Monty Hall Problem (best explanation) - Numberphile
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- čas přidán 6. 09. 2024
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This is the absolute best explanation of this problem. Never before had someone pointed out to me the value in the fact that Monty knows everything.
+João Vítor M It's SUPER-critical (and often overlooked in the introduction to the problem) that Monty KNOWS where the prize is AND will intentionally open the goat-door initially. Once you understand that, the solution becomes nearly simple.
I have a better explanation: you will ALWAYS win if you initially choose a goat and then swap. ALWAYS. What are the odds of choosing a goat initially? 66%. So 66% of the time you choose a goat, he shows the other goat, then you switch to the car and win. Thats it.
+João Vítor M It becomes even more clear why you should always switch, if there were 10 or 1 million doors at the beginning. Then Monty practically gives you the car every time.
@@mjlv3862 ooooooooohhhhhhhhh
Jea me to
I think that if you drew out all the possibilities that would demonstrate the fact better. For example.
Scenario 1:
Car / Goat / Goat
Scenario 2:
Goat / Car / Goat
Scenario 3:
Goat / Goat / Car
Let's say you pick the door on the left and do not switch.
Scenario 1: Win
Scenario 2: Lose
Scenario 3: Lose
Let's say you pick the door on the left and switch doors.
Scenario 1: Lose
Scenario 2: Win
Scenario 3: Win.
Not Switching = 1 win out of 3.
Switching = 2 wins out of 3.
Hard to make it any easier then this.
Thx that helped even more
i still feel like there is something missing that makes your logic not valid
Please don't be mad, it's just that I have a fundamental disagreement with one crucial point of the argument and I'd like to know if someone can enlighten me on my error, if there is one.
If I choose the first door, then the moment the second door is revealed to be a goat, that means scenario 2 is not in play anymore. You can't still count it as a possible scenario. Scenario 2 has a car behind the second door, but that conflicts with the goat that has been revealed to be behind door two.
Therefore, after the first reveal, there are only two scenarios left, scenario 1 and scenario 3. I don't see why the scenario that was just eliminated has its probability merged into the remaining scenarios.
Or, to show an example like Rundvelt has done:
Scenario 1:
Car / Goat / Goat
Scenario 2:
Goat / Car / Goat
Scenario 3:
Goat / Goat / Car
Let's say you pick the door on the *left*.
The *middle* door is revealed to have a *goat*.
You do _not_ switch.
Scenario 1: Win
-Scenario 2- (Eliminated, due to goat behind middle door)
Scenario 3: Lose
Let's say you pick the door on the *left*.
The *middle* door is revealed to have a *goat*.
You _do_ switch.
Scenario 1: Lose
-Scenario 2- (Eliminated, due to goat behind middle door)
Scenario 3: Win.
So after a goat has been revealed to be behind the middle door, there are only two possible scenarios left (#1 & #3), not three scenarios.
Therefore I'm puzzled why this seems to be incorrect. Where is the crucial point I seem to be mistaken about?
Dharma Jannyter You are mistaking Scenario for Door. They are scenarios. I'm showing all the possible combinations for a single door. The reason is because all other doors will be the same. But let's do them all just to understand the issue. I've put (Scenario #) beside the trials that I represented in my previous post.
Let's stick with the same door.
You Pick Door #1 -> Prize is behind Door #1 -> WIN (Scenario 1)
You Pick Door #1 -> Prize is behind Door #2 -> LOSE
You Pick Door #1 -> Prize is behind Door #3 -> LOSE
You Pick Door #2 -> Prize is behind Door #1 -> LOSE (Scenario 2)
You Pick Door #2 -> Prize is behind Door #2 -> WIN
You Pick Door #2 -> Prize is behind Door #3 -> LOSE
You Pick Door # 3 -> Prize is behind Door #1 -> LOSE (Scenario 3)
You Pick Door # 3 -> Prize is behind Door #2 -> LOSE
You Pick Door # 3-> Prize is behind Door #3 -> WIN
So, as you can see, there are 9 possible combinations, and out of those 9 combinations, 3 result in wins.
Now, let's examine what happens if your strategy is to switch.
You Pick Door #1 -> Prize is behind Door #1 -> LOSE (Scenario 1)
You Pick Door #1 -> Prize is behind Door #2 -> WIN
You Pick Door #1 -> Prize is behind Door #3 -> WIN
You Pick Door #2 -> Prize is behind Door #1 -> WIN (Scenario 2)
You Pick Door #2 -> Prize is behind Door #2 -> LOSE
You Pick Door #2 -> Prize is behind Door #3 -> WIN
You Pick Door # 3 -> Prize is behind Door #1 -> WIN (Scenario 3)
You Pick Door # 3 -> Prize is behind Door #2 -> WIN
You Pick Door # 3-> Prize is behind Door #3 -> LOSE
So, as you can see, there are 9 possible combinations here too. And out of those 9 combinations, 6 result in wins.
Above, I was doing scenarios to show what happens for one door (as all other doors will show the same pattern). Hopefully this is a clearer explanation.
Rundvelt
I can understand those possibilities you've enumerated, but I think the moment we gain information (the goat reveal) the probabilities adjust as we now have more knowledge about what could not have been possible from the start.
I've prepared all the possibilities as I see them, but in order not to create more confusion than necessary I'm holding it back for now and instead try to explain the main problem I'm having in understanding this:
Imagine we're at the stage where we've chosen a door and a goat was just revealed.
Imagine further that our memory gets wiped right after that.
Now we see 2 closed doors & a goat.
That means we have the choice between two doors and the car could be behind either one of them.
Choosing the door we had chosen before the memory-wipe would be like sticking to the door.
Choosing the door we haven't chosen before the memory-wipe would be like switching the door.
So, would you say that before the memory-wipe the door I would've _switched_ to had a 2/3 win-chance, but _after_ the memory wipe that same door had only a 1/2 win-chance?
If you do, then how is the pre-memory-wipe situation any different in terms of making a decision to switch than the post-memory-wipe situation?
