Monty Hall Nerd Thread for COOL KIDS

DarkDepths

Your friendly neighbourhood robot overlord
#1
but that is the thing, even with the "if this is true" it is super suepr reactionary and with a complete lack of vision

it frustrates me when people think that the same solution that works for 1 company will naturally work for another

it frustrates me when no thought is put into WHY a company would do something different, and WHY t might not be a bad idea to do so

it all just feels so limited to me

it is like when people debate the old 3 doors thing...

there are 3 doors, behind 2 are hungry tigers, behind the 3rd is an escape... a host, with knowledge of what is behind each door asks you to pick one... when you do, he reveals, for a split second, one of the tigers behind a door you didn't pick... then asks you if you want to change your pick....

this question is debated constantly, and it is the msot idiotic thing in the world IMO... because anybody that says it is statistically more likely to give you better odds by switching doors isn't solving for the other side of the equation.... NOTHING has changed... no matter what door you picked the door revealed would have had a tiger, the hose has knowledge you do not, it is obvious.. and yet there are some very smart people out there that argue otherwise... and those very smart people are solving for only 1 side of the equation... they are re-solving for the door unpicked rather than re-solving for both doors based on new knowledge

this is how I feel when dealing with reactionary people discussing the switch...

they become obsessed with 1 facet or another without considering the whole

to me that is just stupid, I guess

I mean I don't want to call people names, but it seems very unintelligent to me, to not take the broader view
So I agree that Rich is short-sighted, though I think that's separate from his reactionary nature. Alas, you are 1000% wrong about the Monty Hall problem. It is proven theoretically, and plays itself out in simulation. That last real challenges fizzled out the 90's when computer simulation demonstrated it to be true to those who had previously refused to accept it. It's what Quine calls a veridical paradox, the outcome seems bogus and it is true nonetheless.

Your problem is that you are skipping over a key piece of information. After revealing a tiger, the host is not the only one with knowledge, the player has additional knowledge as well. In fact, not only is it better to switch doors than to keep your door, it's also better to switch doors than to choose one of the remaining doors at random. I get that with 3 doors, this is a bit hard to intuit, so let's do the counter-intuitive and scale this problem up to make it more intuitive!

The scenario is much the same, but this time there are 100 doors - for simplicity of typing, we'll say 1/100 doors wins, 99/100 doors lose. After choosing a door, the host will open 98 doors that lose, leaving only your chosen door and one other.

When you initially choose a door, you have a 1/100 chance of winning. After the host opens 98 losing doors, that has not changed. Even though there are only 2 doors remaining, you still have a 1/100 chance. Why?

Because the host is guaranteed to open only losing doors. If the host chooses doors at random and may open the winning door (but not your door), then things change: If the host opens the winning door, you now have a 0% chance of winning. If the host does not open the winning door, then you now have a very good chance of winning with your current door. Why?

If the host opens randomly, then he is effectively playing the same game as you, except with 99 doors (since you've put one out of play). By opening all but one of the remaining doors at random, he has a 98/99 chance of opening the winning door if you had not chosen it, and a 0% chance of opening the winning door if you had. So, there is almost a 100% chance that the host would open the winning door if you had not chosen it, and of course no chance at all if you had.

So as the player, you are watching the host open doors randomly, and the winning door does not get opened. Whereas you previously had a 1/100 chance of having the correct door, you would now have a 98/99 chance of having the correct door, so you would of course keep it. Of course, if the winning door had been opened by the host, you would go home sad.

But that's not the scenario. The scenario is that the host knows the contents, and is guaranteed to only open losing doors. However, the previous example illustrates a point, which is that the player gains knowledge as the host opens doors. When the host knowingly opens a losing door, he is telling you that this particular door is not a winning door. More importantly, he is telling you that the winning door is still in the remaining set.

So you choose 1 of the 100 doors which has a 1/100 chance of winning. That means there is a 99/100 chance that the winning door is in the set of doors that you did not pick - the set of doors which the host must narrow down to only 1 (let's call it the hosts' set). So if there is a 99% chance that the host's set contains the winning door, and the host knows which door is the winning door and is contractually obligated to not remove it from the set, the single door he leaves in his set at the end still has a 99% chance of winning.

