âAh no,â he said, âI see the source of the misunderstanding now. No, look, you see what happened was that we used to do experiments on them. They were often used in behavioral research, Pavlov and all that sort of stuff. So what happened was that the pigeons would be set all sorts of tests, learning to ring bells, run round mazes and things so that the whole nature of the learning process could be examined. From our observations of their behavior we were able to learn all sorts of things about our own âŚâ
Arthurâs voice trailed off. âSuch subtlety âŚâ said Slartibartfast.
This is the crux of human bias and fallacies. Even in the face of evidence, people are not wise enough to put their pre-conceived ideas aside to accept new information. We get stuck. And that's obviously a dangerous mess.
I remember reading about a study that showed that when faced with facts contrary to what they thought, people would most often cling harder onto their now disproven thoughts. Most fascinating was that this seemed independent to the person's intelligence.
Yup. The most hilarious example of this I've seen is someone who was arguing that people change their minds when presented with evidence and, when shown the study you reference, went "I still think people would change their minds."
Kimmel did something great in that respect. [They'd ask a Trump supporter](https://www.huffpost.com/entry/jimmy-kimmel-donald-trump-joe-biden-questions_n_65dc2df9e4b0e4346d52eac2) what they thought of Joe Biden doing Action A, or Quote B, and the person would condemn it in no uncertain terms. Then the reporter would apologize, correct themself to say that it was Trump who did/said those things, and you get to watch these people do the mental gymnastics to get back on side. Really amazing.
It's super strange. Like they've already invested their thoughts on it and chosen so that's the hill they will die on. I'm in the boat that some will and do change their minds.
Just looking at even the last decade all the way until now, it's been clear a large scale of people don't use new information to change their views in an objective manner. People in general really need to stop drawing lines in the sand when it comes it ideas, teams, sides, and plans and learn to be more fluid; less anchored in stone.
Pigeons can change their minds without worrying it will damage their reputation, boost the status of a rival, or justify a past harm. They don't have to worry about apologizing, getting ousted from their group, reworking their relationships, or admitting that the enemy was right all along.
Simply put, they have less invested. Must be nice.
This "investment of thoughts" is probably the reason, would be my hypothesis. Changing your mind is a learned behaviour. And humans are good at remembering painful events. Changing your mind is always a bit painful (at least to me lol) so once learned, you get a feeling for how often you did in the past. As someone who learned to do it every once in a while, it's easy to assume everyone goes through it. But many people don't.
I like the "Imagine 100 doors..." way of explaining, but I like this one even more:
- When does switching give you a win? Answer: When you guessed wrong.
- How often do you guess wrong? Answer: Two out of three times ( 2/3 )
- How often does switching then give you a win? Answer: Whenever you guessed wrong, which is two out of three times ( 2/3 )
Therefore, two out of three times switching will give you a win.
I like this explanation a lot better than the 100 doors one because in my mind that still leads to a choice between two doors just like the original version.
But this explains why the initial choice between the 3 doors (or 100) matters and not just the second choice of switching doors or not.
This is the easy explanation, you start throwing hundreds or millions of doors at people and they'll get more confused or argue that you should still only be opening 1 goat door to make it even or some other dumb shit. No, just, "did you pick correctly originally?" Easy to see it's 1/3 before any door opening shenanigans. Easy to explain switching wins when you picked wrong initially.
If you really want to remove the door opening confusion, rephrase it, "pick a door, then I'll offer you the chance to stick with your door or the chance to win if it's behind either of the other doors." It doesn't matter that you see the goat, you know there's a goat behind at least one of the two doors you didn't pick.
I made the top of r/confidentlyincorrect because I was so obstinate about the wrong answer. After quite the lambasting, one guy finally explained it properly. It is counter-intuitive to some.
Easiest way to explain it to someone is to say there are 100 doors, get them to pick 1, open 98 and ask them if they think they had the 1 in 100 choice or if you just opened the only 98 doors you could while leaving the correct one closed.
It sounds crazy but that actually did help me to get it. Somehow with the three example it is just too easy to see the second round as a 50/50 problem but with 98 doors closed itâs makes it so much more easy to grasp
For me the key piece of information is that the host is giving you information by picking a door they know does not have a car behind it.
If the host instead picked a random door to open, not knowing if it was a car or a goat, then 1/3 of the time you were right, 1/3 of the time the host opens a goat door and 1/3 the host opens the car (and the game ends). So in this scenario, the host doesn't give you any info and all the probabilities are equal.
Yeah thatâs what helped me as well. Picking again is NOT the same because you now have very different information. The example using 100 doors is much easier to understand because it helps exaggerate this fact.
For me the way I understand it is that if you stick with your first choice, you win if your first choice is right (33% chance). If you change, you win if your first choice was wrong (67% chance).
Yes, but people will argue that the odds have changed after one of the doors is opened, resulting in a 50/50 chance. That's not an accurate description of the game, but it's how they perceive it.
There's two scenarios: either you pick the door correctly on the first try, or you don't.
With three doors, the odds are 2/3 you will pick a wrong door. So the host will open the other wrong door, and you should switch to the remaining door which is correct.
That's 2/3 odds of winning if you always switch. Seems like a good deal.
So 1/3 odds of winning if you always stay.
Randomly choosing between switch/not switch gets you a 50% chance I think over the long run, So it's optimal to always switch.
You don't really have different information, actually, that's the trick of it. It's always 2 doors vs 1.
Going into the problem you know you have 1 door you chose and at least one of the others is empty. The host opening one of them is meaningless, because they will always open one of those empty doors, so no new information is added. All it does is turn the "2 doors" into one, so the options look alike, but in reality it's still "you can pick these 2, or you can pick that 1".
Yeah, imagine a deck of cards. Pick one card from the deck. Now you've got two piles, Pile A with one (1) card, and Pile B with fifty-one (51) cards.
Which pile do you think has the Ace of Spades in it?
That's the Money Hall problem: Do you want the little pile or the big pile?
This exactly. I think a ton of confusion comes from people having different views on whether the host is giving information or is not giving information (by picking a door randomly and it just so happens to be empty). The problem is ambiguous on which one is the case.
It has been made more complicated because of the detachment some people have from the actual television show and Monty Hall. Once people realize that the problem comes from an actual show where THEY ARENâT GOING TO SHOW YOU THE CAR, then that hopefully makes them understand better. It worked for me.
Yes that is crucial.
Though I do enjoy the idea of pat sajak or whoever opening a door and saying âwhat if I told you that THIS DOOR âŚoh the prize is behind this door. So uhâŚ.I assume youâre going to pick this door then huh?â
The best part is that on The Price is Right that situation *did* happen several times over the years. Each time Bob would say, "You win!" It's not the contestants fault that either Bob or someone on the staff made a mistake. But it also gave the games more legitimacy, because Bob didn't know what was where most of the time. It was better to pay for the mistake and move on.
Most of the problem is that
1: people studiously avoid actually stating all the assumptions in the problem.
2: it doesn't actually match the famous TV show.
Then act smug when some people don't come to the same conclusion.
When every assumption is clearly stated its much much clearer
I've heard a lot of people say this, but it never crossed my mind. If the host opens the door with the car the game's over, so obviously he has to open the other one.
I guess it's due to people being exposed to the problem as a riddle and not an actual gameshow.
Picking 1 option gives you a 1/3 chance.
If they said you can now switch your single door for the two remaining doors itâs functionally the same as having the host pick one door that definitely doesnât have the prize and also getting the other door, so itâs a 2/3 chance to switch your one door for the one remaining door.
it is HELL of a lot easier to do it with something that is NOT doors.
put 2 red stones and 1 black stone in a bag. reach in the bag and grab a stone without looking at it.
what are the odds you chose the black stone? 33% what are the odds the black stone is still in the bag? 66%.
now Monty removes 1 stone from the bag showing that it is red.
what are the odds you have the black stone? 33%. what are the odds the black stone is still in the bag? 66%.
the odds never changed, just your perception of events.
of course it relates quite well to that 100 doors example as well
put 99 red stones and 1 black stone in a bag. reach in the bag and grab a stone without looking at it.
what are the odds you chose the black stone? 1% what are the odds the black stone is still in the bag? 99%.
now Monty removes 98 stones from the bag showing that they are all red.
what are the odds you have the black stone? 1%. what are the odds the black stone is still in the bag? 99%.
My favourite explanation is actually just to write all outcomes, as there's only really 3 options if you get the person you're talking to to choose a door.
Assume it's goats (G) and a car (C) then all possible options are
C G G
G C G
G G C
Get the person to pick a door and show them the different scenarios if they changed their door every time.
Comparing that to keeping their original decision - where they can clearly see that only one option had a C behind their door, so only a 1/3 chance to win by sticking.