FINALLY.
I think the issue is that when we talk about "concentration of probability", we shouldn't say that there is a 2/3 chance the car is in door 3, we REALLY should be saying there's a 2/3 chance there is a goat behind your door, so you should switch to avoid opening your likely goat door.
Yes! I kinda got it from the 'concentration of probability' idea, but by flipping the problem, i.e. it's not "What's the chance I've picked the car" it's "what's the chance I've picked a goat" I finally get it.
tanukigalpa OOOOH ITS ALL ABOUT THE INITIAL PROBABILITY THAT YOU CHOSE A GOAT OR CAR
+Double Shrekt And that Monty knows whats behind each door, and so indirectly leads you to higher odds of winning, if you switch.
yes that's also how i understand it, i think it's easier that way.
EXACTLY, its better to say you have a 2/3 chance that you get a goat rather than saying there is a 2/3 chance that you will get a car when you switch
This really clicked for me. Much better than the other video with “concentrated probability”.
There are three possibilities. One where your initial guess is correct, and switching makes you lose. Then two different possibilities where your initial guess is wrong, and switching means you win. Boom, 2/3 chance to win if you switch. Great video
That makes so much more sense ty
@@pranavsanthakumar106 Agreed. I was lost, until this explanation.
The concentrated probability explanation is equally valid but more purely mathematical, and that is why a lot of people struggle with it.
You said that much more succinctly than I did.😀
The explanation I posted before reading yours is identical...but you said it way more concisely.
I do think it's worth adding that in the two cases where your initial guess is wrong, Monte has only ONE possible choice, which "gives away" that you MUST switch to win.
Omg, after so many years I finally understand it. I'm find it really odd that nobody could ever just have explained it to me in this way. Thank you.
Or it could've just been explained normally through maths....
Yeah but some people need to understand the consepts or else they won't accept the math to be anything more than "theory", but some do get it that way.
To me, it was pretty simple. You know your first choice has only a 1 in 3 chance of being correct, and it's more likely to be one of the other two doors. It's like Monty is pointing you to where it probably is (2/3 chance) by showing you where it isn't.
My math teacher made a helpful example back in the day. If you do the game with 100 doors and Monty opens 98 of them after your choice, it became clearer to us that your chance of having picked the correct door in the beginning is smaller than the likelihood of the other (last) door being the car.
yes ok . my brain can understand this , thank you ,
make it even clearer: If you had to select from a trillion doors then after opening all but two it becomes very obvious that it’s extremely likely that the car is in the other door than what you picked
This explanation FINALLY made sense to me where every other one sounded ridiculous. Thank you!!
Disagreed with the official explanation until I saw this video. I was wrong all along. Thank you for correcting me.
+strayster2 No you where not wrong. We where not given enough information to fully understand the concept. This man explained that the host knew where the items where. I for instance was under the impression that everything was random
+David Arencibia Ah. That really is the rub, isn't it?
+David Arencibia of course the host knows where the car is... Do you really think he would reveak the car and say " oops lets retake this"...
+Sundqvist And even if the host revealed the car you then should CHANGE your door to the "car door", right?
I faced the exact same situation. I thought I 'solved' the paradox and found out that the probability is, in fact, 50%.
Last video on this I went "aaaaaaah I get it"
Some time passed, forgot some stuff
This video I went "aaaaaaahhhh I get it"
You could keep making these videos for eternity, I'd still be equally amazed each time.
I GET IT! I FINALLY GET IT! THANK YOU!
Great explanation, Monty knows, that's the key. Now suppose there are 10 doors, after you pick, Monty shows you 8 doors with goats.... you better switch to the one door Monty didn't show you.
falcon tinker that’s different as you added in more doors love
Ugh! Then PLEASE explain it to me! I can’t get beyond 50/50 chance.
Yeah the key is this. The player is picking randomly. Monty Hall is NOT picking randomly. So think of it this way: Player picks door 1 (33% chance of being right). Monty picks door 2 (0% chance of being right). Door 3 must have 66% of being right.
You are wrong... if he hads 0% of being right then why you add 33% to the other door insted of adding 16.5% to each door?
trululasoxd scotti Because of the order that the doors are picked, When you pick there are 3 doors. Each has 33% at that point (pure random). Your door will never change from 33% until the final reveal. But by adding information to the system, however, you CAN change the odds of the remaining doors because the remaining picks are non-random. Suppose there are 100 doors. You pick one at random (1% chance). Monty then opens 98 bad doors because he has information of which are bad. Now there are only two doors left (yours and one other). You think your door has 50% now? I say your door still has 1%, and Monty's door has 99%.
raven lord I think your explanation is even better than the video's.
even after watching the vid, i still didn't understand. But reading ur comment, I did :D
Honestly for me when I think of this problem the way I remember it is that Monty isn't being random. That's why the probabilities get skewed. Noticing non-random actions is the most useful technique for dealing with this class of problem, which most other solutions do not help with.
This video is still helping people like me ten years on. This never made sense to me before, and now it at least sort of does. Thank you so much!!
just remember that most likely you will choose a door that does NOT make you win.
There is a greater chance of picking the goat first time so it's better to swap.
+nekogod This one line you wrote is literally the simplest and best way to describe the problem.
+UsuriousCactus8 The problem is that people incorrectly assume that once a goat is revealed, the probability of our originally selected door containing a goat changes instead of remaining fixed. Unfortunately, people often don't understand that the deliberate, non-random nature of the goat elimination is the key, and there doesn't exist a very simple one sentence explanation for that.
+Pine Fe beginner programmer here: may I ask why you would use a times variable (especially as a double) when it is essentially the same number as the count variable?
maxid87 You're right, he could have either initialized count to 1 and then printed (success/count), which would have made for 100,000,000 iterations instead of the 100,000,001 iterations he's getting with the way he set it up, or he could have printed (success/(count+1)) and stopped at < 100000000 instead of
+TedManney yeah I was just wondering why one would do this in some example code that is usually supposed to use as little code as possible. Also it seemed inefficient to initialize a double when it only stores integer values.
