Monty Hall problem revisited
- Thread starter EXTRAPOLATER
- Start date
IMHO, If you are
IMHO, If you are
on a game show dressed up as a giant carrot, you have bigger issues facing you then figuring out the varying odds on a correct curtain choice. :0008
IMHO, If you are
on a game show dressed up as a giant carrot, you have bigger issues facing you then figuring out the varying odds on a correct curtain choice. :0008
As a response to this:
Of course the odds adjust. You are making a new wager. Originally you had a 33% chance and
with the 1/2 option now offered you're odds increase to 50%.
You are not betting if your original pick was correct, but on what might be correct with 2 remaining
options.
---
What is not true, here?
You are basically given a push on your initial play and asked to re-choose. What part of my statement
is untrue?
You can respond with 'nah' and have me conclude that cognitive dissonance plagues you but to
tell me that something I postulated is wrong requires a better argument; if it is wrong then what's
wrong or has the whole world gone crazy?
Believe me, I can change my mind quite readily. I can find no temptation to do so on this matter
either within or with other sources I've checked.
This, actually, is no longer an issue for me.
I'm still curious on other thoughts, being that weirdo that I am (I before E except after C or
whenever else it doesn't apply).
I think you're right but the comment is unclear whether or not you're referring to the 3-door
original or million-door original.
I guess it's the same in either case as your 'adjusted odds' (please consider that phrase, anyone
else crazy enough to endeavour) will be at 50%.
I need a vacation.
Or a Maple Leafs win on...nevermind...
OK, I'll give it a try. This is exactly what jord (and others) already said, but presented in a slightly different way.
We'll call the 3 doors A, B, and C. We all agree that no matter which door you pick, your odds are 1 in 3 of being correct. So let's say you select Door A, then logically this follows:
Odds of it being behind Door A: 1 in 3
Odds of it being behind Doors B or C: 2 in 3
Now Monty chooses a door to open, and how he chooses it is the key.
If he decided which door to open via a random method (like rolling dice) and any of the doors were fair game to open (including yours and also including the one with the prize if they are not the same door) then Extrapolator (and everyone who agrees with him) would be right. If the door he opened didn't have the prize (and wasn't yours), then it would be a pure 50/50 proposition between the remaining 2 doors and your odds would be exactly the same whether you switched or not.
But that's not what Monty does. He knows where the prize is and is using that information in deciding which door to open. And because he does it that way, this ceases to be a purely mathematical problem and becomes a psychological one.
Since the prize can only be behind one of the 3 doors, and Monty always gets to choose which of 2 doors to open, he will always be able to open a door with a goat behind it. Therefore, as counter-intuitive as it seems, Monty's actions don't matter. The simple act of opening a door (judgmentally and not randomly selected) does not alter the original odds that there was a 2 in 3 chance that the car was behind Door 2 or 3. All he's shown you is that one of those doors has a goat behind - which, since there is only 1 prize, you already knew!
Or think of it this way:
Let's say the game is played slightly differently. You select Door A, and then Monty immediately offers to let you have Doors B and C instead. Do you make that switch? Assuming that this is just how the game is played and you know this is not some Monty Hall shenanigans, of course you switch. You've just improved your odds from 1 in 3 to 2 in 3. Now he says, "At least one of your doors has a goat behind it; I'm going to open that door now." You respond by saying "No shit one of my doors has a goat behind it; there's only one car and I picked 2 doors." Again, because he knows where the car is, all he's giving you is information you already knew and it doesn't change the fact that you still stand a 2 in 3 chance of being right - even though there are only 2 doors remaining and it looks on the surface like a 50/50 prop. This is Monty Hall shenanigans.
The two scenarios above are actually identical. Whether Monty gives you the switch option before or after he opens a door has no bearing on the odds - you're still getting 2 doors if you switch, and only 1 if you don't.
We'll call the 3 doors A, B, and C. We all agree that no matter which door you pick, your odds are 1 in 3 of being correct. So let's say you select Door A, then logically this follows:
Odds of it being behind Door A: 1 in 3
Odds of it being behind Doors B or C: 2 in 3
Now Monty chooses a door to open, and how he chooses it is the key.
