The rules of the game mean you can only ever switch from a losing box to a winning box or a winning box to a losing box. You can never switch from a losing box to a losing box. It's more likely you start with a losing box so therefor it's more likely you'll switch from a losing box to a winning box. So assuming the prize is in box 1 (although the paths will be the same for the other boxes) these are all the posibilities if you switch pick 1, switch -> lose pick 2, switch -> win pick 3, switch -> win

@xapie128

4 жыл бұрын

Your simple explanation is concise and perfect. As much as I like Eddie Woo's videos, I think he made his explanation of this one much too complicated. There are only three paths. Simple.

@Mad5am

3 жыл бұрын

Great explanation!

@user-zy8sf7tv2f

2 жыл бұрын

Hey man, you just give me a really great explanation, it's so great !!!

@ezzeldienrashad8374

2 жыл бұрын

Thank you. In fact, I didn't believe that switching has a higher chance after watching the video. I did after seeing this comment. This is much simpler.

@hueiwodh3408

Жыл бұрын

@@ezzeldienrashad8374 same

Anyone else wondering why there’s a whiteboard screwed onto another whiteboard?

@matthewcarl6369

3 жыл бұрын

This mad me laugh out loud. So Thank you.

@derfunkhaus

2 жыл бұрын

Maybe some one ruined the central part of the larger by writing on it with a non-erasable marker? Or maybe there's a portal to Hell that they had to close off.

@soumajitsen1395

2 жыл бұрын

@@derfunkhaus The mathematical probability for both is the same... right?

@derfunkhaus

2 жыл бұрын

@@soumajitsen1395 Maybe the reason you cannot divide by zero is that’s what opens the portal to Hell.

@soumajitsen1395

2 жыл бұрын

@@derfunkhaus A probability indeed.

What a great explanation! Wish I had you teaching this stuff in university. It was clear and concise!

I found it more helpful to break the problem down into a chart of possible decisions the player could make, keeping in mind that the game master will 1. Never reveal the prize 2. Always reveals a door that isn't the one the player picked and 3. Always lets the player switch doors if they want to Since the player first makes a decision with 3 options (choosing a blind door) and makes another decision with 2 options (switch or don't), the total number of possible distinct playthroughs of this game are 6 2 * 3 = 6 For argument's sake, let's say that Door 3 is the door with the prize. We'll start by looking at what happens if the player always stands by their initial choice: Scenario A -> Player chooses Door 1 and loses Scenario B -> Player chooses Door 2 and loses Scenario C -> Player chooses Door 3 and wins Overall odds of winning without switching -> 1/3 Now let's keep Door 3 as the winning door and see the distribution of wins when the player commits to switch ahead of time: Scenario D -> Player chooses Door 1, Game Master reveals Door 2, Player switches to Door 3 and Wins Scenario E -> Player chooses Door 2, Game Master reveals Door 1, Player switches to Door 3 and Wins Scenario F -> Player chooses Door 3, Game Master reveals Door 1 or 2, Player switches to Door 1 or 2 and loses. (Note that it doesn't matter whether Door 1 or Door 2 is revealed here, if you switch you lose either way) Overall odds of winning when you switch -> 2/3 The real bet that someone who always switches makes is that their blind initial choice will be wrong, because they start with a 2/3 chance of being wrong. And if their initial choice is a losing one, then switching makes them win. What I find interesting about this game is that if someone believes that switching makes no difference, and runs through it randomly choosing to switch or stay, then they will win 50% of the time and reinforce their belief that the odds are half and half.

@linsqopiring6816

9 ай бұрын

But if they do it always not switching 90 times for example, they believe they should win about 45 times but they will only end up winning about 30 times. To make it clearer let there be 10 doors and the host always reveals 8. They would still expect to win 50% of the time when it got down to the last two doors. But they would only win 10% of the time. By this time they should be asking themselves how can I expect my ORIGINAL guess to be right 50% of the time when there are 10 doors?

The key to this is that the host will *always* open a non-prize door. The host is who turns the odds in favor of switching. The scenario is easier to illustrate if you set up 10 doors instead, have the contestant choose one door, then the host closes 8 non-prize doors. The one remaining door has a 9/10 chance of being the right door. It's the same principle, just a little more clear to the intuition.

@dalsenov

Жыл бұрын

I think you forgot that the warden did not turn the previously chosen door. So you can consider 10 doors but at the end there are still left unopened 2 doors, not one! You can compare only what is comparable.

@linsqopiring6816

9 ай бұрын

@@dalsenovHe means the one remaining door OF THE 9 that were not chosen by the contestant.

If people still don’t get it I always like to put forward an example. Say you had 1000 doors, picked one from 1000 - odds are it is 1/1000 that you get a prize. Then, the host reveals 998 doors that don’t have a prize. This leaves 2 doors: the one you picked with a 1/1000 chance, or another door. Clearly, you have far better odds of winning if you switch.

@cuffzter

5 жыл бұрын

yep. Ive explained it this way too

@cullenmatthews19

5 жыл бұрын

Katniss Everdeen That doesn’t help me at all haha

@WilliamCacilhas

5 жыл бұрын

Cullen and Friends when you make your choice you have a 1/1000 chance of winning. If all doors except the one with the prize and the one you picked are open this doesn’t change anything. That’s because you made your choice before the other doors were open. So the probability you chose the right door is still 1/1000. The other door however would have a 999/1000 probability because only doors that didn’t hold the prize were open as well as the door you chose. What do you trust more, your ability to guess the correct door out of 1000 or the door that didn’t open after you made your low probability choice?

