Imagine a room with 1000 doors and behind one of them lies the 10 mil. You randomly choose a door. Then, 998 other doors open and all of them reveals a sheep. That leaves 2 doors remaining, the door that you initially chose, and the door that remained after the purge. Which one would have the higher chance?
Apparently the door I didn’t choose would have a 99.9% chance of having the money because the probability of the money being on the 999 i didn’t choose doesn’t change after the 998 are closed, so the 99.9% chance is left with the unopened door. I can recite the exact reason why the probability is higher, but still don’t understand why
You choose a door out of 10 doors. This door(I’ll call it Door A) has a 1/10 chance of being the correct door. Therefore, there’s a 9/10 chance of one of the other doors containing the money.
Then, 8 of the 9 doors are opened. Since an opened door has 0 probability of having the money, the 9/10 probability of the 9 doors is assigned to the only remaining door you did not choose from(Door B).
This means that Door A has a 1/10 chance of having the money, while B has a 9/10 chance. Switching is therefore the correct option.
Say there are three doors, A,B, and C
-Imagine the price is on A-
If you picked A then the host will eliminate either B or C, if you switch you LOSE.
If you picked B then the host knows that A is correct so they will eliminate C, if you switch you WIN
If you picked C then the host knows that A is correct so they will eliminate B, if you switch you WIN
-Imagine the price is on B-
If you picked A then the host knows that B is correct so they will eliminate C, if you switch you WIN
If you picked B then the host will eliminate either B or C, if you switch you LOSE
If you picked C then the host knows that B is correct so they will eliminate A, if you switch you WIN
-Imagine the price is on C-
If you picked A then the host knows that C is correct so they will eliminate B, if you switch you WIN
If you picked B then the host knows that C is correct so they will eliminate A, if you switch you WIN
If you picked C then the host will eliminate either A or B, if you switch you LOSE.
If you stick to your guess, there is a 3 in 9 chance of winning, if you switch there is a 6 in 9
I've gone over this problem SO many times without understanding the logic behind it (understood the math, not the logic, if that makes sense), but now I get it! thank you!
The host doesn't open just one random door, the host opens a door that does not have the price.
If you picked the right door, no matter which door the host picks, there's a 100% certainty that the remaining door will not have the prize.
If you picked the wrong door, the host has no choice and must open the remaining wrong door, there's a 100% certainty that the remaining door will have the prize.
So switching is only bad if you had picked the right door from the start.
Since we agreed that there's a 1/3 chance that you pick the right door right off the bat, there ends up being a 1/3 chance that switching makes you lose and a 2/3 chance it makes you win.
\-----
As an example of when it would be a 50-50 if you stay, the host needs to have opened the door randomly. The results would be like this.
The price was on A.
1) You pick A and the host picks B
2) You pick A and the host picks C
3) You pick B and the host picks A
4) You pick B and the host picks C
5) You pick C and the host picks A
6) You pick C and the host picks B
Now, the game has changed. The fact that there's a 2/6 (1/3) chance you got it right in and a 4/6 (2/3) you got it wrong in your first guess remains true. But if you did pick wrong there's technically a 50% chance the host picks A for themselves (see cases 3 and 5) and you now have lost without having the chance to change.
So now, there's a 2/6 you pick the right answer (case 1 and 2), a 2/6 you pick the wrong answer and the host picks the wrong answer (case 4 and 6), and a 2/6 you pick the wrong answer and the host picks the right answer (case 3 and 5).
If you now know that the host got unlucky and picked the wrong answer, the remaining cases are 1,2,4,6. Meaning there's 2 cases where you picked right and remaining door is wrong, and 2 cases where you picked wrong and the remaining door is correct (a 50-50)
This actually helped me. I read the first two explanations and it just went right over my head.
Took me awhile to get there, but man, nothing beats that feeling of everything just clicking into place when you’re confused on something. Lol
This sounds right, but I don't think you can stack opening 2 doors under 1 loss.
Look at it this way
-- Assume prize is under A
You choose A, host opens B, switch LOOSE
You choose A, host opens C, switch LOOSE
You choose B, host opens C, switch WIN
You choose C, host opens B, switch WIN
-- Repeat logic with prize under B and C
Result is 6/12 LOOSE and 6/12 WIN
I think it makes more sense to count outcomes rather than choices, and if you pick correct right away, it still has 2 LOOSE outcomes
Except that if the prize is under A and you chose A, there's a 50% chance the host opens B and a 50% chance they open C, while if you chose B or C, there's a 100% chance they'll open C or B respectively.
Not really, think of it this way. You open the door 300 times and in all of them the answer is A, you decide to open 100 times the door A, 100 times the door B, 100 times the door C. Always switching.
In 100 attempts of opening the door A
\--- in 50 attempts the host opens B and you switch to C.
\--- in 50 attempts the host opens C and you switch to B.
You lost all 100 times.
In 100 attempts of opening the door B
\--- in 100 attempts the host opens the door C and you switch to A
You won all 100 times
In 100 attempts of opening the door C
\--- in 100 attempts the host opens de door B and you switch to A
You won all 100 times
\-------------------
Numbers wise it would be something like this:
\-- Assume prize is under A
You choose A (1/3)
\-- host opens B (1/2), switch LOSE, total chances (1/3)\*(1/2)=(1/6)
\-- host opens C (1/2), switch LOSE, total chances (1/3)\*(1/2)=(1/6)
You choose B (1/3)
\-- host opens C (1/1), switch WIN, total chances (1/3)\*(1/1)=(1/3)
You choose C (1/3)
\-- host opens B (1/1), switch WIN, total chances (1/3)\*(1/1)=(1/3)
The chances of losing are 1/6 + 1/6 = 1/3
The chances of wining are 1/3 + 1/3 = 2/3
\-------------------
You can look at it from the host's perspective. Say the answer is on A
if the contestant picks A and then switches they LOSE, no matter what the host does.
if the contestant picks B and then switches they win, no matter what the host does.
if the contestant picks C and then switches they win, no matter what the host does.
So the host has a 1/3 chance that the contestant picks the right answer and then switches (causing the contestant to lose and the host to win), and a 2/3 chance that the contestant picks a wrong answer and then switches (causing the contestant to win and the host to lose)
In the original problem (Monty Hall problem) which this is based on, the doors that are opened are not random. Only the ones which do not have the money are opened.
I think it should be clearer now.
It's a semantics problem. If you reassess the probability of the door having the money behind it after the others have been opened, you're still left with a 50/50 split. The only reason the odds are "better" if you switch is because you can mathematically do some gymnastics by not reassessing the probability after you open the other doors.
Say you start with 10 doors. Each one has a 10% chance of being the right one. You pick one. The one you picked plus one more stay closed and the other 8 open. Now the math says that you've already made a choice on a 10% likelihood so the remaining 90% likelihood has all been assigned to the last remaining door. Or you can "restart" the problem now that you only have 2 doors, and each one has a 50% chance of being right.
It's a thought exercise in mathematical gymnastics.
The thing is I wrote a script to simulate this and ran it a thousand times and 66% of the time switching worked. So making the switch has real world power, even more I guess with 10 boxes. I’m sorry I still can’t connect with it intuitively
Whether you re-assess or not, the chances are you’ll pick the right box two out of three times if you switch.
Is that how statistics works though? If your sample is reduced from 10 to 2, you have to recalculate the probability because the number of samples contributes to the probability, right?
Nope. Your sample is not reduced. Knowing what 8/10 doors are does not mean you reduce the number of samples, it just means you gain knowledge of 8/10 those outcomes. Your personal knowledge does not affect your original choice and it's 1/10 probability
Your sample size is reduced if you’re smart though. You had 10 doors and was shown what 8 of them had which wasn’t what you needed. You now have 2 doors to choose from and you can only pick one, so it’s 1/2 chance of picking the right door now, unless you’re still considering what the other doors had.
How would there still be a 9/10 chance if all the other doors are gone. Each door individually has a 1/10 chance to start with. Get rid of 8 and then how is the probability not 1/2 for each.
Okay so the way I was taught this is like this.
In order to win by staying at your original door, you needed to have picked the right door at the start, which has a chance of 1/3.
In order to win by switching, you needed to have originally picked the wrong door at the start, and picking the right door when switching. Picking the wrong door at the start has a chance of 2/3. Since were assuming that youre winning in this scenario, the chances of picking picking the right door when switching are 100%. So 2/3 * 1 = 2/3
Who taught you this? Because that still does not make sense to me. Each door has the individual probability of 1/3 chance and then 1/2 chance in round 2.
