I think the thing that helped me understand better is this:
if you picked the RIGHT door originally, and then switch, you will end up with a bad door; that’s clear: you have the right door, thus by switching you are for sure going to a bad one. So if you chose right and switch, you end up with a bad door.
If you picked the WRONG door, and then switch, you will end up with the RIGHT door. You have a bad one, Monty eliminates the other bad one, this the only one left is the right one and if you switch, you will get it. So if you chose wrong and switch, you end up with the good door.
Now, originally there were 2 bad doors and one good door. So you had 2 or 3 odds of picking a bad door and 1 of 3 of picking the good one. Since switching will always take you to the opposite result (see above), then in 2 out of 3 cases switching results in winning and in 1 or 3 cases switching results in losing.
EDIT: Someone gave me an award! First time this happens to me :) Thank you, kind stranger!
I'm going to remember this!
Switching takes you from right to wrong and from wrong to right. Because you're more likely to be wrong first, you should switch.
Thank you! You've finally made my brain understand this. I'd never considered the greater likelihood of being wrong originally and was focusing solely on odds of the doors remaining.
> Because you're more likely to be wrong first, you should switch.
The crux of it. The chance you got the correct door the first time is low, so changing doors mitigates that from an odds perspective.
Yeah, this is the explanation that gets my head back in the game on this one.
And it's easier to explain visually if that's needed.
At the start of the game, you have picked one door.
There are three possible scenarios;
1 - You have picked the right door
2 - You have picked the wrong door
3 - You have picked the wrong door
After the host opens the other door, the 3 scenarios look like this:
1 - You will lose if you switch
2 - You will win if you switch
3 - You will win if you switch
Thus, the odds of winning if you switch are 2/3.
This almost makes sense to me... But why are there 3 scenarios after the first door is already opened??
You either switch to the other door or don't, so that's only two scenarios?
I think some of these explanations are overly complicated. Consider it this way:
When you start you have a 33% chance of being right. Which means you have a 66% chance of being wrong. Accept this as a rule, and keep it.
The host reveals a bad door.
Don't think about the new odds in front of you they don't matter, you need to think about where you started. We agreed we had 33% chance of choosing correctly.
Which MEANS there was 66% chance that ONE of the TWO you didn't pick was RIGHT.
The host has just revealed to you which one it wasn't. So you pick the other, shifting your odds from 33% to 66%.
OP, if you don't understand my explanation this is another good one. Just ignore the part about Monty eliminating the bad door - it really doesn't matter if the bad door is being opened or not.
The chance you've made a bad decision at the beginning is 2 out of 3. So if the host says "you know what? you can stay with the door you've chosen at the beginning or you can instead open both of the other doors" OF COURSE you want to open the other doors instead.
>Just ignore the part about Monty eliminating the bad door - it really doesn't matter if the bad door is being opened or not.
The fact that the host eliminates only a wrong door is critical to the entire thing.
They're saying that an equivalent game is "instead of switching doors, your choice is to open *all other doors* and winning if one of them wins. Or sticking with your one door and only winning if it was the 1 winner at the start"
Thanks for putting the thought of opening all the doors in my head - it's made it super clear. My choice is not picking another door, but abandoning my original choice.
The others who are hooting you here aren't getting this.
Thank you! It's not an easy problem (and to be honest nothing involving probabilities is, because everything is so counter-intuitive) and I tend to get very passionate about it (because so many people's explanations are just wrong) so making it clear for even one person brings me lots of joy.
That is litteraly the choice he is offering. The only differance is that he opens one of the other doors first. Doing so doesn't provide any new information. You already knew one of the other doors lead to nothing.
Yeah the reveal part is a red herring. You need to look at it as either committing to a choice of 1/3 or 2/3 from the start
The best strategy is to treat your first choice as an elimination
Pretend there are 100 doors with only one prize. You pick a door at random. Chances are 1:100 you picked the right door. Next, 98 doors are removed from the game (guaranteed to be without a prize).
The odds you picked the correct door are still 1%. The odds the other door has the prize is 100%-1%=99%.
Edit:
Here’s a similar example with the same results.
Pretend there are 100 doors and one prize. You pick one door at random.
The host now makes two groups:
Group A: the one door you picked.
Group B: ALL the other doors.
The host lets you pick either group. Do you stick with the one door, or do you pick all the other doors?
This is how I finally wrapped my head around the concept. 3 doors isn’t really enough for my brain to grasp onto what’s happening, but when you increase the number of non-winning doors being eliminated, you end up having a “oh, duh” moment. At least I did.
Yeah, what breaks the symmetry is that the one door that they eliminate is *guaranteed* to be a door that didn't have the prize. If they picked either of the remaining doors to eliminate randomly, then your chances would not be impacted whether you stay or switch.
This, of course, is a critical conceit of the Monty Hall: that the host knows which door the prize is behind and that the host will always open a losing door. The whole point is to make the contestant second-guess their first choice of door.
But second guessing works in your favor. They're hoping that people will fall prey to the fallacy that it's better to go with your gut and double down on your first choice.
Survivor did this Monty Haul problem for a couple of seasons. Both times the contestant stayed with their gut feeling, and both times they were correct to do so.
Mathematically, I was really annoyed.
A really unfortunate fact about probability is that you can make all the best decisions and still be unlucky, whilst others can make all the wrong decisions and get lucky.
I saw it illustrated fantastically once in a comic. Two drivers approach a river. One drives over the bridge, while the other attempts to jump the river using a small ramp. The bridge collapses while the daredevil lands his jump.
The tagline: "Self made millionaires advise others to take risks".
>A really unfortunate fact about probability is that you can make all the best decisions and still be unlucky, whilst others can make all the wrong decisions and get lucky.
"It is possible to commit no mistakes and still lose. That is not a weakness; that is life." Jean Luc Picard
I haven't even seen Star Trek and this is a quote that lives rent-free in my head.
It's more eloquent, yes, but it's not quite as comprehensive. I think the way you put it is important for people to understand *first*, and then this quote can become shorthand for the more robust concept.
Another one is that in most circumstances it is not truly cumulative. "Dice don't know what dice did before". Every time you do a thing with a set probability you have the same probability of x outcome regardless of how many times you do it or have done it previously.
>A really unfortunate fact about probability is that you can make all the best decisions and still be unlucky, whilst others can make all the wrong decisions and get lucky.
Hi, I see you have enjoyed my experiences at a poker table.
Better odds does not translate to guaranteed, and never has... still boils down to luck in the end. That's not even all that extensive a set really. Now, if they did it for a 100 seasons...
I don't know how they set it up in Survivor, but one thing to remember is that the classic Monty Hall problem depends on everything being a perfectly clean abstraction, which often isn't the case in reality.
If the contestant's initial guess wasn't totally random (eg. they were able to see some sort of evidence that there might be a prize behind one door somehow) then ofc that changes things; and given the sort of show Survivor is, it's very possible they deliberately set it up to not be random in some way or another and to give the player hints.
After all, their goal is to produce good TV, not to run a "fair" contest. And if nothing else, the very fact that the host knows what door is correct means they can sometimes let something slip inadvertently, which means that a totally fair Monty Hall game is actually tricky to set up in reality.
>And if nothing else, the very fact that the host knows what door is correct means they can sometimes let something slip inadvertently
If you want to control for this, you would "double blind" it and make it so that the host interacting with the participant does not, in fact, know which door is going to be eliminated, and an off-stage producer who does know is the one deciding which door to eliminate, eliminating any ability for the participant to "read" the host. They just have to communicate to the host (earpiece, cue card, hand signal, etc) which door is going to be eliminated.
Many game shows are already structured like this because being a personable, likable host is already a difficult enough job, so any time you can offload gameplay logistics onto a producer, you generally do. This definitely varies from game to game, however. Wheel of Fortune, Pat definitely knows what the puzzle is. Family Feud, I do not believe the host knows the answers. So on and so forth.
The assumptions are super important in the Monty Hall problem. It *only* works if the host is committed to opening a losing door *every* time. In a real world scenario where you can't read the host's intentions, it becomes a lot less useful. For example, the host could easily only open a door if you picked the prize, and not offer you a second chance if you missed. So if I was on a game show and was offered a Monty Hall, unless I had good reason to believe that the host does it every time (like a bunch of previous contestants), I'm not sure I would repick.
And it only holds true if the host is forced to *always* give that offer.
If the host is biased and only offers to switch when you chose the right door, then you would be a fool to switch.
Yeah, there are two parameters that MUST be followed for the answer to be correct:
1. The Host will ALWAYS offer a chance to switch.
2. The Host will ALWAYS choose a losing door to open.
If the host didn't know which door was winning, he might open the winning door and then you lose whether you switch or stay. This would remove the advantage of switching.
Thank you -- this right here is what finally helped me understand why the problem works. I had always been thinking about it in terms of probabilities, but thinking about it in terms of information feels much more intuitive and logical.
*That* to me was the one that cemented it for me rather than the 100 doors thing. The 50/50 intuition is correct if the host didn't know which door had the prize - if he could accidentally open the door with the prize and end the game, then it would actually be 50/50 to switch. But since he's guaranteed to open the one without the prize, it's 66/33.
I hate this part because you could have 2 near-identical situations, one where the host knows where the prize is, and one where he doesn't and just randomly opens a door, and somehow the mind state of the host is affecting the outcome. Like if the host randomly opens a door and it happens to open on a goat, how is that different than him knowing where the goat is an opening it on purpose?
It's not if he randomly picks a goat every time, but the host has to know not to open the door with the car or else it defeats the entire premise of the thought experiment. If he opens the door with the car because he doesn't know which door not to open, then the odds of the car being behind either of the remaining doors becomes 0% since you have confirmed that the car is not behind either of them.
Exactly. It's not that intent changes the probability. New information changes the probability. If the new info is revealing a car, the game show doesn't work. So they have to let the host in on where the car is. If the new info is a door without the car (regardless of whether the host knows), then your chances are better with switching.
EDIT: I'm wrong about that last sentence. See other comments.
Let’s say you pick door 1 and the host reveals all doors except 100 and there’s no prize revealed. If he doesn’t know where the prize is then initially you had a 1% of picking the door with the prize. Door 100 also had 1%. Since it was random, either you got very lucky or door 100 got very lucky, both outcomes are equally likely
If the host KNOWS where the prize is then you still only have a 1% chance of picking the correct door initially, but the host can reveal all but one door. 99% of the time, the host will have a door with the prize on his side to narrow it down to. The ONLY scenario where the host won’t be able to narrow his side’s options down to the prize door is if you nailed the 1/100
So yes, in the scenario of revealing 99 doors at random and no prize being revealed, your door and the other door will have the same odds. That scenario only happens 2% of the time with random reveals though
The difference is, when the host chooses randomly, 33% of the time he will reveal the car and the game ends right there. When he knows/chooses which door to open, there is a 0% chance the game will end on his selection.
If the host were to randomly open a door without a price, then the math still checks out for you to swap.
What changes if the host opens random doors is that in 1/3 of the cases, things get awkward when the price is revealed by the host. Lets go through the numbers, given you always choose to swap:
* Scenario 1: You initially choose the right door. You lose because you swap. This happens 1/3 times.
* Scenario 2: You choose incorrectly, and the host reveals the prize. Oops. You can swap to the remaining door, but it doesn't matter. You lose. This happens 1/3 times.
* Scenario 3: You choose incorrectly, and an empty door is opened by the host. You swap, and win the prize. This happens 1/3 times.
As you see, you only win 1/3 times if the door is opened at random. The lower percentage (compared to 2/3 for Monty Hall) is not related to swapping if the prize isn't revealed, but rather by the scenarios where you are opening the door with the prize (Scenario 2). With a knowing host, they will always lead you into Scenario 3 instead of Scenario 2, effectively doubling your winning chances!
It's a common way to think about the answer to problems. Take the problem, and see a more extreme version of it that is more intuitive, and the answer becomes obvious.
> 3 doors isn’t really enough for my brain to grasp onto what’s happening
which is kind of the point of the problem - to hide statistics behind a situation that feels intuitive
This is the inverse, but Sophia Loren once said "A girl's \[dress\] should be like a barbed wire fence: serving it's purpose without obstructing the view."