@dalsenov

Жыл бұрын

I think one is compelled to reevaluate once he has new information.I find meaningless that 998 doors were opened if I I didn't choose previously one of those doors.I think the problem to be put is "Should I reevaluate or not?".Suppose after the warden revealed 998 doors (as you supposed) , I stick to the previous choice and then the warden reveals another information ,revealing my choice that indeed contains the car. Shall I choose the losing door just for the sake of switch and bigger theoretical chances?? This would be preposterous!

@linsqopiring6816

9 ай бұрын

@@cullenmatthews19 It might help if you actually did the experiment. You can go to random org and generate several random numbers from 1 to 1000. Always pick number 1 and never switch. You will see that you will only win 1/1000 times. But if you switch you will win 999/1000 times. You can do this experiment in only a couple minute and it should convince you 100% if you still had any doubt.

My 6th grade teacher taught us this story and I thought it was the coolest thing every. She used the example of 64 suitcases and kept eliminated suitcases until there were only two options left. That concept clearer with more options.

To you doubters, you can easily prove this to yourself with 3 playing cards, say two jokers (goats) and one ace (the prize). Get your little brother to be Monty Hall and always reveal one door that HE KNOWS the prize is not behind. Try switching 20 times and sticking 20 times. You will prove that switching doubles your odds. That will convince you to go back and pay closer attention to Eddie's details. :)

@dhruvbaheti4512

5 жыл бұрын

I don't doubt it. I understand in mathematically. But comprehending that in real life is hard for me.

@freddycalabrese319

5 жыл бұрын

Some months ago I was doubting it even after the explanation, so what I did was I made a simple little program code that played the Monty Hall problem for me, and it showed me that not only it was true, but that the probability vastly increases for every door you add (if there are 100 doors, you pick one, and the showman opens all the other ones except one, you have a 99% chance of winning by changing doors). This is a mind-blowing problem!

@freddycalabrese319

5 жыл бұрын

@Pawel I think you might have misread my comment

@lilywest9552

5 жыл бұрын

Bob Uncle we did that in maths class but I kept looking at the scratches rips on the cards so I instantly knew which was which. My partner was really confused, especially when she switched them around behind her back

@doctorf7501

5 жыл бұрын

Dhruv Baheti it only works when someone knows what the answer is and controls what is revealed. The system is not purely random :)

I’ve tried to understand this problem 2 or 3 times before. This was the first time I got it, and it clicked so easy! What a great teacher :)

@philip5940

Жыл бұрын

He may have delivered a better spin , that's all . Trust me; it's always 50/50 .

@linsqopiring6816

9 ай бұрын

@@philip5940So if you decide to always pick door 1 and never switch it will always be 50/50 your chance of winning? That's a remarkable bit of magic since when you pick one of 3 doors your chance of winning is 1/3. How do you have a 1/3 chance of winning and win 1/2 the time?

The reason this is counter intuitive is that it is easy to overlook the fact that there are two "bad" boxes, hence initially selecting them is more likely. There are two ways of selecting a "bad" box, and the host will only reveal a bad one and different from what you picked. Hence, there are two ways in which you will win if you switch (bad to good) since one is already revealed. There is only one way to loose, that is when you picked the "good" box and switched.

The important information in this case is that he never rules out the prize door. If he didn’t knew what was behind the door he ruled out the chances would be 1/3 for each of the remaining doors.

@linsqopiring6816

9 ай бұрын

Nope if the host randomly chose which of the remaining two doors to open and he opened a goat door , in that case it WOULD be 1/2 and 1/2 and it would make no difference if you switch.

@kallek919

9 ай бұрын

Okay, I see you don't get it. I rest my case.

@linsqopiring6816

9 ай бұрын

@@kallek919 I can see that I do get it. So go ahead and rest your case.

I thought “that first explanation made perfect sense, I’m sure this next one can’t be as clear as it”. And then the second seemed even stronger to me! Just started combing through your content, but I’m learning a lot about math and teaching watching it!

A good way to visualize it is going to extremes. Consider there to be a million doors, you pick door #1, the host then opens every other door except for door #583456 "for whatever reason", and shows you they are all empty. Do you stick or switch? Now it seems a lot more intuitive, given that the host knows where the prize is.

I heard a pretty good explanation of how switching is favorable in a book called "The Curious Incident of the Dog in the Night-time." Basically, increase the number of doors to lets say, 10. Pick a door, 1/10 chance you pick the prize door outright. Now the host opens every door except the door you chose and 1 other door. Now you have the option of switching, if you think about it, it makes much more intuitive sense to switch because the odds of you picking the right door the first time is 1/10, but now that 8 other doors have been revealed to have no prize it's much more likely the door the host didn't open has the prize in it because after all, the host knows where the prize is.

Fantastic explanations!

Wow! I'm blown away! Makes so much sense now.

Visual way of the 2nd explanation, Draw a 3x3 table, the rows are what door you choose, the columns what door the prize is in. Thus you have 9 possibilities (3x3), 3 correct guesses, 6 incorrect. Now play the game, with each square and always switch. For the 3 winning squares, you switch and loose, so 3 of the 9 total squares is 1/3. For all 6 of the other squares, the other empty box is removed, and so the only other box you can switch to is the prize box, and so 6 of the 9 results get you the prize, or 2/3

Try not to overthink it. The probability you picked the right door at the beginning is 1/3. That means the probability that you DIDN'T pick the right door at the beginning is 2/3. That is the key. That is why you want to switch. Because there's a 2/3 chance you picked the wrong box at the beginning. In other words, there's a 2/3 chance the prize is behind ONE OF THE OTHER TWO doors that you didn't pick. The game host conveniently shows you which of those doors does NOT have the prize behind it. That means there's a 2/3 chance it's behind the other remaining door. So switch to it.