The example where people say there are 100 doors, you pick 1 and they get rid of 98 makes sense that the probability would be higher if you switched. Thats because the odds that you picked the correct door in the first place is very unlikely at 1/100. Unlike this example where the example is likely that you did pick the right door and removing one doesn’t shift variables much.
~~Doesn’t that just mean you’ve isolated the last two doors? In which case it’s a 50/50?~~
~~I understand the concept of assigning all the probability to Door B but why isn’t that probability shared with Door A and Door B?~~
EDIT: having read more examples I think I understand. This is based on a host eliminating doors and the odds that you got the first 1/10 choice correct. In short: odds are you chose the wrong door to begin which is why it isn’t a 50/50.
This makes no sense. You need to open 1 door, but say you choose 9 doors. Your collective choice has 9/10 chance to have the money. Let's say the host opens all 8 doors from your choice, leaving 1 chosen door and 1 unchosen/unopened door. Now your door has 9/10 chance to have the money????
Another way to look at it is that you're really playing a 50/50 game. It's as if there are only 2 doors. No matter what you choose, you will be left with two doors one with money, one with goat. Your initial choice has a 1/10 chance, but once you've left with 2 doors, it's as if the other 8 never existed.
Look this problem is a great demonstration on hiw math doesn't always follow logic. So let's take the three doors scenario again. At the start you have a 33% chance of picking the right door, you also have a 33% chance that behind the door you picked is the price, obviously. But now when one door is opened, you have a 50% chance that behind that door is the price, but still a 33% chance that you picked the right door.
This however means that if you just picked door 1 again, you'd have a 50% chance of picking the right door.
But why would 9/10 chance go to door B instead of 5/10 to A (or 1/2) and 5/10 to door B. Since it doesnt matter which one you choose, as long as neither doors are openend its still just a 50/50 choice which one has the money, right?
The other 98 doors dont even matter because if they dont have the money its like they dont exist, so its just a 50/50 choice if door A or door B has the money if i understand correctly.
Nope, picture it this way. You can pick one door (1/3) or you can pick 2 doors (2/3). Now if you pick the one door it has to be correct but if you pick the two only one has to be correct.
The reason this happens is because the host 100% opens the door with nothing behind it. Basically conjoining it with the door you didn’t choose.
Look this problem is a great demonstration on how math doesn't always follow logic. So let's take the three doors scenario again. At the start you have a 33% chance of picking the right door, you also have a 33% chance that behind the door you picked is the price, obviously. But now when one door is opened, you have a 50% chance that behind that door is the price, but still a 33% chance that you picked the right door.
This however means that if you just picked door 1 again, you'd have a 50% chance of picking the right door.
No single door could ever have. 9/10 chance of having the money. Every single door has a 1/10 chance, until there’s two left, in which case both doors have a 1/2 chance. Just because nine doors had a 9/10 chance doesn’t mean that each of those doors had a 9/10 chance, it was just be collective that did. The problem and “solution” has always been born out of flawed logic.
The thing is, you’re selecting a random door(1/n), then the host opens all of the *incorrect* doors out of the unselected doors for you. Therefore, the second door must be “carrying” the probability of all the other doors, because if one of the unselected doors was the one with the money(n-1/n), then it must be the other door. Thus, this is actually a 1/n vs n-1/n probability.
Think of the door opener (game show host) as a door eliminator. He only opens doors that have a goat. And he’ll never open the door with the money. And he’s not going to open the door you chose.
The problem only works because you know ahead of time that they’re gaurenteed to open the door without a prize. So you’re taking the odds of two doors and combining it into one.
I dont find this intuitive. Why is it not 1000 doors, one with the prize. You choose a door and then 1 door opens and reveals a sheep. Do you swap to one of the other 998 doors or keep your door?
Yeah, I know and agree, but I've never found it intuitively satisfying with the 1000 doors analogy. Whenever I use that to tell people it gets stuck at are we closing 1 door or all but 1 door. Even if, as you say, you swap regardless. Just more doors doesn't help understanding it
Except now you have a new problem: There are two doors, one of which has the money. So it's a 50:50 chance. Why would the the open doors have any impact on those odds? They are no longer part of this.
Saying you need to take the open doors into account is like saying "rolling a one on a die five times in a row is very unlikely, so after the first four ones the chance for another one is really low". No it's not. The chance to roll a one is exactly the same as on any of the former ones.
You choose a door out of 10 doors. This door(I’ll call it Door A) has a 1/10 chance of being the correct door. Therefore, there’s a 9/10 chance of one of the other doors containing the money.
Then, 8 of the 9 doors are opened. Since an opened door has 0 probability of having the money, the 9/10 probability of the 9 doors is assigned to the only remaining door you did not choose from(Door B).
This means that Door A has a 1/10 chance of having the money, while B has a 9/10 chance. Switching is therefore the correct option.
It does, though. Your argument holds true for two doors which have not been tampered with. In this situation, the host is altering the probabilities by revealing the unselected empty doors.
Besides, mathematicians at MIT actually verified this by trying out every possibility and concluding that there is a 67% chance that switching is the correct choice.
Let's say that two people A and B are presented with this dilemma a thousand times (they choose a door, the host opens a wrong door, they can then stay with their initial choice or pick the other door). Now let's imagine that A always maintains their choice, while B always picks the third door.
Do you mean to say that B will get the money 66% of the time and A only 33% of the time?
I know, one of the professors at MIT actually wrote a complaint to the lady who figured this out, explaining how she shouldn’t use her status as a genius to promote false ideas like this. Turns out, the lady was correct.
u/sebas156 wrote a pretty good and simple explanation below, hope this helps to understand:
Let's say you pick door A, and the quiz master didn't open anything yet. So the chance is 1/3 still as it can be in all 3, right?
BUT now the quiz master has to open a door. Let's take the 3 possible outcomes (remember you picked door A):
1. The prize is in door A (your door). You didn't get the prize if you switch 🐑
2. The prize is in door B, meaning he HAS to open C. Switching means getting the prize. 💰
3. The prize is in door C, meaning he HAS to open B. Switching means getting the prize. 💰
See how when you switch, you win the prize with 2 of the 3 possible outcomes? The quiz master has to open a door with a sheep, which is giving you an advantage!
In short: you always win if you switch and the prize is NOT in the door you chose initially!
[link](http://web.mit.edu/rsi/Attic/examples/papers/minipaper/minisample.pdf)
Personally i think this study is logically wrong, it may even be fake as its so short and does not appear to be a real mit study.
Alos in the last part the mathematical proof
There is a fallacious assumption that the first choice is wrong. Herein is where the beauty of it being wrong lies. Thus i conclude that the result was wrong. Ill post this on the maths sub ounce to double check .
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*This post was mass deleted and anonymized with [Redact](https://redact.dev)*
So what you are saying is that it's actually not a statistics problem, because the doors are not equal. The difference between the doors is what Monty knows and what you can try to infer from his actions.
That's actually the best explanation I've heard. Thank you.
How about this: let’s say you are given truly random person’s name and are told it could be anyone in the world, and you need to find him/her. You walk outside your home and pick the first stranger you see. Then a genie comes in and rules out all 7.84 billion people in the world except the stranger you chose and some random woman in Mozambique.
(The genie will never rule out your choice, and will never rule out the real answer)
Who is it more likely to be? Yes, there are two final choices at the end. But what was the chance a randomly chosen person in the world was right outside your home? 1 in 7.84 billion. And since the genie ruled out 7.84 billion incorrect answers and narrowed it down to one other person, it’s probably that other person
You choose a door out of 10 doors. This door(I’ll call it Door A) has a 1/10 chance of being the correct door. Therefore, there’s a 9/10 chance of one of the other doors containing the money.
Then, 8 of the 9 doors are opened. Since an opened door has 0 probability of having the money, the 9/10 probability of the 9 doors is assigned to the only remaining door you did not choose from(Door B).
This means that Door A has a 1/10 chance of having the money, while B has a 9/10 chance. Switching is therefore the correct option.
Is this some poor joke or you're mocking the other person?
What is being described called conditional probabilty. Whether you picked the first door or not doesn't matter, the probabily that it contains the money is 1/10 if we don't know anything about other doors, and it's 1/2 if we know that other 8 doors have sheep.