So now you get to change your mind, do you want the door that still has a 1% chance of winning, or the one with the 99% chance of winning?

This scales no matter the number of doors. If you have 3 doors, and you choose 1, it is a 1/3 chance of winning. The host's set then has a 2/3 chance of containing the winning door. Since the host must keep the winning door if his set contains it, the remaining door that he leaves also has a 2/3 chance of winning. So you are indeed better of switching doors.
 

Shoulder

Your Resident Beardy Bear
#2
To sum it up, the Monty Hall problem means you are statistically more likely to win if you switch than by keeping your original choice. It's been documented for years, and proves every time that switching your choice gives you a 66% chance of winning over the original 33%.

No need to Rich-ify the answer :p
 

DarkDepths

Your friendly neighbourhood robot overlord
#3
To sum it up, the Monty Hall problem means you are statistically more likely to win if you switch than by keeping your original choice. It's been documented for years, and proves every time that switching your choice gives you a 66% chance of winning over the original 33%.

No need to Rich-ify the answer :p
@theMightyME disagrees with the conclusion. I wanted to make it as clear as I could!
 

theMightyME

Owner of The Total Screen
#4
So I agree that Rich is short-sighted, though I think that's separate from his reactionary nature. Alas, you are 1000% wrong about the Monty Hall problem. It is proven theoretically, and plays itself out in simulation. That last real challenges fizzled out the 90's when computer simulation demonstrated it to be true to those who had previously refused to accept it. It's what Quine calls a veridical paradox, the outcome seems bogus and it is true nonetheless.

Your problem is that you are skipping over a key piece of information. After revealing a tiger, the host is not the only one with knowledge, the player has additional knowledge as well. In fact, not only is it better to switch doors than to keep your door, it's also better to switch doors than to choose one of the remaining doors at random. I get that with 3 doors, this is a bit hard to intuit, so let's do the counter-intuitive and scale this problem up to make it more intuitive!

The scenario is much the same, but this time there are 100 doors - for simplicity of typing, we'll say 1/100 doors wins, 99/100 doors lose. After choosing a door, the host will open 98 doors that lose, leaving only your chosen door and one other.

When you initially choose a door, you have a 1/100 chance of winning. After the host opens 98 losing doors, that has not changed. Even though there are only 2 doors remaining, you still have a 1/100 chance. Why?

Because the host is guaranteed to open only losing doors. If the host chooses doors at random and may open the winning door (but not your door), then things change: If the host opens the winning door, you now have a 0% chance of winning. If the host does not open the winning door, then you now have a very good chance of winning with your current door. Why?

If the host opens randomly, then he is effectively playing the same game as you, except with 99 doors (since you've put one out of play). By opening all but one of the remaining doors at random, he has a 98/99 chance of opening the winning door if you had not chosen it, and a 0% chance of opening the winning door if you had. So, there is almost a 100% chance that the host would open the winning door if you had not chosen it, and of course no chance at all if you had.

So as the player, you are watching the host open doors randomly, and the winning door does not get opened. Whereas you previously had a 1/100 chance of having the correct door, you would now have a 98/99 chance of having the correct door, so you would of course keep it. Of course, if the winning door had been opened by the host, you would go home sad.

But that's not the scenario. The scenario is that the host knows the contents, and is guaranteed to only open losing doors. However, the previous example illustrates a point, which is that the player gains knowledge as the host opens doors. When the host knowingly opens a losing door, he is telling you that this particular door is not a winning door. More importantly, he is telling you that the winning door is still in the remaining set.

So you choose 1 of the 100 doors which has a 1/100 chance of winning. That means there is a 99/100 chance that the winning door is in the set of doors that you did not pick - the set of doors which the host must narrow down to only 1 (let's call it the hosts' set). So if there is a 99% chance that the host's set contains the winning door, and the host knows which door is the winning door and is contractually obligated to not remove it from the set, the single door he leaves in his set at the end still has a 99% chance of winning.