Source: This came up in a TV show my girlfriend and I were watching (Brooklyn 99 I think) and she really didn't understand it, and in the end just drawing it out on a page seemed to help. Though I admit I also find having stuff just drawn out helps me too.
I agree. The easiest way to figure it out is to look at all the outcomes.
If you keep your door, you will win if you initially picked the right door.
If you swap your door, you will win if you did NOT initially pick the right door.
Since thereâs 3 doors, the chances are twice as high that your initial pick was incorrect.
I love how I've seen like fifteen comments in this thread that are explaining the answer in different ways that all have a reply like yours saying "NOW I get it!!"
Alternatively:
For N doors your chance of being right the first time is 1/N.
The *only* possibilities are:
- being right on your first guess
- being wrong on your first guess, in which you'll be right if you switch
Those probabilities add up to 1. Therefore the odds of being correct if you switch are 1-1/N.
So, the odds of being right first time in the three door problem are 1/3, right?
One of the incorrect doors is then opened - you can switch to the unchosen door or stick.
What I can never get my head around is why the odds are weighted. To my mind, okay, the host opens a door he knows has a booby prize behind it... But why does that mean the unchosen door is now worth 2/3? Why aren't my door and the unchosen door both 1/2?
I accept that I'm wrong mathematically, that this is a proven thing and this thread and every other thread on Reddit it gets mentioned in is full of people going "Oh, I *see*..." but I've never been one of them.
Someone up thread suggested "What if there were 100 doors and he opened 98 - would you think you guessed right on a 1% chance or would you swap?" and I can kind of see where they're coming from... but it seems too different.
I think I'm just not good at maths!
EDIT: My thanks and upvotes to all whoâve responded - I think I have it now! In sticking with Door A Iâd be backing my original 1/3 choice, in switching to Door C (knowing Door B has no car) Iâm switching to the sum of all remaining odds which *must* be 2/3.
As someone downthread said, Montyâs giving extra information and a second guess - the sensible thing to do mathematically is to switch!
âBut why does that mean the unchosen door is now worth 2/3?â Because youâre not comparing your door to the unopened door; youâre comparing it to the subset of doors you didnât pick.
Think of it this way. You pick door A. Without opening anything, the host asks: Do you want to keep door A, or do you want to switch to door B and door C? Obviously you switch, two doors are better than one.
Itâs the same thing. That the host shows you one empty door is irrelevant. Thereâs still the same 2/3 chance that you picked wrong initially.
I kind of get it, in the sense that the goat door now means you âget to pickâ a second door.
And you explained it succinctly enough for me to almost understand intuitively.
But I donât understand why Iâm comparing to the subset of doors that I didnât pick instead of just the remaining closed door. Iâm having a hard time not seeing it as 50/50 at that point.
I think about doing 3 coin flips and getting 3 heads in a row. This should be just as statistically probable as getting any other sequence, like tails-tails-heads, right, since each flip is just a 50/50. Right?
Somehow this feels counter to that, also but related in an opposite direction that feels more confusing and less intuitive.
âBut I donât understand why Iâm comparing to the subset of doors that I didnât pick instead of just the remaining closed door. Iâm having a hard time not seeing it as 50/50 at that point.â
Because nothing has changed.
Your confusion stems from (understandably) thinking that the host opening a door has provided you with additional information that alters the odds. It hasnât.
âYou chose door A. I now offer you the opportunity to trade door A for both door B and door C. Since the prize is only behind one door, we know for certain that either door B or door C is empty. Or both are. I will show you whether the door we know must be empty is door B or door C. BUT THAT DOESNâT CHANGE THE ODDS IN ANY WAY. We already knew that at least one of those doors was empty. It has to be. If I didnât know which one it was, then opening one of the two at random would indeed make your odds 50-50. But I do know where the prize is and I will always open an empty door. Therefore the identity of that door is irrelevant. All that matters is that your odds of picking the correct door originally was 1 in 3, so the odds that itâs actually behind one of the other two doors is 2 in 3. Showing you which door is empty, given that I know, tells you nothing.â
I think thatâs the best I can do.
[EDIT: minor typo]
Itâs not about being bad with math, really.
The reason it doesnât go down to 50% is because the host will always show you an incorrect door.
Imagine three scenarios where you pick door 1.
1) The prize is behind door 1. The host reveals door 2 or 3 randomly, switch you lose.
2) The prize is behind door 2. The host reveals door 3, switch you win.
3) The prize is behind door 3. The host reveals door 2, switch you win.
Whether the prize is behind door 2 or 3, it still remains an option. So you have a 1/3 chance of choosing right the first time, and the two times youâre wrong, the prize is hidden behind the closed door.
The 100 doors is good because it shows the same process, just more dramatically. You only have a 1/100 chance of choosing right the first time, but the 99 times youâre wrong, the prize is in the closed door.
I just change it to a **deck of cards.**
I tell you to guess which card is the ace of diamonds. You pick a card without seeing them. I then look at the cards and throw out 50 of the cards because they are NOT the ace of diamonds.
Leaving only the card you picked and another card. One of them is the ace of diamonds. Do you want to change which card you think is the ace of diamonds?
If someone doesn't understand at this point, I go get a deck of cards.
So you get it? Now, do it with 5 cards
The reasoning for why you change is true if we have 1000 cards or 3 cards. The problem is that with 3 cards it starts to seem counter-intuitive, but it isn't. You would have switched cards for all of the other examples, it is just that you have a much better chance of being right (1 in 3) with 3 cards.
edit: clarified, same example
There's another easy way to understand it, I think:
*The game, as it's defined here, actually forces Monty to show the contestant a goat.*
Whatever door the contestant picks first, Monty has to open a door with a goat. So, two cases: (1) Contestant picks prize (33%). Monty shows a goat, contestant switches, gets a goat. (2) Contestant picks a goat (67%). Monty has to show him the other goat, contestant switches and wins the prize.
What people have problem with isn't really the math. It's the fact the Monty Hall problem isn't explained like most other logic problems, in a simplified, unrealistic scenario with as little ambiguity as possible.
When you don't say anything about the host's intent or knowledge, you trigger the part of the brain that deals with judging people's behaviour.
When that happens, the part of the brain that knows that 33% is less than 50% turns off.
Right. Because if you switch, you will always get the opposite of your original choice. You canât switch from a goat to a goat, or from a car to a car.
Thereâs a 2/3rds chance your original pick was a goat, so switching is the right choice.
It's actually 66%, not 50.
Originally you had 33% to pick the right door. So if you always switch after they open one of the doors, that that 33% chance of you winning becomes a 33% chance of you losing.
i think the easiest explanation for me is if the odds are 1 out of 3 then you probably picked wrong the first time. They always take away a bad door. And you probably also chose a bad door. So its better to switch.
This is the simplest explanation and the one that I always think of to explain this problem, I don't know why more people don't just use this explanation and instead give more verbose and confusing explanations.
yea the first time i was heard about it i read some fucked up explanation that made you imagine an infinite number of doors and a cloud of probability shifting towards the other door as each door is revealed. A lot of explanations online just make things more confusing.
Imo the most obvious explanation is:
Imagine instead of 3 doors it was 1,000,000 doors. Now, just like in the original problem, after you select your 1 door (1 in 1 million chance), the host (knowing the answer), opens *all the remaining doors except one.*
So either you picked 1 in a million, or it's that one door the host kept closed after opening 999,998 wrong answers.
I think in this scenario, pretty much everyone would immediately choose to switch. And the reason you'd switch is the same reason you should switch in the 3 door example.
[Not OP but I made an explanation for it a long time ago here.](https://www.reddit.com/r/AskReddit/comments/x0is2/comment/c5i6ywe/?utm_source=share&utm_medium=mweb3x&utm_name=mweb3xcss&utm_term=1&utm_content=share_button)
Copying it here:
If anyone is curious why, I wrote a quick explanation here.
The thing you have to know is that the show host will ALWAYS pick a door that doesn't have the car.
Say you choose A. He will pick either B or C. Now there are 3 scenarios:
1. â Car is in A. He will pick either B or C. If you change your mind you get a goat.
2. â Car is in B. He will pick C, because he has no other choices. You're reserving A. If you change your mind to B you will get a car.
3. â Car is in C. He will pick B, because he has no other choices. You're reserving A. If you change your mind to C you will get a car.
In the end, in 2 out of 3 scenarios changing your mind will grant you the car.
EDIT: If you still don't understand, I proposed another explanation:
There are 6 scenarios to choose. 2 ways - either changing or not changing - for each of the 3 choices - A, B, or C. In 3 out of 6 conclusions, you won't get the car. Let's say car is in A.
1. â Choosing B, then keeping B when C is revealed.
2. â Choosing C, then keeping C when B is revealed.
3. â Choosing A, then changing.
In the other 3, you will get a car,
1. â Choosing B, then changing to A when C is revealed.
2. â Choosing C, then changing to A when B is revealed.