THIS video explained it properly. the last video said that the 1/3 chance of the opened-goat-door was 'added' to the not-chosen-door, which is kind of 'imprecise' and not very easy to understand.
With this video you explained all possible outcomes and showed that 2/3 of the time you get the car, if you switch. Well done
I dont feel it was imprecise. Now that I understand it both feel valid. But definitely for me, reframing the concept as the odds your first pick contains the car cemented the idea--allowing me to accept the other mental model of probability "concentration" which has different intuition benefits.
Kind of like how both analog and digital clocks accurately represent time, but analog quickly gives a rough perception of duration left in an hour or day and digital more quickly gives a precise answer for the exact number of minutes remaining. Both tools accurately represent time, but answer different questions better.
This is the explanation I posted on the other video. The other explanation with 100 doors just makes things too complicated for some people and is unnecessary when this simple explanation does the job.
This problem has been explained to me at length by multiple people, and it’s still been baffling me for years. I skimmed Wikipedia before watching this, and it being pointed out that Monty knows the contents helped a little bit, but I still didn’t totally grasp it. It’s never occurred to me that of course the probability of the INITIAL choice being a dud was actually the key factor. THANK YOU. Truly. I can sleep peacefully now.
OK I didnt get it last time but now I have it and I'll try and give an alternative explanation to the problem. We'll test all the doors with the same configuration so you get a feel for what is happening. Here goes: Lets say that you have a goat behind both doors number 1 and 2 and the car is always behind door number 3. Ok so first lets pick the first door. There is a goat behind it. Monty then reveals the second door because it is the other one that has a goat. If you decide to switch you win the car. Now lets pick the second door. Yet again there is a goat behind it. Monty is then forced to open the first door to reveal a goat. If you switch to the third door, you win the car again. Now lets say you picked the third door. Monty then reveals either door number one or two. If you switch you loose the car in this situation. But so if we add up the numbers you won 2 times out of 3 and only lost one. So the probability of winning if you switch is 2/3. Hope that helped
I commented on the first video pretty much explaining "If you switched while on a goat, you get the car (2/3), and if you switch when on the car, you get a goat (1/3)", similar to how this video explained it. Even then, I recall at least a couple of responses still not understanding that the third door plays a part xD
The third door doesn't play a part because he is always going to open a door with a goat no matter what anyone says.
I just say "With the always-switch strategy, you only lose when you guessed right, and that only happens one out of three times. The rest of the time you win".
I read an entire Wikipedia article on this and didn’t understand where the 2/3 chance was but by 2:19 I understood. Great video and very clear explanation.
This is my favorite explanation, but presented better than I usually do.
I usually put it [not quite this briefly] ..
The game plays out the same every time: You pick, Monty shows a goat, you get tyo decide if you switch.
You know it's going to go this way.
You can decide to switch or not BEFORE Monty does anything, and nothing would be different.
If you decide to not switch, you only win if you initially pick the car: 1/3 chance.
If you decide TO switch, you win if you initially pick EITHER goat: 2/3 chance
I made this too - it is on Numberphile2 but in case you don't see it --- Monty Hall Problem Express Explanation
Still totally wrong.. I mean I dont get it.. no Im kidding :D This is even better an explanation then the other ones, ohnestly. But they were good too :)
Those who don't get it with this explanation ... Please do yourself a favour and give it up to try :-)
This absolutely kills you, doesn't it? An explainable logic (well, statistics) problem that some people refuses to believe is true. While well maintained, I could feel the absolute rage in your gestures and complete disbelief people couldn't or wouldn't understand.
Thank you brady! this is the key piece of information many people seem to miss, and many describers seem to leave out, im very happy you clearly stated how one of the key factors is the hosts obligation and knowledge to open a zonk, and showing the odds before switching
I don't think this problem could be explained any better than you have here, cheers!
*****
I tried with code, using random function in Java after 1 million experiments I get about 66% (like 66.6331% , 66.7482% and etc. ) chance of winning. Or someone could try rolling dice like few hundred times if they do not trust computers :)
Who says I want the car? Maybe I want the goat.
OK, so then don't switch, because there's a 2/3 chance you have a goat.
You can sell the car and buy more goats. You have greater +EV of goats by switching.
😂, smart man, goats are a great investment.
@@euming got me there
@@driziiD dude, that's great.
I fail to see how anyone will disagree with your logic here Brady, but I'm pretty sure they still will!
I can disagree. The 2 in three is correct, but the "if the host didn't know is NOT correct in the second half of the video. Let's grant thatbthe host doesn't know, and 1/3 times opens the door with the car...that changes the odds to 0....bit if he opens a goat door the odds to switch are 2 in 3, we BOTH now know it was a goat door, and the prior knowledge of the host is irrelevant...he got that bit wrong
@@75ur15 No. If the host opens one of the remaining doors by random, the chance is 50/50.
In 1/3 of cases you choose a car, host opens a random goat and you lose if you switch.
In 1/3 of cases you choose a goat, host opens a car and game is over.
In 1/3 of cases you choose a goat, host opens the other goat and you win if you switch.
The two cases when you get the opportunity to switch has equal probability.
@@sverkeren I've already corrected this on my own comment, and you are correct
I want to post to admit I was wrong. I did my test using cards, two jokers and one ace. Shuffling at random and picking a card. The key is I think that the odds are 2 of 3 that I picked a joker. Removing the other joker still means that the two of three chance is on your side if you switch. I think it adult to admit my first reaction was totally wrong, but the proof is in a test.
That way of explaining it was really helpful to me. The biggest thing (for me), is that you have 2/3 chances of getting it wrong the first time.
Thanks! :)
I feel like Brady explained it the best.
I thought I understood it after the first video. Now, I've re-imagined the problem ... and now I'm having issues.
Presume that door number 1 is the door first chosen by us -- the contestant. Initially, we know that there are only three possibilities:
(a) 1.car - 2.goat - 3.goat;
(b) 1.goat - 2.car - 3.goat; and
(c) 1.goat - 2.goat - 3.car.