@evaahh9584

5 жыл бұрын

ssoundasleep ok, but if you switch there is a 2 third chance switching was wrong.

@wictor1992

5 жыл бұрын

​@@evaahh9584 No, the chance of switching being wrong is only 1/3. The only case where you don't want to switch is when you were lucky enough to pick the door with the prize at the very beginning. And the chances of this happening are only 1/3. That's why switching only goes wrong in 1/3 of the cases. In every other situation switching is better.

@_barrel6901

5 жыл бұрын

what if u happen to pick the prize door first tho? if switching means u have a 2/3 chance of winning u should win right? but u switched FROM the prize dooe

@stevewatson2383

5 жыл бұрын

@ssoundasleep Wonderfully explained..

@XxStuart96xX

4 жыл бұрын

@@_barrel6901 'if switching means u have a 2/3 chance of winning u should win right?' Using 'should' when talking about probabilities is a bad word. It leads people on; if something has a 99% chance of happening, but doesn't, people say 'how is that possible? It should have happened!'. But even 99% is not a guarantee, there is no reason why the 1% chance event can't happen. You say switching has a 2/3 chance of winning; that means switching has a 1/3 chance of not winning. That 1/3 chance happens precisely when you pick the prize door first.

Years ago, I wrote a program to run this 50,000 times, and yes, it did prove that switching doors (boxes) would get the prize 2/3 of the time. But, as you said, it's still a game of chance.

@philip5940

Жыл бұрын

There's programmes and then there's programmes . The programmes showing a 50/50 outcome do exist.

I had trouble learning this and it wasn't until someone explained the problem with 100 doors that it finally made sense. With the 100 door problem, you pick one door and the game show host reveals 98 of the empty ones. Your option is choosing your original door that had a 1% chance to win or choosing 99 doors, 98 of which have been opened.

Hey I've replied to most of these but here's the easiest explanation I can think of (sorry if you're reading this again lol) (By the way: when this problem was first solved, there were heaps of very well-regarded mathematicians who rejected the solution - however, all of them eventually realised their error. Because Eddie's solution does make sense; the game is just constructed in a way that is inherently confusing.) Say a robot with no biases plays the show, and say, for simplicity's sake, the correct door is door 1. (Of course, this can be repeated with the other two doors each being the correct; I'm sure we can all agree that they will have the same result, it is merely that result that we are arguing!!) Here is a probability tree. Line 1: choice A. Line 2: opened (no prize) door. Line 3: choice B. 1. 2. 3. / \ | | 2. 3. 3. 2. / \ / \ / \ / \ 1. 3. 1. 2. 1. 2. 1. 3. In all of these, your original chance of picking the correct door is 1/3. If you originally chose door 1 and you switch, you lose. If you originally chose either of the other two and you switch, you win. It is the fact that the same action gives the same result in two cases but gives a different result in one that skews the probability. Hope this helps!

@sophiebobb4010

5 жыл бұрын

Lol correct, I did hope that would be apparent from the diagram/terms of the game but you're right, I should have added that information in. But I honestly was not at all bothered to draw a probability tree diagram using text in a KZread comment; I care about mathematical accuracy, but not quite *that* much 😂 Thanks for pointing it out though :)

I already understand this. Why am I watching this? And why am I enjoying it?

you should also give an example with a 1000 doors such that when you choose a door the host opens up the rest 998 doors where the prize is not. this connects the point you're trying to teach to regular intuition.

How were there many mathematicians that didnt think of the 100 box example, when thinking about the 3 boxes?

I think I have a better explanation: There are 3 locked garages (G1,G2,G3) and I will play this game 1000 times.I will always choose firstly G1 and I know that I will win at least 333 cars if I stick at G1 permanently,and if I never switch. The most important remark is that the G1 is the garage that is NEVER disclosed until the very end. .If I switch half of the time and half of the time I don't switch then I will get 333/2=166 cars from G1 and 666/2=333 out of G2 and G3 combined so 333+166=499 cars. But if I never bet G1 and I stick of G2 and G3,I always benefit from the disclosed garage and I get altogether 666 cars.

To make it less complicate i would mention 1 million doors. You pick one random door, and have 1/1000000, of course. Then the host opens every door except yours and the prized one. Now you have to ask yourself: “How many chances i had one the first pick?” Very low, so it has to be the other door

@timeless_trends932

27 күн бұрын

may be your door have a prize,and i think it doesn't relate to math even it relates to human cleverness,or whatever i say i think i can't never understand this proble m i have so many questions in it,and i have n't that much IQ,as a 14 year old i gave menssa test which for 16's i get my iq as 120

@-Ifrit

27 күн бұрын

@@timeless_trends932 IQ is just a social convention. You have a “low iq”? Based on what standard? Screw that shit, we are better than this. Now, think about it. You have 100 doors. Only one has a prize behind. Pick one. You have 1% chance of getting it right. Then, the conductor opens 98 doors besides yours and another one. Behind any of the 98 doors there isn’t the prize. Now think about your last situation. You had only 1% of success and, most importantly, 99% of failure. Now he opened every one of the other doors. Do you think it was more likely that your door was one of the 99 to have nothing behind, or the winning one? Of course it is not the winning one. Math is in everything, sometimes is just a little bit harder to spot.

Good explanation

The best demonstration of the monte hall problem is to take a whole deck of cards with one winning card, say the Ace of Spades. Lay them all out on the table face down. Only the dealer knows where the winning card is. Same scenario. Player picks a card. The dealer then shows every other card except the players initial choice and the winning card. Do you change your choice at this point? All of a sudden, if you change the card you have increased your odds from 1/52 to 51/52 and you’d be a mug not to change. Anyone who says not to because it’s 50:50, come over my place and we’ll start playing for money. :). Increase the number of doors and reveals and the choices become more obvious. I’m amazed that people still get this wrong today.