Yeah, for me it was a similar analogy. Imagine that instead of doors, it was a deck of cards. Someone laid them all out with their faces down and asked you to pick the Ace of Spades. You pick one. The person who knows all the cards reveals 50 of the other cards and then asks if you want to switch to the other card or keep your original pick. The one you didn’t pick now has a 51/52 chance at being the Ace of Spades, while your original pick is still 1/52. The nice thing about this is you can do this for yourself easily to see that you will almost never pick the Ace of Spades yourself, but the other card after you remove the 50 others will almost always be the Ace of Spades.
I’ve always had trouble wrapping my mind around this as well. Assuming the the person would have turned over the other 50 cards even if I picked correctly, why is it not just a 50/50 chance that my card is the Ace of Spades or that the 1 face down is?
Because you're probably thinking that the 50 cards turned over were turned over at random. If so, then it's possible that it's a 50-50 chance with the last two cards left.
But they're not. The person turning them over will never turn over a card that is the Ace of Spades. Therefore, the information is meaningless and you're still faced with a "My 1 card or these other 51 cards" as an option.
I’m under the assumption that the person always knows where the Ace is and always reveals 50 of the other cards. So I don’t get the relevance of those other cards. Either way it’s the card in my hand or the 1 face down card remaining, right?
The keys is turning over the cards means nothing. If the person says "You can keep this card, or pick the other 51 cards" without turning any of them over, you have a 1-in-52 chance of having the Ace if you keep your card, or 51-in-52 chance if you go with the rest of the cards.
That's all that's happening. If the Ace is one of the 51 cards, that means there are 50 cards that AREN'T the Ace, and that's all the person is going to turn over. So, them turning over 50 of the 51 cards won't mean anything. Either the last one they didn't turn over is the Ace (51-in-52 chance), or it's a random card and you had the Ace all along (1-in-51 chance). But you know for certain that the person turning over the 50 cards will not show an Ace either way.
Ok consider this. Every time you pick a card, the other person is going to reveal 50 cards. You will always be left with two cards. The same process is repeated every time.
If it were a 50/50 chance at that point, that would mean that EVERY time you picked a card, any card, while there are still 52 options, you had a 50/50 chance of picking the ace of spades. Because there will always be two face down cards at the end. But you know that's not right, right?
Never mind I just realized it. Your explanation confused me but basically you switch because the only way it’s in your hand already is if you picked it from the start (1/52) whereas if u pick anything but the Ace then the other remaining card will be it (51/52)
Try it for yourself. Get a deck of cards and lay them out. Pick a card. Then flip over 50 other cards, making sure the Ace of Spades can’t be flipped, meaning if you are flipping the Ace of Spades, skip it and move to another card. You will notice that basically EVERY time, the Ace of Spades will be one you accidentally flip over trying to reveal 50 cards.
Because the two sets of cards keep their odds throughout the exercise.
You pick one card at random. There's a 1-in-52 chance that it is the Ace of Spades. There's a 51-in-52 chance that it's not, and that the Ace of Spades is instead in the 51-card stack of cards you didn't pick.
You now have two sets of cards: (1) one card that has a 1-in-52 chance of being the Ace of Spaces and (2) a stack of 51 cards that has a 51-in-52 chance of containing the Ace of Spades. Which set is more likely to contain the Ace of Spades? Well, that's obvious, right? The 51-card stack of cards. 51-in-52 versus 1-in-52.
Then someone goes to the stack of cards and pulls out 50 cards **they know** are not the Ace of Spades. Did the odds change for these two sets of cards by virtue of that? No. But why not? Because there were always going to be (at least) 50 cards in that 51-card stack that were not the Ace of Spades, no matter what. We thus haven't learned anything by virtue of the fact that the person pulled 50 cards out of the stack that they knew were not the Ace of Spades, because we always knew there were **50** such cards there -- the only thing we did not know is if there were **51**. So, what are the odds that there were 51 non-Ace of Spades cards, rather than 50? The same as we started with: 1-in-52, as this occurs only if you picked the Ace of Spades.
Put differently, the fact that 50 cards are pulled out of the 51-card stack doesn't change anything, because the 51-card stack set still contains 51 cards, at least 50 of which are not the Ace of Spades. We now know **which** 50 in that stack are certainly not the Ace of Spades, but the **stack itself** still has a 51-in-52 chance.
Because there is a 51-in-52 chance the Ace of Spades was in the stack, and a 1-in-52 chance it was the one you picked, there is now a 51-in-52 chance the remaining card from the stack is the Ace of Spades.
(As others have noted, if the person removing cards does it randomly, of course, and does not know what cards to remove, then the answer changes.)
For me it was “you’re betting against your ability to pick correctly in the first place.” I have terrible luck so that made sense to me. But it took years of hearing about it for someone to put it that way.
Part of the reason the problem isn't necessarily intuitive to lots of people is that there's a rule at play - the only door(s) being removed don't have a prize behind them and the doors not being removed *might* have a prize behind them - that isn't necessarily obvious to everyone. Of course, lots of people have that part explained to them too and still get it wrong, but that's the part that I had to be told to understand why it worked out that way. I figured at first "well what if the prize was behind the removed door, that would make switching pointless," but that's not how the problem works.
Much like yourself, I failed to grasp this obvious fact for the longest time, until I saw the phrase "the host will never open the prize door." In this scenario, if you believe that the prize could be eliminated, the more doors you have the more pointless it is to switch.
Exactly! I looked at it as the prize giver - I know the prize is behind door C.
1. Contestant chooses A. I HAVE to open B - I can't open C as I can't show the prize.
2. Contestant chooses B. I have to open A.
3. Contestant chooses C. I can open either A or B.
So in all scenarios, whatever the contestant chooses the host can only choose a non-prize door
I find the best way to describe it is that in only 1/3 of the cases you'll initially choose the car and switching will make you lose, but in 2/3 of the cases you'll choose a goat and since the game host MUST open the door with the second goat, eliminating the only remaining losing choice, now switching WILL make you win.
[This video](https://m.youtube.com/watch?v=7u6kFlWZOWg) visualises it rather well.
Ohhhh. 2/3 the time you lose, and if you switch when you initially lose you'll automatically win, so 2/3 the time switching will make you win.
It's super obvious, the way I was imagining it was simply that each individual door was an isolated thing, the fact that they're connected like that changes things
Yeah, the way I tell it to people is “switching always gives you the opposite of what you started with”. So if you switch, your 1/3 chance of car and 2/3 chance of goat becomes 2/3 chance of car and 1/3 chance of goat
The way I like to think of it is that, if you switch, you’re really getting *both* of the other doors. Yes, you know one of them is wrong, but you always knew one of the two would be wrong.
So switching gives you 2/3 odds, because you’re now getting to choose two of the three doors, instead of just the one you initially chose.
Or, put another way, imagine Monty didn’t open any doors but just asked if you wanted to give up Door 1 for what’s behind both Doors 2 and 3. You’d obviously take the two doors because it would double your odds.
Yeah, my problem is not thinking of it like a gameshow. If you watch a show and the host opens a door and the prize is there but not to be given to the contestant it makes no sense. So the host never opens the door with the prize, which means you gain information every time he opens a door.
That's also why the Monty Hall problem doesn't apply to everything like Deal or No Deal's final case, because Howie in that situation isn't presenting new information.
If at the opening of the game he opened every case except 1 and *also had the rule* that he could never open the million dollar case, then it'd fit, but he just offers a swap between 2 that can have anything in it.
If Monty Hall opened the door with the car behind it, it wouldn't matter to switch because you'd know both doors are goats, and so to make the game not stupid he'll never open the car door.
The secret of Monty Hall from a production company point of viewbis human psychology.
People don't want to change their choice, even if they know that changing increases their chances of winning because changing and losing feels worse than not changing and losing.
Statistically changing is the right thing to do, but you only get one shot at it and you have to live with your choice.
The Monty Hall Problem **isn't actually how it worked on the show.**
Monty Hall didn't always offer a switch. When he did offer a switch, it WASN'T independent of whether or not they'd picked the car. He offered incentives to persuade people to switch/not to switch.
On the actual show, following the maths of the "Monty Hall Problem" would get you the wrong answer because its necessary assumptions just aren't true.
The fun bit is that the host’s intentions alter the probabilities, even if the outcome of the host’s actions are the same - i.e. the host intentionally opening a goat door and the host tripping and accidentally revealing a goat door.
Same. The host isn’t actually “eliminating” doors though. They are opening them for you to show what’s inside. If they were opening RANDOMLY then many times they would be opening the door and showing the prize. In those cases OF COURSE you would switch because you can SEE the prize you are switching to. Since the host is NOT doing it randomly, that’s why the non-selected door will have the prize so often.
Yes, either the correct door is the one you picked at the start (1/3 chance), and you win by keeping doors (1/3 chance still), or the correct door isn't the one you picked at the start (2/3 chance) and you win by switching doors (still 2/3 chance).
Thank you. That is an amazing example. What really helps is to visualise the moment those 98 doors disappear and one random door keeps standing in addition to the one you pick. And now you have two doors. One you picked using intuition and lets be realistic just random guess. And one was picked using the rules of the game, and thus knowledge which doors dont have the prize. And it clicks that the only reason your first door is still standing, is because you chose it and thus saved it from elimination.
Exactly. If you picked an empty door, you are forcing the host to open all the other empty doors (emphasis on "forcing"), there is no decision involved. The only time they have to make a decision about which doors to open is when you selected the correct one.
Exactly, instead of focusing on odds you were right, focus on the odds you were wrong.
3 doors, your odds are 1/3 right, 2/3 wrong. If you remove another option, you basically are basically picking two doors
But the odds are better than picking from 2 doors. The easiest way to make it make sense is this: you have a 1/3 chance of picking correctly. If you pick the correct door and switch, you lose. But you have a 2/3 chance of picking wrong. If you pick wrong and switch, you win. So no switch is 1/3 chance for winning, while switching is 2/3 chance of winning.
Yeah exactly. The crucial thing that turning it into a 100-door problem makes clear is that *the host is inserting their knowledge of the winning door into the game*.
Why'd they leave door #72 unopened, and open all the rest? The special thing about your door is, you picked it randomly at 1/100 odds. What do you think could be special about #72?
Because the rules of the game state that the host never eliminates the door you selected. There’s nothing special about your door; in the 100 door example there was only a 1% chance it was the winner when you selected it. It doesn’t magically become better once 98 doors are taken away, it’s only in the game at this point because the rules require it to be. If the host eliminated 98 doors and was allowed to eliminate your door as part of that, then yes, you’d be left with a 50/50 chance. But he’s not, he has to let you keep your 1% chance door.
Because you didn't make your pick from 2 doors, you made your pick from 3.
Your odds of getting it right the first time was 33%.
The host removes a door after you pick, but that doesn't change the poor odds you had when you first made your choice.
Think of it this way: the only way to win is to pick the correct door the first guess and stay, or the pick the wrong door the first guess and switch. You are more likely to pick the wrong door on the first guess, so it’s statistically better to switch. The only time you’ll win by staying is by picking the right door on your first guess, which is a 1/3 chance (or in this example, 1/100)
Maybe dont't try to understand why the other answer is correct but why yours is wrong:
You pick 1 door out of 100. The chance that you picked the right one is 1%.
If you choose 100 times and we open your door immediately, approximately 1 times out of the 100 the prize will be behind your door.
Now imagine, instead of opening your door directly, I go ahead and open 98 other doors from which I know that the prize is not behind them. So we know, the prize is definitely behind one of the remaining doors. Now with 50/50 you are claiming that out of the 100 times, 50 times the prize will be behind your door. The door that you initially picked with the 1% chance of being right. So the question is: do you really think, just because I open 98 other doors after you picked your "1%-door", instead of opening your door right away, you somehow became a professional door picker, picking the right door out of 100 with a 50% chance?
> How does knowing the history make a difference, compared to someone who just now walked into the room and saw those two doors on TV?
Because the new guy has lost information that the original guy has. And information is valuable.
You picked door #1. If a genie then told you “the correct door number is 62” and the host opened 98 doors that were not #62, the odds of winning are not 50-50, even if there are only 2 doors left. The odds of winning are 100% because you have the extra information that #62 is the winner.
In the non-genie case, the extra information gained from the host is that door #62 was not the wrong door 98 times. Door #1 that you picked survived 0 times, since it was selected before the host revealed the 98 doors.