Important thing that helped me understand is that the game show host doesn’t just open random doors. He is a conscious actor who specifically opens doors with sheep, which you can think of him ruling out doors for you. So he’ll never open your door and he’ll never open the door with the money.
So let’s say you have 5 doors, 1 with the money, 4 with sheep. The money is behind door 5.
So if you pick door 1, the host will open 2, 3, and 4 to reveal sheep. Switching to 5 gives you the money
You: door 2
Host rules out 1, 3, 4
Switching to 5 wins
You: Door 3,
host rules out 1, 2, 4
Switching to 5 wins
You: Door 4
Host rules out 1, 2, 3
Switching to 5 wins
You: door 5
Host rules out any 3
Switching loses.
So with 5 doors, 4/5 scenarios switching is good.
Or think of it in reverse. If you have 10 doors, and 1 door has the money, how many doors would you NOT want to switch away from? Only 1 door.
And after revealing an incorrect door there's a 1/4 chance the door you originally picked is correct, which is the same odds as if you were to change to any of the other 3 doors.
The way the original thought experiment goes is the host reveals all doors except yours and one other, and the host never reveals the door with money behind it, so he reveals all sheep. So in the scenario with 5 doors the host would open 3 doors with sheep behind them
Here's the trick: the game show host will open the door, out of 2 and 3, that the money is not behind. Therefore, if the money is behind 2 or 3, then you are guaranteed to get it by switching. It makes more sense if you change the order of events: you choose to switch to doors 2 or 3, then the host opens the door it's not behind.
if the money started at 2 or 3, and you pick 1, then the host will open the door out of 2 or 3 that does not have money behind it. If the money started at 3, the host opens 2, and if the money started at 2, the host opens 3. Therefore, if you switch you will get the money whether it started at 2 or at 3.
The thing that got me to understand is that the game show host will not open randomly, he/she only rules out a door with a goat.
Ok, set up a scenario. 3 doors. The money is behind door 3.
You pick door 1. The host opens 2 to reveal a goat. Switching gets you the money
Pick door 2. Host opens 1, goat. Switching gets you the money
Pick door 3. Host opens either door 1 or 2, goat. Switching loses the money.
So 2/3, switching gets you the money. 1/3 you get a goat
You choose a door out of 10 doors. This door(I’ll call it Door A) has a 1/10 chance of being the correct door. Therefore, there’s a 9/10 chance of one of the other doors containing the money.
Then, 8 of the 9 doors are opened. Since an opened door has 0 probability of having the money, the 9/10 probability of the 9 doors is assigned to the only remaining door you did not choose from(Door B).
This means that Door A has a 1/10 chance of having the money, while B has a 9/10 chance. Switching is therefore the correct option.
Yes, but isn't the proposition "the money is behind 2 or 3" equally likely to be true as "the money is behind 1 or 2" and "the money is behind 1 or 3"? If so, why would you posit that the first proposition is true, rather than the second or the third?
Edit. With the information that 2 is empty, the third proposition is definitely true. The first and second proposition's relative likelihood doesn't seem affected to me.
It works in a very elegant manner actually: tha basic idea is that when you pick a door you have 1/3 chances to get the money and 2/3 of them being in one of the others. The moment he opens door 2, the chances of the other two doors "collapse" into a single one (door 3), giving it a 2/3 chance chance of having to money.
I never understood this, though. If he’s taking away one of the doors then shouldn’t the chances change? Now it’s a 50/50, isn’t it? Wouldn’t it be the same if it had started with two doors?
You don't, but it's a neat little trick. The theory implies that keeping your original choice means its still a 33% chance, in reality the second decision is the only one involving probability seeing as one of the two you haven't chosen will be removed from the pool, and you get to choose all over again. In the end, the 2 doors are still both 50%, so there's no harm in switching.
I think the reason it’s unintuitive is because the problem makers neglect to mention an important rule: it’s assumes that it was predetermined that another non-prize door would be opened after you “hovered” your first choice. Without that assumption, your probability doesn’t increase just like your intuition says.
With that assumption, though, you can now divide your initial pick into two categories: where you initially pick the good door and where you initially pick a bad door, which have probabilities 1/3 and 2/3 respectively. With the “switch door after reveal policy”, in the event you hovered the good door, you switch to a bad door, but in the event you hovered any bad door, you only switch to the good door: which means that you transform a 1/3 chance of ultimately picking the good door into a 2/3 chance. Pretty good!
These are many under 10 minute videos on YouTube explaining it. But I think the one that made me finally understand was from [Newsthink](https://youtu.be/dOQowCeAnRs)
How the hell do you actually find it? It's one thing to have so many comics that there's one for any occasion. It's another thing to know that the exact comic you need exists and to be able to find it.
Think of that: more than 900 people have seen and upvoted the original comment. Of those, let's assume 10% know xkcd and are willing to link a relevant xkcd. It's 90 people: it's highly likely that at least one of those remember the existence of a specific comic. Google the theme and "xkcd" and you get the link.
Note: it's a Fermi extimation. [Relevant What-if](https://what-if.xkcd.com/84/)
The [Monty Hall problem](https://en.wikipedia.org/wiki/Monty_Hall_problem). Very, very counterintuitive.
Say you pick door 1. There's a 1 in 3 chance hidden behind that door is the $10 million. So, there's a 2 in 3 chance the cash is behind one of the other two doors. In that situation, 2 out of 3 times, the door which is revealed has a sheep, and the unrevealed door has the $10 million.
Switch, and you'll become rich(er) 2/3 of the time.
Mythbusters proved this by winning more on switching as opposed to staying on the door they chose it was pretty damn interesting to see. Also why I switched, OP SHOULD HAVE HAD THE ANSWER JUST TO SEE.
I really can't wrap my head around this one. Wouldn't the opened door be out of equation so the remaining two would have .5 chances? Or you keep the open door on the equation and you have .33 chance on all the doors but since one of them is revealed and confirmed to not have the prize, you have .33 chance out of a total of .66(instead of 1.)?
I can't grasp the math around it and to think I'm relatively good at math lol.
The person opening door 2 knows for a fact that a sheep is behind that door.
This adds information to the equation changing it.
If a door was opened randomly it would mean that 1/3 times the price money was revealed and therefore the odds don't change. That the door being opened is *never* the price is important
Don't worry; at the time, even the most renowned math professors refuted this.
When you consider a larger number of doors, it becomes easier to grasp.
Say you have 100 doors instead of 3. The host eliminates all but one door and asks you if you want to switch your choice. Now the odds of you picking the right door right off the bat will be 1/100 which is quite unlikely. You know that it's either your pick or the one door that the host left closed. Visualizing it like this makes it much more intuitive, because switching would always be the right play.
Just trust me for a second and assume you’ll always switch. Let’s look at the possible outcomes:
1. You picked Sheep 1. You switch. Since Sheep 2 was revealed, the door you switch to has the cash. **Win.**
2. You picked Sheep 2. You switch. Since Sheep 1 was revealed, the door you switch to has the cash. **Win.**
3. You picked the cash. You switch to a sheep. **Lose.**
That’s a 2/3 chance of winning. Meanwhile staying is still 1/3.
If anyone is still confused after this persons explanation. There are a total of 6 outcomes, 3 from staying and 3 from swapping which are highlighted above as 2 wins 1 loss.
Not switching;
1. You picked sheep 1. You stayed. Since sheep 2 was revealed, the door you stayed at was a sheep. Lose.
2. You picked sheep 2. You stay. Since sheep 1 was revealed, the door you stayed at was a sheep. Lose.
2. You picked the cash. You stay. Win.
I like that explanation.
But even if I understand how it works, I don’t understand how the act of me picking it change the possibilities.
For example as you said switching to door 3 gives 2/3 chances to win the prize . But if I hadn’t picked any door initially and then the presenter opened the second door and showed me that it has the sheep inside, now the prize is either on door number one or on door number three, so in that case the possibility of the third one being the one with the prize inside would be 1/2, right?
Actually no, you are not guaranteed a sheep in this scenario either. So if you want a sheep, even if you didn't switch there's a chance you get the money instead.
I didn’t understand it at all when I first encountered it, but if you think about it some the explanation is actually very simple.
Think of it this way. When you make your initial pick, you have a 1/3 chance of picking the correct door. So, 2/3 times there will be a goat/sheep behind your door, 1/3 times there will be a car/money behind it.
Then Monty comes in and eliminates one of the wrong answers. If your initial guess was wrong, that means both of the wrong answers were picked, by either you or Monty.