It's very simple. If you stick then in order to win you had to have picked the winner when the odds were 1/3. That being the case, then logically you have 2/3 chance to win if you switch. To see why, let's examine what would happen if Monty DIDN'T know where the prize was. You pick one, Monty picks one. Your odds are 1/3, and so are his, and 1/3 chance neither of you win. Now what if you get to pick 2 doors and Monty just gets one door? Your odds become 2/3 and Monty only gets 1/3. And that is what the odds end up for you. 2/3 if you pick 2 doors (switching). BECAUSE, since Monty knows he has to pick a loser then your odds are 2/3: the same odds you would have had if Monty didn't know where the prize was but where you got 2 doors.

That is a pretty good explanation, but there is another way that I think about it: When you pick a box your probability of losing is 2/3, and that probability can only be altered if you change the amount of boxes BEFORE the choice. The host then removes one of the empty boxes, leaving you with only 2 boxes. This means that if you chose an empty box from the start the other remaining box is the correct one. What this boils down to is that if you always switch box your chances of winning becomes the same as your chances of picking an empty box from the start. This can be tested with more than 3 options as well, as long as you only leave 1 other option to switch to. What that will do is increase the chance of winning even more when switching, which can make it easier to understand.

@aspiringcloudexpert5127

5 жыл бұрын

losing*

I have to admit that your videos are great and took me back to my math days. Probability depends on the presence of number of choices. When one door was opened and prize was not there.. Then the 1/3 chance of that door having prize shifted to both other doors making it 2/3 each or we can say 50-50 chances.. nothing less nothing more.

@KalOrtPor

6 жыл бұрын

It didn't shift to both others, it wasn't random choice since he knows where the prize is and will never open it first, the prize is *always* never there. For what you say to be true would only apply if he randomly opened the door and saw the prize was not there. But playing the game many times that way would mean there's a chance he could accidentally reveal the prize, but by definition that can't happen. Think of it like the host picking 2 doors at the beginning and letting you peek at one, then offering you to swap your door for both of his. Or, guess what the lottery numbers are, and I'm going to remove all possibilities except your numbers and 4, 8, 15, 16, 23, 42, and offer you to switch.

@yulisgold7201

5 жыл бұрын

I agree

FINALLY I UNDERSTAND!!!

I think I have a better explanation but I don't know how to put it. I'll try the first box that the host opens has to be a box other than the one which was selected by the player. Now the host is left with two boxes Even on those two he only can open an empty box. That leaves him with only one choice two third of the time i.e. he cannot open the one that the player selected and not the one which has the price, but an empty one. Let's go iterative I'm going in as a player I'm fixing my decision as, at first I will choose box one and after the host open a box I will choose the other box, i.e I will switch. 1) first box has the price other boxes empty. I choose first box, then the host opens third box, then I switch to the second box and finally I lose 2) second box has the price other boxes empty Now I like before choose first box, the host has no choice but to open the third box, as the first one was already selected and the second one has the price. Host opens the third box and I switch to the second box, I win. 3) third box has the prize others are empty I select the first box, and this time also the host has no choice but to open the second box as the third box has the price, so he opens the second box. I as usual switch to the third box and I win. So you can see by switching I won two rounds, but if I had stuck with the original, i only would have won the first round. We can conclude that switching makes as more likely to win i.e 2/3 of the time. Hope I'm clear and not wrong as I've just understood this by myself. Please like if I'm correct Mr. Woo.

the key part people miss in this problem is that the host never opens the door you've chosen, that's what makes all the difference.

@peterboris3765

3 жыл бұрын

And never opens the winning door

It’s the fact that these probabilities are so similar. Avg minds make 1/3 and 2/3 as “close” to one another. If you do the same problem with 100 doors, it’s immediately apparent because the host would open 98 doors except for your choice and another door. Did you really just beat 1:99 odds? Or is it clear then that you should switch. It only remains constant if the host doesn’t know either

Let me have a take at this.. I choose a random door from the three. Therefore I have chosen either Winning (1/3 chance) or Losing (2/3 chance). The game show host reveals a losing door because I couldn't choose 2 doors. Now, a normal person would see that, on the circumstances seen with no perspective, he has 1/2 chance of winning and losing. There's no point in choosing.. But as I have a perspective, I can change the probability. Now, I must establish 2 categories. [WINNING] and [LOSING] I had a 1/3 chance of choosing [WINNING], so I most likely had chosen [LOSING]. This establishes that the unchosen doors hold a 2/3 chance..now this is what confuses most people. Then, let me add 2 more categories. [CHOSEN] and [UNCHOSEN]. The [LOSING] was most likely [CHOSEN]. The [UNCHOSEN] have a higher chance of containing a [WINNING]. Why? Because of the 1/3 chance of me obtaining [WINNING], it had a higher chance of being in the [UNCHOSEN]...ok that didn't explain much but I think establishing these categories might help

@AnshulRaman

5 жыл бұрын

Anthony Nguyen This is how I like to think of it. Instead of looking at the perspective of the contestant, look at it from the host's side. Once you make your initial choice, the host, Eddie, must show you a box. When he chooses which box to show you, the box must fulfill 2 requirements. 1, it must be empty, and 2, you can't have chosen that box. If the prize is in box 1 and you choose box 1 initially, Eddie can choose to show you either box 2 or box 3. However, if the prize is in box 1 and you choose box 2, Eddie must show you box 3. Similarly, if the prize is in box 1 and you choose box 3, Eddie must show you box 2. There is a 1/3 chance that you picked the right box, and this means that there is a 1/3 chance that Eddie gets to choose which box to show you. Otherwise, there is a 2/3 chance that you chose the wrong box, which means there is a 2/3 chance that Eddie is forced to show you a certain box. Once Eddie reveals that box, the question becomes, what are the chances that Eddie was forced to show me the box that he did? As established above, there is a 2/3 chance that he had to show you the box that he did. What is the only way that he is forced to show you a box? That can only happen if you initially chose the wrong box. That means that there is a 2/3 shot that Eddie was forced to show you a certain box, and if he was forced to, that means the prize has to be behind the box you didn't choose. Let me know if that makes sense! That is the way I best understood it.