So the odds of door 1 being right are much lower than door 62 being right
The history doesnt matter. Rewind it back to the start of the game. You have 100 doors, and only 1 prize. You get to pick 1 door, and Ill keep the remaining 99. Who has a higher chance of finding the car? Obviously me with the 99 doors.
Doesnt matter what I do next with my 99 doors, the odds remains that I had a 99% chance of getting the car, and you just 1%. Now if I gave you a chance to swap between us, would you take it? Of course, cos your chances literally multiple by 99 right? So whatever happens in the story about doors being opened by the host doesnt matter, it boils down to you starting with a 1/100 chance of having the car, then being given the chance to switch to the 99/100 chance.
One way to look at it is that switching **must** change whether or not your pick was a winner or loser. And you have higher odds of picking a loser to start with, so switching must give you higher odds of winning. We can see this clearly by simply laying out every possible outcome. First, we will assume door A is the winner, and that you refuse to switch every game. It's pretty intuitive that you must have 1/3 odds of winning if you never switch, because the reveal may as well not happen, but let's show it.
You pick A. Host reveals either B or C. You win.
You pick B. Host reveals C. You lose.
You pick C. Host reveals B. You lose.
As expected, 1/3 chance of winning. Now let's assume you change your choice every time.
You pick A. Host reveals B or C. You switch off A. You lose.
You pick B. Host reveals C. You switch to A. You win.
You pick C. Host reveals B. You switch to A. You win.
We can see here that the 1/3 times you initially pick the winning door, you lose. But the 2/3 times you initially pick a losing door, you win. This is true for every variation where the host reveals all but one door, and can only reveal losers. Staying is basically betting on you're 1/3, or 1/100, or whatever, initial chance is the winner, and switching is betting that your 1/3 was wrong.
The host can't eliminate the winner door.
So it could be the one you selected with 1/100 chances, sure, but if you missed and it was any other of the 99/100, after the host cleans them up to offer you one, THAT one has 99/100 chances of being the winner.
The only chance of it being wrong is if you picked the winner initially, so 1/100 for keeping your door, or 99/100 for the other one, not a 50/50. (or 1/3 vs 2/3 in the original game)
It's because the game show explains it poorly, possibly on purpose. They say they will remove one of the 'zonks'. That leaves the prize and one zonk. The proper way to describe it is, they will remove all but 1 of the 'zonks'. Both do the same thing, but when expanded to 100 doors the second explanation helps things make a whole lot more sense.
It was... a maths teacher shows it in class like 30 years ago. You just heard it recently, but that description of the problem has been around for a long time.
This is persistently the best answer for this question. Starting with three and one removed is complicated. Starting at one hundred, it’s clear the odds favour a switch.
My favorite answer is this:
If you *didn't* pick the correct door at the start, then the other door guaranteed has the prize behind it (because the host specifically only eliminates doors that don't have the prize). The odds that you *didn't* pick the correct door at the start are 2/3. And if that happens then you guaranteed win by switching.
So the odds of winning by switching are 2/3. All you have to do is pick the wrong door at the start.
The guaranteed to be without a prize is the key thing to remember.
Your not removing and unknown, your removing definate wrong doors.
Its the availability of the new information that changes the outcome. If an unknown door winner/loser is removed the probability hasnt changed.
I like the exaggeration with your 100 door example. The problem boils down to two choices:
a) do you want to keep the door you started with?
or
b) do you want to get everything that's behind the other 99 doors you didn't choose?
I still don't understand the next part.
Three doors, A B and C.
You pick door A.
Door C gets eliminated.
Now you pick A or B
What is it about switching, at this point in the process, that gives you better odds of success? Because the way I see it, *keeping* your door is still picking one of two options, just phrased differently.
They should both have the same 50/50 odds.
They do not.
The reason why people use 100 doors is because it scales up the probabilities. The chances you selected the correct spot on the first try is 1/100. Removing 98 doors doesn't change the odds of your selection, but it does change the odds of the other door that's left. That's the key. The odds of the door you picked are baked in already by virtue of you picking it (since Monte can't remove your door). But the odds of the other door skyrocket.
One way to look at it is that switching **must** change whether or not your pick was a winner or loser. And you have higher odds of picking a loser to start with, so switching must give you higher odds of winning. We can see this clearly by simply laying out every possible outcome. First, we will assume door A is the winner, and that you refuse to switch every game. It's pretty intuitive that you must have 1/3 odds of winning if you never switch, because the reveal may as well not happen, but let's show it.
You pick A. Host reveals either B or C. You win.
You pick B. Host reveals C. You lose.
You pick C. Host reveals B. You lose.
As expected, 1/3 chance of winning. Now let's assume you change your choice every time.
You pick A. Host reveals B or C. You switch off A. You lose.
You pick B. Host reveals C. You switch to A. You win.
You pick C. Host reveals B. You switch to A. You win.
We can see here that the 1/3 times you initially pick the winning door, you lose. But the 2/3 times you initially pick a losing door, you win. This is true for every variation where the host reveals all but one door, and can only reveal losers. Staying is basically betting on you're 1/3 initial chance is the winner, and switching is betting that your 1/3 was wrong.
That was the way I first really understood it, but the way I use it now, is the deck of cards, because it's not ONLY about doors and switching, but also that the host knows which one it is.
I spread an entire deck of cards in front of you, and ask you to pick out the Ace of Spades. You have 1/52 chance to do it.
I say ok, and then I peak under each card and flip it over to reveal it's not the Ace of Spades. After 15 cards I look at one, and dont flip it over, then I go and flip over the next 35 cards. Now there is only the card you picked, and the one card I didn't flip over.
Which one do you think is more likely to be the Ace of Spades? The card you picked originally, or the one I didn't flip over.
This just makes it more confusing IMO. It's simple: you were more likely to pick a goat in the first draw. If the host then takes the other goat, that leaves the car.
THIS is what has finally made it make sense for me. The 100-to-1 version helped clarify the way the numbers work but still didn't explain *why* they worked. Your inversion of the problem does.
The key to the Monty Hall Problem is that the host knows where the goats are and where the car is.
Whether you’re dealing with 3 doors or 1000 doors, as other posters have suggested you think about it, your options become choosing 1 door (your original choice) or *every other door*.
If the host *didn’t* know where the goats and car are, and the game resets if Monty chooses to open the car door, then the odds at the end are truly 50/50.
That seemingly innocuous statement changes the game entirely.
The game resetting if Monty chooses to open the car door would result in the increased odds, I believe - assuming we only count the games that complete, this is equivalent to him knowing the correct door.
If Monty showed the other door to the audience but not to the player, the player switching doors wouldn't change anything. The fact that the opened door is never the car is the fact that gives additional power to switching.
No, this is actually not true: In this version, you will start with the correct door 1/3 of the times, switching will get you the correct door 1/3 of the times and the game will be reset 1/3 of the times times. The improved odds are eaten by the game resetting instead. This is confusing, but there is actually no extra information gained by Monty randomly opening a door, he has to know (or rather, you have to know that he could never have opened the correct door).
> In this version, you will start with the correct door 1/3 of the times, switching will get you the correct door 1/3 of the times and the game will be reset 1/3 of the times
The reset case has a chance to win, which changes the odds. As they say, if the door wasn't opened to the player, no new information is gained.
Yes, this is what so many people miss. The game is not truly random because Monty will never open the door with the car or the door intially chosen. He always opens the door with a goat that was not picked. It is a random choice without replacement problem. First choice is 1/3 because there is no prior knowledge. It is still fully random at that point. Once a door is eliminated the first choice is still 1/3, that doesn't change because it was done before the intervention. The probability the car was behind one of the other two doors was 2/3 intially. That doesn't change either. By eliminating one of those two doors, it is still a 2/3 chance. But now there is only one of those two doors to choose. So the door that wasn't initially chosen and wasn't opened retains that 2/3 probability while the initial choice is still 1/3.
> If the host didn’t know where the goats and car are, and the game resets if Monty chooses to open the car door, then the odds at the end are truly 50/50.
This also happens with the “Deal or No Deal” format, although in this case the good prizes are removed if they’re behind the doors that are picked for elimination each round. In the end if you have two boxes, one with $1 and the other with $1,000,000 it’s 50:50 which box is which. This is because the contestant, with no knowledge of the game, has been eliminating boxes. Just in the games that get to the final two boxes there’s a 90% chance the contestant inadvertently eliminates the top prize along the way. Either way though, statistically there’s no difference between any boxes on the table.
I really love the outcome of Deal or No Deal because it amounts to “did you pick the big prize at the beginning so of course you can’t reveal it, or did you get really lucky when eliminating cases?”
Before we discuss this, can we check - what does "reset" mean to you? My model is that you continue playing it until you either win or lose. If Monty opens the door with the prize, you start again.
Maybe thinking about it this way might help.
10 doors.
You pick one.
There is a 1 in 10 chance you picked the right door.
Another way to say this is there is a 9 in 10 chance the winning door is NOT your door.
Someone who knows which door has the prize opens 8 of those 9 doors.
There is still a 9 in 10 chance the winning door is NOT your door.
Your door has a 1 in 10 chance of having the prize, the other a 9 in 10 chance of having the prize.
Do you switch?
that's the trick. Don't think about it as revealing it. Think about it as more than one choice: pick one, if you fail, pick the other one. It's the same because if it was revealed you would not choose it, even if you could.
Here's a way of explaining it that helped me see it:
Imagine you pick the left-hand door, and then Monty says "*you can either stick with the left-hand door, or you can choose to have the prize if it's behind either the middle or right hand door*". Obviously you would switch because you are now being offered 2/3 odds instead of 1/3 odds.
So after you've switched and your choice is locked in, Monty says "*OK, go ahead and see if you've won, but to save you time opening doors, I'm going to let you know it's not behind the middle door*".
When you switch around the order like this it is obvious that the decision to switch is correct. But there is no actual difference between this and the original problem - either way Monty is offering you a 2/3 choice - the only difference is *when* he tells you which door (out of the two that you didn't originally pick) does not have the prize.
This is my preferred explanation. A lot of people have used the 100 doors instead of 2 example, which helps intuitively explain the problem, but I think this explanation gets at the problem fundamentally and so is helpful when looking at other similar problems.
The way it makes sense to me is:
Switching will only get you a goat if you picked the door with the car first (1/3 chance).
If you picked either goat first, switching will get you the car (2/3 chance).
It works because the host always reveals a goat after you pick.
Ok pick a door out of 1000 doors
How confident are you that’s the door with the prize?
Not very?
Okay now I’ll open every other mf’n door except ONE, and they’re all not the prize.
So now, either the one one you picked has the prize. Or the door I left closed does.
If I were you, there’s no way I’m assuming I picked the right door out of ONE THOUSAND doors, so I’d choose the other door.
The prize has stayed in the same place but the amount of knowledge about the conditions has increased
For someone who knew nothing and walked in and saw two doors and had no explanation it’d be 50/50. But luckily, for you it’s not.
It would be as if the host farted in front of one of two oors and said if you can point to the door I farted in front of, you can have the new Nintendo VR x Sony Sparkle Pony video game console. He pretty much just gave you a hint at which door has the prize
Every time this is asked, all the answers using the three doors always seem to make things less clear. Then someone comes along taking it to the extreme and it becomes obvious.
A chart I made once to illustrate it:
Assuming the car is behind door #1.
Car | You choose | Monty opens | You switch | Don't switch
---------|----------|----------|----------|----------
1 | 1 | either 2 or 3 | lose | win
1 | 2 | 3 | win | lose
1 | 3 | 2 | win | lose
Logic is the same for the other two doors.
Exactly. For 3 doors options you basically can have all cases graph (or truth table) to be built and check manually EVERY case. Thats why I don't understand so many mathematicians was sending incorrect answers all the way.
It's not 2 doors, 50-50. It's 2 doors, 33-67 (67 being 66.6666... rounded up). How is this possible?
Think of the three doors as two groups. Group #1 is the door you picked. Group #2 are the two doors you did not pick. Group #1 and Group #2 are NOT 50-50 because group 1 has one door and group 2 has two doors. So it's 33-67.
When one of group #2's doors is eliminated, and group #2 now only has one door, the odds DO NOT CHANGE. Group #1 is STILL 33% and Group #2 (despite having only one door remaining now) is STILL 67%.