If he decided which door to open via a random method (like rolling dice) and any of the doors were fair game to open (including yours and also including the one with the prize if they are not the same door) then Extrapolator (and everyone who agrees with him) would be right. If the door he opened didn't have the prize (and wasn't yours), then it would be a pure 50/50 proposition between the remaining 2 doors and your odds would be exactly the same whether you switched or not.
But that's not what Monty does. He knows where the prize is and is using that information in deciding which door to open. And because he does it that way, this ceases to be a purely mathematical problem and becomes a psychological one.
Since the prize can only be behind one of the 3 doors, and Monty always gets to choose which of 2 doors to open, he will always be able to open a door with a goat behind it. Therefore, as counter-intuitive as it seems, Monty's actions don't matter. The simple act of opening a door (judgmentally and not randomly selected) does not alter the original odds that there was a 2 in 3 chance that the car was behind Door 2 or 3. All he's shown you is that one of those doors has a goat behind - which, since there is only 1 prize, you already knew!
Or think of it this way:
Let's say the game is played slightly differently. You select Door A, and then Monty immediately offers to let you have Doors B and C instead. Do you make that switch? Assuming that this is just how the game is played and you know this is not some Monty Hall shenanigans, of course you switch. You've just improved your odds from 1 in 3 to 2 in 3. Now he says, "At least one of your doors has a goat behind it; I'm going to open that door now." You respond by saying "No shit one of my doors has a goat behind it; there's only one car and I picked 2 doors." Again, because he knows where the car is, all he's giving you is information you already knew and it doesn't change the fact that you still stand a 2 in 3 chance of being right - even though there are only 2 doors remaining and it looks on the surface like a 50/50 prop. This is Monty Hall shenanigans.
The two scenarios above are actually identical. Whether Monty gives you the switch option before or after he opens a door has no bearing on the odds - you're still getting 2 doors if you switch, and only 1 if you don't.
People argue this all over the interwebs.
I like this guy's explanation.
Here's the logical reasoning method of proving that switching is better:
1. At the beginning of the game, when you select a door, it is more likely that you will guess a wrong door than the right one. (Specifically, you will be wrong 2 out of 3 times, but "more likely to be wrong" is close enough for a logic-based non-numerical explanation.)
2. If you guess the wrong door, then Monty MUST open the only other incorrect door, since he cannot open (a) the door you picked, or (b) the prize door. Another way of saying this is that if your first guess is wrong, then the door that neither you nor Monty pick MUST be the right door.
3. The only situation in which switching your guess would cause you to lose is if your first guess had been correct.
4. However, as stated in point 1, you were more likely to be wrong in your first choice than correct.
5. Since it is more likely that your original guess was wrong, then changing your guess to the only other available door MUST improve your chances of winning.
I like this guy's explanation.
Here's the logical reasoning method of proving that switching is better:
1. At the beginning of the game, when you select a door, it is more likely that you will guess a wrong door than the right one. (Specifically, you will be wrong 2 out of 3 times, but "more likely to be wrong" is close enough for a logic-based non-numerical explanation.)
2. If you guess the wrong door, then Monty MUST open the only other incorrect door, since he cannot open (a) the door you picked, or (b) the prize door. Another way of saying this is that if your first guess is wrong, then the door that neither you nor Monty pick MUST be the right door.
3. The only situation in which switching your guess would cause you to lose is if your first guess had been correct.
4. However, as stated in point 1, you were more likely to be wrong in your first choice than correct.
5. Since it is more likely that your original guess was wrong, then changing your guess to the only other available door MUST improve your chances of winning.
Another way to look at it.
It's 1/3-2/3
Monty revealing an empty door tells you nothing.
EVERYONE knew one of the unchosen doors was empty.
This simply squeezes the probability of both unchosen doors into the one left. ( 2/3)
Try this.
Assume you choose door A.
Monty reveals nothing and offers to let you trade for
BOTH B AND C.
This is essentially the same as the original question.
Revealing the empty door won't change it.
It's 1/3-2/3
Monty revealing an empty door tells you nothing.
EVERYONE knew one of the unchosen doors was empty.
This simply squeezes the probability of both unchosen doors into the one left. ( 2/3)
Try this.
Assume you choose door A.
Monty reveals nothing and offers to let you trade for
BOTH B AND C.