@Bluedragon2513

5 жыл бұрын

Anshul Raman it makes sense.

It is far easier via Bayes and you recast your hypothesis with he additional information - there are many complex models that you can't brute-force your way through

If you still aren't convinced, make it a problem with 10 doors, 1 with the prize, 9 without. What you do as a game host is to open 8 doors, and then ask people to switch. (General: N doors, 1 prize, N-1 no prize, make a choice, open N-2 doors with no prize, the chance of getting the prize is 1/N for the originally choses door and (N-1)/N for switching doors. For N large, winning with the original choice approaches 0 and switching approaches 1.) If you stay with the first choice, you would have a 1/10 chance of winning. If you switch, you have a 9/10 of winning.

You have a 1/3 chance of winning, you have a 2/3 chance of losing. If you chose the losing option (which is a 2/3 chance), when they take away the other other losing option, switching always guarantees a win.

@KuroTora-

2 жыл бұрын

Moral of the story, always be pessimistic (~said sarcasticly).

The easiest way to comprehend, is imagine having 100 doors, and 98 doors are revealed, its quit obvious that u switch, xd

The way I've always explained Monty Hall to people is to suppose a much more extreme case. Suppose you have 100 doors, and there's a prize in one of them. Then the host reveals 98 bad doors and asks if you want to switch. Obviously you're vastly more likely to win if you switch in this case.

@sereya666

Жыл бұрын

Well 50-50% would still be vastly more likely so this doesn't help us directly

@ALeafOnTheWind42

Жыл бұрын

@@sereya666 The point is it's easier to intuit what's going on in this hypothetical situation

Ohhhh now I get it

It's even easier if you see it as: Do you want to go with 1. the one door you first chose 2. ALL the other doors

@klaus7443

2 жыл бұрын

That is incorrect. If you are offered the other doors then the host CANNOT know where the car is in order for you to double your chances of winning.

@Cardium

2 жыл бұрын

@@klaus7443 The choice essentially always comes down to either staying with the one door you originally picked or essentially taking all the other doors. The fact that the host opens a wrong door before you do it just concentrates the chance into the remaining door, but it's the same as taking them both if they hadn't. Easier to imagine with a 100 door version: you either stick with the one door you picked or switch to the other 99. The fact that the host opens 98 wrong doors first just concentrates the chance into the remaining other one, but you're still essentially looking at "is the car behind the one door I picked or behind the other 99"

@klaus7443

2 жыл бұрын

@@Cardium Your explanation is the worst one possible...here's why. If the contestant is offered the other two doors after he has picked one then there is no way of knowing if the host would have opened another door that must reveal a goat and regardless as to whether or not the car was picked. So if the host knows where the car is then you are at the mercy of the host, he could have offered a goat if you picked the car, or if he wanted you to win he could of offered the car if you picked a goat. Therefore the only way for certain you would double your chances of winning is when the host does NOT know where the car is which is the opposite reason switching doubles the chances in the MHP. And if the host offers the other two doors before the contestant picks one then he doesn't even need to know where the car is at all. And in the MHP if the host doesn't know where the car is and revealed a goat from another door then switching does NOT double the chances of winning. You cannot take an unconditional probability problem and disguise it to look like a conditional one, it doesn't work.

@klaus7443

2 жыл бұрын

@@Cardium "The fact that the host opens a wrong door before you do it just concentrates the chance into the remaining door" No it doesn't. The probabilities of the two doors the host has have changed the moment the contestant has picked one. Again you are treating this problem as if it was an unconditional one.

@Cardium

2 жыл бұрын

@@klaus7443 of course the host is going to open all unchosen doors that contain goats except one that contains either a goat or a car, those are the rules, but before they even do so you know that there's 1/n chance that you picked the car and (n-1)/n chance it's in the remaining doors. Whether the host says "do you want to switch to all remaining doors" or says "do you want to switch to all remaining doors, and by the way I'll open n-2 losing doors" is irrelevant. If it wasn't the case, the chance with switching wouldn't be 2/3 (chance when picking all other doors). The chance in switching in a 100 door game wouldn't be 99/100 (chance when picking all other doors). Yet that's what it is, because that's all you're really doing when you switch: you're saying "it's not behind the one I picked" - well if it's not behind the one you picked, what does that leave?

*jaws open, silence*

If the way to win is to choose the non-prize door in the first place, the problem is, just like you don’t know which door has the prize behind it, you don’t know which two doors don’t have the prize behind them either. So, is there is way to win?

@stevenh9413

5 жыл бұрын

no but you have a greater chance of picking a non prize door in the first place right? so its not guarenteed win but you have a higher chance to pick a non prize door and siwtching off that means you get the prize

what prevents the host from picking your choice and opening it in the first place?

@RonaldABG

5 жыл бұрын

That's a rule of the game. He knows the positions and cannot reveal your choice and neither the option that hides the prize. Those are the conditions that make switching better.

