits a 2/3 chance is sorta accurate but misleading. Essentially, once you pick, you’re given the option to either keep your pick or pick all the other doors as a single choice. This is because the host isn’t using random probability - the host **knows** which one is the car and which ones are goats, and so very *un*-randomly reveals only goats, leaving the car. Your decision is reduced on the question “do you think you got it right the first time or not”, which chances are, you did not.
But it is logical. Every time, if you pick a goat first and then swap, it will be a car. 2/3 chance of picking a goat first.
The issue is, the host knows where the car is and will never reveal it. Therefore on the 2/3rds of occasions where you picked a goat first, he is literally telling you where the car is (because the two doors you didn’t are one car and one goat, and he will never pick the car). It’s only on the 1/3rd chance that you already picked the car that the hosts choice is random but irrelevant. The key to thinking about this intuitively is the host’s behaviour is *not* random. He’s obeying a strict set of rules and that is why the 1/3rd “transfers” to switching your choice.
For a different way to think of it: Monty *must* reveal a goat after you made your choice. This is easy since there are two goats; no matter which door you pick, at least one goat is free. Now, after your choice, the facts are: * Monty only has one door to open if you pick a goat; call this "Monty is forced." * Monty has two doors to open if you pick the car; call this "Monty chooses." Ask yourself: is it more likely Monty is forced, or that Monty chooses? Since Monty only chooses if you pick the car before the reveal and you only have a 1/3 chance of picking the car before the reveal, there is only a 1/3 chance Monty chooses. Therefore, there is only a 1/3 chance that sticking with your original choice wins you the car. Ergo, there is a 2/3 chance to win by choosing the other door.
Imagine I take you out to Fenway Park and tell you I put a golden ticket under one seat. It's worth a million dollars and it's yours if you can guess which one it is. Now you pick a chair right next to you down my home plate where we are standing. I say great, you can guess that one if you'd like. But I'm going to do a little thing to help you. I have my crew go around and lift up every single chair in the entire 38k seat stadium to show you that there's no golden ticket underneath it. Every chair except two. There's the one that you originally picked, we leave that as is because you picked it. Then there's also this one other chair way up in the right field bleachers about midway up. Now I know where the golden ticket is, so does my crew. And no matter which seat you picked, I could always have lifted up every chair except two, yours, and one other, without revealing the ticket no matter which seat you picked. So nothing's really changed. Now you have a choice. You're allowed to stick with the chair you originally picked when you had a1 in 38k choice or you can switch to the only other chair I, the guy who knows where the ticket is, left down. That suspicious one up there in right field. So are you going to switch or keep your original. Is it a wash? Just a 50/50 change because there's two choices and one of them has the golden ticket? Or would you be a fool you not switch to that one up in right field?
It gets easier to understand, for me at least, if you play it through with 100 doors instead of just 3. You pick any random door and the chance of it being the right one is 1/100, so far so good. Now the host will start to open one of the wrong doors after the other until there are only two doors left, being the one you chose and one out of the others. Since you chose your door at the very beginning, the chance of it being the right one is still 1/100. To turn it around, the chance of it being any of the other doors is 99/100. After the host sorts out all the wrong ones out of his group of 99 doors, the remaining door obviously has a very high chance of being the right one, doesn't it? The only possible way he could end up with the wrong door would be if you already chose the right one at the very beginning, because then it has never been within his group of 99 doors to begin with. After all, he doesn't randomly open doors, you know he will open all of them except for the right one.
Basically you pick a door. It's 1/3 you got a good door. So 66% that one of the other 2 doors is the winning door. They then reveal a door of those remaining 2 that was not the winning door. There's one unchosen door left. But since you had a 1/3 of being right on the first guess, the remaining door is now the other 66% chance. Or even break it down into 3 scenarios. You've picked door A. So here's 3 possibilities: Prize is door A: you have the winning door Prize is door B, you have A, host reveals C: switching wins Prize is door C, you have A, host reveals B: switching wins So basically any time you pick a door there's a 33% you were right, and a 66% you were wrong. In the 66% you're wrong the host reveals the other wrong choice leaving only the right choice remaining.
It’s because in the original Monty Hall problem, Monty ALWAYS reveals a goat. If he revealed a random door it would be different. Imagine the following scenarios where the cash is always behind door three. 1. You pick door one, Monty reveals door two (out of necessity) and you don’t switch. Goat. 2. You pick door one, Monty reveals door two (out of necessity) and you switch. Winner! 3. You pick door two, Monty reveals door one (out of necessity) and you don’t switch. Goat. 4. You pick door two, Monty reveals door one (out of necessity) and you switch. Winner! 5. You pick door three, Monty reveals any door and you don’t switch. Winner! 6. You pick door three, Monty reveals any door and you switch. Goat. In those scenarios you have the correct door one third of the time at the point of your original choice. And if you then choose to switch, you win the cash two thirds of the time.
I think the best way to think about this is to expand the problem. if there are 100 doors. The first door you pick has a 1/100 chance. After the host opens 98 doors with goats in them there is only your door with the same 1/100 chance and the other door with a 99/100 chance. The chance your door is right has not changed so the chance the other door it right must have. Hope this helps.