This is essentially the same as the original question.
Revealing the empty door won't change it.
I think that johnnyonthespot might closest to a reasonable explanation that I have seen within;
if I'm right about that then, as expected, factors beyond objective probability totally apply.
Your response, Jack, suggests the same thing..look at #2 in your first go:
2. If you guess the wrong door, then Monty MUST open the only other incorrect door, since he cannot open (a) the door you picked, or (b) the prize door. Another way of saying this is that if your first guess is wrong, then the door that neither you nor Monty pick MUST be the right door.
---
That is true if you guess the wrong door. There is a still a 50% chance that such is not the case.
If you picked the right door then--assuming the mind-game issue is excluded, say that Monty wants
you to change to avoid winning or whatever--then Monty has opened one of two remaining losers
and, therefore, would NOT have opened the ONLY other incorrect door; barring a motivational
factor from Monty gives no relevant information, which is why I mentioned that johnnyonthespot
might be more relevant here but I'm going to have to re-read his response after more sleep than
the sox and injuns are likely getting.
Your second response, Jack, leaves me with the obvious conclusion that taking 2 of 3 doors--other
factors aside (i.e. Monty is coercive)--is obviously the right move. This is certainly not the same
as the original question, as you mentioned; if given the 2/3 option (66%) then you would still
be above the 50% option if a losing door is revealed.
I'm still confused as to why this is an issue but am fascinated by the controversy and intrigued
enough to delve deeper into insanity.
Over and out for tonight, Hendrix willing.
if I'm right about that then, as expected, factors beyond objective probability totally apply.
Your response, Jack, suggests the same thing..look at #2 in your first go:
2. If you guess the wrong door, then Monty MUST open the only other incorrect door, since he cannot open (a) the door you picked, or (b) the prize door. Another way of saying this is that if your first guess is wrong, then the door that neither you nor Monty pick MUST be the right door.
---
That is true if you guess the wrong door. There is a still a 50% chance that such is not the case.
If you picked the right door then--assuming the mind-game issue is excluded, say that Monty wants
you to change to avoid winning or whatever--then Monty has opened one of two remaining losers
and, therefore, would NOT have opened the ONLY other incorrect door; barring a motivational
factor from Monty gives no relevant information, which is why I mentioned that johnnyonthespot
might be more relevant here but I'm going to have to re-read his response after more sleep than
the sox and injuns are likely getting.
Your second response, Jack, leaves me with the obvious conclusion that taking 2 of 3 doors--other
factors aside (i.e. Monty is coercive)--is obviously the right move. This is certainly not the same
as the original question, as you mentioned; if given the 2/3 option (66%) then you would still
be above the 50% option if a losing door is revealed.
I'm still confused as to why this is an issue but am fascinated by the controversy and intrigued
enough to delve deeper into insanity.
Over and out for tonight, Hendrix willing.
Sorry; I simply didn't have the time for discourse.
Maybe this page will help you see your errors in thought. If not, I'm happy to do my best to help you see the light tomorrow.
http://betterexplained.com/articles/understanding-the-monty-hall-problem/
This is simply not true. The odds of your original pick being correct NEVER change. IN my scenario with a million doors - every single time you pick initially you are one in a million to guess correct. MOnty, on the other hand, gets to see the correct answer and the 999,999 you guess wrong he is 100% guaranteed to be correct. So, the ONLY way the car isn't in the OTHER door, is if you guessed correctly initially. That, again was a one in a million shot. SO, if you played this game one million times, you will be correct 999,999 if you take the other door. You will be correct only once if you stick with your original choice. There isn't a reset button in the game, that re-randomized between the ultimate 2 doors.
Or put another way, I will play this game with you where you keep your door every time, and I change every time... I will lay you 5-1 for every penny I have,
BTW, Extrapolater... I'm just messing around with that clip... not being a dick. You seem like a cool guy.
Cheers :toast:
Cheers :toast:
Alright, I had to go away and think about this for a couple of days.
Not non-stop but in between baseball, music, and my checkers and go-fish competitions.
I get it now. It was a mind-fuck.
BTW, there are way simpler ways to explain it.
Not non-stop but in between baseball, music, and my checkers and go-fish competitions.
I get it now. It was a mind-fuck.