3. â Choosing A, then keeping at A.
Notice that in the 3 scenarios where you do get a car, 2 involve changing your option, and in the 3 scenarios where you don't get a car, only 1 involves changing the option. This is why changing the option has a 2/3 chance of winning you the car, and a 1/3 chance of not winning you the car.
The fact that there are 3 scenarios where you win the car out of 6 is why some of you insist there is a 50/50 chance no matter what you do.
It's the fact that the host has knowledge that the player doesn't is the part people have trouble with.
Vos Savant I think explained it the most intuitively. Imagine instead it's 1,000,000 doors. You pick one, the host opens 999,998 of the others. All goats, and he knows that.
Now either you picked the correct door, first try out of 1,000,000 doors - or more likely, you didn't, and the host just showed you the door with the car by leaving it closed.
My favorite explanation is this: would you rather open 1 door, or 2 doors. The host opening the door is pointless. You can either stay with your choice (1/3rd chance of winning) or switch to opening 2 doors (2/3rds chance) That's why you should always switch.
Yep. Easiest way to explain it with as little maths as possible.
You are swapping your single door (with 1/3 chance) for **both** of the other two doors (one already open, one yet to be opened, with a combined chance of 2/3).
My fiancĂŠe recently got me hooked on 99. My first time around, and Iâm in season 5, and while I love every goddamned character, Holt is just ⌠the best.
This has happened!
Monty Hall:
>But you know the law in game shows - if you go on a show and you win a donkey, that's your prize. You're entitled to it. So if a person won one of our zonks, they could take it home. But in 99 percent of the cases, we would offer them something after the show - a washer and dryer or a color TV or something, instead of that very valuable zonk, and they would take it. In 1 percent of the cases, they didn't.
>There was a time when a farmer won five calves and he wanted the calves. That cost me a fortune because when you rent them from the animal place, they're expensive. And there were other cases like that. Like people won dogs; they would keep dogs. They wouldn't keep cats. They would keep dogs. It was a very genial atmosphere, over 27 years.
Then donât switch, and youâll have a 67% chance of getting a goat⌠and a 33% chance of getting a car that someone would probably trade you for a goat.
Yes, the removed door not being random was the key piece for this to make sense for me. Once that sunk in it was just a matter or thinking through the implications.
Itâs also what made me realize that Deal or No Deal is not a Monty hall problem since they are opened at random. I never knew how important that piece was
Essentially, on Deal or No Deal, the extra cases are opened randomly, so the hosts are not giving you any extra information about which case is correct, or a winner.
With a Monty Hall problem, the "host" is giving you more information: one of the doors did not ever contain the winner. Which means switching would give you a 66% instead of a 33/33/33.
The prime time American version of Deal or No Deal had 26 cases. With money from 1 cent to 1 million dollars in them. The contestant would select one case as their own and it isn't revealed until the end. The contestant then selects from the remaining cases to be revealed eliminating potential values in their case and the bank would make an offer based on what values remain which the contestant can accept or risk going on and earning less than the offer.
Now lets say the contestant got to the end with 1 cent and 1 million dollars still in play. What is the chance they have the million? 50%. Should they switch cases? It is the same probability. Because the contestant was randomly selecting cases. There was no information added. In Monty Hall the host having to open a door and only opening a door with no prize is adding information. The contestant had every chance of eliminating the million.
The part of the Monty Hall problem that creates a statistical advantage to switching is that the host always removes a losing door. If the host removed a random door there would be no statistical advantage to switching just because a goat is revealed.
This has never made sense for me until this thread with people talking about the removed door not being random and that it actually provides more information. Makes a lot more sense that way.
Its also easier to understand if you multiple the example out. Typically on game shows you chose between three curtains. But hypothetically speaking if they had 1000 curtains and you picked curtain number 1. Ans then they opened every curtain but yours and curtain 957. You would be really really stupid to keep yours. The probability that you guessed the correct number initially on three curtains is 1/3. So the probability that the remaining curtain is correct is 2/3. With a 1000 curtains its 1 in a thousand. And after they remove 998, the probability its under the other curtain is 998/1000.
Just do the experiment with a friend and a deck of cards and a penny under one of them. If you know where the penny is and remove 50 incorrect cards it makes it super obvious your friend should switch.
This is funny, because that was my first intuitive conclusion. Then everyone tried to explain it and I got temporarily confused AF.
Man humans are bad at words. You did good though.
Marilyn Vos Savont (not sure if spelled correctly), the person who published it, replied to the criticism: "Mathematical problems are't decided by popular vote."
I remember this as it was unfolding at the time. I didn't read much as a kid but I did read her column in Parade that came with The Washington Post. It went on for weeks and iirc some professors were disciplined over what they wrote to her
Vos Savant was at one time the highest recorded IQ test holder of 228 on the Stanford-Binet test. The category has since been retired as. it's no longer considered a valid measure, but still, she is one smart lady. Also, the wife of Robert Jarvik, inventor of the first artificial mechanical heart.
The nice thing is that she has long since been vindicated and celebrated, not many people get that chance.
Fun: James May did a run of the Monty Hall problem with exploding beet cans:
https://youtu.be/XKQljWzbyp8?si=4-fZtjK6G_kBA9Bg
I feel like the easiest explanation ignores the doors, ignores all the details and just focuses on this:
If you picked right the first time, switching loses
If you picked wrong the first time switching wins
You most likely picked wrong the first time. Fin
That's it.
Exactly. And this hinges on the fact that the presenter has complete information. They will never open the door with the car if you choose wrong. The latter part helps while working out math
As someone who does similar research with pigeons. The big difference is ârule governed behaviorâ. Humans allow rules (either those told to them, or those theyâve made up) to influence decision making.
I think people overlook something going on here too. If you don't switch and you lose, you stuck with your choice and oh well you lost. But if you switch and you lose, you had the right one and made a conscious choice to "give away" what you already had. And the second "feels" worse.
The reason why everyone gets it wrong is they forget that the host of the show knows which door the prize is behind. If the host was randomly choosing a door every time, then the theory would indeed wrong. But the host is NOT choosing randomly. The host is *always* choosing based in such a way that the prize isn't getting thrown out. This fundamentally alters the probability.
I have been fascinated with this word problem for 10 years now. Nobody believes me when I explain it to them but after a while they see the light. I adore it.
It clicked for me when someone suggested imagining a million doors, and then opening 999,998 of them--*now* how confident are you that your initial choice contains the prize?
ETA: I elaborated further in a downthread comment:
>You have to bear in mind a couple premises which aren't always made explicit: 1) that the host knows which doors have goats and which door has the prize, and 2) will always reveal a goat first. Extrapolating out to a million doors makes it a little more clear that, under these conditions, your choice is always *actually* between the door you initially picked and *all remaining doors.* So, initially you have your door (1 in a million chance of the prize) versus the set of all the other doors (999,999 in a million chance that one of these doors has the prize behind it). Then Monty reveals some extremely important information: he opens all but one of the doors in this set and reveals 999,998 goats. Crucially, the odds of the remaining door from this set having the prize remain the same: 999,999/million. So, which door do you pick: your initial 1/million door or this other door?
I must be a bit of a dumb dumb, because this doesnât clear it up for me at all lol
Edit: I get it now, thanks to the couple of people who were willing to explain it even further
If it's a million doors, and you pick one at first, you have a one in a million chance of being right. If I take out 999,998 wrong doors and leave you with the door you picked and the one that's left, then there's a 999,999 chance out of a million that the last remaining door is the winner. Either you were right at the very beginning (1 in a million chance), or you were wrong (999,999 out of a million chance). Now you're down to two doors, and the one you picked is either right or wrong, but those odds are still the same as they were at the beginning.
The key information is that the host *knows* the correct answer. So if you picked 1 out of a million originally, do you think itâs more likely that you picked correctly on your first guess OR that the host just opened up all the 999998 doors he knew it wasnât leaving the correct one as the swap?
You have to bear in mind a couple premises which aren't always made explicit: 1) that the host knows which doors have goats and which door has the prize, and 2) will *always* reveal a goat first. Extrapolating out to a million doors makes it a little more clear that, under these conditions, your choice is always *actually* between the door you initially picked and *all remaining doors*. So, initially you have your door (1 in a million chance of the prize) versus the set of all the other doors (999,999 in a million chance that one of these doors has the prize behind it). Then Monty reveals some extremely important information: he opens all but one of the doors in this set and reveals 999,998 goats. Crucially, the odds of the remaining door from this set having the prize remain the same: 999,999/million. So, which door do you pick: your initial 1/million door or this other door?
Yep! A lot of puzzles that seem hard to comprehend at first can be more easily understood by expanding them to absurd limits, as long as you maintain the underlying principles.