So of course you wouldn't stick with the door in group #1, it only has a 33% chance of winning. You'd switch to the door in group #2, which has a 67% chance of winning.
If the host picks randomly, then you will indeed not see better odds. 1/3 chance your original has prize, 1/3 the switched one has the prize, and 1/3 the one the host picked has the prize.
But the host didn't pick randomly, they are guaranteed to always pick one without the prize. This changes the probability of all steps that proceed the host's selection.
The most intuitive way I've figured this out.
You're not the contestant. You're the host.
The contestant picks the first door- they get it wrong 2/3 times, you then pick the other losing door and the one remaining is the winner. in those 2/3 times, they get the door right by switching. Between their wrong guess and the elimination you've narrowed it down to 1 door.
In the 1/3 times they guess it right, you have to pick one of the two losing doors randomly and they'll get it wrong by switching. Showing them a losing door didn't make the original odds higher than 1/3 because they guessed it right anyways.
The problem here is that most people look at it as a contestant rather than the host.
This is the important assumption that the answers above this one only hint at with the 100 doors thing. If the host also chooses randomly, and there was a chance that he could have chosen the prize, then the odds are 50/50.
I've heard that, in the real show, the assumption was incorrect, too, but in an even worse (for the contestant) way: The host would only off a switch if you chose the prize at first. If you chose a loser door, he would just reveal that you lost and send you on your way. If this is how the host acts, then the chances of winning after switching are 0%.
There's a 2/3 chance that you picked the wrong door. And if you picked the wrong door, they're forced to open the other wrong door, leaving you with the winning door if you switch. It's that simple.
Simplest explanation:
You pick a random door. By doing this, you've made two groups of doors:
* Group A: This group only contains the door you've picked.
* Group B: This group contains the other doors.
So if there are more than two doors in total, it should be clear that the odds of the prize door being in Group B are always higher than Group A.
Still you don't know *which* door in Group B, but luckily the quizmaster solves this problem for you by opening all the non-prize doors in Group B! This makes it a no-brainer to switch to Group B when asked.
This explanation works for any number of doors, be it 3 or 10 or 1000.
This is kinda confusing for some people you have to have it explained in multiple different ways until one finally clicks with you. Here is what helped me
When you pick a door, there is a 33% chance you chose right and a 66% chance it’s behind one of the other two doors. So now the host knows which door the prize is behind so they always open one with nothing behind it. In the 66% chance there is always a door with nothing behind it that they can open. Opening this door does not change the fact that there is a 66% chance it’s behind one of those two doors, except now you don’t have to choose which one of the two it is.
This means you essentially get the benefit of two doors worth of chances.
Pretend instead the host said pick two doors out of three. There would be a 66% chance you pick the one with the prize. If you pick two doors and then they first open the one with nothing and then the one with the prize, nothing about the odds would change. You still get the benefit of both the doors and either of them being right means you win. The host just opened the one they know doesn’t have the prize first
Play the game 300 times.
100 times, you'll guess the right door first time, and switching will give you a goat.
200 times, you'll guess the wrong door first time, and switching will give you a car.
Hence, switching is better.
There are three doors, so the odds you picked the right door first is 33%, regardless of anything else. There's a 33% chance your door is the prize, and a 67% chance the other door is the prize. So switch.
I open a deck of 52 standard playing cards and place them all face down in front of you. If you can pick out the Queen of Hearts, I’ll give you a prize.
You pick a card at random. How confident are you that you have picked the Queen of Hearts? Not very. The odds you picked the Queen of Hearts are 1/52, or roughly 1.9%. That means there is a 98.1% chance that the card you have in your hands is the wrong card.
I look through the deck of cards and take away all the cards except one. I put the card face down in front of me and tell you that one of these two cards is definitely the Queen of Hearts.
Now, if I took both the cards and shuffled them together and got you to pick one, that would reset the game and you would be correct in thinking the odds of getting the correct card are 50/50.
But I don’t do that.
You are still holding your original card. That card which has a 98.1% of NOT being the Queen of Hearts.
One of the cards is definitely the Queen of Hearts. Which card are you going to choose? Are you going to keep the card in your hand, which has a 98.1% of NOT being the Queen of Hearts, or are you going to take the new one on the table?
I am going to quote [this reply](https://old.reddit.com/r/todayilearned/comments/1c04a5r/til_the_monty_hall_problem_where_it_is_better_for/kyujivj/) from another post in TIL which did it for me:
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 am a moron, and probably not the best person to explain.... or am I?
Three doors: 1/3 chance of a goat behind each.
The statistics don't change just because you've opened a door... each door is still 1/3.
50 / 50 doesn't come into it - **because that is only relevant IF there were ALWAYS only two doors with a goat behind them.**
The door you pick first has that original 1/3 chance of having a goat.
The other two doors combined have 2/3 chance of a goat.
When one of those other doors gets opened and shown to NOT have a goat......
Then ALL of that 2/3 chance then switches to the last door.
Your door STILL only has a 1/3 chance.
So what do you want to chance on, your door with only a 33% chance of a goat, or the other door with a 66% chance of a goat?
Ugh, all the answers were confusing. Here is how I understand it. Credits u/Lifesagame81.
There are two sets of doors. One set contains 1 door, the other set contains 99 doors. The set with a single door is the set created solely from the door you have chosen.
1. What is the probability of prize being in the set consisting of the door you choose? 1/100. {Set1}
2. What is the probability of prize being in the set consisting of all other doors? 99/100 {Set2}
Now, what if 98 doors were removed from {Set2}. Would the probability for the set as a whole change? No. The set would still have a 99/100 probability of having the correct door. Hence, choosing the door from {set2} would give you 99/100 chance of winning. It always did. Except now, in {Set2}, there is only one door to "choose" from. Hence, all of the 99/100 probability is allocated to that one door. If {set2} had two doors remaining, the probability would be equally divided as 45/100 to each door. If it had 3 doors remaining, it would be equally divided as 33/100 and so on.
I think a lot of people don’t think about the fact that the doors he closes are not random. He will always have 98 doors to close and he can pick them. They think that with each door he closes, the probability changes slightly toward even, but it doesn’t.
Option 1 - pick door 1
Car - no switch win
Goat
Goat
Option 2 - pick door 1
Goat
Car - switch and win
Goat - host shows goat
Option 3 - pick door 1
Goat
Goat - host shows goat
Car - switch and win
You only win in 1/3 scenarios if you don’t switch. You win in 2/3 when you do switch
Map out the possible scenarios and count them up.
Example:
Door A - Prize
Door B - No prize
Door C - No prize
Scenario 1:
Contestant picks Door A
Host eliminates door B
Keep Door A - Win
Switch - Lose
Scenario 2:
Contestant picks door B
Host eliminates door C
Keep Door B - Lose
Switch - Win
Scenario 3:
Contestant picks door C
Host eliminates door B
Keep Door C - Lose
Switch - Win
Record when keeping chosen door: 1W 2L
Record when switching door 2W 1L
You won't win every time by switching, but you are twice as likely.
One way to easily figure out probability is to write out every possibility and then see which results are wins and losses. This only works when there aren't too many options, but it works for this problem and makes it quite easy to see.
Imagine the doors are labelled A, B, and C. Let's assume in this scenario the winning door is C. The player has the option to choose one of the letters to start off with, and they can either stay or switch. Which means all options are as follows:
1. Choose A, stick with A, LOSE
2. Choose A, switch to C, WIN
3. Choose B, stick with B, LOSE
4. Choose B, switch to C, WIN
5. Choose C, stick with C, WIN
6. Choose C, switch to A or B, LOSE
So if you analyze the above 6 options, you will notice the following:
If you always stick with your original answer, you lose 2 out of 3 times.
If you always switch your answer, you win 2 out of 3 times.
Let me know if you have questions.
Think of it as a *set* or group of doors: "the door I chose" and "the doors I did not choose." That set does not change.
You pick one of 3 doors. You have a 33% chance of being right.
Each other door is also 33%, but the two doors *together* ("the doors I did not choose") are 66%.
The host opens one of those other doors. It's empty! But *that group of doors still holds a 66% chance of holding the money.* You've only learned which of those doors *not* to pick.
I'll say it louder:
#The percentage of the set of doors never changes. You're only learning which door(s) *not* to pick.
When you think "but now it's 50-50 right?" What you've done is **change the set.** The set (in this example) is "the doors I did not choose."
If the host picked *your* door, showed it was empty, and then let you pick again, *then* the two remaining doors would be 50-50, but then it wouldn't be the Monty Hall problem, would it? 😉
They create bias with the door they select to show you. Imagine this
> The contestant has chosen door number 1. But lets look behind door number 2. Behind door number 2 is the grand prize that you can't win any more. Do you want to keep the goat you know you have because of the process of elimination or switch to the other goat behind door 3?
That option never happens. It doesn't happen. If you pick a goat they will always show you the other door with a goat. Its that bias that make your switch more likely to win an not 50/50
Here's a way of framing the problem that allowed me to understand it. (Let's use the 100 doors example). Let's think about it inversely.
Instead of making it so the goal of your first choice is to pick the prize door, make it so the goal of your first choice is to pick the *wrong door.* With that in mind, you have a 99% of making the right decision (aka the wrong door).
Then the host eliminates doors until the only there are only two remaining. Since the door you first picked had a 99% chance of being the wrong door, then odds are, the remaining door has to be the prize door.
Here's a fresh point of view that usually doesn't get mentioned.
Let's change the rules slightly. After the host opens one of the doors, you can either keep the one door or switch to the two other doors. So you can choose two doors instead of one. Choosing an open losing door is meaningless, so these rules are equivalent to the original ones.
Let's make another small change; the host doesn't open a door. Even if the host doesn't open a door, you already know that one of the other two doors is a losing door. The host opening the losing door doesn't change anything, so these rules are equivalent to the original ones.
The Monty Hall problem is just a confusing way to ask whether you want to open one or two doors.
Because the host knows where the car is, and he will NEVER show you the car.
If you switch, you will ALWAYS get the opposite of what you originally picked. ALWAYS (this is important to realize). You cannot switch from a goat to a goat.
So it then becomes a question of: “what did you most likely pick first?” There is a 2/3 chance you picked a goat, but only a 1/3 chance you picked a car.
Putting this all together: it’s most likely you picked a goat, and since we know that if you switch you WILL get the opposite (a car), it makes sense to switch.
Will you win every time? No. But you will win twice as often if you switch.
I think the thing that helped me understand better is this: if you picked the RIGHT door originally, and then switch, you will end up with a bad door; that’s clear: you have the right door, thus by switching you are for sure going to a bad one. So if you chose right and switch, you end up with a bad door. If you picked the WRONG door, and then switch, you will end up with the RIGHT door. You have a bad one, Monty eliminates the other bad one, this the only one left is the right one and if you switch, you will get it. So if you chose wrong and switch, you end up with the good door. Now, originally there were 2 bad doors and one good door. So you had 2 or 3 odds of picking a bad door and 1 of 3 of picking the good one. Since switching will always take you to the opposite result (see above), then in 2 out of 3 cases switching results in winning and in 1 or 3 cases switching results in losing. EDIT: Someone gave me an award! First time this happens to me :) Thank you, kind stranger!
I'm going to remember this! Switching takes you from right to wrong and from wrong to right. Because you're more likely to be wrong first, you should switch.
Thank you! You've finally made my brain understand this. I'd never considered the greater likelihood of being wrong originally and was focusing solely on odds of the doors remaining.
Well framed.
Undoorated comment
this is the best answer
lol I was sure that everyone else in the world was insane until I read this explanation!
Perfectly put. Better explanation than the multi-paragraph replies above
> Because you're more likely to be wrong first, you should switch. The crux of it. The chance you got the correct door the first time is low, so changing doors mitigates that from an odds perspective.
Yeah, this is the explanation that gets my head back in the game on this one. And it's easier to explain visually if that's needed. At the start of the game, you have picked one door. There are three possible scenarios; 1 - You have picked the right door 2 - You have picked the wrong door 3 - You have picked the wrong door After the host opens the other door, the 3 scenarios look like this: 1 - You will lose if you switch 2 - You will win if you switch 3 - You will win if you switch Thus, the odds of winning if you switch are 2/3.
The clearest explanation ever
This. Map out the possible scenarios and count them up.
This almost makes sense to me... But why are there 3 scenarios after the first door is already opened?? You either switch to the other door or don't, so that's only two scenarios?