Can I go back in time and have you as a teacher, please?

This problem hurt my brain when I first heard of it. Made charts, still couldn't accept it. Finally, I took some playing cards and did it myself one hundred times. Mix them up, choose one, flip another. If the target card was flipped, disregard that round. If it wasn't, switch. After a hundred repetitions, switching was 59-41. Tolerably close to the expected 67-33.

@klaus7443

5 жыл бұрын

If the target card was flipped that round should have been counted as a win by switching because you already have a losing card selected first.

@RonaldABG

5 жыл бұрын

@Jonathan Lovelace As you were told, the discarded rounds should be count as winning by switching because the host, that knows the postions, would have not revealed the target card (that's a rule of the game). Discarding them alters the results, and to show why, let's divide the three kinds of results in three groups: 1) 1/3 of the time you pick the target card. The other revealed will necessarily be a wrong card, You win by staying. 2) 1/3 of the time you pick a wrong one. The other revealed is the other wrong one. You win by switching. 3) 1/3 of the time you pick a wrong one, but then you reveal the correct card and discard that game. You repeat it. If the game was made as in Monty Hall, always revealing wrong cards, you should win by switching in cases 2 and 3 -> 2/3 of the time. But you are keeping only with cases 1 and 2, both with the same probability, so in the valid cases staying and switching win the same amount of cases (on average). With the repetitions will occur exactly the same. A third part will be discarded and so you will consider as valid only the same amount of winnings for each strategy. The point is to see that the repetitions don't have necessarily the same result as the original ones, so you are replacing some of the winnings by switching with winnings by staying, which is what gives the incorrect result 50%. In the extreme case, if everytime you got a wrong card you discarded the game, you would win staying 100%.

@jonathanlovelace521

5 жыл бұрын

I am, as is so often the case, confused. As you stated, one of the rules of the game is that the host never reveals the target card. So, when I discarded the ones where the target had been initially flipped, that was my way of imitating that by myself. Because the situation we're interested in is solely "if a non-target card is flipped, should I switch?" I felt justified in discarding times when a target card was flipped. If I understand what you're saying... The times that I discarded were necessarily times when I did not pick the target card, and so would be wins for switching in the true Monty Hall scenario with non-random flipping. So I replaced half of the switching wins with redos, where switching wins 2/3, except it doesn't win 2/3 of the time, because I replace half the wins with redos, where switching wins 2/3 of the time, except it doesn't because I replace half the wins with redos... It seems intuitive then that in that scenario, switching should win half the time, and my 59-41 advantage for switching was pure luck. But I'm not sure I buy this? If I chart out including the discards, I'm not eliminating one of the 3 possibilities. I'm including a fourth and fifth. Pick one of two non-targets, and flip the target. Any time this happens, whether it happens only once or five times in a row, I replace it with one of the three original possibilities. Pick right, switch, lose. Pick wrong, switch, win. Pick wrong, switch, win. The reason I am watching high school math videos is because I didn't learn it so well in high school. IE, If I'm wrong here, I want to know that and know why.

@klaus7443

5 жыл бұрын

In the MHP you need to break the problem down when you picked a door from three available, and not when a door was revealed...that's too late. Pretend you picked Door 1. If the car is behind Door 2 the host reveals Door 3. If the car is behind Door 3 he reveals Door 2. So you will win by switching if the car is behind either Door 2 or Door 3. Now when the host reveals one of them it is the other one that has the 2/3 chance of winning. So all you want to do if you experiment with cards is to cover one of the three, that is your first pick, then turn both the other cards over and if the target card is among them then you record that as a win by switching. You would have ended up with the target card anyways when the host had to reveal the other one. So if the target card is among the other two that's a win by switching, and if it isn't among the other two it's a win by staying.

@philip5940

Жыл бұрын

Thankyou for that honest trial. It's actually closer to 50/50 . But you are still seeing ⅓ to ⅔ and that's interesting. The correct answer is actually 50/50 .

There's a better explanation in The Curious Incident of the Dog in the Nighttime This was still good though

@Xperim

5 жыл бұрын

Mythbusters still did it the best.

I like to see it like this keep your door or get both the other doors (pretend one was never opened) this is because the host only ever opens a dud door and there is ALWAYS a dud door for him to select. So ask yourself do you want two doors or one?

@unbound26z

6 жыл бұрын

The Raven Black thats a nice way to explain it to the people that dont get eddie , good comment

@klaus7443

6 жыл бұрын

It's a bad explanation because the host does not need to know where the car is in that case. In the MHP the chances to win by switching double because he knows where it is and must reveal a goat. It's 50/50 if he didn't know where the car is and randomly revealed a door with a goat.

@klaus7443

5 жыл бұрын

You don't know what you are talking about. The car randomly placed will be behind your door 1/3 of the time, 1/3 of the time behind the door you can switch to, and 1/3 of the time behind the host's door.

@klaus7443

5 жыл бұрын

Try it in your head 99 times....in the MHP you picked the car 33 times and the goat 66 times. Switching wins 66 times. Randomly you STILL picked the car 33 times AND SO DID THE HOST. So you have the car 33 times, host has the car 33 times (which we don't have to count because a goat wasn't revealed) , and the door you can switch to has the car 33 times.

@klaus7443

5 жыл бұрын

In the MHP twice as many contestants picked goats then cars and they all continue. When the host reveals doors at random there aren't twice as many contestants with goats that can continue because half of them already lost when the host revealed the car. So 1/2 of those that picked goats have lost early but none of those that picked the car have because the host cannot reveal it the same time that you have picked it.