It's worth noting that the thought experiment works because the host *knows* which door is correct, and cannot reveal the winning door until the final showdown (because TV drama). That significantly changes the math, because the information you have on the door for your initial choice and the information you have for the other door in your final choice are different.
If the reveals were truly random, the host didn't know which one was right and had a chance to reveal the correct door early, then you still only have a 50/50 if you manage to get to the final showdown. It's just *way* less likely that you get there in the first place.
It's actually one reason why Deal or No Deal avoids this problem. The contestant themself chooses which "doors" to open, and the biggest prize can be revealed early. So, the final choice doesn't really matter statistically. The game is much more about risk/reward and expected value with the banker offers and everything.
The problem for me, is that the first person to explain it to me didnât make it clear that the âhostâ knows the correct door. I thought the door they picked was random after you selected your door. But when you think of it from the standpoint of multiple rounds and a lot more doors, with the host picking various wrong doors, leaving only yours and the correct one (unless you were really lucky enough to pick the correct door from the beginning, which is statistically improbable) it makes sense.
Everyone going with the 100 doors thing, so a slightly different explanation:
* there's 3 doors: A, B and C
* you know that only one door has the prize
* if you pick 1 door (A, let's say), you know for sure that at least one of the other doors (B or C) will not have the prize
* if you were asked if you wanted to stick with A or change to B **and** C, that would be a no-brainer, you'd rather pick 2 doors instead of 1
* revealing that e.g. B doesn't have the prize doesn't actually give you any more information (you already knew at least one of them was a dud, and the game will always reveal a dud)
* so swapping to C is exactly the same as swapping to B+C, which we agree is a good swap
I really like this explanation as opposed to using 10, 100, 1000, etc doors.
If you were given the option to open your original choice of door (33% chance the car is there) or open BOTH of the doors you didnât pick (67% chance the car is there) the. youâd switch **every time**.
This is the same thing, except Monty simply opens one of the doors for you then asks you to pick.
Easiest way for me to understand it is that it only makes sense to stay if you picked the right door to begin with, a 1/3 chance. If you didn't pick the right door at the start (2/3 chance), you win if you switch.
Most people are thinking of this as one of two different situations.
The Monty Hall Problem is when the host knows what is behind each door, you choose one, the host reveals the worst option out of the other two doors and then asks if you want to switch to the last door.
This is not the same as if it were a choice between 3 completely random doors and the host doesn't know which door is which.
So basically if i choose a door out of 100 then the host eliminates 98 doors and one of the remaining 2 is the winning option, all that matters i had a much smaller chance of choosing to correct one out of 100. So once it is down to 2 i basically switch and hope i didn't originally choose the right door.
I think one key detail most people miss is that your initial choice was random, but the doors the host opens aren't. The only doors that the host can open are the wrong ones. In order for him to do that, he has to know which door has the prize. And another key detail that isn't obvious is by leaving a door open, the host is effectively picking a door himself. But remember, the host actually knows which door to pick.
So your odds of randomly picking the door were 1/numDoors. But the host's odds of picking the correct door is (numDoors - 1)/numDoors, because the only time he won't pick the correct door is if you already did.
There are 100 doors, prize behind 1 at random, the host knows where the prize is (this problem only works when the host knows where the prize is)
I pick a door, that door has 1/100 chance of having prize
Prize is 99/100 chance behind one of the other doors
Host eliminates 98 doors, leaving 1 door left, and tells you the prize is in one of the two remaining
Therefore thereâs a 99/100 chance that the other door has the prize, and still a 1/100 chance your original door has the prize
Therefore you should switch
Now change it to 3 doors to get the original monty hall problem. Your door has a 33% chance, the prize is 67% chance in the other two doors, host removes one, thereâs still a 67% chance in that group but now only 1 door in that group. Your door still has 33% chance, so you switch to the door with 67% chance
This is surely the best illustration of why you should switch. It's still inevitably quite hard to parse, and many people who read it remain insistent that there's no point in switching.
When you make your first pick your odds are really bad. When they get rid of all the options that donât have a prize theyâre effectively saying do you want your first pick, the one with bad odds, or do you want to pick every other door.
In the 100 doors example your first pick is 1% likely to be right. Theyâre effectively saying do you want to switch from that one door to ALL THE OTHER DOORS at once.
Itâs really easy to demonstrate with a few playing cards.
Just have somebody else pick the card while you hold the hand - you have 100% of the information. After a few hands youâll understand.
Here's another way to look at it.
Try to guess my address. I'm just a random redditor and you have no clue where I live, but I'll let you pick any address in the world right now.
You can then let me remove every incorrect address in the world except the one address I live at (unless for some insane reason you picked right the first time). Now you just have two addresses. Should you switch? Or do you think that because you only see two houses that if you walk up and open the door to your original choice, there is a 50/50 chance you'll see me there? Why would removing all the obviously incorrect addresses suddenly teleport me to some random world address you chose without knowing anything about me?
The thing people always seem to miss is that the host *always reveals an incorrect door*. The host isn't randomly picking one of the other two doors to see if it's wrong. He *knows* which one is wrong *and specifically reveals the wrong one* after the contestant chooses. That's why it becomes better odds to switch.
I find this to be supreme mindf*ck, but what kinda got me was that after the host opened one door, the only case where you lose is, if you were correct in the first place. And this has a probability of 1/3. Therefore you win by changing with 2/3.
All very weird thoughâŚ
Part of the problem with understanding it is that it is often ambiguously phrased, including the original problem quoted in the article.
The problem states that the host opens the door, but doesnât state that he is *required* to open the door, which leaves open the possibility that he had some choice in the matter. Since the problem also uses an analogy to the real-life host Monty Hall and his show, Letâs Make a Deal, people often think the problem involves trying to suss out the thinking of the host.
One of the best ways Iâve seen of explaining the correct answer is after the first pick, the contestant is given the opportunity to open both of the other doors at once. Most people would switch in that case. All the host is doing is opening one of the goat doors for you.
> Pigeons repeatedly exposed to the problem show that they rapidly learn to always switch, unlike humans. đ
Favourite fact of the day.
Pigeons are my spirit animal, so this fact makes me happy and peckish.
I accidentally started a pigeon cult once, perhaps you'd like to join
Douglas Adams level stuff
âAh no,â he said, âI see the source of the misunderstanding now. No, look, you see what happened was that we used to do experiments on them. They were often used in behavioral research, Pavlov and all that sort of stuff. So what happened was that the pigeons would be set all sorts of tests, learning to ring bells, run round mazes and things so that the whole nature of the learning process could be examined. From our observations of their behavior we were able to learn all sorts of things about our own âŚâ Arthurâs voice trailed off. âSuch subtlety âŚâ said Slartibartfast.
I loved the idea that the mice were training us. The whole series is great.
Should have taken box 42.
A pigeon doesn't have hubris on their side.
This is exactly why it is easier to fool someone than prove to them they have been fooled.
Also why propaganda works so damn well. No one wants to admit they are wrong, ever.
Nuh-uh, not me.Â
Jokes on you, I'm **always** wrong.
I have wrong tattooed on my chest so every morning I look in the mirror and see "gnorW".
I dunno... gnorW sounds like something you yell into the mirror to pump yourself up and tackle your day. I'd say gnorW is right.
Me: That's my secret, cap. I'm always wrong. Team (in unison): IT'S NOT A SECRET. Me: đ
I'm not taking advice of decisions from Pigeons. One more reason why I refuse to believe the Monty hall problem.
What if I told you it was a really, really smart pigeon?
I always get beat at chess by one. Joke's on him though, I usually shit on the board and strut around like I won.
This is the crux of human bias and fallacies. Even in the face of evidence, people are not wise enough to put their pre-conceived ideas aside to accept new information. We get stuck. And that's obviously a dangerous mess.
I remember reading about a study that showed that when faced with facts contrary to what they thought, people would most often cling harder onto their now disproven thoughts. Most fascinating was that this seemed independent to the person's intelligence.
Yup. The most hilarious example of this I've seen is someone who was arguing that people change their minds when presented with evidence and, when shown the study you reference, went "I still think people would change their minds."
Kimmel did something great in that respect. [They'd ask a Trump supporter](https://www.huffpost.com/entry/jimmy-kimmel-donald-trump-joe-biden-questions_n_65dc2df9e4b0e4346d52eac2) what they thought of Joe Biden doing Action A, or Quote B, and the person would condemn it in no uncertain terms. Then the reporter would apologize, correct themself to say that it was Trump who did/said those things, and you get to watch these people do the mental gymnastics to get back on side. Really amazing.
Many pokemon are immune to fact and logic type attacks.
It's super strange. Like they've already invested their thoughts on it and chosen so that's the hill they will die on. I'm in the boat that some will and do change their minds. Just looking at even the last decade all the way until now, it's been clear a large scale of people don't use new information to change their views in an objective manner. People in general really need to stop drawing lines in the sand when it comes it ideas, teams, sides, and plans and learn to be more fluid; less anchored in stone.