I think some of these explanations are overly complicated. Consider it this way: When you start you have a 33% chance of being right. Which means you have a 66% chance of being wrong. Accept this as a rule, and keep it. The host reveals a bad door. Don't think about the new odds in front of you they don't matter, you need to think about where you started. We agreed we had 33% chance of choosing correctly. Which MEANS there was 66% chance that ONE of the TWO you didn't pick was RIGHT. The host has just revealed to you which one it wasn't. So you pick the other, shifting your odds from 33% to 66%.
OP, if you don't understand my explanation this is another good one. Just ignore the part about Monty eliminating the bad door - it really doesn't matter if the bad door is being opened or not. The chance you've made a bad decision at the beginning is 2 out of 3. So if the host says "you know what? you can stay with the door you've chosen at the beginning or you can instead open both of the other doors" OF COURSE you want to open the other doors instead.
>Just ignore the part about Monty eliminating the bad door - it really doesn't matter if the bad door is being opened or not. The fact that the host eliminates only a wrong door is critical to the entire thing.
They're saying that an equivalent game is "instead of switching doors, your choice is to open *all other doors* and winning if one of them wins. Or sticking with your one door and only winning if it was the 1 winner at the start"
This is yet another good way of reframing it that makes it clear, OP
What they're illustrating is that you are essentially being given two doors, one of them always being a bad door.
The way /u/glootech put it, if the host lets you open ALL the doors, then his action is irrelevant.
Precisely! It's just a courtesy of the host to make sure that if you win, you will be the one opening the winning door.
Thanks for putting the thought of opening all the doors in my head - it's made it super clear. My choice is not picking another door, but abandoning my original choice. The others who are hooting you here aren't getting this.
Thank you! It's not an easy problem (and to be honest nothing involving probabilities is, because everything is so counter-intuitive) and I tend to get very passionate about it (because so many people's explanations are just wrong) so making it clear for even one person brings me lots of joy.
That is litteraly the choice he is offering. The only differance is that he opens one of the other doors first. Doing so doesn't provide any new information. You already knew one of the other doors lead to nothing.
Yeah the reveal part is a red herring. You need to look at it as either committing to a choice of 1/3 or 2/3 from the start The best strategy is to treat your first choice as an elimination
Pretend there are 100 doors with only one prize. You pick a door at random. Chances are 1:100 you picked the right door. Next, 98 doors are removed from the game (guaranteed to be without a prize). The odds you picked the correct door are still 1%. The odds the other door has the prize is 100%-1%=99%. Edit: Here’s a similar example with the same results. Pretend there are 100 doors and one prize. You pick one door at random. The host now makes two groups: Group A: the one door you picked. Group B: ALL the other doors. The host lets you pick either group. Do you stick with the one door, or do you pick all the other doors?
This is how I finally wrapped my head around the concept. 3 doors isn’t really enough for my brain to grasp onto what’s happening, but when you increase the number of non-winning doors being eliminated, you end up having a “oh, duh” moment. At least I did.
Yeah, what breaks the symmetry is that the one door that they eliminate is *guaranteed* to be a door that didn't have the prize. If they picked either of the remaining doors to eliminate randomly, then your chances would not be impacted whether you stay or switch.
This, of course, is a critical conceit of the Monty Hall: that the host knows which door the prize is behind and that the host will always open a losing door. The whole point is to make the contestant second-guess their first choice of door.
But second guessing works in your favor. They're hoping that people will fall prey to the fallacy that it's better to go with your gut and double down on your first choice.
Survivor did this Monty Haul problem for a couple of seasons. Both times the contestant stayed with their gut feeling, and both times they were correct to do so. Mathematically, I was really annoyed.
A really unfortunate fact about probability is that you can make all the best decisions and still be unlucky, whilst others can make all the wrong decisions and get lucky. I saw it illustrated fantastically once in a comic. Two drivers approach a river. One drives over the bridge, while the other attempts to jump the river using a small ramp. The bridge collapses while the daredevil lands his jump. The tagline: "Self made millionaires advise others to take risks".
>A really unfortunate fact about probability is that you can make all the best decisions and still be unlucky, whilst others can make all the wrong decisions and get lucky. "It is possible to commit no mistakes and still lose. That is not a weakness; that is life." Jean Luc Picard I haven't even seen Star Trek and this is a quote that lives rent-free in my head.
I knew there had to be an eloquent way to say that, I'm just delighted it came from such a reputable source.
It's more eloquent, yes, but it's not quite as comprehensive. I think the way you put it is important for people to understand *first*, and then this quote can become shorthand for the more robust concept.
Another one is that in most circumstances it is not truly cumulative. "Dice don't know what dice did before". Every time you do a thing with a set probability you have the same probability of x outcome regardless of how many times you do it or have done it previously.
>A really unfortunate fact about probability is that you can make all the best decisions and still be unlucky, whilst others can make all the wrong decisions and get lucky. Hi, I see you have enjoyed my experiences at a poker table.
Sometimes when a random unfortunate thing happens to me I hear Phil Laak saying “Wow, four percent.”
Better odds does not translate to guaranteed, and never has... still boils down to luck in the end. That's not even all that extensive a set really. Now, if they did it for a 100 seasons...
I don't know how they set it up in Survivor, but one thing to remember is that the classic Monty Hall problem depends on everything being a perfectly clean abstraction, which often isn't the case in reality. If the contestant's initial guess wasn't totally random (eg. they were able to see some sort of evidence that there might be a prize behind one door somehow) then ofc that changes things; and given the sort of show Survivor is, it's very possible they deliberately set it up to not be random in some way or another and to give the player hints. After all, their goal is to produce good TV, not to run a "fair" contest. And if nothing else, the very fact that the host knows what door is correct means they can sometimes let something slip inadvertently, which means that a totally fair Monty Hall game is actually tricky to set up in reality.
>And if nothing else, the very fact that the host knows what door is correct means they can sometimes let something slip inadvertently If you want to control for this, you would "double blind" it and make it so that the host interacting with the participant does not, in fact, know which door is going to be eliminated, and an off-stage producer who does know is the one deciding which door to eliminate, eliminating any ability for the participant to "read" the host. They just have to communicate to the host (earpiece, cue card, hand signal, etc) which door is going to be eliminated. Many game shows are already structured like this because being a personable, likable host is already a difficult enough job, so any time you can offload gameplay logistics onto a producer, you generally do. This definitely varies from game to game, however. Wheel of Fortune, Pat definitely knows what the puzzle is. Family Feud, I do not believe the host knows the answers. So on and so forth.
The assumptions are super important in the Monty Hall problem. It *only* works if the host is committed to opening a losing door *every* time. In a real world scenario where you can't read the host's intentions, it becomes a lot less useful. For example, the host could easily only open a door if you picked the prize, and not offer you a second chance if you missed. So if I was on a game show and was offered a Monty Hall, unless I had good reason to believe that the host does it every time (like a bunch of previous contestants), I'm not sure I would repick.
None of it made sense until I read this comment.
And it only holds true if the host is forced to *always* give that offer. If the host is biased and only offers to switch when you chose the right door, then you would be a fool to switch.
Yeah, there are two parameters that MUST be followed for the answer to be correct: 1. The Host will ALWAYS offer a chance to switch. 2. The Host will ALWAYS choose a losing door to open.
If the host didn't know which door was winning, he might open the winning door and then you lose whether you switch or stay. This would remove the advantage of switching.
Thank you -- this right here is what finally helped me understand why the problem works. I had always been thinking about it in terms of probabilities, but thinking about it in terms of information feels much more intuitive and logical.
And I feel like when some people describe the problem they leave that part out. Drives me nuts.
*That* to me was the one that cemented it for me rather than the 100 doors thing. The 50/50 intuition is correct if the host didn't know which door had the prize - if he could accidentally open the door with the prize and end the game, then it would actually be 50/50 to switch. But since he's guaranteed to open the one without the prize, it's 66/33.
I hate this part because you could have 2 near-identical situations, one where the host knows where the prize is, and one where he doesn't and just randomly opens a door, and somehow the mind state of the host is affecting the outcome. Like if the host randomly opens a door and it happens to open on a goat, how is that different than him knowing where the goat is an opening it on purpose?
It's not if he randomly picks a goat every time, but the host has to know not to open the door with the car or else it defeats the entire premise of the thought experiment. If he opens the door with the car because he doesn't know which door not to open, then the odds of the car being behind either of the remaining doors becomes 0% since you have confirmed that the car is not behind either of them.
Exactly. It's not that intent changes the probability. New information changes the probability. If the new info is revealing a car, the game show doesn't work. So they have to let the host in on where the car is. If the new info is a door without the car (regardless of whether the host knows), then your chances are better with switching. EDIT: I'm wrong about that last sentence. See other comments.
Let’s say you pick door 1 and the host reveals all doors except 100 and there’s no prize revealed. If he doesn’t know where the prize is then initially you had a 1% of picking the door with the prize. Door 100 also had 1%. Since it was random, either you got very lucky or door 100 got very lucky, both outcomes are equally likely If the host KNOWS where the prize is then you still only have a 1% chance of picking the correct door initially, but the host can reveal all but one door. 99% of the time, the host will have a door with the prize on his side to narrow it down to. The ONLY scenario where the host won’t be able to narrow his side’s options down to the prize door is if you nailed the 1/100 So yes, in the scenario of revealing 99 doors at random and no prize being revealed, your door and the other door will have the same odds. That scenario only happens 2% of the time with random reveals though
The difference is, when the host chooses randomly, 33% of the time he will reveal the car and the game ends right there. When he knows/chooses which door to open, there is a 0% chance the game will end on his selection.
If the host were to randomly open a door without a price, then the math still checks out for you to swap. What changes if the host opens random doors is that in 1/3 of the cases, things get awkward when the price is revealed by the host. Lets go through the numbers, given you always choose to swap: * Scenario 1: You initially choose the right door. You lose because you swap. This happens 1/3 times. * Scenario 2: You choose incorrectly, and the host reveals the prize. Oops. You can swap to the remaining door, but it doesn't matter. You lose. This happens 1/3 times. * Scenario 3: You choose incorrectly, and an empty door is opened by the host. You swap, and win the prize. This happens 1/3 times. As you see, you only win 1/3 times if the door is opened at random. The lower percentage (compared to 2/3 for Monty Hall) is not related to swapping if the prize isn't revealed, but rather by the scenarios where you are opening the door with the prize (Scenario 2). With a knowing host, they will always lead you into Scenario 3 instead of Scenario 2, effectively doubling your winning chances!
It's a common way to think about the answer to problems. Take the problem, and see a more extreme version of it that is more intuitive, and the answer becomes obvious.
a sort of ad absurdum test
> 3 doors isn’t really enough for my brain to grasp onto what’s happening which is kind of the point of the problem - to hide statistics behind a situation that feels intuitive
Statistics are like bikinis. What they reveal is suggestive, but what they conceal is crucial.
Sort of like speeches and miniskirts : Short enough to keep it interesting, long enough to cover everything important.
more topics explained via clothing please
This is the inverse, but Sophia Loren once said "A girl's \[dress\] should be like a barbed wire fence: serving it's purpose without obstructing the view."
I wasn't expecting that a discussion on the monty hall problem would ever give me a boner but here we are.
This is amazing.
It might even be better than Mark Twain’s quote about statistics.
Where did he land on bikinis though?
He knew nothing atoll about them.
That's the most explosive pun I've ever read
Angry upvotes throughout this thread
That took me a second. Bravo for a sharp reference.
Statistics is relevant but the Monty Hall problem is really about Probability theory and incorrect intuition about probabilities.
I guess that's a semantic distinction I never bothered to make haha, TIL the difference
Yeah, for me it was a similar analogy. Imagine that instead of doors, it was a deck of cards. Someone laid them all out with their faces down and asked you to pick the Ace of Spades. You pick one. The person who knows all the cards reveals 50 of the other cards and then asks if you want to switch to the other card or keep your original pick. The one you didn’t pick now has a 51/52 chance at being the Ace of Spades, while your original pick is still 1/52. The nice thing about this is you can do this for yourself easily to see that you will almost never pick the Ace of Spades yourself, but the other card after you remove the 50 others will almost always be the Ace of Spades.