That was the best explanation of the Monty Hall problem that I've seen but I still respectfully disagree. I consider the second stick/switch question is completely independent of the first choice since after eliminating one of the choices, you now have free will to choose box 1 (switch) nor box 2 (stay). But, I'm neither a mathematician or a statistician.

@ianfoote

4 жыл бұрын

It's not completely independent because the elimination is forced to be one particular door when switching wins.

Andrew Church does not understand the Monty Hall Problem as evidenced in a debate with, of all things, a cat. He failed to convince the cat that a randomly placed object would be behind one specifically marked door of the three twice as often as compared to another specifically marked door.

@klaus7443

5 жыл бұрын

...lol...well he did say he was 60 and pretty dumb, no argument there.

@georgcantor3603

5 жыл бұрын

…….it sure worked for yourself.

Is the reality of quantum mechanics more logical than what the math? The player never knows which goat is the one revealed.。.. If life is an endless choice, then if we face it with this way of dealing with things, then what aspect of human nature is eternal? Some people always want to control the world and act as the so-called "God"?

Could have explained it a lot better. Door 1 Door 2 Door 3 car goat goat If you pick door 1, then switch to door 2, you get nothing Door 1 Door 2 Door 3 Goat car goat If you picked door 1 and switch to door 2, you win the prize Door 1 Door 2 Door 3 goat goat car If you pick door 1 and switch to door 3 (because door 2 would have been opened) then you win the prize. So out of every possible scenario you will win the prize 2 out of the 3 times. If you simply would have stuck to the original, you would only win 1 out of 3 times.

It is 1/2 because if you gain information, the information at the beginning isnt useful anymore. It is true that in the beginning you have a 2/3 chance of being wrong, and 1/3 chance of being right, but that is when you dont have the information that one of those 2/3 is wrong. when you gain that information, the problem stops being about three doors -- its about two doors. then, you have a 1/2 chance of being right and 1/2 chance of being wrong. If we go back to the beginning, if were to pick one of the two other doors, there would still be a 1/3 chance of being wrong, because you can not pick two doors. Thats where the problem with this paradox stands: you cant really pick two doors at once, which seems to be what you do when you assume that because the first door in the initial scenario was a 1/3 chance, that the second door left in the second scenario (which has added information that was previously inherited) somehow carries over the leftover 2/3 chance. You start by picking one door, and then you are given information and can pick one of the two doors again. You are not picking the first door (1/3) *and* the second door (1/3) which you will switch to right at the beginning of the situation (if so, it really would be a 1/3 chance). You pick the second door when there has been a new development in the problem, which would make it a different problem itself.

@RonaldABG

5 жыл бұрын

You are not gaining information about your option, because the host does not open a door randomly; he knows the positions and is forced by the rules to always reveal a goat from the other two doors. We already knew that in the other two it was always possible to find at least one goat, because they could have two goats (1/3 of probability) or a car and a goat (2/3 of probability). So the fact that other person deliberately takes it out, knowing where to find it, does not say anything about what you already have in your door. To make an analogy, we can agree that the chances of winning the lottery are very few. Suppose you buy a ticket and its number is 456432. You don't see the results on the day of the contest but you tell a friend to do it for you. You tell him that if your number was not the winner, write yours and the winner together on a piece of paper. For example, if the winner is 989341, he would have to write: /////456432,,,989341/////. On the other hand, if by chance of life yours was the winner, then he would have to write yours and any other that he could think of. For example, he writes: /////456432,,,278226///// Note that with these conditions we have managed to be in the same situation as in Monty Hall problem: despite what your first selection is, there will be always two possible options remaining, one of them is necessarily the winner and your option is also one of them. All the rest is discarded. But do you think yours is 50% likely to be right and so you will start winning the lottery 1/2 of the time? If someone knows the results, obviously that person can write your number next to the winner as many times as you play, but that does not mean that your option will be correct half of those times. The probability that the winner is the other number he wrote and not yours is very very high. Now, note that your friend acts exactly as the host in Monty Hall. The difference here is that you don't have the opportunity to switch; you are forced to stay.

I am very disappointed with this teacher. After the first choice is made, the “game show host” will ALWAYS show the contestant the empty box. You know right from the beginning that the empty box will Be revealed after your choice. If I had a random number generator in my pocket and made it make the first choice then showed it to the Host. I, however, did not look at the generator, but I tell the host this is what I’m choosing. Now the host knows what is being chosen but the contestant does not, yet. So now the host will reveal an empty box. So now the contestant suddenly has more information without actually making a choice. What is the probability now? The contestant never made a first choice.

@klaus7443

6 жыл бұрын

The contestant still doesn't have additional information of the door he picked. If he did he would know his door has a 1/3 chance and by switching a 2/3 chance. When you have an equal amount of information of two doors, in this case none, then it's 50/50.

Souldn't a proper explanation involve Conditional Probability ? Bayes Theorem and P ( Box 2 given Box 1 empty ) = P ( 2 | not 1 ).