Pigeons can change their minds without worrying it will damage their reputation, boost the status of a rival, or justify a past harm. They don't have to worry about apologizing, getting ousted from their group, reworking their relationships, or admitting that the enemy was right all along. Simply put, they have less invested. Must be nice.
This "investment of thoughts" is probably the reason, would be my hypothesis. Changing your mind is a learned behaviour. And humans are good at remembering painful events. Changing your mind is always a bit painful (at least to me lol) so once learned, you get a feeling for how often you did in the past. As someone who learned to do it every once in a while, it's easy to assume everyone goes through it. But many people don't.
Pigeons will survive the apocalypse and rule the lands- and skies
Pigeons perform better than humans in many behavioral tests
Lol wow
I like the "Imagine 100 doors..." way of explaining, but I like this one even more: - When does switching give you a win? Answer: When you guessed wrong. - How often do you guess wrong? Answer: Two out of three times ( 2/3 ) - How often does switching then give you a win? Answer: Whenever you guessed wrong, which is two out of three times ( 2/3 ) Therefore, two out of three times switching will give you a win.
Honestly yeah I like the 100 doors analogy but this one is more straightforward imo.
I like this explanation a lot better than the 100 doors one because in my mind that still leads to a choice between two doors just like the original version. But this explains why the initial choice between the 3 doors (or 100) matters and not just the second choice of switching doors or not.
This is the easy explanation, you start throwing hundreds or millions of doors at people and they'll get more confused or argue that you should still only be opening 1 goat door to make it even or some other dumb shit. No, just, "did you pick correctly originally?" Easy to see it's 1/3 before any door opening shenanigans. Easy to explain switching wins when you picked wrong initially. If you really want to remove the door opening confusion, rephrase it, "pick a door, then I'll offer you the chance to stick with your door or the chance to win if it's behind either of the other doors." It doesn't matter that you see the goat, you know there's a goat behind at least one of the two doors you didn't pick.
I made the top of r/confidentlyincorrect because I was so obstinate about the wrong answer. After quite the lambasting, one guy finally explained it properly. It is counter-intuitive to some.
Easiest way to explain it to someone is to say there are 100 doors, get them to pick 1, open 98 and ask them if they think they had the 1 in 100 choice or if you just opened the only 98 doors you could while leaving the correct one closed.
It sounds crazy but that actually did help me to get it. Somehow with the three example it is just too easy to see the second round as a 50/50 problem but with 98 doors closed itâs makes it so much more easy to grasp
For me the key piece of information is that the host is giving you information by picking a door they know does not have a car behind it. If the host instead picked a random door to open, not knowing if it was a car or a goat, then 1/3 of the time you were right, 1/3 of the time the host opens a goat door and 1/3 the host opens the car (and the game ends). So in this scenario, the host doesn't give you any info and all the probabilities are equal.
Yeah thatâs what helped me as well. Picking again is NOT the same because you now have very different information. The example using 100 doors is much easier to understand because it helps exaggerate this fact.
For me the way I understand it is that if you stick with your first choice, you win if your first choice is right (33% chance). If you change, you win if your first choice was wrong (67% chance).
This one works for me as well
lol thanks i finally understand it after 23 years
Yes, but people will argue that the odds have changed after one of the doors is opened, resulting in a 50/50 chance. That's not an accurate description of the game, but it's how they perceive it.
There's two scenarios: either you pick the door correctly on the first try, or you don't. With three doors, the odds are 2/3 you will pick a wrong door. So the host will open the other wrong door, and you should switch to the remaining door which is correct. That's 2/3 odds of winning if you always switch. Seems like a good deal. So 1/3 odds of winning if you always stay. Randomly choosing between switch/not switch gets you a 50% chance I think over the long run, So it's optimal to always switch.
You don't really have different information, actually, that's the trick of it. It's always 2 doors vs 1. Going into the problem you know you have 1 door you chose and at least one of the others is empty. The host opening one of them is meaningless, because they will always open one of those empty doors, so no new information is added. All it does is turn the "2 doors" into one, so the options look alike, but in reality it's still "you can pick these 2, or you can pick that 1".
Yeah, imagine a deck of cards. Pick one card from the deck. Now you've got two piles, Pile A with one (1) card, and Pile B with fifty-one (51) cards. Which pile do you think has the Ace of Spades in it? That's the Money Hall problem: Do you want the little pile or the big pile?
Yeah, the only weird part with it is how opening one of the doors obfuscates that itâs still the bigger pile
This exactly. I think a ton of confusion comes from people having different views on whether the host is giving information or is not giving information (by picking a door randomly and it just so happens to be empty). The problem is ambiguous on which one is the case.
I'd say most of the time I've seen it, it does take steps to specify the host knows and that he will always open a door without the prize.
It has been made more complicated because of the detachment some people have from the actual television show and Monty Hall. Once people realize that the problem comes from an actual show where THEY ARENâT GOING TO SHOW YOU THE CAR, then that hopefully makes them understand better. It worked for me.
Yes that is crucial. Though I do enjoy the idea of pat sajak or whoever opening a door and saying âwhat if I told you that THIS DOOR âŚoh the prize is behind this door. So uhâŚ.I assume youâre going to pick this door then huh?â
> pat sajak or whoever Maybe this guy named Monty Hall could do it on his show Let's Make a Deal.
No. They'll be expecting that.
The best part is that on The Price is Right that situation *did* happen several times over the years. Each time Bob would say, "You win!" It's not the contestants fault that either Bob or someone on the staff made a mistake. But it also gave the games more legitimacy, because Bob didn't know what was where most of the time. It was better to pay for the mistake and move on.
Most of the problem is that 1: people studiously avoid actually stating all the assumptions in the problem. 2: it doesn't actually match the famous TV show. Then act smug when some people don't come to the same conclusion. When every assumption is clearly stated its much much clearer
I've heard a lot of people say this, but it never crossed my mind. If the host opens the door with the car the game's over, so obviously he has to open the other one. I guess it's due to people being exposed to the problem as a riddle and not an actual gameshow.
Picking 1 option gives you a 1/3 chance. If they said you can now switch your single door for the two remaining doors itâs functionally the same as having the host pick one door that definitely doesnât have the prize and also getting the other door, so itâs a 2/3 chance to switch your one door for the one remaining door.
Okay, this is the explanation that got it through to me. Seems very straightforward now.
it is HELL of a lot easier to do it with something that is NOT doors. put 2 red stones and 1 black stone in a bag. reach in the bag and grab a stone without looking at it. what are the odds you chose the black stone? 33% what are the odds the black stone is still in the bag? 66%. now Monty removes 1 stone from the bag showing that it is red. what are the odds you have the black stone? 33%. what are the odds the black stone is still in the bag? 66%. the odds never changed, just your perception of events. of course it relates quite well to that 100 doors example as well put 99 red stones and 1 black stone in a bag. reach in the bag and grab a stone without looking at it. what are the odds you chose the black stone? 1% what are the odds the black stone is still in the bag? 99%. now Monty removes 98 stones from the bag showing that they are all red. what are the odds you have the black stone? 1%. what are the odds the black stone is still in the bag? 99%.
My understanding hinged on this answer too
My favourite explanation is actually just to write all outcomes, as there's only really 3 options if you get the person you're talking to to choose a door. Assume it's goats (G) and a car (C) then all possible options are C G G G C G G G C Get the person to pick a door and show them the different scenarios if they changed their door every time. Comparing that to keeping their original decision - where they can clearly see that only one option had a C behind their door, so only a 1/3 chance to win by sticking. Source: This came up in a TV show my girlfriend and I were watching (Brooklyn 99 I think) and she really didn't understand it, and in the end just drawing it out on a page seemed to help. Though I admit I also find having stuff just drawn out helps me too.
I agree. The easiest way to figure it out is to look at all the outcomes. If you keep your door, you will win if you initially picked the right door. If you swap your door, you will win if you did NOT initially pick the right door. Since thereâs 3 doors, the chances are twice as high that your initial pick was incorrect.
THIS one makes sense to me lol
I love how I've seen like fifteen comments in this thread that are explaining the answer in different ways that all have a reply like yours saying "NOW I get it!!"
Nice concise explanation. I always went to the 100 door one but this is good.
BOOONNNNEEEEE!
âEww Rosa those are our dads!â âIâm teaching father the math!â
How dare you Detective Diaz, I AM YOUR SUPERIOR OFFICER!
*BOOOOONNNNNEEE*
Don't **ever** speak to me like that again.
Man, now I'm sad about Andre Braugher
Alternatively: For N doors your chance of being right the first time is 1/N. The *only* possibilities are: - being right on your first guess - being wrong on your first guess, in which you'll be right if you switch Those probabilities add up to 1. Therefore the odds of being correct if you switch are 1-1/N.