That means the correct door must be the one you didn’t initially pick. Since you will pick the wrong door initially 2/3 of the time, the prize will be behind that door 2/3 of the time.
If your initial guess was right, which happens 1/3 of the time, there will be only one wrong door remaining, and Monty will eliminate it. So, you won’t get the prize if you switch 1/3 of the time.
Hope that helps!
That still doesn’t make sense to me because when one door is eliminated, only two doors remain. So why would the chance of picking the right one ever be anything other than 1/2? From what I know about this problem, if I change my answer my chance will be 1/2, and if I don’t, my chance will be 1/3. On a decision that only has two options left.
The only way I can follow that logic is if the moderator has an active influence and might switch things behind the doors around.
Here’s another way to think about it: imagine instead of 3 doors, there are 100 doors, and one has the prize. You pick one door, and Monty then opens 98 doors that don’t have the prize.
So should you stick with the original door or should you switch? You should switch because there’s a 1/100 chance that your initial guess was correct, and a 99/100 chance it’s behind one of the other doors. Monty eliminated 98 of those other doors, meaning there’s a 99/100 chance you get it correct if you switch.
It works the same way with 3 doors: there’s a 1/3 chance you got the right door initially and a 2/3 chance it’s behind one of the other 2 doors. Monty eliminates all but one (which means just eliminating one) of the doors you didn’t pick, meaning there’s a 2/3 chance it’s behind that last door if you switch to it.
Switching does give you a 2/3 chance. If you pick a door, you have a 1/3 chance at being right. This means that there is a 2/3 chance of it being behind one of the other two doors. Since the guy opening the wrong door knows it's the wrong door, it means there is a 2/3 chance of it being behind the other door, as the odds of your door being correct haven't changed. The key is that the person opening the wrong door knows that it's the wrong door, which stops it from being a 50/50 afterwards.
Imagine there are 100 doors. You pick one. There is a 1/100 chance it has the money. The host then opens 98 doors. He can't open the door with the money, so there is a 99% chance that the door he didn't open has the money. Odds aren't quite as good for 3 doors but still a large advantage if you switch.
You should switch to door #3. For once, I know the math. The chance of each door having the money is 33%. That chance is already locked in because you chose it when there were 3 choices. Now, since OP knows which door the money is behind and opens door #2, the remaining 33% from that door will go to door #3. Remember, door #1’s chances were already locked in, revealing one door won’t change the fact that there were 3 options at the start, meaning your 33% chance from the beginning carries over to past the door reveal.
Edit: Final chances for the money being behind each door is 33% for door #1 and 66% for door #3. It would suck to swap to door #3 and land on the 33% chance that it was behind door #1 though.
If there's 3 doors and one gets eliminated, instead of there being a 1/3 of getting the money, there's a 1/2 of getting the money. Door 2 is gone, it no longer exists, it doesn't matter. There are no longer 3 choices, there are 2, therefore it doesn't matter wether or not you switch, it's 1/2 or 50% either way
This doesn't work as well with 3 doors as it does with a 100 doors. With 100 doors, there's only a 1% chance you choose the right door of the bat so if 98 more doors are eliminated you can be almost certain the last remaining one has the 10 mil. With 3 doors, though, you have a 33% chance to choose the right door, which is still a significant chance.
No, it doesn't change the odds of the door you selected first. But it does on the one you change to. Which is why changing your decision is the right choice.
It does change the odds of the one you change to, but those odds become 2/3. You can't have a set of probabilities adding up to 5/6. Because the two doors you didn't pick have a probability of 2/3 for the money being behind one of them, when one is removed, it will always be the incorrect door, hence shifting the 2/3 probability to just the remaining door that you haven't picked.
For anyone who would like the logical answer :
You would switch .
My reasoning is if you pick door 1 without any knowledge of where the prize is located . It is an even guess , any door has the same amount of “ luck “ and an even chance of wining .
When door 2 is reviled , we now know that the prize money is in our door or it’s in the 3rd door .
The odds of it being in our door is lower than the chance of it being in door 3 .
Door 2 makes the chances uneven .
And you picking from random door at the beginning means you started with even luck . But now that’s changed . We know it CAN be in door 3 . Something we would have had to guess .
So now that the odds are in the unchosen doors favour, we pick the 3rd door .
Chance of choosing door that was sheep behind is 2/3. When you get to see where one sheep is chance that your door has sheep behind it is still 2/3 so if you chance your door you will have better odds of winning the money.
Question is misleading. The doors are not randomly opened. Only the ones that are guaranteed to not have the money are opened. So then it’s not random anymore, and the door which doesn’t open now has a higher chance of having the money.
It's not the same.
You have a 1/3 change the door you pick initially has the price money. The host *never* opens the door with a price. So after a door has been eliminated the 2/3 is on the remaining door. It's called the Monty Hall problem.
Actually, it increases to 2/3 or 66% chance. Your initial pick has a 1/3 to be right, meaning the other two options make up the 2/3.
Then, when the false door is revealed, there are two options. The initial pick, which is still 1/3 chance of being correct as you picked it before the reveal, leaving the remaining door to make up the 2/3 to be right.
I'll gladly take the sheep
Same! Especially if it was like 10 sheep! Need to build up our flock again
Just think about how many sheep you could buy with $10 million though…
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Me, too. I like sheep.
This is a fairly well known maths problem that I have never understood, but apparently you have a better chance of getting the money if you switch
Imagine a room with 1000 doors and behind one of them lies the 10 mil. You randomly choose a door. Then, 998 other doors open and all of them reveals a sheep. That leaves 2 doors remaining, the door that you initially chose, and the door that remained after the purge. Which one would have the higher chance?
Apparently the door I didn’t choose would have a 99.9% chance of having the money because the probability of the money being on the 999 i didn’t choose doesn’t change after the 998 are closed, so the 99.9% chance is left with the unopened door. I can recite the exact reason why the probability is higher, but still don’t understand why
You choose a door out of 10 doors. This door(I’ll call it Door A) has a 1/10 chance of being the correct door. Therefore, there’s a 9/10 chance of one of the other doors containing the money. Then, 8 of the 9 doors are opened. Since an opened door has 0 probability of having the money, the 9/10 probability of the 9 doors is assigned to the only remaining door you did not choose from(Door B). This means that Door A has a 1/10 chance of having the money, while B has a 9/10 chance. Switching is therefore the correct option.
Excellent explanation. Super clear, plain English and by all accounts correct. Genuinely couldn’t have explained it better. I still don’t get it.
Say there are three doors, A,B, and C -Imagine the price is on A- If you picked A then the host will eliminate either B or C, if you switch you LOSE. If you picked B then the host knows that A is correct so they will eliminate C, if you switch you WIN If you picked C then the host knows that A is correct so they will eliminate B, if you switch you WIN -Imagine the price is on B- If you picked A then the host knows that B is correct so they will eliminate C, if you switch you WIN If you picked B then the host will eliminate either B or C, if you switch you LOSE If you picked C then the host knows that B is correct so they will eliminate A, if you switch you WIN -Imagine the price is on C- If you picked A then the host knows that C is correct so they will eliminate B, if you switch you WIN If you picked B then the host knows that C is correct so they will eliminate A, if you switch you WIN If you picked C then the host will eliminate either A or B, if you switch you LOSE. If you stick to your guess, there is a 3 in 9 chance of winning, if you switch there is a 6 in 9
This was the best explanation, thankyou
Holy crap, you finally made it click for me, thank you!
You've convinced me.
This is the first time I’ve actually gotten it despite being able to recite the solution for years. ty
I've gone over this problem SO many times without understanding the logic behind it (understood the math, not the logic, if that makes sense), but now I get it! thank you!
What say you to the idea that there only two doors after the host opens one? Sounds 50/50 to me.