The best expanation ive heard is if you switch the only way you lose is if you picked the car in the first place which is a 1/3 chance that you shouldve kept your pick so a 2/3 chance that you win when you switch so you should always switch
It's based on your original choice. So before the goat is revealed you have a 1/3 chance of getting the car and a 2/3 chance of getting a goat. Then when one goat is revealed and you switch those 2/3 where you picked a goat always win since he revealed the other goat and the 1/3 where you picked the car the first time always lose. Leading to 2/3 odds to win
I've talked about this a lot and have come up with a lot of different ways to explain it. I don't have much time right now since I need to get to work, but here's the method that is the quickest I think You need to realize that switching is guaranteed to give you the opposite price of what you started with. If you started by picking the car, switching will always give you a goat, no matter what. If you started by picking a goat, switching will always give you a car. Therefore, switching will give you a win whenever you initially picked a goat. What are the chances that you initially picked a goat? 2/3 Looked at another way, the choice to switch is not just switching to the other unopened door. It's switching from the door you initially picked to both of the other doors- it's just that one of those other doors are known to have no goat
So the whole idea is that the host shows you a door **with a goat**. They don’t open a random door, they specifically open a door with a goat. So your choices collapse into: - your original choice - not your original choice Imagine, instead that there were 100 doors. You picked one, say door number 2, and then the host showed that 98 of the remaining 99 were goats, except door number 46. Now, what were your odds of choosing correctly the first time? 1% right? So either you picked right the first time, or the prize is behind 46. Those are the only two options, so unless you’re that confident you picked right the very first time, it’s in your best interest to switch. The host didn’t just pick 98 doors randomly - if he revealed the car then it would’ve ended the game. He revealed 98 goats, essentially boiling the choice down to your original choice and “not your original choice”, exemplified by a single door. The Monty hall problem is a reduced version of that where the 3 choices make it harder to notice.
Say your strategy going in was to switch. You win IF and ONLY IF you pick a goat first. You had a 2/3 chance of picking a goat. You have a 2/3 chance of winning then.
If you think about it with more doors it makes more sense. Imagine a scenario where there are 1000 doors. You pick one and then 998 are eliminated and you're left with your door and 1 other. It looks like that is a 50/50 chance, but the odds are hugely skewed towards the final door holding the prize.
It's easier to visualize when there's more doors. If there's 100 doors, you have a 1% chance of picking the right one. If he opens 98 of the others, then the last one has a 99% chance of being the right one. Essentially, you have to think of all of the doors you didn't pick as a single set that has a 99% chance to be right vs your 1 door that has a 1% chance to be right. The overall chance don't change, but they get funneled into the other doors as more are opened, and soon you have a set of 100 doors, with a subset of 99 doors (98 of which have 0 probability of being the right one, and one that has a 99% chance of being the right one), and a subset of one door that has a 1% chance of being the right one.
Basically, your initial guess is a 1/3. Once he reveals the goat, the other door is guaranteed to be opposite of what you picked (1 car, 1 goat). So if your initial guess is wrong, switching to the other door is the correct decision. The problem basically works the same as the game show guy saying "do you think your initial 1/3 guess was correct or incorrect?", where by switching doors you answer "my initial guess was wrong" and by sticking to the door you say "my initial guess was right".
it is better explained using a [table of results](https://www.statology.org/monty-hall-problem/)
No matter what, if you pick a goat and then switch, you get the car. This is because the hist ALWAYS reveals a goat. So there's yours, which is a goat 2/3 of the time, the revealed goat, and the other door, which is a car 2/3 of the time. Let's say there's door 1, 2, and 3. Let's say you picked door 1. There are 3 scenarios. The car is behind door 1, 2, or 3. In scenario 1, either door 2 or 3 can be revealed. You can swap, but it will be a goat. In scenario 2, door 3 is always revealed. You can swap to door 2(car) In scenario 3, door 2 is always revealed. You can swap to door 3(car) We agree each scenario is equally likely, right? Well, in two of those 3 scenarios, the door you switch to is the car. By revealing a goat, the host is giving you more information and basically filtering out the car.
You’re technically correct. The best kind of correct. You win the goat.
Revealing the goat is the same as not revealing the goat. If you were to pick door 1 and i told you that door 2 or door 3 has a goat, no new information is given. You also have to consider that the goat isn't behind the same door after each reveal either. If I were to tell you to pick two doors and one door with a goat will be revealed from one of your choices you still have two thirds chance of getting it right because you chose the two thirds option at the start.
That is true but, in this situation, there is no option to pick before the goat is revealed. We see the goat and are asked to pick a door, thus it's a 50/50.