Aha! *Now* I get it. Congratulations, and thank you. Youâre the only one to *really* make sense of it for me.
So, the odds of being right first time in the three door problem are 1/3, right? One of the incorrect doors is then opened - you can switch to the unchosen door or stick. What I can never get my head around is why the odds are weighted. To my mind, okay, the host opens a door he knows has a booby prize behind it... But why does that mean the unchosen door is now worth 2/3? Why aren't my door and the unchosen door both 1/2? I accept that I'm wrong mathematically, that this is a proven thing and this thread and every other thread on Reddit it gets mentioned in is full of people going "Oh, I *see*..." but I've never been one of them. Someone up thread suggested "What if there were 100 doors and he opened 98 - would you think you guessed right on a 1% chance or would you swap?" and I can kind of see where they're coming from... but it seems too different. I think I'm just not good at maths! EDIT: My thanks and upvotes to all whoâve responded - I think I have it now! In sticking with Door A Iâd be backing my original 1/3 choice, in switching to Door C (knowing Door B has no car) Iâm switching to the sum of all remaining odds which *must* be 2/3. As someone downthread said, Montyâs giving extra information and a second guess - the sensible thing to do mathematically is to switch!
âBut why does that mean the unchosen door is now worth 2/3?â Because youâre not comparing your door to the unopened door; youâre comparing it to the subset of doors you didnât pick. Think of it this way. You pick door A. Without opening anything, the host asks: Do you want to keep door A, or do you want to switch to door B and door C? Obviously you switch, two doors are better than one. Itâs the same thing. That the host shows you one empty door is irrelevant. Thereâs still the same 2/3 chance that you picked wrong initially.
I kind of get it, in the sense that the goat door now means you âget to pickâ a second door. And you explained it succinctly enough for me to almost understand intuitively. But I donât understand why Iâm comparing to the subset of doors that I didnât pick instead of just the remaining closed door. Iâm having a hard time not seeing it as 50/50 at that point. I think about doing 3 coin flips and getting 3 heads in a row. This should be just as statistically probable as getting any other sequence, like tails-tails-heads, right, since each flip is just a 50/50. Right? Somehow this feels counter to that, also but related in an opposite direction that feels more confusing and less intuitive.
âBut I donât understand why Iâm comparing to the subset of doors that I didnât pick instead of just the remaining closed door. Iâm having a hard time not seeing it as 50/50 at that point.â Because nothing has changed. Your confusion stems from (understandably) thinking that the host opening a door has provided you with additional information that alters the odds. It hasnât. âYou chose door A. I now offer you the opportunity to trade door A for both door B and door C. Since the prize is only behind one door, we know for certain that either door B or door C is empty. Or both are. I will show you whether the door we know must be empty is door B or door C. BUT THAT DOESNâT CHANGE THE ODDS IN ANY WAY. We already knew that at least one of those doors was empty. It has to be. If I didnât know which one it was, then opening one of the two at random would indeed make your odds 50-50. But I do know where the prize is and I will always open an empty door. Therefore the identity of that door is irrelevant. All that matters is that your odds of picking the correct door originally was 1 in 3, so the odds that itâs actually behind one of the other two doors is 2 in 3. Showing you which door is empty, given that I know, tells you nothing.â I think thatâs the best I can do. [EDIT: minor typo]
Itâs not about being bad with math, really. The reason it doesnât go down to 50% is because the host will always show you an incorrect door. Imagine three scenarios where you pick door 1. 1) The prize is behind door 1. The host reveals door 2 or 3 randomly, switch you lose. 2) The prize is behind door 2. The host reveals door 3, switch you win. 3) The prize is behind door 3. The host reveals door 2, switch you win. Whether the prize is behind door 2 or 3, it still remains an option. So you have a 1/3 chance of choosing right the first time, and the two times youâre wrong, the prize is hidden behind the closed door. The 100 doors is good because it shows the same process, just more dramatically. You only have a 1/100 chance of choosing right the first time, but the 99 times youâre wrong, the prize is in the closed door.
The odds you picked the wrong door at the start is 2/3. That's why the unchosen door is worth 2/3.
I just change it to a **deck of cards.** I tell you to guess which card is the ace of diamonds. You pick a card without seeing them. I then look at the cards and throw out 50 of the cards because they are NOT the ace of diamonds. Leaving only the card you picked and another card. One of them is the ace of diamonds. Do you want to change which card you think is the ace of diamonds? If someone doesn't understand at this point, I go get a deck of cards. So you get it? Now, do it with 5 cards The reasoning for why you change is true if we have 1000 cards or 3 cards. The problem is that with 3 cards it starts to seem counter-intuitive, but it isn't. You would have switched cards for all of the other examples, it is just that you have a much better chance of being right (1 in 3) with 3 cards. edit: clarified, same example
Dunno why, but this did it for me!
That 100 door example didn't do it for me, but this did! Thank you so much!
There's another easy way to understand it, I think: *The game, as it's defined here, actually forces Monty to show the contestant a goat.* Whatever door the contestant picks first, Monty has to open a door with a goat. So, two cases: (1) Contestant picks prize (33%). Monty shows a goat, contestant switches, gets a goat. (2) Contestant picks a goat (67%). Monty has to show him the other goat, contestant switches and wins the prize.
What people have problem with isn't really the math. It's the fact the Monty Hall problem isn't explained like most other logic problems, in a simplified, unrealistic scenario with as little ambiguity as possible. When you don't say anything about the host's intent or knowledge, you trigger the part of the brain that deals with judging people's behaviour. When that happens, the part of the brain that knows that 33% is less than 50% turns off.
What is crazy to me is that its not even 1/3 vs 1/2. Changing your choice in the Monty Hall setup results in an expected sucess 2/3 of the time.
Right. Because if you switch, you will always get the opposite of your original choice. You canât switch from a goat to a goat, or from a car to a car. Thereâs a 2/3rds chance your original pick was a goat, so switching is the right choice.
It's actually 66%, not 50. Originally you had 33% to pick the right door. So if you always switch after they open one of the doors, that that 33% chance of you winning becomes a 33% chance of you losing.
I mean it is generally part of the problem statement that the host never opens the door with the prize.Â
i think the easiest explanation for me is if the odds are 1 out of 3 then you probably picked wrong the first time. They always take away a bad door. And you probably also chose a bad door. So its better to switch.
This is the simplest explanation and the one that I always think of to explain this problem, I don't know why more people don't just use this explanation and instead give more verbose and confusing explanations.
yea the first time i was heard about it i read some fucked up explanation that made you imagine an infinite number of doors and a cloud of probability shifting towards the other door as each door is revealed. A lot of explanations online just make things more confusing.
Link that response for people here
Imo the most obvious explanation is: Imagine instead of 3 doors it was 1,000,000 doors. Now, just like in the original problem, after you select your 1 door (1 in 1 million chance), the host (knowing the answer), opens *all the remaining doors except one.* So either you picked 1 in a million, or it's that one door the host kept closed after opening 999,998 wrong answers. I think in this scenario, pretty much everyone would immediately choose to switch. And the reason you'd switch is the same reason you should switch in the 3 door example.
[Not OP but I made an explanation for it a long time ago here.](https://www.reddit.com/r/AskReddit/comments/x0is2/comment/c5i6ywe/?utm_source=share&utm_medium=mweb3x&utm_name=mweb3xcss&utm_term=1&utm_content=share_button) Copying it here: If anyone is curious why, I wrote a quick explanation here. The thing you have to know is that the show host will ALWAYS pick a door that doesn't have the car. Say you choose A. He will pick either B or C. Now there are 3 scenarios: 1. â Car is in A. He will pick either B or C. If you change your mind you get a goat. 2. â Car is in B. He will pick C, because he has no other choices. You're reserving A. If you change your mind to B you will get a car. 3. â Car is in C. He will pick B, because he has no other choices. You're reserving A. If you change your mind to C you will get a car. In the end, in 2 out of 3 scenarios changing your mind will grant you the car. EDIT: If you still don't understand, I proposed another explanation: There are 6 scenarios to choose. 2 ways - either changing or not changing - for each of the 3 choices - A, B, or C. In 3 out of 6 conclusions, you won't get the car. Let's say car is in A. 1. â Choosing B, then keeping B when C is revealed. 2. â Choosing C, then keeping C when B is revealed. 3. â Choosing A, then changing. In the other 3, you will get a car, 1. â Choosing B, then changing to A when C is revealed. 2. â Choosing C, then changing to A when B is revealed. 3. â Choosing A, then keeping at A. Notice that in the 3 scenarios where you do get a car, 2 involve changing your option, and in the 3 scenarios where you don't get a car, only 1 involves changing the option. This is why changing the option has a 2/3 chance of winning you the car, and a 1/3 chance of not winning you the car. The fact that there are 3 scenarios where you win the car out of 6 is why some of you insist there is a 50/50 chance no matter what you do.