In scenarios (b) and (c), Monty must open up a specific door -- door number 3 in (b) and door number 2 in (c). However, in (a), Monty could choose to open up door number 2 or 3 ... because there's a goat behind both doors.
Thus, here are all the possible scenarios:
(a) ... Monty opens 2 ... SWITCH = GOAT and STAY = CAR ;
(a) ... Monty opens 3 ... SWITCH = GOAT and STAY = CAR ;
(b) ... Monty opens 3 ... SWITCH = CAR and STAY = GOAT;
(c) ... Monty opens 2 ... SWITCH = CAR and STAY = GOAT.
So, if all scenarios are listed, IF YOU SWITCH, you will win the car in two instances and get the goat in two instances; IF YOU STAY, you will win the car in two instances and get the goat in two instances. This returns me to the 50% chance.
Please, somebody, explain why I'm wrong.
mickavellian It is not a "50/50 choice". You have failed to understand the argument he gave here based on the principles of probability. Always switch, like he says, that doubles your chance of winning.
mickavellian You are the incoherent one here. Your other i-word might apply too;) You have missed the point of the whole Monty Hall probllem. Once the contestant sees one opened door with no prize behind it, it is no longer 50/50.
Sure, that is counterintutiive. But real math and real science are full of results that are counterintuitive to those who have not learned the concepts.
Here, the main concept that is so importans is "conditional probability". It is notoriously difficult to learn.
mickavellian There are many things you got wrong in your long screed, but the key one is this: "With two doors you are going for heads or tails (50/50 .. correct?)"
No, not correct. Once the game show host opens a door revealing no prize behind it, he has leaked information about where the prize is, so it is no longer 50/50; you now know (if you understand conditional probabilities) that your odds of winning are 2/3 if you switch, 1/3 if you stay.
mickavellian You still haven't figured out your objection is without merit. What happens if the door is taken off and another contestant brought on it entirely irrelevant. You have changed the conditions, so the conditional probabilities are different.
mickavellian No, he is wrong and so are you. You simply do not understand probability at all -- just like the many people Monty Hall took advantage of on the game show.
I have been grappling with this and could not get past believing it comes down to a 50/50 choice after revealing where the goat is. I believe this explanation helped me properly understand the paradox by focusing on the fact that Monty knows what is behind all the doors.
In this case, knowing is 2/3 the battle. GI JOE!
So basically, if you choose a door with a goat behind and switch you win, which happens 2 out of 3 times. Understood
Thanks for helping me understand
I couldn't understand this problem for days but this explanation really helped me visualize it. So let's say that instead of 3 doors, there are 100 doors. There's a car behind one of the 100 doors and a goat behind the other 99 doors. The rules are the same. You pick a door and then, the host opens 98 other doors, leaving only the door you picked and one other door unopened. Now, would you rather stay with the door you picked randomly amongst 100 doors and switch to the one door that the host suspiciously left unopened? The problem works the same way with 3 doors but the only difference is that this time, the probability of winning from switching is 67% instead of 99%. Hope this helps and please leave a like if this helped you understand the problem.
It makes so much more sense when there’s more doors
Okay I was upset that I still didn’t understand but you putting it this way made it click instantly and I feel silly for not understanding it from the video. Thanks!
Your explanation is clear and compelling. My explanation is very similar to yours, but may be even a smidge easier to understand...
Scenario 1: I go to the door with goat #1.There are two doors left, but given his knowledge Monte has only ONE choice. He has to open the only remaining goat door, and therefore I SHOULD DEFINITELY SWITCH to the door he didn't pick, which has the car.
Scenario 2: I go to the door with goat #2. Identical logic as above, I SHOULD DEFINITELY SWITCH.
Scenario 3: I go to the door with the car. Now Monty does in fact have two choices - goat 1 or goat 2 (50/50 chance each). This is the only scenario where I WIN BY STAYING.
Each of the three scenarios is equally likely, but switching wins in two out of the three of them.
Yeah, your going through each step makes it more clear to me, but I've found by reading comments that the solution seems to click into place for people in slightly different ways, which is interesting. Most other comments make no sense to me.
Edit: this was explained in the first video..
This is how I explain it:
There's 1,000,000 doors.
You pick one of them.
Monty opens all but 2 doors - the one you picked (#2 maybe?), and another one (lets say #3647).
All the doors Monty opened have goats behind them, and now he asks if you'd like to stick with your door (#2) or change to this seemingly random door (#3647).
The chances that you picked the correct door the first time are 1 in a million, therefore it will nearly always be that other remaining door.
Just because there's only 2 doors remaining does not mean there is a 50/50 chance.
Eetarsaurus That only works when the moderator knows where the prize is. If he opens random doors with the chance of revealing the prize early, the chances split up evenly for the doors.
The original video made the example with 100 doors and people still didn't understand it, even though it was clearly obvious. Going higher should make it even more obvious, but I think some will still not understand it xD
Eetarsaurus Shouls have watched the video first. He said just that. Sorry mate. :D
Eetarsaurus Troll post is Troll :D
There is a very simple answer to this. There is a 2/3 chance that you pick a goat, and Monty always picks a goat. This means there is also a 2/3 chance that you both pick a goat, meaning the last door has a 1/3 chance of NOT being a goat (and 2/3 chance of being the car).
Yup, that is exactly what it is
NOW I GET IT. God I feel dumb.
When I was introduced to this problem I also thought it has to be 50/50. What opened my eyes was the same problem with 1000 doors.
1 car, 999 goats. You pick one door, the host opens 998 goat doors.
If you stick you have a 1/1000 chance to win.
If you switch you have a 999/1000 chance.
That was eye opening to me.
I recently discovered this problem. I've watched several videos and yours are by far the best ones. This specific video clarifies any doubt that I could had remaining. Great job.