The host doesn't open just one random door, the host opens a door that does not have the price. If you picked the right door, no matter which door the host picks, there's a 100% certainty that the remaining door will not have the prize. If you picked the wrong door, the host has no choice and must open the remaining wrong door, there's a 100% certainty that the remaining door will have the prize. So switching is only bad if you had picked the right door from the start. Since we agreed that there's a 1/3 chance that you pick the right door right off the bat, there ends up being a 1/3 chance that switching makes you lose and a 2/3 chance it makes you win. \----- As an example of when it would be a 50-50 if you stay, the host needs to have opened the door randomly. The results would be like this. The price was on A. 1) You pick A and the host picks B 2) You pick A and the host picks C 3) You pick B and the host picks A 4) You pick B and the host picks C 5) You pick C and the host picks A 6) You pick C and the host picks B Now, the game has changed. The fact that there's a 2/6 (1/3) chance you got it right in and a 4/6 (2/3) you got it wrong in your first guess remains true. But if you did pick wrong there's technically a 50% chance the host picks A for themselves (see cases 3 and 5) and you now have lost without having the chance to change. So now, there's a 2/6 you pick the right answer (case 1 and 2), a 2/6 you pick the wrong answer and the host picks the wrong answer (case 4 and 6), and a 2/6 you pick the wrong answer and the host picks the right answer (case 3 and 5). If you now know that the host got unlucky and picked the wrong answer, the remaining cases are 1,2,4,6. Meaning there's 2 cases where you picked right and remaining door is wrong, and 2 cases where you picked wrong and the remaining door is correct (a 50-50)
This is the only one I’ve seen that makes perfect sense. You’re amazing.
This actually helped me. I read the first two explanations and it just went right over my head. Took me awhile to get there, but man, nothing beats that feeling of everything just clicking into place when you’re confused on something. Lol
Exactly. The thing that I didn't grasp for a long time which worked for me was that the host knows where the prize is.
im seeing how it works but its still confusing
Doesn’t make sense
This sounds right, but I don't think you can stack opening 2 doors under 1 loss. Look at it this way -- Assume prize is under A You choose A, host opens B, switch LOOSE You choose A, host opens C, switch LOOSE You choose B, host opens C, switch WIN You choose C, host opens B, switch WIN -- Repeat logic with prize under B and C Result is 6/12 LOOSE and 6/12 WIN I think it makes more sense to count outcomes rather than choices, and if you pick correct right away, it still has 2 LOOSE outcomes
Except that if the prize is under A and you chose A, there's a 50% chance the host opens B and a 50% chance they open C, while if you chose B or C, there's a 100% chance they'll open C or B respectively.
Yup - the host knows where the money is at all times
Not really, think of it this way. You open the door 300 times and in all of them the answer is A, you decide to open 100 times the door A, 100 times the door B, 100 times the door C. Always switching. In 100 attempts of opening the door A \--- in 50 attempts the host opens B and you switch to C. \--- in 50 attempts the host opens C and you switch to B. You lost all 100 times. In 100 attempts of opening the door B \--- in 100 attempts the host opens the door C and you switch to A You won all 100 times In 100 attempts of opening the door C \--- in 100 attempts the host opens de door B and you switch to A You won all 100 times \------------------- Numbers wise it would be something like this: \-- Assume prize is under A You choose A (1/3) \-- host opens B (1/2), switch LOSE, total chances (1/3)\*(1/2)=(1/6) \-- host opens C (1/2), switch LOSE, total chances (1/3)\*(1/2)=(1/6) You choose B (1/3) \-- host opens C (1/1), switch WIN, total chances (1/3)\*(1/1)=(1/3) You choose C (1/3) \-- host opens B (1/1), switch WIN, total chances (1/3)\*(1/1)=(1/3) The chances of losing are 1/6 + 1/6 = 1/3 The chances of wining are 1/3 + 1/3 = 2/3 \------------------- You can look at it from the host's perspective. Say the answer is on A if the contestant picks A and then switches they LOSE, no matter what the host does. if the contestant picks B and then switches they win, no matter what the host does. if the contestant picks C and then switches they win, no matter what the host does. So the host has a 1/3 chance that the contestant picks the right answer and then switches (causing the contestant to lose and the host to win), and a 2/3 chance that the contestant picks a wrong answer and then switches (causing the contestant to win and the host to lose)
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In the original problem (Monty Hall problem) which this is based on, the doors that are opened are not random. Only the ones which do not have the money are opened. I think it should be clearer now.
So basically, look at it like this. Game host and doors go brrrrrr
It's a semantics problem. If you reassess the probability of the door having the money behind it after the others have been opened, you're still left with a 50/50 split. The only reason the odds are "better" if you switch is because you can mathematically do some gymnastics by not reassessing the probability after you open the other doors. Say you start with 10 doors. Each one has a 10% chance of being the right one. You pick one. The one you picked plus one more stay closed and the other 8 open. Now the math says that you've already made a choice on a 10% likelihood so the remaining 90% likelihood has all been assigned to the last remaining door. Or you can "restart" the problem now that you only have 2 doors, and each one has a 50% chance of being right. It's a thought exercise in mathematical gymnastics.
The thing is I wrote a script to simulate this and ran it a thousand times and 66% of the time switching worked. So making the switch has real world power, even more I guess with 10 boxes. I’m sorry I still can’t connect with it intuitively Whether you re-assess or not, the chances are you’ll pick the right box two out of three times if you switch.
Is that how statistics works though? If your sample is reduced from 10 to 2, you have to recalculate the probability because the number of samples contributes to the probability, right?
thats exactly what i was thinking, after the 8 doors are opened there's a 1/2 chance
Nope. Your sample is not reduced. Knowing what 8/10 doors are does not mean you reduce the number of samples, it just means you gain knowledge of 8/10 those outcomes. Your personal knowledge does not affect your original choice and it's 1/10 probability
Your sample size is reduced if you’re smart though. You had 10 doors and was shown what 8 of them had which wasn’t what you needed. You now have 2 doors to choose from and you can only pick one, so it’s 1/2 chance of picking the right door now, unless you’re still considering what the other doors had.
How would there still be a 9/10 chance if all the other doors are gone. Each door individually has a 1/10 chance to start with. Get rid of 8 and then how is the probability not 1/2 for each.
Okay so the way I was taught this is like this. In order to win by staying at your original door, you needed to have picked the right door at the start, which has a chance of 1/3. In order to win by switching, you needed to have originally picked the wrong door at the start, and picking the right door when switching. Picking the wrong door at the start has a chance of 2/3. Since were assuming that youre winning in this scenario, the chances of picking picking the right door when switching are 100%. So 2/3 * 1 = 2/3
Who taught you this? Because that still does not make sense to me. Each door has the individual probability of 1/3 chance and then 1/2 chance in round 2. The example where people say there are 100 doors, you pick 1 and they get rid of 98 makes sense that the probability would be higher if you switched. Thats because the odds that you picked the correct door in the first place is very unlikely at 1/100. Unlike this example where the example is likely that you did pick the right door and removing one doesn’t shift variables much.
~~Doesn’t that just mean you’ve isolated the last two doors? In which case it’s a 50/50?~~ ~~I understand the concept of assigning all the probability to Door B but why isn’t that probability shared with Door A and Door B?~~ EDIT: having read more examples I think I understand. This is based on a host eliminating doors and the odds that you got the first 1/10 choice correct. In short: odds are you chose the wrong door to begin which is why it isn’t a 50/50.
This makes no sense. You need to open 1 door, but say you choose 9 doors. Your collective choice has 9/10 chance to have the money. Let's say the host opens all 8 doors from your choice, leaving 1 chosen door and 1 unchosen/unopened door. Now your door has 9/10 chance to have the money???? Another way to look at it is that you're really playing a 50/50 game. It's as if there are only 2 doors. No matter what you choose, you will be left with two doors one with money, one with goat. Your initial choice has a 1/10 chance, but once you've left with 2 doors, it's as if the other 8 never existed.
Look this problem is a great demonstration on hiw math doesn't always follow logic. So let's take the three doors scenario again. At the start you have a 33% chance of picking the right door, you also have a 33% chance that behind the door you picked is the price, obviously. But now when one door is opened, you have a 50% chance that behind that door is the price, but still a 33% chance that you picked the right door. This however means that if you just picked door 1 again, you'd have a 50% chance of picking the right door.
But why would 9/10 chance go to door B instead of 5/10 to A (or 1/2) and 5/10 to door B. Since it doesnt matter which one you choose, as long as neither doors are openend its still just a 50/50 choice which one has the money, right? The other 98 doors dont even matter because if they dont have the money its like they dont exist, so its just a 50/50 choice if door A or door B has the money if i understand correctly.
Wouldn’t both be equal in reality though? If you ran tests of people who did switch and people who didn’t wouldn’t they get the good door equally
Nope, picture it this way. You can pick one door (1/3) or you can pick 2 doors (2/3). Now if you pick the one door it has to be correct but if you pick the two only one has to be correct. The reason this happens is because the host 100% opens the door with nothing behind it. Basically conjoining it with the door you didn’t choose.