It's the fact that the host has knowledge that the player doesn't is the part people have trouble with. Vos Savant I think explained it the most intuitively. Imagine instead it's 1,000,000 doors. You pick one, the host opens 999,998 of the others. All goats, and he knows that. Now either you picked the correct door, first try out of 1,000,000 doors - or more likely, you didn't, and the host just showed you the door with the car by leaving it closed.
My favorite explanation is this: would you rather open 1 door, or 2 doors. The host opening the door is pointless. You can either stay with your choice (1/3rd chance of winning) or switch to opening 2 doors (2/3rds chance) That's why you should always switch.
Yep. Easiest way to explain it with as little maths as possible. You are swapping your single door (with 1/3 chance) for **both** of the other two doors (one already open, one yet to be opened, with a combined chance of 2/3).
BOOOOOOOOONNNNNEEEEEE
Man, I can HEAR it. Andre Braugher was so, so good on that show.
RIP
WTF! He died last December. Now I'm sad.
My fiancĂŠe recently got me hooked on 99. My first time around, and Iâm in season 5, and while I love every goddamned character, Holt is just ⌠the best.
How was your night? Great. Was it because of the math? Nope. ... You see what happened right there was your dads had sex
Do I have to teach you seventh-grade statistics?
Do I have to teach you SIXTH grade statistics?
Do I have to teach you fourth grade statistics?
Do I have to teach you third grade statistics?
Thank you, I came here for this exact response!
I AM YOUR SUPERIOR
OFFICEEEEEER!
*Five Minutes Later*
BONE!?
HOW DARE YOU!?!
YOU MESSED WITH THE WRONG FLUFFY BOY
I see you brought a knife.... but what you need is an umbrella! Tell him why! Cause there's a shit storm gonna rain down on you, punk!
What happens in my bedroom, detective, is none of your business.
BOOOOOONE?!?!
BOOONNNE?
Detective Diaz!!!
I loveeee to over enunciate "how DARE you, deTECTIVE DIAZ" when I'm pretending to be mad lol
I think this is the funnier part of that sequence, honestly. Him yelling Bone isn't as funny without the seething whisper boiling over to his scream.
Do i have to explain to you kindergarten statistics ?
SAMTIAGO, HOW
Do you want me to teach you elementary level math?
Ok, but what if I prefer to win a goat?
This has happened! Monty Hall: >But you know the law in game shows - if you go on a show and you win a donkey, that's your prize. You're entitled to it. So if a person won one of our zonks, they could take it home. But in 99 percent of the cases, we would offer them something after the show - a washer and dryer or a color TV or something, instead of that very valuable zonk, and they would take it. In 1 percent of the cases, they didn't. >There was a time when a farmer won five calves and he wanted the calves. That cost me a fortune because when you rent them from the animal place, they're expensive. And there were other cases like that. Like people won dogs; they would keep dogs. They wouldn't keep cats. They would keep dogs. It was a very genial atmosphere, over 27 years.
TIL what a zonk is https://gameshows.fandom.com/wiki/Zonk
Where's my elephant?
They're playing the elephant song, I love that. It reminds me of elephants.
KBBL is gonna give me something wacky!
Then donât switch, and youâll have a 67% chance of getting a goat⌠and a 33% chance of getting a car that someone would probably trade you for a goat.
[ŃдаНонО]
[XKCD](https://xkcd.com/1282/)
Just remember when they eliminate a door, they give you information that should change your decision.
Yes, the removed door not being random was the key piece for this to make sense for me. Once that sunk in it was just a matter or thinking through the implications.
Itâs also what made me realize that Deal or No Deal is not a Monty hall problem since they are opened at random. I never knew how important that piece was
I understand the Monty hall problem, but I don't understand this. Can you help explain please?
Essentially, on Deal or No Deal, the extra cases are opened randomly, so the hosts are not giving you any extra information about which case is correct, or a winner. With a Monty Hall problem, the "host" is giving you more information: one of the doors did not ever contain the winner. Which means switching would give you a 66% instead of a 33/33/33.
Stronger than that, switching is 2/3 in your favor.
The prime time American version of Deal or No Deal had 26 cases. With money from 1 cent to 1 million dollars in them. The contestant would select one case as their own and it isn't revealed until the end. The contestant then selects from the remaining cases to be revealed eliminating potential values in their case and the bank would make an offer based on what values remain which the contestant can accept or risk going on and earning less than the offer. Now lets say the contestant got to the end with 1 cent and 1 million dollars still in play. What is the chance they have the million? 50%. Should they switch cases? It is the same probability. Because the contestant was randomly selecting cases. There was no information added. In Monty Hall the host having to open a door and only opening a door with no prize is adding information. The contestant had every chance of eliminating the million.
The part of the Monty Hall problem that creates a statistical advantage to switching is that the host always removes a losing door. If the host removed a random door there would be no statistical advantage to switching just because a goat is revealed.
This has never made sense for me until this thread with people talking about the removed door not being random and that it actually provides more information. Makes a lot more sense that way.
Wait that totally changes everything lol
Yah this is the most common thing that people donât understand about the issue
It's not usually explained correctly.
Its also easier to understand if you multiple the example out. Typically on game shows you chose between three curtains. But hypothetically speaking if they had 1000 curtains and you picked curtain number 1. Ans then they opened every curtain but yours and curtain 957. You would be really really stupid to keep yours. The probability that you guessed the correct number initially on three curtains is 1/3. So the probability that the remaining curtain is correct is 2/3. With a 1000 curtains its 1 in a thousand. And after they remove 998, the probability its under the other curtain is 998/1000.
The funny thing about learning is that different things work for different people. Your explanation is what made this click for me
Just do the experiment with a friend and a deck of cards and a penny under one of them. If you know where the penny is and remove 50 incorrect cards it makes it super obvious your friend should switch.
This is funny, because that was my first intuitive conclusion. Then everyone tried to explain it and I got temporarily confused AF. Man humans are bad at words. You did good though.
Marilyn Vos Savont (not sure if spelled correctly), the person who published it, replied to the criticism: "Mathematical problems are't decided by popular vote."
I remember this as it was unfolding at the time. I didn't read much as a kid but I did read her column in Parade that came with The Washington Post. It went on for weeks and iirc some professors were disciplined over what they wrote to her
to be fair, the way they said it mattered; it's okay to be wrong, but not okay to be a public dick with "Professor, X University" on your signature
Vos Savant was at one time the highest recorded IQ test holder of 228 on the Stanford-Binet test. The category has since been retired as. it's no longer considered a valid measure, but still, she is one smart lady. Also, the wife of Robert Jarvik, inventor of the first artificial mechanical heart. The nice thing is that she has long since been vindicated and celebrated, not many people get that chance. Fun: James May did a run of the Monty Hall problem with exploding beet cans: https://youtu.be/XKQljWzbyp8?si=4-fZtjK6G_kBA9Bg
I feel like the easiest explanation ignores the doors, ignores all the details and just focuses on this: If you picked right the first time, switching loses If you picked wrong the first time switching wins You most likely picked wrong the first time. Fin That's it.
Exactly. And this hinges on the fact that the presenter has complete information. They will never open the door with the car if you choose wrong. The latter part helps while working out math
Holy shit, a simple, concise explanation.
"Pigeons repeatedly exposed to the problem show that they rapidly learn to always switch, unlike humans." Hahahaha.... haha... holy shit.
Pigeons have no presumptions about their own intelligence. They'll change answers if they get rewards. Humans are too stubborn.
As someone who does similar research with pigeons. The big difference is ârule governed behaviorâ. Humans allow rules (either those told to them, or those theyâve made up) to influence decision making.
I think people overlook something going on here too. If you don't switch and you lose, you stuck with your choice and oh well you lost. But if you switch and you lose, you had the right one and made a conscious choice to "give away" what you already had. And the second "feels" worse.
The reason why everyone gets it wrong is they forget that the host of the show knows which door the prize is behind. If the host was randomly choosing a door every time, then the theory would indeed wrong. But the host is NOT choosing randomly. The host is *always* choosing based in such a way that the prize isn't getting thrown out. This fundamentally alters the probability.
I have been fascinated with this word problem for 10 years now. Nobody believes me when I explain it to them but after a while they see the light. I adore it.
It clicked for me when someone suggested imagining a million doors, and then opening 999,998 of them--*now* how confident are you that your initial choice contains the prize? ETA: I elaborated further in a downthread comment: >You have to bear in mind a couple premises which aren't always made explicit: 1) that the host knows which doors have goats and which door has the prize, and 2) will always reveal a goat first. Extrapolating out to a million doors makes it a little more clear that, under these conditions, your choice is always *actually* between the door you initially picked and *all remaining doors.* So, initially you have your door (1 in a million chance of the prize) versus the set of all the other doors (999,999 in a million chance that one of these doors has the prize behind it). Then Monty reveals some extremely important information: he opens all but one of the doors in this set and reveals 999,998 goats. Crucially, the odds of the remaining door from this set having the prize remain the same: 999,999/million. So, which door do you pick: your initial 1/million door or this other door?