I am back again. This time to counter my own comment that I made a couple days ago. Damn, I would have swore and bet my life that the odds in the Monty problem were 50/50 But I drew it out on paper and realized that Monty DOES change the odds. Before, I thought he made no difference but he does. Do this: Draw three doors three times. Put the car behind #1 in the first row of doors, behind #2 in the second row and behind #3 in the third. First row if you pick #1 Monty will pick #2 if you switch you loose. Next row if you pick #2 Monty picks #3 (because the car is behind #1). If you switch you win. Third row you pick #3 Monty picks #2 you win. Yes, you have twice the chance of winning if you switch. I have been humbled.
Neat explanation Dean. Please stick around - there are harder nuts to crack, some persevering with their own 50/50 blogs for over six years and not worked it out!
For me the visualizing the 100-door version did the magic. 99 goats, 1 car, the chance of you picking a goat is 99%. If you do pick a goat then the other door MUST have the car because Monty has to reveal 98 goats no matter what. In other words, the only case you lose after switching is when you have initially picked the car. Which has the probability of 1% (or 33% with 3 doors). This is somehow much easier for me to grasp than the original 3-door problem
your'e wrong, but the answer is still switch. say you always pick door A, you would think the odds would be
1. W, L, L
2. L, W, L
3. L, L, W
but option 2 and 3 are the exact same, it makes no difference what order the doors are in.
Morgan Bartholomew Well spotted. I'm sure the OP meant: "Draw three doors three times. Put the car behind #1 in the first row of doors, behind #2 in the second row and behind #3 in the third. First row if you pick #1 Monty will pick #2 or #3; if you switch you lose. Next row if you pick #1 ... etc". That's what I saw (by not reading it properly).
I've never actually seen someone on the internet admit they were wrong. Bravo
Well explained. Emphasing that Monty's knowledge forces his hand and therefore increases your probability of winning.
Fantastic video!
My take - you have a two in three chance of picking a goat at the beginning and therefore it’s more likely that the car is behind one of the doors you did not pick. Monty HAS to pick a goat out of the remaining two doors thus eliminating the door you don’t want out of the the 2/3 or “better” side. Monty is the key.
To make it simple:
The choice is "Would you like to stick? Or choose BOTH of the other doors instead?"
We already KNOW that there HAS to be a goat behind one of them since there is only one car. So showing it to us doesn't matter in the slightest...
reading these comments is depressing... here I tought your average numberphile viewer was somewhat intelligent. And while initially being wrong by choosing the intuitive 50/50 answer isnt a sign of low intelligence, refusing to alter your position once presented with such crystal clear evidence surely is.
You can NEVER go wrong when you use an exhaustive solution. Fortunately, in this case, it's ridiculously easy to use an exhaustive solution. There are 3 conditions: oxx, xox, xxo, where the valued pick is the "o" door. And there are three strategies: pick door 1, 2, or 3. If you apply all three strategies to all three conditions, and always take the switching choice, you will see that you will always win 2 out of 3 times under all three conditions. Therefore you KNOW that switching gives you 2/3 probability since you've tried ALL combinations. If you don't switch you will win only 1/3 of the time with each strategy under all three conditions.
This explanation is the best I've seen. In other explanations the fact that "Monty knows" seems to get lost (when this is crucial) and so you only focus on the odds (because you're assuming it's a random environment).
the disbelievers need to have their belief shaken by empirically demonstrating the phenomena !
I have my own explanation of this problem. Might help you.
Step one: There is a 100% possibility that the car is behind the 3 doors.
If you select one of the three doors (lets say door A) that gives you a 33% chance of the car being behind that door.
Which also mean there is a 67% chance of the car being behind door B and C. (33% of door B and 33% of door C)
Step two:
The Host now has door B and C and a 67% chance to have a car behind his doors. (Still the 33% of door B and the 33% of door C combined).
The host opens door B, which has a goat. That still means that Door B and door C have a combined chance of 67%. So door C gains the 33% of door B and has 67% now.
Step three:
You switch to the Combined doors B and C and get a 67% chance.
Thanks, it's more clear now!
Yeah I prefer the explanation in the video but this is valid too.
Oooh, that makes much more sense. The previous video was framing it as if there was somehow probabilities being 'concentrated' which is a silly idea. It makes more sense to frame it as the probability of your initial choice.
You pick a door. 2/3 times you will pick a goat. Monty Hall shows you where a goat is. If you picked a goat (which you will do 2/3 times) Monty is showing you where the other goat is, therefore if you switch, you will be switching to the car, 2/3 times. People seem to think this is a fool proof plan, it isn't. This is a strategy that, statistically, works more often than it doesn't.
THANKS ! I finally understand the correct answer!
you're welcome
Shunya No, it does affect the decision, it merely does so in a trivial way.
When I first was learning about the Monty Hall problem, I had some trouble, but I eventually figured it out, like, a few years ago.
I think this is a very good explanation of it, and one that can be understood by pretty much anyone.
I mean though, there are only 3 possible configurations in this system. Either you've picked the car, you've ;picked goat 1 or you've picked goat 2.
When in doubt, think through every possibility.
I initially chose...
Car..... remaining after Monty reveals a goat has to be a goat. -> Stick preferable
Goat1..remaining after Monty reveals a goat has to be a car -> Switch preferable
Goat2..remaining after Monty reveals a goat has to be a car -> Switch preferable
***** This. Going through every possible state is a pretty irrefutable way of coming to an answer.
I cannot explain how wrong i thought people were who said you should switch but after seeing this i finally understand it. The fact that monty knows it was very well betrayed and that we could see all the things behind the doors at one point brought the idea home. I think a lot of the other videos the people saying it may not know what it all means so are just explaining it from a script but this really helped.
Btw not saying i'm smarter than the people in other videos. Like i'm doing my GCSE'S but i finally understand the meaning and its due to the way that you laid it out which meant that i got it 2 mins in.
Do you mean betrayed or portrayed?
The fact that you have a 2/3 of selecting the goat initially, and knowing Monty will always show you a goat, you have a better chance of winning the car by switching. Not always, though
@@kuperlilu5340im sure it was portrayed
Thank you for stating the problem correctly.