So now you get to change your mind, do you want the door that still has a 1% chance of winning, or the one with the 99% chance of winning?

This scales no matter the number of doors. If you have 3 doors, and you choose 1, it is a 1/3 chance of winning. The host's set then has a 2/3 chance of containing the winning door. Since the host must keep the winning door if his set contains it, the remaining door that he leaves also has a 2/3 chance of winning. So you are indeed better of switching doors.
Yes I am aware that everyone had new information, but you then need to solve for both sides of the equation.... The site you picked didn't remain a 33% chance while the other becomes a 50% chance... Because deciding to stay on the original door resets that choice to a 50/50 as well... The whole problem resets to a choice between 2 doors
 

DarkDepths

Your friendly neighbourhood robot overlord
#5
Yes I am aware that everyone had new information, but you then need to solve for both sides of the equation.... The site you picked didn't remain a 33% chance while the other becomes a 50% chance... Because deciding to stay on the original door resets that choice to a 50/50 as well... The whole problem resets to a choice between 2 doors
No, it really doesn't. The original door retains it's initial probability while the other gets the remainder. So the final choice is either 33% or 66%. Neither choice is 50%.

In order to have a 50/50 situation, you would need a random friend who hasn't been following along to come in and choose one.

Keeping the original doesn't give you a 50% chance. Here's an experiment to do with a friend. Get a deck of cards and shuffle them up. Choose one at random, and don't look at it. The goal is to choose the Ace of Hearts. Have your friend go through the remainder of the deck and discard all but one card - but he's not allowed to discard the Ace of Hearts. Reveal the cards.

Do this several times. Statistically, if you do this 52 times, you will end up with the Ace of Hearts in your hands just once, while your friend will have it all the rest of the times. It's exactly the same.

There is no debate here, the experts aren't failing to "solve the other side of the equation". You're simply wrong on this one.


@FriedShoes
:(
 

theMightyME

Owner of The Total Screen
#6
No, it really doesn't. The original door retains it's initial probability while the other gets the remainder. So the final choice is either 33% or 66%. Neither choice is 50%.

In order to have a 50/50 situation, you would need a random friend who hasn't been following along to come in and choose one.

Keeping the original doesn't give you a 50% chance. Here's an experiment to do with a friend. Get a deck of cards and shuffle them up. Choose one at random, and don't look at it. The goal is to choose the Ace of Hearts. Have your friend go through the remainder of the deck and discard all but one card - but he's not allowed to discard the Ace of Hearts. Reveal the cards.

Do this several times. Statistically, if you do this 52 times, you will end up with the Ace of Hearts in your hands just once, while your friend will have it all the rest of the times. It's exactly the same.

There is no debate here, the experts aren't failing to "solve the other side of the equation". You're simply wrong on this one.


@FriedShoes
:(
it frustrates me to hell, but you are right... you shouldn't be... as if the player had never chosen at first and the door was revealed then it would be 50/50 blarghhh I am rewritting physics... so Now I am right... all hail your new god
 

Juegos

All mods go to heaven.
Moderator
#7
@DarkDepths I don't understand why the host opening doors matters at all. The host is going to open doors regardless of whether you've already picked the right door or not, aren't they? The host is a liar and a cheater, and I refuse to pay attention to their actions.

Edit: Never mind, I got it. So there is always an advantage for switching over sticking with your original pick, but it becomes less significant with higher numbers. That said, there's pretty much never any reason not to switch.

Maybe that's why it's called the Nintendo Switch. Because as your second console it will always be better than your first one.
 
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DarkDepths

Your friendly neighbourhood robot overlord
#8
@DarkDepths I don't understand why the host opening doors matters at all. The host is going to open doors regardless of whether you've already picked the right door or not, aren't they? The host is a liar and a cheater, and I refuse to pay attention to their actions.
After you choose a door, the host opens all other doors except one, and all open doors are losers. The player is then asked if they want their original choice, or this new door.

It adds drama, a second choice, a chance to change your mind. It just so happens that you have better odds of winning if you change your mind. With a 3 door setup it's less pronounced than with 100 doors, but the effect is still there.