BTW, there are way simpler ways to explain it.
Just read this today, and it reminded me of this thread. I had never read this way to think about it, but thought maybe this could help some better get it...
The Jelly Bean Problem
Tim Urban | 03/07/16
Welcome to Mini Week, a week where I?m writing a new mini-post for each weekday. Good idea? Terrible idea? I?m not sure. But based on feedback and whether or not doing five posts in a week makes me die, it can be a one-time thing or something that happens again. Either way, off we go.
___________
You?re wandering through a faraway land one day when you pass a plum tree. Excited, you pick a plum from the tree, but just as you?re about to take your first bite, you?re confronted by a nearby man.
?You stole one of my plums,? he says.
You try to give the plum back and explain that you didn?t mean to steal from him, you just really like plums so you picked one.
?Those aren?t mutually exclusive,? he says. ?It sounds like you both really like plums and meant to steal one from me.?
?I guess, yeah, but like?? you say.
?The penalty for stealing plums in our faraway land is death,? he says.
?Shit,? you say.
He takes you with him over to a tree stump and asks you to sit down next to the stump. He reaches in his pocket and takes out three jelly beans?a green one, a red one, and a blue one. He sits down on the other side of the stump and puts the three jelly beans on the stump in a row.
3 beans
He says, ?Here?s some good news. The way we do the death penalty here is that you have a chance to get out of it. Here?s the deal: Two of the jelly beans on the stump are poisonous?you?ll die within 30 seconds of eating either one of them. But one of the jelly beans isn?t poisonous and won?t harm you at all. All three of the jelly beans are delicious. The situation works like this: You pick one of the jelly beans and eat it, and if you happen to pick the non-poisonous one, you?re free to go. Cool??
?Cool,? you say. He tells you it?s time to choose a jelly bean.
You choose the green one.
2 beans
Before you put it in your mouth, the man stops you and says, ?Wait a minute?there?s one other little tradition we have that we do with each prisoner. Hold on to your jelly bean. I?m going to remove one of the other two jelly beans and put it back in my pocket, and I?m going to remove a poisonous one. I know which colors are poisonous and which aren?t, and one thing I can tell you is that blue jelly beans are poisonous.?
He takes the blue jelly bean and puts it back in his pocket. That leaves the red jelly bean still on the stump and the green jelly bean still in your hand.
1 bean
?Okay, time to eat your jelly bean,? he says. ?Oh and we also always allow prisoners to change their mind up until the last second, so if you want to switch to the red one, you can. I don?t care. But you need to eat one of them in the next 10 seconds.?
So the question is?which jelly bean do you eat? Does it even matter?
Stop and think about what you?d do in this situation before moving on.
___________
Here?s the weird thing about that question?yes, it matters. It matters a lot. Knowing only what you know in that situation, there is one clear choice: the red jelly bean.
My guess is that when you thought about it, you decided to stick with your original decision and eat the green one. Because with two jelly beans left, you had a 50/50 shot, and it feels better to stick with your original choice than to switch. Right?
Wrong. Very wrong.
The green jelly bean is double as likely to kill you as the red jelly bean. Here?s why:
When you initially picked the green jelly bean, there was a 1/3 chance that it was the safe one to eat, and a 2/3 chance that it was poisonous and the safe one was still on the stump. When the man removed a poisonous blue jelly bean from the stump, it told you no new info about the green jelly bean in your hand?that still had a 1/3 chance of being safe. But removing the blue jelly bean told you a lot about the red jelly bean?it told you that if the safe jelly bean had been on the stump, the red one is safe.
Put another way, if you picked a poisonous jelly bean?which you would do two-thirds of the time?then choosing to switch after he removes one will save you every time. If you picked the safe one to start off with?which happens one-third of the time?then switching will kill you. So switching is a good choice two-thirds of the time.
You could run a pretty quick simulation: If you ran this process 300 times and picked the green jelly bean each time, about 100 of the times, it would be safe to eat and switching to the red bean would kill you. The other 200 or so times, the green jelly bean would be poisonous and switching to the red bean would save you.