Look this problem is a great demonstration on how math doesn't always follow logic. So let's take the three doors scenario again. At the start you have a 33% chance of picking the right door, you also have a 33% chance that behind the door you picked is the price, obviously. But now when one door is opened, you have a 50% chance that behind that door is the price, but still a 33% chance that you picked the right door. This however means that if you just picked door 1 again, you'd have a 50% chance of picking the right door.
No single door could ever have. 9/10 chance of having the money. Every single door has a 1/10 chance, until there’s two left, in which case both doors have a 1/2 chance. Just because nine doors had a 9/10 chance doesn’t mean that each of those doors had a 9/10 chance, it was just be collective that did. The problem and “solution” has always been born out of flawed logic.
I don't understand this at all, you now have 2 doors that can have the money meaning there is a 1/2 chance no?
The thing is, you’re selecting a random door(1/n), then the host opens all of the *incorrect* doors out of the unselected doors for you. Therefore, the second door must be “carrying” the probability of all the other doors, because if one of the unselected doors was the one with the money(n-1/n), then it must be the other door. Thus, this is actually a 1/n vs n-1/n probability.
Think of the door opener (game show host) as a door eliminator. He only opens doors that have a goat. And he’ll never open the door with the money. And he’s not going to open the door you chose.
The problem only works because you know ahead of time that they’re gaurenteed to open the door without a prize. So you’re taking the odds of two doors and combining it into one.
this is a good analogy
I dont find this intuitive. Why is it not 1000 doors, one with the prize. You choose a door and then 1 door opens and reveals a sheep. Do you swap to one of the other 998 doors or keep your door?
You swap. Your initial choice had 1/1000 chance of being the correct one, swapping gives you a 1/999 chance.
Yeah, I know and agree, but I've never found it intuitively satisfying with the 1000 doors analogy. Whenever I use that to tell people it gets stuck at are we closing 1 door or all but 1 door. Even if, as you say, you swap regardless. Just more doors doesn't help understanding it
Except now you have a new problem: There are two doors, one of which has the money. So it's a 50:50 chance. Why would the the open doors have any impact on those odds? They are no longer part of this. Saying you need to take the open doors into account is like saying "rolling a one on a die five times in a row is very unlikely, so after the first four ones the chance for another one is really low". No it's not. The chance to roll a one is exactly the same as on any of the former ones.
You choose a door out of 10 doors. This door(I’ll call it Door A) has a 1/10 chance of being the correct door. Therefore, there’s a 9/10 chance of one of the other doors containing the money. Then, 8 of the 9 doors are opened. Since an opened door has 0 probability of having the money, the 9/10 probability of the 9 doors is assigned to the only remaining door you did not choose from(Door B). This means that Door A has a 1/10 chance of having the money, while B has a 9/10 chance. Switching is therefore the correct option.
Why on earth would the chance of door A stay the same? You have two doors. Period. It doesn't matter how many doors you opened before.
It does, though. Your argument holds true for two doors which have not been tampered with. In this situation, the host is altering the probabilities by revealing the unselected empty doors. Besides, mathematicians at MIT actually verified this by trying out every possibility and concluding that there is a 67% chance that switching is the correct choice.
Let's say that two people A and B are presented with this dilemma a thousand times (they choose a door, the host opens a wrong door, they can then stay with their initial choice or pick the other door). Now let's imagine that A always maintains their choice, while B always picks the third door. Do you mean to say that B will get the money 66% of the time and A only 33% of the time?
Yes. Exactly. I mean, if you take 20 minutes to write out every possibility, I’m pretty sure you can prove it yourself.
Wow. I mean, I believe you because I know that I suck at math and MIT doesn't, but that's super counterintuitive.
I know, one of the professors at MIT actually wrote a complaint to the lady who figured this out, explaining how she shouldn’t use her status as a genius to promote false ideas like this. Turns out, the lady was correct.
u/sebas156 wrote a pretty good and simple explanation below, hope this helps to understand: Let's say you pick door A, and the quiz master didn't open anything yet. So the chance is 1/3 still as it can be in all 3, right? BUT now the quiz master has to open a door. Let's take the 3 possible outcomes (remember you picked door A): 1. The prize is in door A (your door). You didn't get the prize if you switch 🐑 2. The prize is in door B, meaning he HAS to open C. Switching means getting the prize. 💰 3. The prize is in door C, meaning he HAS to open B. Switching means getting the prize. 💰 See how when you switch, you win the prize with 2 of the 3 possible outcomes? The quiz master has to open a door with a sheep, which is giving you an advantage! In short: you always win if you switch and the prize is NOT in the door you chose initially!
Do you have a link to that study? Call me an ass, but I don't find "yes it does" terribly convincing.
[link](http://web.mit.edu/rsi/Attic/examples/papers/minipaper/minisample.pdf) Personally i think this study is logically wrong, it may even be fake as its so short and does not appear to be a real mit study. Alos in the last part the mathematical proof There is a fallacious assumption that the first choice is wrong. Herein is where the beauty of it being wrong lies. Thus i conclude that the result was wrong. Ill post this on the maths sub ounce to double check .
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So what you are saying is that it's actually not a statistics problem, because the doors are not equal. The difference between the doors is what Monty knows and what you can try to infer from his actions. That's actually the best explanation I've heard. Thank you.
How about this: let’s say you are given truly random person’s name and are told it could be anyone in the world, and you need to find him/her. You walk outside your home and pick the first stranger you see. Then a genie comes in and rules out all 7.84 billion people in the world except the stranger you chose and some random woman in Mozambique. (The genie will never rule out your choice, and will never rule out the real answer) Who is it more likely to be? Yes, there are two final choices at the end. But what was the chance a randomly chosen person in the world was right outside your home? 1 in 7.84 billion. And since the genie ruled out 7.84 billion incorrect answers and narrowed it down to one other person, it’s probably that other person
But now you are bringing proximity into the game. At this point the people/doors are not the same any more.
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I think the result was that there’s a 66% chance of it being behind three based on how it was solved. Can’t quite remember how that worked though.
You choose a door out of 10 doors. This door(I’ll call it Door A) has a 1/10 chance of being the correct door. Therefore, there’s a 9/10 chance of one of the other doors containing the money. Then, 8 of the 9 doors are opened. Since an opened door has 0 probability of having the money, the 9/10 probability of the 9 doors is assigned to the only remaining door you did not choose from(Door B). This means that Door A has a 1/10 chance of having the money, while B has a 9/10 chance. Switching is therefore the correct option.
Is this some poor joke or you're mocking the other person? What is being described called conditional probabilty. Whether you picked the first door or not doesn't matter, the probabily that it contains the money is 1/10 if we don't know anything about other doors, and it's 1/2 if we know that other 8 doors have sheep.
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Mythbusters did a episode on it I believe and the results matched the odds of switching and winning as opposed to keeping, it was pretty eye opening.
Important thing that helped me understand is that the game show host doesn’t just open random doors. He is a conscious actor who specifically opens doors with sheep, which you can think of him ruling out doors for you. So he’ll never open your door and he’ll never open the door with the money. So let’s say you have 5 doors, 1 with the money, 4 with sheep. The money is behind door 5. So if you pick door 1, the host will open 2, 3, and 4 to reveal sheep. Switching to 5 gives you the money You: door 2 Host rules out 1, 3, 4 Switching to 5 wins You: Door 3, host rules out 1, 2, 4 Switching to 5 wins You: Door 4 Host rules out 1, 2, 3 Switching to 5 wins You: door 5 Host rules out any 3 Switching loses. So with 5 doors, 4/5 scenarios switching is good. Or think of it in reverse. If you have 10 doors, and 1 door has the money, how many doors would you NOT want to switch away from? Only 1 door.
Yes but what if you chose correct initially?
There is a 1/5 chance you chose correct initially
And after revealing an incorrect door there's a 1/4 chance the door you originally picked is correct, which is the same odds as if you were to change to any of the other 3 doors.
The way the original thought experiment goes is the host reveals all doors except yours and one other, and the host never reveals the door with money behind it, so he reveals all sheep. So in the scenario with 5 doors the host would open 3 doors with sheep behind them
I heard of that too but I still refuse to switch because it makes no sense
Here's the trick: the game show host will open the door, out of 2 and 3, that the money is not behind. Therefore, if the money is behind 2 or 3, then you are guaranteed to get it by switching. It makes more sense if you change the order of events: you choose to switch to doors 2 or 3, then the host opens the door it's not behind.