Yes, if you had already chosen the right door, then switching is not ideal. But statistically, you are more likely to win by switching than you are to choose the right door initially.
 

theMightyME

Owner of The Total Screen
#9
After you choose a door, the host opens all other doors except one, and all open doors are losers. The player is then asked if they want their original choice, or this new door.

It adds drama, a second choice, a chance to change your mind. It just so happens that you have better odds of winning if you change your mind. With a 3 door setup it's less pronounced than with 100 doors, but the effect is still there.

Yes, if you had already chosen the right door, then switching is not ideal. But statistically, you are more likely to win by switching than you are to choose the right door initially.
Why does it make me so angry that this is true, and who should be angry with?
 

Juegos

All mods go to heaven.
Moderator
#10
Why does it make me so angry that this is true, and who should be angry with?
RNGesus.

With a 3 door setup it's less pronounced than with 100 doors, but the effect is still there.
I had just convinced myself that it was the opposite of this. The comparison I'm making is between what are your chances of winning if you stick with your choice (always 1/X, where X is the number of possible choices), versus the chances of winning if you switch to a different choice (seems to be [X-1] / [X*(X-2)]). With 3 choices, your chances or winning by sticking with your choice are 1/3, while your chances of winning by switching your choice are 2/3; with 100 choices, your chances of winning by sticking with your choice are 1/100, but by switching they turn into 99/9800, which is just above 1/100.
 

DarkDepths

Your friendly neighbourhood robot overlord
#11
RNGesus.


I had just convinced myself that it was the opposite of this. The comparison I'm making is between what are your chances of winning if you stick with your choice (always 1/X, where X is the number of possible choices), versus the chances of winning if you switch to a different choice (seems to be [X-1] / [X*(X-2)]). With 3 choices, your chances or winning by sticking with your choice are 1/3, while your chances of winning by switching your choice are 2/3; with 100 choices, your chances of winning by sticking with your choice are 1/100, but by switching they turn into 99/9800, which is just above 1/100.
I have no idea how you concluded a 99/9800 probability of winning by switching! But I'm pretty sure it should by 99/100.
 

Juegos

All mods go to heaven.
Moderator
#12
I have no idea how you concluded a 99/9800 probability of winning by switching! But I'm pretty sure it should by 99/100.
Are you taking into account that you can switch from a losing choice to another losing choice, rather than to a winning one? The way I saw it, in each of the 99 or (X-1) cases in which your first guess is incorrect, you have a 1/98 or 1/(X-2) chance to switch to the correct one; in the case that you first guess was correct, you have a 0/98 chance of switching to the correct one. So there are a total of 99 cases in which you switch from a losing choice to a winning choice, out of a total of 9800 possible switches into either a winning or losing choice.

Isn't that right?
 

DarkDepths

Your friendly neighbourhood robot overlord
#13
Are you taking into account that you can switch from a losing choice to another losing choice, rather than to a winning one? The way I saw it, in each of the 99 or (X-1) cases in which your first guess is incorrect, you have a 1/98 or 1/(X-2) chance to switch to the correct one; in the case that you first guess was correct, you have a 0/98 chance of switching to the correct one. So there are a total of 99 cases in which you switch from a losing choice to a winning choice, out of a total of 9800 possible switches into either a winning or losing choice.

Isn't that right?
Oh, I see what you are saying. But in the real scenario, you only get the option to choose when all but one other door has been eliminated. So in fact you don't have the option of switching from losing to losing - one must be a winner. So you don't really end up with a cumulative probability.
 

Juegos

All mods go to heaven.
Moderator
#14
Oh, I see what you are saying. But in the real scenario, you only get the option to choose when all but one other door has been eliminated. So in fact you don't have the option of switching from losing to losing - one must be a winner. So you don't really end up with a cumulative probability.
Oh I see, I hadn't paid attention to that before. So the host opens the other (X-2) doors and then asks you to choose between staying with yours or switching to the other one. That's nuts. Carry on.
 

Shoulder

Your Resident Beardy Bear
#16
As a supplemental video, here's another video with numberphile (great Maths channel on YT) that also goes into the 100 door scenario as well.

 
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