One other way to look at it is this?say instead of three jelly beans, the guy puts down 1,000 jelly beans and tells you again that only one is safe. So this time, instead of two poisonous jelly beans to contend with, there are 999. He labels them each with a number, 1-1,000, and asks you to choose. You choose jelly bean 267 and hold it in your hand, leaving 999 jelly beans on the stump. Then, while you hold onto jelly bean 267, the man removes 998 jelly beans from the stump and tells you that all 998 of them are poisonous. All that?s left on the stump is jelly bean 749. He tells you you can stick with jelly bean 267 or switch to jelly bean 749?what would you do?
You?d switch. Jelly bean 267 has a 1/1,000 chance of being safe. There?s nothing special about it. You picked it and it sat on the sidelines as 998 poisonous jelly beans were removed from the 999 on the stump. Jelly bean 749 is special?it?s the survivor. There?s a tiny, 1/1,000 chance that you happened to pick the non-poisonous jelly bean, and if that?s the case, then the man just took 998 jelly beans at random from 999 that were all poisonous. But far more likely is that the jelly bean you picked is one of the poisonous ones, meaning the stump only has 998 poisonous jelly beans on it. So when the man says he?s going to take 998 poisonous ones off the stump, he has no choice?he has to leave the safe one there, so jelly bean 749 is left there very intentionally. Jelly bean 749 is the one you want to eat?it?s been through the tough test and survived, while the one in your hand survived nothing and is super dangerous to eat.
Back to our three jelly bean situation. Same deal. The green jelly bean in your hand and its 1/3 odds of being safe were kept in isolation during the jelly bean removal process, so the 1/3 odds are still valid. But the red jelly bean survived the removal process and all the safe choices that might have been on the stump are now consolidated into the red jelly bean. That?s why you switch.
___________
As many of you have figured out, I didn?t make this up. Well, I made the jelly bean story up?but it?s just my own version of the famous Monty Hall problem, which was popularized by Marilyn vos Savant in her Ask Marilyn column in 1990.
Monty Hall was the host of the game show Let?s Make a Deal, and one of the games went like this:
There are three doors, labeled 1, 2, and 3.
3 doors
Behind one of them is a new car. Behind each of the other two is a goat. You get to pick a door, and if you pick the door with the car behind it, you win the car.1
You pick Door 2, but before you have a chance to open the door, Monty, who knows which door contains the car, tells you he?s going to release a goat from behind one of the two doors you didn?t pick. He opens Door 3 and a goat comes running out.
2 doors
He then gives you an option to switch from your original choice before opening a door.
Marilyn explained the problem and then explained why it was definitely the right choice to switch to Door 1?because if you were wrong in your original choice of Door 2, then the car will be behind Door 1. And there?s a 2/3 chance that you were wrong.
But over the next week, Marilyn received over 10,000 letters, including over 1,000 from PhD?s?and almost all of the letters were berating her for her huge mistake. One PhD put it like this:1
?You blew it, and you blew it big! Since you seem to have difficulty grasping the basic principle at work here, I?ll explain. After the host reveals a goat, you now have a one-in-two chance of being correct. Whether you change your selection or not, the odds are the same. There is enough mathematical illiteracy in this country, and we don?t need the world?s highest IQ propagating more. Shame!?
When people first hear the Monty Hall three-doors problem (or the Tim Urban inane jelly bean story), it doesn?t matter how smart they are?they almost always get it wrong and fervently believe that the final two options have an equal probability of being correct. Which is why almost everyone chooses not to switch their decision when given the opportunity.
One note is that this problem only works with three key conditions in place:
1) The host must always open a door, and it has to be one that was not picked by the contestant.
2) The host must know where the car is and always open a door with a goat behind it.
3) The host will always give you the option to switch your choice.
If any of those conditions isn?t there, the problem doesn?t work. For example, if the host doesn?t know where the car is and takes a guess by opening one of the other two doors, and a goat runs out, then it?s no longer better to switch your choice?it?s 50/50.
A third version of this problem is the Three Prisoners problem, which I?ll explain in this footnote2 for those who are interested.
So?
Has it clicked? You with me? Or still in 50/50 land?
:toast::toast:
The bank of elevators I take up to my office has 3 doors. I constantly think about this problem, even though my guess as to which door will open is not really similar. This one and the birthday paradox are great ways to see probability and statistics in a way that is betrayed by the knee jerk common sense of the averagely intelligent person.