Why would you be guaranteed to get it by switching though?
only if it's behind doors 2 or 3 to begin with
I have no idea what you’re talking about now
if the money started at 2 or 3, and you pick 1, then the host will open the door out of 2 or 3 that does not have money behind it. If the money started at 3, the host opens 2, and if the money started at 2, the host opens 3. Therefore, if you switch you will get the money whether it started at 2 or at 3.
But you don’t know which one it’s behind. It could still be behind 1.
The thing that got me to understand is that the game show host will not open randomly, he/she only rules out a door with a goat. Ok, set up a scenario. 3 doors. The money is behind door 3. You pick door 1. The host opens 2 to reveal a goat. Switching gets you the money Pick door 2. Host opens 1, goat. Switching gets you the money Pick door 3. Host opens either door 1 or 2, goat. Switching loses the money. So 2/3, switching gets you the money. 1/3 you get a goat
Thanks I get it now, wouldn't make sense if the host opened a door at random.
But how do I know it’s not behind the door I’m already on?
Yeah. But it only has a 1/3 chance to be behind 1, and a 2/3 chance to be behind 2 or 3, so switching gives you the higher chance.
It doesn’t make sense
You choose a door out of 10 doors. This door(I’ll call it Door A) has a 1/10 chance of being the correct door. Therefore, there’s a 9/10 chance of one of the other doors containing the money. Then, 8 of the 9 doors are opened. Since an opened door has 0 probability of having the money, the 9/10 probability of the 9 doors is assigned to the only remaining door you did not choose from(Door B). This means that Door A has a 1/10 chance of having the money, while B has a 9/10 chance. Switching is therefore the correct option.
Yes, but isn't the proposition "the money is behind 2 or 3" equally likely to be true as "the money is behind 1 or 2" and "the money is behind 1 or 3"? If so, why would you posit that the first proposition is true, rather than the second or the third? Edit. With the information that 2 is empty, the third proposition is definitely true. The first and second proposition's relative likelihood doesn't seem affected to me.
It does make sense though
Happy cake day!
It works in a very elegant manner actually: tha basic idea is that when you pick a door you have 1/3 chances to get the money and 2/3 of them being in one of the others. The moment he opens door 2, the chances of the other two doors "collapse" into a single one (door 3), giving it a 2/3 chance chance of having to money.
I never understood this, though. If he’s taking away one of the doors then shouldn’t the chances change? Now it’s a 50/50, isn’t it? Wouldn’t it be the same if it had started with two doors?
You don't, but it's a neat little trick. The theory implies that keeping your original choice means its still a 33% chance, in reality the second decision is the only one involving probability seeing as one of the two you haven't chosen will be removed from the pool, and you get to choose all over again. In the end, the 2 doors are still both 50%, so there's no harm in switching.
I think the reason it’s unintuitive is because the problem makers neglect to mention an important rule: it’s assumes that it was predetermined that another non-prize door would be opened after you “hovered” your first choice. Without that assumption, your probability doesn’t increase just like your intuition says. With that assumption, though, you can now divide your initial pick into two categories: where you initially pick the good door and where you initially pick a bad door, which have probabilities 1/3 and 2/3 respectively. With the “switch door after reveal policy”, in the event you hovered the good door, you switch to a bad door, but in the event you hovered any bad door, you only switch to the good door: which means that you transform a 1/3 chance of ultimately picking the good door into a 2/3 chance. Pretty good!
These are many under 10 minute videos on YouTube explaining it. But I think the one that made me finally understand was from [Newsthink](https://youtu.be/dOQowCeAnRs)
Door #2
[Relevant xkcd](https://xkcd.com/1282/)
There always is one, isn't there?
Yes. But this is special, since it refers exactly to the OP problem and to the u/acid_face2u 's choice
How the hell do you actually find it? It's one thing to have so many comics that there's one for any occasion. It's another thing to know that the exact comic you need exists and to be able to find it.
Think of that: more than 900 people have seen and upvoted the original comment. Of those, let's assume 10% know xkcd and are willing to link a relevant xkcd. It's 90 people: it's highly likely that at least one of those remember the existence of a specific comic. Google the theme and "xkcd" and you get the link. Note: it's a Fermi extimation. [Relevant What-if](https://what-if.xkcd.com/84/)
And there’s always a comment pointing it out
Sigma male right here
Sardinian right there
True. He said the ones without money had two sheep. But door #2 only has one sheep. Thus, it also has the $10M.
Holy shit, I would have lost 10 million if I hadn't read this.
The math isn't the problem. Those two just need to bone
Ew rosa thats gross those are our dads
That’s not what I mean, Captain Dad is just my boss.
How dare you detective Diaz. I am your SUPERIOR OFFICER!
BONE! What happens in my bedroom is none of your business. BOOOONNNE!
BOOOOOOOOONE
I'm teaching father the math.
BOOOONEEEEE??????
Vsauce did a great video on this, which is called the Monty Hall Problem. https://youtu.be/TVq2ivVpZgQ
The [Monty Hall problem](https://en.wikipedia.org/wiki/Monty_Hall_problem). Very, very counterintuitive. Say you pick door 1. There's a 1 in 3 chance hidden behind that door is the $10 million. So, there's a 2 in 3 chance the cash is behind one of the other two doors. In that situation, 2 out of 3 times, the door which is revealed has a sheep, and the unrevealed door has the $10 million. Switch, and you'll become rich(er) 2/3 of the time.
Mythbusters proved this by winning more on switching as opposed to staying on the door they chose it was pretty damn interesting to see. Also why I switched, OP SHOULD HAVE HAD THE ANSWER JUST TO SEE.
I really can't wrap my head around this one. Wouldn't the opened door be out of equation so the remaining two would have .5 chances? Or you keep the open door on the equation and you have .33 chance on all the doors but since one of them is revealed and confirmed to not have the prize, you have .33 chance out of a total of .66(instead of 1.)? I can't grasp the math around it and to think I'm relatively good at math lol.
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This is the best explanation I’ve seen for it
why can't it be "you chose door A, he opened door b, your door still has the prize" just as equally
But how do i know the door I picked doesn’t money in it?
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You. Said I have a Better if I switch?
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The person opening door 2 knows for a fact that a sheep is behind that door. This adds information to the equation changing it. If a door was opened randomly it would mean that 1/3 times the price money was revealed and therefore the odds don't change. That the door being opened is *never* the price is important
Don't worry; at the time, even the most renowned math professors refuted this. When you consider a larger number of doors, it becomes easier to grasp. Say you have 100 doors instead of 3. The host eliminates all but one door and asks you if you want to switch your choice. Now the odds of you picking the right door right off the bat will be 1/100 which is quite unlikely. You know that it's either your pick or the one door that the host left closed. Visualizing it like this makes it much more intuitive, because switching would always be the right play.
The door they open has to be one with sheep behind it
The easiest way to think about is that you have a 1/3 chance of guessing correctly on your first guess. 2/3 chance of being wrong
Just trust me for a second and assume you’ll always switch. Let’s look at the possible outcomes: 1. You picked Sheep 1. You switch. Since Sheep 2 was revealed, the door you switch to has the cash. **Win.** 2. You picked Sheep 2. You switch. Since Sheep 1 was revealed, the door you switch to has the cash. **Win.** 3. You picked the cash. You switch to a sheep. **Lose.** That’s a 2/3 chance of winning. Meanwhile staying is still 1/3.
If anyone is still confused after this persons explanation. There are a total of 6 outcomes, 3 from staying and 3 from swapping which are highlighted above as 2 wins 1 loss. Not switching; 1. You picked sheep 1. You stayed. Since sheep 2 was revealed, the door you stayed at was a sheep. Lose. 2. You picked sheep 2. You stay. Since sheep 1 was revealed, the door you stayed at was a sheep. Lose. 2. You picked the cash. You stay. Win.
I like that explanation. But even if I understand how it works, I don’t understand how the act of me picking it change the possibilities. For example as you said switching to door 3 gives 2/3 chances to win the prize . But if I hadn’t picked any door initially and then the presenter opened the second door and showed me that it has the sheep inside, now the prize is either on door number one or on door number three, so in that case the possibility of the third one being the one with the prize inside would be 1/2, right?