There are 4 important things to emphasise:
- the host knows what is behind all three doors.
- the host asks the contestant to randomly select one of the three doors.
- the host must then open one of the two doors you have not selected.
- the door the host opens must have a goat behind it.
- the host must then offer you a switch to the other unopened door.
And there is also the implied rule where the host cannot open the door the contestant has chosen, or there is no game
@@jmak9376: "there is also the implied rule where the host cannot open the door the contestant has chosen"
It can't be implied. It must be spelled out.
That was the 3rd of my list of 4 important things to emphasise:
- the host must then open one of the two doors you have not selected.
I got this the first time but I'm glad you were able to make it even clearer for people.
I've watched like 4 explanations of this, but this one is the first one which finally males sense even to a layman :D
I first saw this in OMNI Magazine back in the 70s. They got letters from math professors from very prestigious schools swearing that either way it was 50/50. It must be hard to see past the idea that there are two doors and you are choosing one. It is like saying either big foot exists or big foot doesn't exist, so it's 50/50. I just want to find one person willing to play this game for money giving those odds.
Summary is what you chose is probably wrong, plus the fact the host will always reveal another wrong one, further confirming your wrong initial choice, so the remaining is most probably the right one so always switch to that to maximize winning.
Best non visual breakdown & explaining further, you only have 1/3 chance of choosing the car door, so switching has a bigger winning rate of 2/3. the host will ALWAYS remove a goat door which gives the change of choice (switching) an additional 1/3 (total of 2/3) compared to your initial choice of 1/3. this solution only works if the host ALWAYS removes a goat door. if the host doesn't open any doors then this will truly be a 1/3 chance of winning regardless if the host asks you to change your choice or not.
Fun fact: Monty Hall is still alive, he's turning 95 next month!
maybe he will (or already did) no guarantee. What is the probability he will?
Ro Jay His birthday was 2 days ago (and he's still alive) :P
So 100%...
How about now?
Just to keep you guys in check about ten months later. Monty Hall is still kickin'.
RIP Monty Hall, September 30, 2017
Thanks. You clearly emphasized the major factor that the probability of choosing a goat or a car in the very first door that you are opening. This important factor surely affect the calculation. Plus, we should never omit another important factor that Monty Hall has the knowledge of where the car is and he should open a door that has a goat behind, probably this important factor is always being omitted when stating the scenario...
Ahh this makes sense now. If you pick the door with the car behind, the car is behind your door (obviously). If you pick a door with a goat, the car is always behind the other door. There is a 2/3 chance of picking a goat, therefore a 2/3 chance the car will be behind the other door
It really is that simple. Funny how some people twist themselves into pretzels trying to over-complicate this.
Boom, got it! For me it clicked knowing I had a 2/3 chance of getting it wrong. The odds are I did. THEN Monty reveals where the car can't be. So having a greater chance of getting it wrong first combined with 100% chance of getting it right next (If I indeed DID pick wrong first) - winner, winner, chicken dinner.
9 possibilities in this whole situation:
Pick 1 - Prize 1
Pick 1 - Prize 2
Pick 1 - Prize 3
Pick 2 - Prize 1
Pick 2 - Prize 2
Pick 2 - Prize 3
Pick 3 - Prize 1
Pick 3 - Prize 2
Pick 3 - Prize 3
You win 33% of the time if you stay. 66% of the time if you switch. Any questions?
A player's beliefs are self=fulfilling prophecies.
Player 1) Clearly 1 in 3, no need to change, so I won't.
Player 2) Clearly 50-50, so I will flip coin.
Player 3) Numberfile says 2 in 3, so I switch.
Each of these players will get confirming results!
I kind of got it in the first one, but this one explains it REALLY clearly
Stating the fact that Monty knowing what's behind all the doors like no one ever did in the explanations that I read about this problem made me understand it. Thank you
Monty wouldn't be able to open the door with a goat every single time, if he didn't have information about contents of the doors.
For those still scratching their heads, consider this variation:
Monty takes you into a room of 100 doors. wow. He says 99 of them have goats, and one of them has a brand new sports car. He tells you to pick one. after much hemming and hawing, you pick one.
Monty then waves his hand and 98 doors fly open, and 98 goats spill out into the hallway!
Once the goats have all cleared out, we are left with two doors: The one you originally chose, and one other one.
Then Monty says, "Before we reveal your door, I am going to give you a choice... you can keep the door you have already chosen, or you can switch!
So would you switch?
I really would like to understand this but I don’t. Why would I switch? I chose and he’s now told me I didn’t choose one of those 98. I don’t get how if I had picked the other one he hasn’t yet opened it would change anything. I’m truly not getting it from a practical standpoint. What makes it more likely that the car is behind the door I did not choose if I chose at random to begin with?
You are an absolute lifesaver. It’s 11pm and this was driving me crazy. Finally someone explained it in a way my simple brain could understand. Thank you so much
The biggest mistake of people who don't understand the Monty Hall Problem is that they accidentally see 2 empty boxes as one box. So when they choose a box, they think there are only TWO different scenarios in front of them: The choosen box is empty, or the box has prize in it. It's WRONG.
In fact, there are THREE different scenarios about choosen box: The prize box, the 1st empty box and the 2nd empty box. Remember, there are 2 empty boxes in the game, not one.
WTF? No boxes, just doors and none are empty. If you don't switch doors, you will always only win 1/3 of the time, so switch!
Thanks! The key realisation for me was "Don't focus on what you wanted, focus on what you probably got."
You have a 66% chance at picking a goat on your first try, if you do, your swap will ALWAYS be the car since the remaining goat is then eliminated by the host.
That's it. There's no more trick to it. You had a higher chance of picking the goat the first time therefore the swap will win you the game.
This video is correct.
Well there is actually a trick to it. It's that Monty knows and his hand was forced. Had you arrived at the same scenario where you pick a door, the a goat is revealed, however Monty didn't know where the car was. Your chances are actually 50/50. This happens because rather then converging two option into one, we completely eliminate what was a possibility, that he could have eliminated the car. If Monty knew, odds are 33/66, if he didn't they are 50/50.
reread please: "There's no MORE trick to it."