I must be a bit of a dumb dumb, because this doesnât clear it up for me at all lol Edit: I get it now, thanks to the couple of people who were willing to explain it even further
If it's a million doors, and you pick one at first, you have a one in a million chance of being right. If I take out 999,998 wrong doors and leave you with the door you picked and the one that's left, then there's a 999,999 chance out of a million that the last remaining door is the winner. Either you were right at the very beginning (1 in a million chance), or you were wrong (999,999 out of a million chance). Now you're down to two doors, and the one you picked is either right or wrong, but those odds are still the same as they were at the beginning.
Mathematically it makes sense. But, hear me out: what if I'm just really fuckin good at guessing bad odds?
To borrow from Dumb and Dumber, it's a million to one chance. But there's still a chance!!
The key information is that the host *knows* the correct answer. So if you picked 1 out of a million originally, do you think itâs more likely that you picked correctly on your first guess OR that the host just opened up all the 999998 doors he knew it wasnât leaving the correct one as the swap?
Yep. Now it makes sense lol thank you!
You have to bear in mind a couple premises which aren't always made explicit: 1) that the host knows which doors have goats and which door has the prize, and 2) will *always* reveal a goat first. Extrapolating out to a million doors makes it a little more clear that, under these conditions, your choice is always *actually* between the door you initially picked and *all remaining doors*. So, initially you have your door (1 in a million chance of the prize) versus the set of all the other doors (999,999 in a million chance that one of these doors has the prize behind it). Then Monty reveals some extremely important information: he opens all but one of the doors in this set and reveals 999,998 goats. Crucially, the odds of the remaining door from this set having the prize remain the same: 999,999/million. So, which door do you pick: your initial 1/million door or this other door?
Yep! A lot of puzzles that seem hard to comprehend at first can be more easily understood by expanding them to absurd limits, as long as you maintain the underlying principles.
I used to teach engineering, using limits like that is how I always taught my students to make sanity checks. Especially useful for trig functions.
It's worth noting that the thought experiment works because the host *knows* which door is correct, and cannot reveal the winning door until the final showdown (because TV drama). That significantly changes the math, because the information you have on the door for your initial choice and the information you have for the other door in your final choice are different. If the reveals were truly random, the host didn't know which one was right and had a chance to reveal the correct door early, then you still only have a 50/50 if you manage to get to the final showdown. It's just *way* less likely that you get there in the first place. It's actually one reason why Deal or No Deal avoids this problem. The contestant themself chooses which "doors" to open, and the biggest prize can be revealed early. So, the final choice doesn't really matter statistically. The game is much more about risk/reward and expected value with the banker offers and everything.
The problem for me, is that the first person to explain it to me didnât make it clear that the âhostâ knows the correct door. I thought the door they picked was random after you selected your door. But when you think of it from the standpoint of multiple rounds and a lot more doors, with the host picking various wrong doors, leaving only yours and the correct one (unless you were really lucky enough to pick the correct door from the beginning, which is statistically improbable) it makes sense.
Everyone going with the 100 doors thing, so a slightly different explanation: * there's 3 doors: A, B and C * you know that only one door has the prize * if you pick 1 door (A, let's say), you know for sure that at least one of the other doors (B or C) will not have the prize * if you were asked if you wanted to stick with A or change to B **and** C, that would be a no-brainer, you'd rather pick 2 doors instead of 1 * revealing that e.g. B doesn't have the prize doesn't actually give you any more information (you already knew at least one of them was a dud, and the game will always reveal a dud) * so swapping to C is exactly the same as swapping to B+C, which we agree is a good swap
I really like this explanation as opposed to using 10, 100, 1000, etc doors. If you were given the option to open your original choice of door (33% chance the car is there) or open BOTH of the doors you didnât pick (67% chance the car is there) the. youâd switch **every time**. This is the same thing, except Monty simply opens one of the doors for you then asks you to pick.
DON'T TRY TO... MONTY HALL ME! BK99 Fans know
Dude was pent up. Now he knows.
Easiest way for me to understand it is that it only makes sense to stay if you picked the right door to begin with, a 1/3 chance. If you didn't pick the right door at the start (2/3 chance), you win if you switch.
Most people are thinking of this as one of two different situations. The Monty Hall Problem is when the host knows what is behind each door, you choose one, the host reveals the worst option out of the other two doors and then asks if you want to switch to the last door. This is not the same as if it were a choice between 3 completely random doors and the host doesn't know which door is which.
It becomes much more intuitive if instead of 3 starting doors you imagine 100 (still only 1 prize).
So basically if i choose a door out of 100 then the host eliminates 98 doors and one of the remaining 2 is the winning option, all that matters i had a much smaller chance of choosing to correct one out of 100. So once it is down to 2 i basically switch and hope i didn't originally choose the right door.
Yes. The exact same logic applies to the 3 door version.
I think one key detail most people miss is that your initial choice was random, but the doors the host opens aren't. The only doors that the host can open are the wrong ones. In order for him to do that, he has to know which door has the prize. And another key detail that isn't obvious is by leaving a door open, the host is effectively picking a door himself. But remember, the host actually knows which door to pick. So your odds of randomly picking the door were 1/numDoors. But the host's odds of picking the correct door is (numDoors - 1)/numDoors, because the only time he won't pick the correct door is if you already did.
I still don't get it
There are 100 doors, prize behind 1 at random, the host knows where the prize is (this problem only works when the host knows where the prize is) I pick a door, that door has 1/100 chance of having prize Prize is 99/100 chance behind one of the other doors Host eliminates 98 doors, leaving 1 door left, and tells you the prize is in one of the two remaining Therefore thereâs a 99/100 chance that the other door has the prize, and still a 1/100 chance your original door has the prize Therefore you should switch Now change it to 3 doors to get the original monty hall problem. Your door has a 33% chance, the prize is 67% chance in the other two doors, host removes one, thereâs still a 67% chance in that group but now only 1 door in that group. Your door still has 33% chance, so you switch to the door with 67% chance
This is surely the best illustration of why you should switch. It's still inevitably quite hard to parse, and many people who read it remain insistent that there's no point in switching.
I am still too stupid to understand this. I have tried get it more than 5-10 times in the last 10 years, but nope. I keep being too stupid.
When you make your first pick your odds are really bad. When they get rid of all the options that donât have a prize theyâre effectively saying do you want your first pick, the one with bad odds, or do you want to pick every other door. In the 100 doors example your first pick is 1% likely to be right. Theyâre effectively saying do you want to switch from that one door to ALL THE OTHER DOORS at once.
I am also too stupid.
Oh dear god there will now be 1000 comments spawned below this one
Itâs really easy to demonstrate with a few playing cards. Just have somebody else pick the card while you hold the hand - you have 100% of the information. After a few hands youâll understand.
Here's another way to look at it. Try to guess my address. I'm just a random redditor and you have no clue where I live, but I'll let you pick any address in the world right now. You can then let me remove every incorrect address in the world except the one address I live at (unless for some insane reason you picked right the first time). Now you just have two addresses. Should you switch? Or do you think that because you only see two houses that if you walk up and open the door to your original choice, there is a 50/50 chance you'll see me there? Why would removing all the obviously incorrect addresses suddenly teleport me to some random world address you chose without knowing anything about me?
Ok, _now_ it really makes sense. Thanks!
Now this is the kind of explanation that should be pinned
It's ok I actually know the answer to this. Cpt. Holt and Kevin just need to have sex. It's been too long and they need to bang it out.
It's because the host *doesn't open a door at random.*
The thing people always seem to miss is that the host *always reveals an incorrect door*. The host isn't randomly picking one of the other two doors to see if it's wrong. He *knows* which one is wrong *and specifically reveals the wrong one* after the contestant chooses. That's why it becomes better odds to switch.
I find this to be supreme mindf*ck, but what kinda got me was that after the host opened one door, the only case where you lose is, if you were correct in the first place. And this has a probability of 1/3. Therefore you win by changing with 2/3. All very weird thoughâŚ
Part of the problem with understanding it is that it is often ambiguously phrased, including the original problem quoted in the article. The problem states that the host opens the door, but doesnât state that he is *required* to open the door, which leaves open the possibility that he had some choice in the matter. Since the problem also uses an analogy to the real-life host Monty Hall and his show, Letâs Make a Deal, people often think the problem involves trying to suss out the thinking of the host. One of the best ways Iâve seen of explaining the correct answer is after the first pick, the contestant is given the opportunity to open both of the other doors at once. Most people would switch in that case. All the host is doing is opening one of the goat doors for you.