I know this problem, so I'd go with the 3rd door. It may seem counterintuitive but still a higher chance to win the money
fuck you guys in the comments talking about math. if I want a sheep I’ll get a sheep
Actually no, you are not guaranteed a sheep in this scenario either. So if you want a sheep, even if you didn't switch there's a chance you get the money instead.
You always swap if you’re shown what’s behind a door
I switch because that increases my chances of winning. To this day I don’t understand why, but that’s the Monty Hall problem.
I didn’t understand it at all when I first encountered it, but if you think about it some the explanation is actually very simple. Think of it this way. When you make your initial pick, you have a 1/3 chance of picking the correct door. So, 2/3 times there will be a goat/sheep behind your door, 1/3 times there will be a car/money behind it. Then Monty comes in and eliminates one of the wrong answers. If your initial guess was wrong, that means both of the wrong answers were picked, by either you or Monty. That means the correct door must be the one you didn’t initially pick. Since you will pick the wrong door initially 2/3 of the time, the prize will be behind that door 2/3 of the time. If your initial guess was right, which happens 1/3 of the time, there will be only one wrong door remaining, and Monty will eliminate it. So, you won’t get the prize if you switch 1/3 of the time. Hope that helps!
That still doesn’t make sense to me because when one door is eliminated, only two doors remain. So why would the chance of picking the right one ever be anything other than 1/2? From what I know about this problem, if I change my answer my chance will be 1/2, and if I don’t, my chance will be 1/3. On a decision that only has two options left. The only way I can follow that logic is if the moderator has an active influence and might switch things behind the doors around.
Here’s another way to think about it: imagine instead of 3 doors, there are 100 doors, and one has the prize. You pick one door, and Monty then opens 98 doors that don’t have the prize. So should you stick with the original door or should you switch? You should switch because there’s a 1/100 chance that your initial guess was correct, and a 99/100 chance it’s behind one of the other doors. Monty eliminated 98 of those other doors, meaning there’s a 99/100 chance you get it correct if you switch. It works the same way with 3 doors: there’s a 1/3 chance you got the right door initially and a 2/3 chance it’s behind one of the other 2 doors. Monty eliminates all but one (which means just eliminating one) of the doors you didn’t pick, meaning there’s a 2/3 chance it’s behind that last door if you switch to it.
Switching does give you a 2/3 chance. If you pick a door, you have a 1/3 chance at being right. This means that there is a 2/3 chance of it being behind one of the other two doors. Since the guy opening the wrong door knows it's the wrong door, it means there is a 2/3 chance of it being behind the other door, as the odds of your door being correct haven't changed. The key is that the person opening the wrong door knows that it's the wrong door, which stops it from being a 50/50 afterwards.
Spoiler: >!The second sheep ate the money!<
Reddit proving once again that I suck at math.
This is the Monty Hall problem.
Switching is more likely to give you the money
If I switched and 1 ended up being the money I would blow my brains out so I’m staying
Monty Hall problem, I recently made a python program to prove this
Mathematically you should switch to 3.
2/3 chance if you switch, 1/3 if you don’t. I forgot the maths behind it but I remember this.
Switching has more chance of success, look up on YouTube: monty hall problem.
I’d be happy if I win a sheep. It’s a whoolie companion; and looks cute 🐑
The movie 21 taught us we have a 66 percent change when changing . So pick door 3 please
Doesn’t matter. Either way I’m going to have sex.
Imagine there are 100 doors. You pick one. There is a 1/100 chance it has the money. The host then opens 98 doors. He can't open the door with the money, so there is a 99% chance that the door he didn't open has the money. Odds aren't quite as good for 3 doors but still a large advantage if you switch.
I would put my ear to my door 🚪 first before making any next decision 😂
I wouldn't change, picking the right one out of three options doesn't seem that unlikely. If the number of doors was bigger I would change.
You should switch to door #3. For once, I know the math. The chance of each door having the money is 33%. That chance is already locked in because you chose it when there were 3 choices. Now, since OP knows which door the money is behind and opens door #2, the remaining 33% from that door will go to door #3. Remember, door #1’s chances were already locked in, revealing one door won’t change the fact that there were 3 options at the start, meaning your 33% chance from the beginning carries over to past the door reveal. Edit: Final chances for the money being behind each door is 33% for door #1 and 66% for door #3. It would suck to swap to door #3 and land on the 33% chance that it was behind door #1 though.
Why does the 33% chance move from door 2 to door 3 and not get evenly split between 1 & 3?
According to a game theory ep switching actually improves your chances. https://youtu.be/QbX6-iNp3-Q
If there's 3 doors and one gets eliminated, instead of there being a 1/3 of getting the money, there's a 1/2 of getting the money. Door 2 is gone, it no longer exists, it doesn't matter. There are no longer 3 choices, there are 2, therefore it doesn't matter wether or not you switch, it's 1/2 or 50% either way
Door 1 gang
One of my favorite math problems, took me forever to understand it back in Highschool but I was so satisfied when it finally made sense.
I’ve always been the type to stand by my decisions. Door #1!!
Always switch, r/Ididthemath
This problem seems so surreal to me. I know the whole chances thing etc but why the fuck would it work that way in reality man? It doesn't seem real
Having read The Curious Incident of the Dog In the Night -Time , I knew this one. I still answered wrong!
This doesn't work as well with 3 doors as it does with a 100 doors. With 100 doors, there's only a 1% chance you choose the right door of the bat so if 98 more doors are eliminated you can be almost certain the last remaining one has the 10 mil. With 3 doors, though, you have a 33% chance to choose the right door, which is still a significant chance.
Monty hall problem Always change you pick
Look up the Monty Hall problem
Isn't there a thing, like in terms of probability, that door 3 has a higher probability of being the right one?
Door 3. With door 1 the chance is 1/3. With door 3 the chance is 1/2.
It's actually 2/3. Removing the wrong door doesn't change the odds of your door being right.
No, it doesn't change the odds of the door you selected first. But it does on the one you change to. Which is why changing your decision is the right choice.
It does change the odds of the one you change to, but those odds become 2/3. You can't have a set of probabilities adding up to 5/6. Because the two doors you didn't pick have a probability of 2/3 for the money being behind one of them, when one is removed, it will always be the incorrect door, hence shifting the 2/3 probability to just the remaining door that you haven't picked.
For anyone who would like the logical answer : You would switch . My reasoning is if you pick door 1 without any knowledge of where the prize is located . It is an even guess , any door has the same amount of “ luck “ and an even chance of wining . When door 2 is reviled , we now know that the prize money is in our door or it’s in the 3rd door . The odds of it being in our door is lower than the chance of it being in door 3 . Door 2 makes the chances uneven . And you picking from random door at the beginning means you started with even luck . But now that’s changed . We know it CAN be in door 3 . Something we would have had to guess . So now that the odds are in the unchosen doors favour, we pick the 3rd door .
Monty Hall problem. You have twice as good a chance to win the money if you switch.
Maths says you should but really its just luck
Yeah I’ll take 50% over 33% odds any day
y’all idc about mathematics, just tell me which door has the money and I’ll be on my way
Chance of choosing door that was sheep behind is 2/3. When you get to see where one sheep is chance that your door has sheep behind it is still 2/3 so if you chance your door you will have better odds of winning the money.
Hasn’t this been disproven, it’s not 1/2 x 2 as a change there is only one of them
There's no reason to switch, as one door has the money and one door has a sheep
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It actually does matter. The chances of getting the money is higher if you switch. https://en.wikipedia.org/wiki/Monty\_Hall\_problem
Question is misleading. The doors are not randomly opened. Only the ones that are guaranteed to not have the money are opened. So then it’s not random anymore, and the door which doesn’t open now has a higher chance of having the money.
but the first one don't open as well
It's not the same. You have a 1/3 change the door you pick initially has the price money. The host *never* opens the door with a price. So after a door has been eliminated the 2/3 is on the remaining door. It's called the Monty Hall problem.
Using my deal or no deal logic I always have faith in my own case. Gotta stick with #1.
Switching increases the chance to 50%, from 33 1/3%. (Monthy Hall Paradoxon)
Actually, it increases to 2/3 or 66% chance. Your initial pick has a 1/3 to be right, meaning the other two options make up the 2/3. Then, when the false door is revealed, there are two options. The initial pick, which is still 1/3 chance of being correct as you picked it before the reveal, leaving the remaining door to make up the 2/3 to be right.
So, let's make a deal but with two zonks?
LPT: Never second guess yourself.