Your swap will not ALWAYS be a car. 1/3 your swap will be a goat. 2/3 it will be a car.
@@jpg7616 I did say "If you do" in my first statement. Perhaps I should have written it in all caps to clarify?
To be honest I did not understand this after the last video and my friends who claimed to understand could not explain it well. However, after playing out the scenarios with two salt packets and a sugar packet on a Waffle House table, the whole concept became very clear. This video only reinforces the knowledge I came to understand on my own, which is how most people learn. So, if you are still having trouble with the concept, act it out for yourself! It worked marvelously for me.
Jeez... the video is wrong!!! All of this quite 'frighteningly' shows how people are so easily influenced to believe something, which is clearly incorrect. No wonder that propaganda and fake news are having such a dire consequence on our society.
'Think' for yourselves people... don't let those in authority or with slick videos do that for you!
@@sebastianwrites You need to do some thinking yourself. The host doesn't have a door that has a 1/2 chance of anything to leave for you to switch with.
Ok I think I get it now we should focus on the probability of the outcomes not the focus on the doors themselves. 2 outta 3 times you'll have picked the goat so you're likely to get the car by switching.
''Goat is in 2/3 of cases'' and ''Host knows where the car is and won't open that door''- These two cleared everything.
Someone should make a crossbreed between the Monty Hall and Schrodinger's Cat. Where... I don't know there's a 50/50 chance the goat has died on your indecisiveness to either switch or stay.
How about trolly problem.
Monty hall trolly problem. Two people are stuck to one of three tracks each and one track is clear but you don’t know which one. After picking a track one of the other two is revealed to have someone stuck to it. Do you change the track?
The pain on his face at 4:11 when he says to leave a comment...
lmao
I'm amazed that there are still people who just don't believe the simple trutth when it's handed to them on a silver platter.
If you pick a goat, which you have 2/3 chances of doing, Monty indirectly shows you where the car is and you win it by switching.
There's only 1/3 chances that you don't pick a goat the first try, and lose the car by switching, hence by always switching you have 2/3 chances of winning the car.
It's really not that hard, once you get past the counter-intuitiveness.
how does montys knowledge change this as long as the one he chooses is the goat even if it was by accident in that scenario the chances are 2/3 if you change
So, the key is that your first choice was probably wrong, so when Monty reveals the other wrong choice, you know which one is probably right. This video finally explained it. Thank you!
0:37 HOW RUDE
I like Brady's explanation better, but I also have my own explanation, and it involves the 100 doors. Let's say you chose door #1, just like in the first video. Monty then reveals 98 doors, but leaves one door closed. Now, the car can either be behind your door, or the door that Monty didn't open. Think about it: what are the chances that, on your very first try, out of a 100 doors, you picked exactly the one containing the car? The probability of that happening is only 1/100. But Monty left one door closed, and the difference is what makes it obvious to switch. You prooobably didn't get the car with your first choice, but Monty left one door closed, whilst all other doors contain the goats, so the one he didn't open prooobably contains the car.
+Niker107 best explanation
because your chances of picking the right door is 1/100%. all the other doors are OBVIOUSLY the wrong door because he opened it. you get the benefit of opening 2 doors when you switch, therefore, more doors for you, more chances. The 98 doors opened tells you that the probability favours that one single door monty did not open
even my dad got this
This is a great way to explain the problem, and probably the best for the non technical crowd. But for the technical folks, the easiest way to prove this to them is just to make them demonstrate it to themselves. Programmers can write a trivial simulation, mathematicians can draw out the state table. It's trivial to demonstrate this to yourself.
To add to my previous comment (and after looking at the video again), I further understand that considering that are two goats and one car, that chances of picking a goat are 2/3. For that reason, I should assume I have chosen a goat. After Monty shows where one of the goats is, assuming I have chosen a goat, probability says I should switch as that is more likely the car.
PLEASE, PLEASE, PLEASE! Numberphile, explain the "Bertrand's box paradox" and the "Three prisoners problem"... At first I had trouble believing the Monty Hall problem, but now it's really clear to me!... I tried to use the same logic to solve those two, but it just doesn't feel like it's explanied as this. D:
Ha, I get it now!
Thank you very much!!
thanks for watching
It's good, isn't it?
But then you get annoyed at those who don't get it.
I probably can't explain the logic behind it any better than anyone else, but I can lay out the bare probabilities
If the car is in door one, and you pick door one, switching will lose you the game.
-The car was already in the door you picked, so clearly you lost
-Switching loses the Game
If the car is in door two, and you pick door one, switching will win you the game.
-The car was not in the door you picked. Monty Hall opens the other door without a car. Therefore switching puts you on the door with the car
-Switching wins the Game
If the car is in door three, and you pick door one, switching will win you the game.
-The car was not in the door you picked. Monty Hall opens the other door without a car. Therefore switching puts you on the door with the car
-Switching wins the Game
If the car is in door one, and you pick door two, switching will win you the game.
-Switching wins the Game
If the car is in door two, and you pick door two, switching will lose you the game.
-Switching loses the Game
If the car is in door three, and you pick door two, switching will win you the game.
-Switching wins the Game
If the car is in door one, and you pick door three, switching will win you the game.
-Switching wins the Game
If the car is in door two, and you pick door three, switching will win you the game.
-Switching wins the Game
If the car is in door three, and you pick door three, switching will win you the game.
-Switching loses the Game
Tallying up the results, you'll see that I won 2/3 times by switching doors.
I think ASAPScience's video on the Monty Hall Problem may help too.
I have a friend who had a hard time with this. How did I convince him? We sat at the table and played the game with two jokers and an ace. 20 times with no switch, 20 times with always switch. Once it was in his head that it was true, it was easier to show him the explanation of WHY it was true.
Way better explained than in the previous video :)
Thank you so much for explaining this. I was pretty confused about it but the way you explained it made perfect sense.