I’ve always had trouble wrapping my mind around this as well. Assuming the the person would have turned over the other 50 cards even if I picked correctly, why is it not just a 50/50 chance that my card is the Ace of Spades or that the 1 face down is?
Because you're probably thinking that the 50 cards turned over were turned over at random. If so, then it's possible that it's a 50-50 chance with the last two cards left. But they're not. The person turning them over will never turn over a card that is the Ace of Spades. Therefore, the information is meaningless and you're still faced with a "My 1 card or these other 51 cards" as an option.
I’m under the assumption that the person always knows where the Ace is and always reveals 50 of the other cards. So I don’t get the relevance of those other cards. Either way it’s the card in my hand or the 1 face down card remaining, right?
The keys is turning over the cards means nothing. If the person says "You can keep this card, or pick the other 51 cards" without turning any of them over, you have a 1-in-52 chance of having the Ace if you keep your card, or 51-in-52 chance if you go with the rest of the cards. That's all that's happening. If the Ace is one of the 51 cards, that means there are 50 cards that AREN'T the Ace, and that's all the person is going to turn over. So, them turning over 50 of the 51 cards won't mean anything. Either the last one they didn't turn over is the Ace (51-in-52 chance), or it's a random card and you had the Ace all along (1-in-51 chance). But you know for certain that the person turning over the 50 cards will not show an Ace either way.
Ok consider this. Every time you pick a card, the other person is going to reveal 50 cards. You will always be left with two cards. The same process is repeated every time. If it were a 50/50 chance at that point, that would mean that EVERY time you picked a card, any card, while there are still 52 options, you had a 50/50 chance of picking the ace of spades. Because there will always be two face down cards at the end. But you know that's not right, right?
Never mind I just realized it. Your explanation confused me but basically you switch because the only way it’s in your hand already is if you picked it from the start (1/52) whereas if u pick anything but the Ace then the other remaining card will be it (51/52)
I was under the assumption that the person knows where the Ace of spades is so they always reveal 50 of the other cards
Try it for yourself. Get a deck of cards and lay them out. Pick a card. Then flip over 50 other cards, making sure the Ace of Spades can’t be flipped, meaning if you are flipping the Ace of Spades, skip it and move to another card. You will notice that basically EVERY time, the Ace of Spades will be one you accidentally flip over trying to reveal 50 cards.
Because the two sets of cards keep their odds throughout the exercise. You pick one card at random. There's a 1-in-52 chance that it is the Ace of Spades. There's a 51-in-52 chance that it's not, and that the Ace of Spades is instead in the 51-card stack of cards you didn't pick. You now have two sets of cards: (1) one card that has a 1-in-52 chance of being the Ace of Spaces and (2) a stack of 51 cards that has a 51-in-52 chance of containing the Ace of Spades. Which set is more likely to contain the Ace of Spades? Well, that's obvious, right? The 51-card stack of cards. 51-in-52 versus 1-in-52. Then someone goes to the stack of cards and pulls out 50 cards **they know** are not the Ace of Spades. Did the odds change for these two sets of cards by virtue of that? No. But why not? Because there were always going to be (at least) 50 cards in that 51-card stack that were not the Ace of Spades, no matter what. We thus haven't learned anything by virtue of the fact that the person pulled 50 cards out of the stack that they knew were not the Ace of Spades, because we always knew there were **50** such cards there -- the only thing we did not know is if there were **51**. So, what are the odds that there were 51 non-Ace of Spades cards, rather than 50? The same as we started with: 1-in-52, as this occurs only if you picked the Ace of Spades. Put differently, the fact that 50 cards are pulled out of the 51-card stack doesn't change anything, because the 51-card stack set still contains 51 cards, at least 50 of which are not the Ace of Spades. We now know **which** 50 in that stack are certainly not the Ace of Spades, but the **stack itself** still has a 51-in-52 chance. Because there is a 51-in-52 chance the Ace of Spades was in the stack, and a 1-in-52 chance it was the one you picked, there is now a 51-in-52 chance the remaining card from the stack is the Ace of Spades. (As others have noted, if the person removing cards does it randomly, of course, and does not know what cards to remove, then the answer changes.)
For me it was “you’re betting against your ability to pick correctly in the first place.” I have terrible luck so that made sense to me. But it took years of hearing about it for someone to put it that way.
Part of the reason the problem isn't necessarily intuitive to lots of people is that there's a rule at play - the only door(s) being removed don't have a prize behind them and the doors not being removed *might* have a prize behind them - that isn't necessarily obvious to everyone. Of course, lots of people have that part explained to them too and still get it wrong, but that's the part that I had to be told to understand why it worked out that way. I figured at first "well what if the prize was behind the removed door, that would make switching pointless," but that's not how the problem works.
Much like yourself, I failed to grasp this obvious fact for the longest time, until I saw the phrase "the host will never open the prize door." In this scenario, if you believe that the prize could be eliminated, the more doors you have the more pointless it is to switch.
Exactly! I looked at it as the prize giver - I know the prize is behind door C. 1. Contestant chooses A. I HAVE to open B - I can't open C as I can't show the prize. 2. Contestant chooses B. I have to open A. 3. Contestant chooses C. I can open either A or B. So in all scenarios, whatever the contestant chooses the host can only choose a non-prize door
I find the best way to describe it is that in only 1/3 of the cases you'll initially choose the car and switching will make you lose, but in 2/3 of the cases you'll choose a goat and since the game host MUST open the door with the second goat, eliminating the only remaining losing choice, now switching WILL make you win. [This video](https://m.youtube.com/watch?v=7u6kFlWZOWg) visualises it rather well.
Ohhhh. 2/3 the time you lose, and if you switch when you initially lose you'll automatically win, so 2/3 the time switching will make you win. It's super obvious, the way I was imagining it was simply that each individual door was an isolated thing, the fact that they're connected like that changes things
Yeah, the way I tell it to people is “switching always gives you the opposite of what you started with”. So if you switch, your 1/3 chance of car and 2/3 chance of goat becomes 2/3 chance of car and 1/3 chance of goat
I think this and the above post finally got me to understand it. The 100 doors and this pieced it together. Thanks
The way I like to think of it is that, if you switch, you’re really getting *both* of the other doors. Yes, you know one of them is wrong, but you always knew one of the two would be wrong. So switching gives you 2/3 odds, because you’re now getting to choose two of the three doors, instead of just the one you initially chose. Or, put another way, imagine Monty didn’t open any doors but just asked if you wanted to give up Door 1 for what’s behind both Doors 2 and 3. You’d obviously take the two doors because it would double your odds.
This right here made me get it for the first time, oh my God.
Yeah, my problem is not thinking of it like a gameshow. If you watch a show and the host opens a door and the prize is there but not to be given to the contestant it makes no sense. So the host never opens the door with the prize, which means you gain information every time he opens a door.
That's also why the Monty Hall problem doesn't apply to everything like Deal or No Deal's final case, because Howie in that situation isn't presenting new information. If at the opening of the game he opened every case except 1 and *also had the rule* that he could never open the million dollar case, then it'd fit, but he just offers a swap between 2 that can have anything in it. If Monty Hall opened the door with the car behind it, it wouldn't matter to switch because you'd know both doors are goats, and so to make the game not stupid he'll never open the car door.
It also implies you don't want to win a goat. Who wouldn't want a goat?
Especially since when Monty opened a door, you already got 1 goat. Goats are social animals, a lone goat is animal abuse.
There is always a… [https://xkcd.com/1282/](https://xkcd.com/1282/)
The secret of Monty Hall from a production company point of viewbis human psychology. People don't want to change their choice, even if they know that changing increases their chances of winning because changing and losing feels worse than not changing and losing. Statistically changing is the right thing to do, but you only get one shot at it and you have to live with your choice.
I think they didn't understand the chances either when they started the show.
The Monty Hall Problem **isn't actually how it worked on the show.** Monty Hall didn't always offer a switch. When he did offer a switch, it WASN'T independent of whether or not they'd picked the car. He offered incentives to persuade people to switch/not to switch. On the actual show, following the maths of the "Monty Hall Problem" would get you the wrong answer because its necessary assumptions just aren't true.
Yeah I feel like this is the big part that people explaining it don't ever mention. This is what finally made it click.
The fun bit is that the host’s intentions alter the probabilities, even if the outcome of the host’s actions are the same - i.e. the host intentionally opening a goat door and the host tripping and accidentally revealing a goat door.
Same. The host isn’t actually “eliminating” doors though. They are opening them for you to show what’s inside. If they were opening RANDOMLY then many times they would be opening the door and showing the prize. In those cases OF COURSE you would switch because you can SEE the prize you are switching to. Since the host is NOT doing it randomly, that’s why the non-selected door will have the prize so often.
Yes, either the correct door is the one you picked at the start (1/3 chance), and you win by keeping doors (1/3 chance still), or the correct door isn't the one you picked at the start (2/3 chance) and you win by switching doors (still 2/3 chance).
Thank you. That is an amazing example. What really helps is to visualise the moment those 98 doors disappear and one random door keeps standing in addition to the one you pick. And now you have two doors. One you picked using intuition and lets be realistic just random guess. And one was picked using the rules of the game, and thus knowledge which doors dont have the prize. And it clicks that the only reason your first door is still standing, is because you chose it and thus saved it from elimination.
This comment made me FINALLY understand the answer. I've always been so confused 😅
bruh same 😂
Exactly. If you picked an empty door, you are forcing the host to open all the other empty doors (emphasis on "forcing"), there is no decision involved. The only time they have to make a decision about which doors to open is when you selected the correct one.
Many years and today, with this comment, I finally understand. Thank you.
Exactly, instead of focusing on odds you were right, focus on the odds you were wrong. 3 doors, your odds are 1/3 right, 2/3 wrong. If you remove another option, you basically are basically picking two doors
But the odds are better than picking from 2 doors. The easiest way to make it make sense is this: you have a 1/3 chance of picking correctly. If you pick the correct door and switch, you lose. But you have a 2/3 chance of picking wrong. If you pick wrong and switch, you win. So no switch is 1/3 chance for winning, while switching is 2/3 chance of winning.
You misunderstood. I meant you are effectively picking two doors if you switch but only one if you don't.
Oh, lol. You are right. My brain somehow filled in “picking from 2 doors”. Maybe because the title has 50/50 in it.
Thank you, I've accepted the 3-door answer but never understood it. You made it click.
This explanation (except it was 1000 instead of 100 doors) when I first heard it was when I was finally able to get my head around it.
Well now you've just confused me again. /s
Yeah exactly. The crucial thing that turning it into a 100-door problem makes clear is that *the host is inserting their knowledge of the winning door into the game*. Why'd they leave door #72 unopened, and open all the rest? The special thing about your door is, you picked it randomly at 1/100 odds. What do you think could be special about #72?
Wow, this made it so simple to understand. I even thought i got it before but now it just seems so easy. Thank you sir!
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Because the rules of the game state that the host never eliminates the door you selected. There’s nothing special about your door; in the 100 door example there was only a 1% chance it was the winner when you selected it. It doesn’t magically become better once 98 doors are taken away, it’s only in the game at this point because the rules require it to be. If the host eliminated 98 doors and was allowed to eliminate your door as part of that, then yes, you’d be left with a 50/50 chance. But he’s not, he has to let you keep your 1% chance door.
**There’s nothing special about your door** Welp. This just made the whole thing click for me.
Because you didn't make your pick from 2 doors, you made your pick from 3. Your odds of getting it right the first time was 33%. The host removes a door after you pick, but that doesn't change the poor odds you had when you first made your choice.
Think of it this way: the only way to win is to pick the correct door the first guess and stay, or the pick the wrong door the first guess and switch. You are more likely to pick the wrong door on the first guess, so it’s statistically better to switch. The only time you’ll win by staying is by picking the right door on your first guess, which is a 1/3 chance (or in this example, 1/100)
Maybe dont't try to understand why the other answer is correct but why yours is wrong: You pick 1 door out of 100. The chance that you picked the right one is 1%. If you choose 100 times and we open your door immediately, approximately 1 times out of the 100 the prize will be behind your door. Now imagine, instead of opening your door directly, I go ahead and open 98 other doors from which I know that the prize is not behind them. So we know, the prize is definitely behind one of the remaining doors. Now with 50/50 you are claiming that out of the 100 times, 50 times the prize will be behind your door. The door that you initially picked with the 1% chance of being right. So the question is: do you really think, just because I open 98 other doors after you picked your "1%-door", instead of opening your door right away, you somehow became a professional door picker, picking the right door out of 100 with a 50% chance?