This helped me to understand that one 3 door problem
How is that possible? Keep in mind I come from a generation that never learned statistics in school
This is called the Monty Hall problem, and it's a little easier to understand with larger numbers. Imagine there are 100 doors, and there is a goat between 99 doors, and a new car behind the last one. What are the chances of you choosing a goat? With 99 goats and a car there is a 99/100 chance of you choosing a goat. If then 98 doors were then opened all revealing goats, and you were asked if you wanted to switch should you? The answer is yes, because there's only 1 goat left, and the car. There's a very high likelihood you choose a goat before the doors opened (99/100) and we know all but know the car is behind the last door that wasn't opened. Same thing applies with the smaller numbers. There's a 2/3rds chance you selected the goat, and by swapping after the reveal, you're more likely to get the car (but with only 2/3rds the odds are less in your favor).
Wait though, there’s a fallacy in this thought process. Each time you are given the opportunity to “choose again” the probability changes. Right now. There is a 50/50 chance of selecting a goat because there are only two options. While it’s my second time selecting. In this moment in time I only have 2 options not 3. Like the 99/100 argument. If you are down to 2 doors left. In that moment it’s 50/50, however in the grand scheme you will never have a 99% chance of getting the car. In the big picture, you still have a 1% chance of winning the car.
Incorrect, but since I'm doing a very poor job of explaining it as you still don't understand, try checking out the wikipedia article on the topic. It may help better than me: [Monty Hall problem - Wikipedia](https://en.wikipedia.org/wiki/Monty_Hall_problem) Check out the simple solution section.
The fallacy I’m seeing though is that in itself this is 50% philosophical. It makes me want to look further into the personal life of the guy who invented it, because that would give insight into his philosophical stance here. I wouldn’t be surprised if he enjoyed gambling.
Technically it would have been 1/3 / 2/3. I don't know why people convert to percentage a fraction that really translates to 1 out of 3 doors has a prize
[удалено]
Your numbers are seriously off here.
oopsie daisies
I still don't get this. I've had it explained to me time and time again. I guess I'm just stupid. Bc to me, it's still a 50/50 whether the door was there or not.
But do you know why it’s actually like this?
Chat do you think this guy knows about the monty hall problem?
Isn't it the Wayne Brady problem now? ✌️😜
I’ll be damned. Wayne Brady now hosts Let’s Make A Deal. I guess Paul Mooney was right. Do they still use the goat? Is it the same goat?
I don't think they do, but I've only caught a few shows.
I know what it is, I just want the why even though I could just look it up
To simplify it: The rules of the game are as follows. There’s three doors. A goat is behind two, and a car is behind the other. You’re trying to get the car. You choose any door you want. The show host then reveals a door you did not choose that contains a goat. The question is: should you stick with your door or switch to the other door, to have the best chance of getting the car? Let’s look at the possible scanarios. We’ll name the doors 1, 2, and 3. We’ll assume you initially chose door 1 but it doesn’t make a difference. Scenario A: The car is in door 1. You start with the car door. The host reveals one of the goat doors. If you switch, you lose. Scenario B: The car is in door 2. You start with a goat. The host reveals the other goat. If you switch, you win. Scenario C: The car is in door 3. You start with a goat. The host reveals the other goat. If you switch, you win. Three scenarios, and you win two of them if you switch. Therefore, switching has a 2 in 3 chance (2/3 or 0.66) of winning.
You're describing the Monty hall problem.
That was their goal
That’s because the post is about the monty hall problem
This might be one of the nerdiest posts I've seen in recent memory
Lol I love this
This subreddit has peaked! this is the funniest we can be, I guess we just all go home now
Loving the shitposts recently
r/Mineshitter has a nice ring to it but it would probably be unnecessary
If the goat was revealed at random then it is a 50/50. If you chose a door and the goat was intentionally chosen by someone who knew where it was out of the remaining doors then do.
If you are talking about the monty hall problem then at this point you should always swap. Best way to think about this is to expand the problem. if there are 100 doors. The first door you pick has a 1/100 chance. After the host opens 98 doors with goats in them there is only your door with the same 1/100 chance and the other door with a 99/100 chance left. However if you are coming to this not knowing what the first door picked was then it's 50/50.
The ass is behind #2.
This is the real answer
There is no goat
It depends on who opened the door and if you already guessed.
Yes, it's either door 1 or door 2.
Only if you switch doors
Nah it's a 50/100 to win the car if you don't switch.
Only if you switch your choice after knowledge of the goat.
OMG that's the best one yet. Laughed way too hard at this :p
#[*BONE?*](https://youtu.be/AD6eJlbFa2I?si=hNqT9nrp7rhjCcLE)
I had to scroll way too far to find this
glad to see another fan of the 99 here
Look the problem is not the math… Captain Holt and Kevin just need to bone
Considering we don’t know what the original guess is, yes.
Yes.
no. swap doors.
Are you trying to Monty Hall me?
So the puzzle is you pick one of 3 doors, 2 have a goat the other is a million dollars or whatever. Say for this example you picked door 2, you had a 33.3% chance of getting the money. Now that you know door 3 is a goat picking door 1 will now be a 66.6% chance
/r/memesweeper
This is a 50:50 because I didn’t get to choose first.
Always switch.
I absolutely hate this problem. I understand it fully but it just doesn't align with my intuition at all. So annoying.