You are either asking if switching is better BEFORE the third door is opened, or after. If the third door is a variable then yes it favors the player no doubt, but once that door is revealed and your answer is dependent upon this information, then the third door being NOT opened or being the CAR are no longer variables so yeah.
*Play Monty Hall: Comments Edition!*
Pick a door!
[1] [2] [3]
The gameshow host opens a different door, revealing a goat.
Now, do you want to stick with your original door or switch to the other one? You must decide STICK or SWITCH before reading the next paragraph.
...
Ok. You chosen stick or swtich. If the current number of likes on this video is a multiple of 3 then the door you selected at the start was the one with the car behind it.
(an easy check to see if a number is a multiple of 3 is to add its digits. If the sum of the digits is a multiple of 3, the original number is also a multiple of 3. e.g. 732 is a multiple of 3 because 7+3+2 = 12)
If you stuck with your original door and won, congratulations - you beat the odds!
If however you switched and won, you were sensible and trusted in the mathematics.
:D
This finally clicked :D Instead of thinking about this as pure chances, I needed to think about this as permutations, and that all possible outcomes are goat -> car, goat -> car and car -> goat, and it is only because the host only opens the goat doors. Brilliant explanation
I believe it's true, but it still feels so strange, because all of my "math instinct" keeps telling me that the two doors should be 50/50.
This is why it's dangerous to do math on hunches and "gut reactions" they can be highly misleading.
Even if monty wouldn't have known where the car is and wouldn't have opened the door With the car, you should switch.
I tried a simulation on my ti83 let it run for 100 games and I won 64,978% of the time
It will creep ever closer to 66.66... the more iterations you do :)
after two days I FINALLY GET IT thank you. all the other videos do not emphasize enough that monty KNOWS what is behind every door, and what that means. the scenario for this brain game is more complex than most people understand it to be, and i think that makes people, me included, "get" but not get the answer.
Now try explaining to someone that swapping does *not* confer an advantage in Deal or No Deal because the eliminations of the cases are blind and random. It can be literally impossible to get some people to accept this, they insist that DoND is a special application of the Monty Hall problem.
tl/dr: You will *LOSE* two-thirds of the time if you *DON'T* switch, because naturally two-thirds of the time your initial choice will be a goat. You will *WIN* two-thirds of the time if you *DO* switch, because two-thirds of the time your initial pick will be a goat, and then the other goat will be revealed/eliminated/avoided behind one of the other doors, and then you will switch to the remaining door which will have the car! Two-thirds of the time. So, switch every time!!!
DEMONSTRATION:
================
Only THREE possible scenarios/combinations:
----------
1 2 3
a truly infinitely better explanation.
It doesn't matter what he reveals. All that matters is the second choice. Forget about the first event. At the end of it, the odds change to 1/2. There is a goat behind one door, and a car behind the other. 50/50.
Now just focus on what we have at this point. There's no higher likelihood by this point, or no more reason to assume that the other door has a sports car. When there's door 1 and door 3 left, each one has its own thing. When he asks, "Do you want to switch?" He's really just asking, "Door 1 or 3?" Simple as that. Just because you've chosen one, it doesn't have an impact on likelihood. You have to think about it in terms of real world chance.
Two choices doesn't imply 50/50. Take two people. One is a 90 year old woman from Fargo, N. Dakota named Mary Sue Lovejoy. The other is a 21 year old man from Bakalakastan named Ali Abdul Kaboom. You are told that one of them is wearing a suicide bomber vest. You have time to disarm one of them, but if you pick wrong, then the other one will detonate the bomb. Which one do you try to disarm? If you think it is 50/50, the you'll be wrong half the time. Dead wrong. This is a case were two choices is like 0.0000001% vs 99.9999999%.
The thing is you chose before he opened one of the doors. Your logic says by him opening a door your 1/3 chance of choosing the correct door on you first try became a 1/2 magically because he opened a door. After he opens a door it's still a 1/3 chance that you chose the right door. Ignore the fact he eliminated a door. You get 1/3 because WHEN YOU CHOSE THE DOOR THERE WERE 3 OPTIONS. The number of options after he eliminates one doesn't affect how likely you were to choose the correct one BEFORE he eliminated said door. So that 1/3 magically becoming 1/2 makes no sense because you are changing a independent variable based off another dependent variable, it's dependent because it's based off the door you choose first.
Finally I understand the problem, no video/person has ever before explained that Monty is FORCED to show a GOAT, thanks Numberphile
Well damn. I would just stick either way. I made my choice and i'd stick with it.
That's what most people did, in fact.
IronSoldier I think everyone can be happy with that, provided that you accept that you are taking reduced probability of winning by sticking.
Pick with your head, not your heart
Well I picked with my head the first time. When the odds were equal and guaranteed.
IronSoldier Then you picked what you've had to, and your head is probably not working very well.
I think I finally get it. Seems fairly obvious now lol. So when this was discovered were all the rules of game shows changed to get around it?
Thank you! I love this!
+Mami Oso you're welcome
After musing over this probability puzzle for a while, I arrived at a very simple explanation:
When you pick one out of three closed doors, you're more likely to lose. In other words, the car is more likely to be behind one of the other two doors. - This is the crucial point that many people fail to consider. - no matter which door you initially pick, the car is more likely to be behind some other door. ... So, after you picked a door, and the show host opens another door to reveal a goat, the best probability of getting the car shifts to the closed door that you didn't pick.
I would beg to slightly differ (about which part confuses people). People understand well enough that picking the car is less likely than picking a goat, but the issue is that people incorrectly think that seeing a deliberate elimination of a goat changes the probability of the initially-selected door containing the car, which it doesn't.
It *would* if and only if the host blindly, randomly eliminated a door which happened to contain a goat by chance. The difference that blind, random eliminations and deliberate eliminations make is the single hardest concept for people to grasp. It seems intuitive that swapping would still be more likely to win even if the host blindly selected a door that happened to contain a goat, but this is simply not true.
well put! You don't even need arithmetic to see that.