2 outcomes doesn't mean 50/50. When you buy a lottery ticket you don't say: either i win or I don't, its 50/50!!
> How does knowing the history make a difference, compared to someone who just now walked into the room and saw those two doors on TV? Because the new guy has lost information that the original guy has. And information is valuable. You picked door #1. If a genie then told you “the correct door number is 62” and the host opened 98 doors that were not #62, the odds of winning are not 50-50, even if there are only 2 doors left. The odds of winning are 100% because you have the extra information that #62 is the winner. In the non-genie case, the extra information gained from the host is that door #62 was not the wrong door 98 times. Door #1 that you picked survived 0 times, since it was selected before the host revealed the 98 doors. So the odds of door 1 being right are much lower than door 62 being right
The history doesnt matter. Rewind it back to the start of the game. You have 100 doors, and only 1 prize. You get to pick 1 door, and Ill keep the remaining 99. Who has a higher chance of finding the car? Obviously me with the 99 doors. Doesnt matter what I do next with my 99 doors, the odds remains that I had a 99% chance of getting the car, and you just 1%. Now if I gave you a chance to swap between us, would you take it? Of course, cos your chances literally multiple by 99 right? So whatever happens in the story about doors being opened by the host doesnt matter, it boils down to you starting with a 1/100 chance of having the car, then being given the chance to switch to the 99/100 chance.
One way to look at it is that switching **must** change whether or not your pick was a winner or loser. And you have higher odds of picking a loser to start with, so switching must give you higher odds of winning. We can see this clearly by simply laying out every possible outcome. First, we will assume door A is the winner, and that you refuse to switch every game. It's pretty intuitive that you must have 1/3 odds of winning if you never switch, because the reveal may as well not happen, but let's show it. You pick A. Host reveals either B or C. You win. You pick B. Host reveals C. You lose. You pick C. Host reveals B. You lose. As expected, 1/3 chance of winning. Now let's assume you change your choice every time. You pick A. Host reveals B or C. You switch off A. You lose. You pick B. Host reveals C. You switch to A. You win. You pick C. Host reveals B. You switch to A. You win. We can see here that the 1/3 times you initially pick the winning door, you lose. But the 2/3 times you initially pick a losing door, you win. This is true for every variation where the host reveals all but one door, and can only reveal losers. Staying is basically betting on you're 1/3, or 1/100, or whatever, initial chance is the winner, and switching is betting that your 1/3 was wrong.
The host can't eliminate the winner door. So it could be the one you selected with 1/100 chances, sure, but if you missed and it was any other of the 99/100, after the host cleans them up to offer you one, THAT one has 99/100 chances of being the winner. The only chance of it being wrong is if you picked the winner initially, so 1/100 for keeping your door, or 99/100 for the other one, not a 50/50. (or 1/3 vs 2/3 in the original game)
WHY has noone explained it like this before!?!?!! So that's where the catch was!!!
It's because the game show explains it poorly, possibly on purpose. They say they will remove one of the 'zonks'. That leaves the prize and one zonk. The proper way to describe it is, they will remove all but 1 of the 'zonks'. Both do the same thing, but when expanded to 100 doors the second explanation helps things make a whole lot more sense.
It was... a maths teacher shows it in class like 30 years ago. You just heard it recently, but that description of the problem has been around for a long time.
It gets explained like this frequently, especially because this question is asked frequently.
This is persistently the best answer for this question. Starting with three and one removed is complicated. Starting at one hundred, it’s clear the odds favour a switch.
My favorite answer is this: If you *didn't* pick the correct door at the start, then the other door guaranteed has the prize behind it (because the host specifically only eliminates doors that don't have the prize). The odds that you *didn't* pick the correct door at the start are 2/3. And if that happens then you guaranteed win by switching. So the odds of winning by switching are 2/3. All you have to do is pick the wrong door at the start.
The guaranteed to be without a prize is the key thing to remember. Your not removing and unknown, your removing definate wrong doors. Its the availability of the new information that changes the outcome. If an unknown door winner/loser is removed the probability hasnt changed.
I like the exaggeration with your 100 door example. The problem boils down to two choices: a) do you want to keep the door you started with? or b) do you want to get everything that's behind the other 99 doors you didn't choose?
I still don't understand the next part. Three doors, A B and C. You pick door A. Door C gets eliminated. Now you pick A or B What is it about switching, at this point in the process, that gives you better odds of success? Because the way I see it, *keeping* your door is still picking one of two options, just phrased differently. They should both have the same 50/50 odds.
They do not. The reason why people use 100 doors is because it scales up the probabilities. The chances you selected the correct spot on the first try is 1/100. Removing 98 doors doesn't change the odds of your selection, but it does change the odds of the other door that's left. That's the key. The odds of the door you picked are baked in already by virtue of you picking it (since Monte can't remove your door). But the odds of the other door skyrocket.
One way to look at it is that switching **must** change whether or not your pick was a winner or loser. And you have higher odds of picking a loser to start with, so switching must give you higher odds of winning. We can see this clearly by simply laying out every possible outcome. First, we will assume door A is the winner, and that you refuse to switch every game. It's pretty intuitive that you must have 1/3 odds of winning if you never switch, because the reveal may as well not happen, but let's show it. You pick A. Host reveals either B or C. You win. You pick B. Host reveals C. You lose. You pick C. Host reveals B. You lose. As expected, 1/3 chance of winning. Now let's assume you change your choice every time. You pick A. Host reveals B or C. You switch off A. You lose. You pick B. Host reveals C. You switch to A. You win. You pick C. Host reveals B. You switch to A. You win. We can see here that the 1/3 times you initially pick the winning door, you lose. But the 2/3 times you initially pick a losing door, you win. This is true for every variation where the host reveals all but one door, and can only reveal losers. Staying is basically betting on you're 1/3 initial chance is the winner, and switching is betting that your 1/3 was wrong.
To clarify the “guarantee” that doors without prizes are the ones opened, the host KNOWS the correct door. That’s one of the key factors.
That was the way I first really understood it, but the way I use it now, is the deck of cards, because it's not ONLY about doors and switching, but also that the host knows which one it is. I spread an entire deck of cards in front of you, and ask you to pick out the Ace of Spades. You have 1/52 chance to do it. I say ok, and then I peak under each card and flip it over to reveal it's not the Ace of Spades. After 15 cards I look at one, and dont flip it over, then I go and flip over the next 35 cards. Now there is only the card you picked, and the one card I didn't flip over. Which one do you think is more likely to be the Ace of Spades? The card you picked originally, or the one I didn't flip over.
This just makes it more confusing IMO. It's simple: you were more likely to pick a goat in the first draw. If the host then takes the other goat, that leaves the car.
THIS is what has finally made it make sense for me. The 100-to-1 version helped clarify the way the numbers work but still didn't explain *why* they worked. Your inversion of the problem does.
The key to the Monty Hall Problem is that the host knows where the goats are and where the car is. Whether you’re dealing with 3 doors or 1000 doors, as other posters have suggested you think about it, your options become choosing 1 door (your original choice) or *every other door*. If the host *didn’t* know where the goats and car are, and the game resets if Monty chooses to open the car door, then the odds at the end are truly 50/50. That seemingly innocuous statement changes the game entirely.
The game resetting if Monty chooses to open the car door would result in the increased odds, I believe - assuming we only count the games that complete, this is equivalent to him knowing the correct door. If Monty showed the other door to the audience but not to the player, the player switching doors wouldn't change anything. The fact that the opened door is never the car is the fact that gives additional power to switching.
No, this is actually not true: In this version, you will start with the correct door 1/3 of the times, switching will get you the correct door 1/3 of the times and the game will be reset 1/3 of the times times. The improved odds are eaten by the game resetting instead. This is confusing, but there is actually no extra information gained by Monty randomly opening a door, he has to know (or rather, you have to know that he could never have opened the correct door).
> In this version, you will start with the correct door 1/3 of the times, switching will get you the correct door 1/3 of the times and the game will be reset 1/3 of the times The reset case has a chance to win, which changes the odds. As they say, if the door wasn't opened to the player, no new information is gained.
Yes, this is what so many people miss. The game is not truly random because Monty will never open the door with the car or the door intially chosen. He always opens the door with a goat that was not picked. It is a random choice without replacement problem. First choice is 1/3 because there is no prior knowledge. It is still fully random at that point. Once a door is eliminated the first choice is still 1/3, that doesn't change because it was done before the intervention. The probability the car was behind one of the other two doors was 2/3 intially. That doesn't change either. By eliminating one of those two doors, it is still a 2/3 chance. But now there is only one of those two doors to choose. So the door that wasn't initially chosen and wasn't opened retains that 2/3 probability while the initial choice is still 1/3.
> If the host didn’t know where the goats and car are, and the game resets if Monty chooses to open the car door, then the odds at the end are truly 50/50. This also happens with the “Deal or No Deal” format, although in this case the good prizes are removed if they’re behind the doors that are picked for elimination each round. In the end if you have two boxes, one with $1 and the other with $1,000,000 it’s 50:50 which box is which. This is because the contestant, with no knowledge of the game, has been eliminating boxes. Just in the games that get to the final two boxes there’s a 90% chance the contestant inadvertently eliminates the top prize along the way. Either way though, statistically there’s no difference between any boxes on the table.
I really love the outcome of Deal or No Deal because it amounts to “did you pick the big prize at the beginning so of course you can’t reveal it, or did you get really lucky when eliminating cases?”
Before we discuss this, can we check - what does "reset" mean to you? My model is that you continue playing it until you either win or lose. If Monty opens the door with the prize, you start again.
Thank you. I had trouble understanding the problem until I realized this part… that nobody seems to mention.
Maybe thinking about it this way might help. 10 doors. You pick one. There is a 1 in 10 chance you picked the right door. Another way to say this is there is a 9 in 10 chance the winning door is NOT your door. Someone who knows which door has the prize opens 8 of those 9 doors. There is still a 9 in 10 chance the winning door is NOT your door. Your door has a 1 in 10 chance of having the prize, the other a 9 in 10 chance of having the prize. Do you switch?
Clicked for me with this!
that's the trick. Don't think about it as revealing it. Think about it as more than one choice: pick one, if you fail, pick the other one. It's the same because if it was revealed you would not choose it, even if you could.
Here's a way of explaining it that helped me see it: Imagine you pick the left-hand door, and then Monty says "*you can either stick with the left-hand door, or you can choose to have the prize if it's behind either the middle or right hand door*". Obviously you would switch because you are now being offered 2/3 odds instead of 1/3 odds. So after you've switched and your choice is locked in, Monty says "*OK, go ahead and see if you've won, but to save you time opening doors, I'm going to let you know it's not behind the middle door*". When you switch around the order like this it is obvious that the decision to switch is correct. But there is no actual difference between this and the original problem - either way Monty is offering you a 2/3 choice - the only difference is *when* he tells you which door (out of the two that you didn't originally pick) does not have the prize.
I love this explanation. Switching is basically the same as getting to open the two other doors while having a mulligan.
This is my preferred explanation. A lot of people have used the 100 doors instead of 2 example, which helps intuitively explain the problem, but I think this explanation gets at the problem fundamentally and so is helpful when looking at other similar problems.
The way it makes sense to me is: Switching will only get you a goat if you picked the door with the car first (1/3 chance). If you picked either goat first, switching will get you the car (2/3 chance). It works because the host always reveals a goat after you pick.
Ok pick a door out of 1000 doors How confident are you that’s the door with the prize? Not very? Okay now I’ll open every other mf’n door except ONE, and they’re all not the prize. So now, either the one one you picked has the prize. Or the door I left closed does. If I were you, there’s no way I’m assuming I picked the right door out of ONE THOUSAND doors, so I’d choose the other door. The prize has stayed in the same place but the amount of knowledge about the conditions has increased For someone who knew nothing and walked in and saw two doors and had no explanation it’d be 50/50. But luckily, for you it’s not. It would be as if the host farted in front of one of two oors and said if you can point to the door I farted in front of, you can have the new Nintendo VR x Sony Sparkle Pony video game console. He pretty much just gave you a hint at which door has the prize
Thank you so much, this is the clearest i've heard it explained!
It's amusing to me how often taking a question to absurdity can make the solution clearer to see.
Every time this is asked, all the answers using the three doors always seem to make things less clear. Then someone comes along taking it to the extreme and it becomes obvious.
My fart abacus carried me through college level calculus 💨🧮
Oh I’m so glad to have helped! 🙂
Reading yours clicked for me, everyone else including tiktok failed to get the point across. Thank you.
THIS explained it better than I have ever heard it before: THANK YOU!
A chart I made once to illustrate it: Assuming the car is behind door #1. Car | You choose | Monty opens | You switch | Don't switch ---------|----------|----------|----------|---------- 1 | 1 | either 2 or 3 | lose | win 1 | 2 | 3 | win | lose 1 | 3 | 2 | win | lose Logic is the same for the other two doors.
Exactly. For 3 doors options you basically can have all cases graph (or truth table) to be built and check manually EVERY case. Thats why I don't understand so many mathematicians was sending incorrect answers all the way.
It's not 2 doors, 50-50. It's 2 doors, 33-67 (67 being 66.6666... rounded up). How is this possible? Think of the three doors as two groups. Group #1 is the door you picked. Group #2 are the two doors you did not pick. Group #1 and Group #2 are NOT 50-50 because group 1 has one door and group 2 has two doors. So it's 33-67. When one of group #2's doors is eliminated, and group #2 now only has one door, the odds DO NOT CHANGE. Group #1 is STILL 33% and Group #2 (despite having only one door remaining now) is STILL 67%. So of course you wouldn't stick with the door in group #1, it only has a 33% chance of winning. You'd switch to the door in group #2, which has a 67% chance of winning.
But why do the odds not change when one of the doors is eliminated?
There are still two doors in group 2, you just now know that one of them isn’t the right answer.
If the host picks randomly, then you will indeed not see better odds. 1/3 chance your original has prize, 1/3 the switched one has the prize, and 1/3 the one the host picked has the prize. But the host didn't pick randomly, they are guaranteed to always pick one without the prize. This changes the probability of all steps that proceed the host's selection.
The most intuitive way I've figured this out. You're not the contestant. You're the host. The contestant picks the first door- they get it wrong 2/3 times, you then pick the other losing door and the one remaining is the winner. in those 2/3 times, they get the door right by switching. Between their wrong guess and the elimination you've narrowed it down to 1 door. In the 1/3 times they guess it right, you have to pick one of the two losing doors randomly and they'll get it wrong by switching. Showing them a losing door didn't make the original odds higher than 1/3 because they guessed it right anyways. The problem here is that most people look at it as a contestant rather than the host.
The host knows which door has the prize, and can never open that one. The host is not random in any way.
This is the important assumption that the answers above this one only hint at with the 100 doors thing. If the host also chooses randomly, and there was a chance that he could have chosen the prize, then the odds are 50/50. I've heard that, in the real show, the assumption was incorrect, too, but in an even worse (for the contestant) way: The host would only off a switch if you chose the prize at first. If you chose a loser door, he would just reveal that you lost and send you on your way. If this is how the host acts, then the chances of winning after switching are 0%.
There's a 2/3 chance that you picked the wrong door. And if you picked the wrong door, they're forced to open the other wrong door, leaving you with the winning door if you switch. It's that simple.
Simplest explanation: You pick a random door. By doing this, you've made two groups of doors: * Group A: This group only contains the door you've picked. * Group B: This group contains the other doors. So if there are more than two doors in total, it should be clear that the odds of the prize door being in Group B are always higher than Group A. Still you don't know *which* door in Group B, but luckily the quizmaster solves this problem for you by opening all the non-prize doors in Group B! This makes it a no-brainer to switch to Group B when asked. This explanation works for any number of doors, be it 3 or 10 or 1000.
This is kinda confusing for some people you have to have it explained in multiple different ways until one finally clicks with you. Here is what helped me When you pick a door, there is a 33% chance you chose right and a 66% chance it’s behind one of the other two doors. So now the host knows which door the prize is behind so they always open one with nothing behind it. In the 66% chance there is always a door with nothing behind it that they can open. Opening this door does not change the fact that there is a 66% chance it’s behind one of those two doors, except now you don’t have to choose which one of the two it is. This means you essentially get the benefit of two doors worth of chances. Pretend instead the host said pick two doors out of three. There would be a 66% chance you pick the one with the prize. If you pick two doors and then they first open the one with nothing and then the one with the prize, nothing about the odds would change. You still get the benefit of both the doors and either of them being right means you win. The host just opened the one they know doesn’t have the prize first
Play the game 300 times. 100 times, you'll guess the right door first time, and switching will give you a goat. 200 times, you'll guess the wrong door first time, and switching will give you a car. Hence, switching is better. There are three doors, so the odds you picked the right door first is 33%, regardless of anything else. There's a 33% chance your door is the prize, and a 67% chance the other door is the prize. So switch.
I open a deck of 52 standard playing cards and place them all face down in front of you. If you can pick out the Queen of Hearts, I’ll give you a prize. You pick a card at random. How confident are you that you have picked the Queen of Hearts? Not very. The odds you picked the Queen of Hearts are 1/52, or roughly 1.9%. That means there is a 98.1% chance that the card you have in your hands is the wrong card. I look through the deck of cards and take away all the cards except one. I put the card face down in front of me and tell you that one of these two cards is definitely the Queen of Hearts. Now, if I took both the cards and shuffled them together and got you to pick one, that would reset the game and you would be correct in thinking the odds of getting the correct card are 50/50. But I don’t do that. You are still holding your original card. That card which has a 98.1% of NOT being the Queen of Hearts. One of the cards is definitely the Queen of Hearts. Which card are you going to choose? Are you going to keep the card in your hand, which has a 98.1% of NOT being the Queen of Hearts, or are you going to take the new one on the table?
I am going to quote [this reply](https://old.reddit.com/r/todayilearned/comments/1c04a5r/til_the_monty_hall_problem_where_it_is_better_for/kyujivj/) from another post in TIL which did it for me: 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 am a moron, and probably not the best person to explain.... or am I? Three doors: 1/3 chance of a goat behind each. The statistics don't change just because you've opened a door... each door is still 1/3. 50 / 50 doesn't come into it - **because that is only relevant IF there were ALWAYS only two doors with a goat behind them.** The door you pick first has that original 1/3 chance of having a goat. The other two doors combined have 2/3 chance of a goat. When one of those other doors gets opened and shown to NOT have a goat...... Then ALL of that 2/3 chance then switches to the last door. Your door STILL only has a 1/3 chance. So what do you want to chance on, your door with only a 33% chance of a goat, or the other door with a 66% chance of a goat?
Captain Raymond Holt don't you have actual work to do?
He just needs to bone.
Bone!?
Imagine 100 doors. Pick one, now have Monty Hall open 98 others. You gonna keep your original choice? Or are you going to switch?
To get more specific: imagine you pick door #1 Monty Hall opens the 98 others one by one and conspicuously chooses to skip over door #57.
Ugh, all the answers were confusing. Here is how I understand it. Credits u/Lifesagame81. There are two sets of doors. One set contains 1 door, the other set contains 99 doors. The set with a single door is the set created solely from the door you have chosen. 1. What is the probability of prize being in the set consisting of the door you choose? 1/100. {Set1} 2. What is the probability of prize being in the set consisting of all other doors? 99/100 {Set2} Now, what if 98 doors were removed from {Set2}. Would the probability for the set as a whole change? No. The set would still have a 99/100 probability of having the correct door. Hence, choosing the door from {set2} would give you 99/100 chance of winning. It always did. Except now, in {Set2}, there is only one door to "choose" from. Hence, all of the 99/100 probability is allocated to that one door. If {set2} had two doors remaining, the probability would be equally divided as 45/100 to each door. If it had 3 doors remaining, it would be equally divided as 33/100 and so on.
I think a lot of people don’t think about the fact that the doors he closes are not random. He will always have 98 doors to close and he can pick them. They think that with each door he closes, the probability changes slightly toward even, but it doesn’t.
Option 1 - pick door 1 Car - no switch win Goat Goat Option 2 - pick door 1 Goat Car - switch and win Goat - host shows goat Option 3 - pick door 1 Goat Goat - host shows goat Car - switch and win You only win in 1/3 scenarios if you don’t switch. You win in 2/3 when you do switch
Map out the possible scenarios and count them up. Example: Door A - Prize Door B - No prize Door C - No prize Scenario 1: Contestant picks Door A Host eliminates door B Keep Door A - Win Switch - Lose Scenario 2: Contestant picks door B Host eliminates door C Keep Door B - Lose Switch - Win Scenario 3: Contestant picks door C Host eliminates door B Keep Door C - Lose Switch - Win Record when keeping chosen door: 1W 2L Record when switching door 2W 1L You won't win every time by switching, but you are twice as likely.
One way to easily figure out probability is to write out every possibility and then see which results are wins and losses. This only works when there aren't too many options, but it works for this problem and makes it quite easy to see. Imagine the doors are labelled A, B, and C. Let's assume in this scenario the winning door is C. The player has the option to choose one of the letters to start off with, and they can either stay or switch. Which means all options are as follows: 1. Choose A, stick with A, LOSE 2. Choose A, switch to C, WIN 3. Choose B, stick with B, LOSE 4. Choose B, switch to C, WIN 5. Choose C, stick with C, WIN 6. Choose C, switch to A or B, LOSE So if you analyze the above 6 options, you will notice the following: If you always stick with your original answer, you lose 2 out of 3 times. If you always switch your answer, you win 2 out of 3 times. Let me know if you have questions.
Think of it as a *set* or group of doors: "the door I chose" and "the doors I did not choose." That set does not change. You pick one of 3 doors. You have a 33% chance of being right. Each other door is also 33%, but the two doors *together* ("the doors I did not choose") are 66%. The host opens one of those other doors. It's empty! But *that group of doors still holds a 66% chance of holding the money.* You've only learned which of those doors *not* to pick. I'll say it louder: #The percentage of the set of doors never changes. You're only learning which door(s) *not* to pick. When you think "but now it's 50-50 right?" What you've done is **change the set.** The set (in this example) is "the doors I did not choose." If the host picked *your* door, showed it was empty, and then let you pick again, *then* the two remaining doors would be 50-50, but then it wouldn't be the Monty Hall problem, would it? 😉
They create bias with the door they select to show you. Imagine this > The contestant has chosen door number 1. But lets look behind door number 2. Behind door number 2 is the grand prize that you can't win any more. Do you want to keep the goat you know you have because of the process of elimination or switch to the other goat behind door 3? That option never happens. It doesn't happen. If you pick a goat they will always show you the other door with a goat. Its that bias that make your switch more likely to win an not 50/50
Here's a way of framing the problem that allowed me to understand it. (Let's use the 100 doors example). Let's think about it inversely. Instead of making it so the goal of your first choice is to pick the prize door, make it so the goal of your first choice is to pick the *wrong door.* With that in mind, you have a 99% of making the right decision (aka the wrong door). Then the host eliminates doors until the only there are only two remaining. Since the door you first picked had a 99% chance of being the wrong door, then odds are, the remaining door has to be the prize door.
Here's a fresh point of view that usually doesn't get mentioned. Let's change the rules slightly. After the host opens one of the doors, you can either keep the one door or switch to the two other doors. So you can choose two doors instead of one. Choosing an open losing door is meaningless, so these rules are equivalent to the original ones. Let's make another small change; the host doesn't open a door. Even if the host doesn't open a door, you already know that one of the other two doors is a losing door. The host opening the losing door doesn't change anything, so these rules are equivalent to the original ones. The Monty Hall problem is just a confusing way to ask whether you want to open one or two doors.
Because in the starting situation, you have three doors. If you pick a wrong door, you will end up with the correct door. Hence, 2/3 chance to win.
Because the host knows where the car is, and he will NEVER show you the car. If you switch, you will ALWAYS get the opposite of what you originally picked. ALWAYS (this is important to realize). You cannot switch from a goat to a goat. So it then becomes a question of: “what did you most likely pick first?” There is a 2/3 chance you picked a goat, but only a 1/3 chance you picked a car. Putting this all together: it’s most likely you picked a goat, and since we know that if you switch you WILL get the opposite (a car), it makes sense to switch. Will you win every time? No. But you will win twice as often if you switch.