If there were a 101 doors, you choose onw and open other 99, then according to u there should be just 1% chance of losing. Is it even logical? Plz explain

@RonaldABG

5 жыл бұрын

The probability of losing by switching wouldn't be 1/100 but 1/101. It is logical since the revelation is not made by random. The host reveals goats not because your selection was good, but because he is forced by the rules to always discard goats from the other doors until two are remaining: yours and any other one. He can do it because he knows the positions (he avoids revealing the car) and despite what your first selection was, it was possible for him to find at least 99 goats. Now note this means that everytime you failed, which happens more times, the other he leaves closed is the correct. To make an analogy, we can agree that the chances of winning the lottery are very few. Suppose you buy a ticket and its number is 456432. You don't see the results on the day of the contest but you tell a friend to do it for you. You tell him that if your number was not the winner, write yours and the winner together on a piece of paper. For example, if the winner is 989341, he would have to write: /////456432,,,989341/////. On the other hand, if by chance of life yours was the winner, then he would have to write yours and any other that he could think of. For example, he writes: /////456432,,,278226///// Note that with these conditions we have managed to be in the same situation as in Monty Hall problem: despite what your first selection is, there will be always two possible options remaining, one of them is necessarily the winner and your option is also one of them. All the rest is discarded. But do you think yours is 50% likely to be right and so you will start winning the lottery 1/2 of the time? If someone knows the results, obviously that person can write your number next to the winner as many times as you play, but that does not mean that your option will be correct half of those times. The probability that the winner is the other number he wrote and not yours is very very high. Now, note that your friend acts exactly as the host in Monty Hall. The difference here is that you don't have the opportunity to switch; you are forced to stay.

@dheersavla3542

5 жыл бұрын

@@RonaldABG my bad. Understood. Instead, it is much more relevant if we take 100 boxes

@RonaldABG

5 жыл бұрын

I am glad it was not difficult for you and the example was clear enough.

The game is all wrong, watch Let's Make a Deal. The contestant is either take the money $1000 or the door. Sometimes a prize is open and and another door may be chosen. All cases give the member has a 50/50 choice between TWO Doors not THREE.

@morbideddie

3 жыл бұрын

The Monty Hall Problem is a mathematical problem inspired by the game show. In the problem Monty always picks a dud to reveal and always offers the switch, hence why your odds double.

I utterly despise the first explanation. It's how I always hear people describe it and it is fundamentally stupid. Probability doesn't work that way and it's no wonder people considered this a paradox as a result - the 1/3 chance doesn't "magically" transfer to the other door once you opened it, everyone knows that's not how probability works and it's no wonder you get resistance when it's described that way. The second explanation is much better, much more elegant and actually mathematically accurate.

@robharwood3538

5 жыл бұрын

Probability actually does work that way. The key concept to understand is called 'conditional probability'. Probability is actually more about our state of knowledge than it is about the state of the universe. Consider Monty Hall's (or Eddie's) perspective at the very beginning of the game. He *knows* which door contains the prize (let's say it's door 2), so for *him* the probability that it's behind door #2 is 1, and the probability behind the other doors is 0. If that's true for Monty, why isn't it true for the contestant (who thinks the probability for each door is 1/3, not 1 or 0)? Answer: Because Monty's state of knowledge is more informed than the contestant's. Each time you learn something new, the probability distribution *does* change, not magically, but logically. After learning that door 3 is empty (and learning nothing else), the only logical probability distribution is 2/3 for door 1, 1/3 for door 2, and 0 for door 3. Do you agree that the probability for door 3 changes from 1/3 to 0 when you learn it's empty? Was that a magical change or a logical change? Logical, right? Same for door 1.

@evaahh9584

5 жыл бұрын

Rob Harwood no it’s not. That’s just wrong. You knowing something does not change the chances.

@robharwood3538

5 жыл бұрын

@@evaahh9584 If that's the case, then what does Eddie think the probability of the prize being in box 2 is? What do the students think it is? How can these two answers be different?

@evaahh9584

5 жыл бұрын

Rob Harwood ok lemme give you an example 10 balls in a bag, 5 red and 5 blue, so there is an equal chance of you getting red or black but you do not know how many of each colour there is. you pick one out, it’s black, you put it back in. You pick another. Red. Black. Black. Black. From this you would guess that there is a 4/5 chance the next ball you pick will be black, but that’s just not true. This is why conditional chances are flawed, this problem works not because of conditional chances but because monty hall will always open a non prize door

@klaus7443

5 жыл бұрын

Colobrinus…..Rob Harwood is entirely correct. The MHP is a conditional probability problem so the probabilities are recalculated BEFORE the reveal, not after. Each door starts off with a 1/3 chance of having the prize. After you pick one the two doors remaining for the host have probabilities of 0 and 2/3 because he CANNOT open one with a 1/3 chance of having the prize.

I like the explanation of choosing to switch before the first door gets revealed. You could also go so far as to choosing to switch after choosing your door but before the host reveals. Only problem is you need to know that the hosts always reveals a door. What if he only does it when you guessed right?

@francisgiovannicarizon5752

6 жыл бұрын

Alex Burns note that this is a problem of probability , chances , and it states that there is a 1/3 chance of losing if you switch , not 0 chance .

@ddebenedictis

Жыл бұрын

I also like the explanation of choosing to switch before the first door gets revealed. Monty's offer would be this: "Do you want to keep the ONE door you selected, or would you like to own BOTH of the other two doors?" This approach seems intuitively obvious, you would rather have two doors than one.

It is totally on Host, He can make you look like a Dumb by playing with your emotion of sticking. Or He can make you a fool by playing with your match instincts. Why we ignoring the fact that Host might have opened any of other door, because GIFT was behind the chosen one. Well, 50-50 chance of that that ? That's why the probability of of opened door should be distributed to the other doors equally. Unless this thing is all a blunder to teach everyone that SWITCHING is best thing so that People can manipulate the public someday. :D just had a thought. haha

kde spadnú planéty

after 7 years, i still don't get it. 🤣

@philip5940

Жыл бұрын

That's good. There is hope still for the species. For some reason to do with spin many folk choose to believe it ain't 50/50 . It's 50/50 pure and simple .

Each sheep is also an independent and distinct individual. This game cannot mislead players by treating every sheep as the same thing. There are three options, and two of the sheep also exist independently. If the player can distinguish them, there will be no chance of winning like 2/3. Is the reality of quantum mechanics more logical than what the math? The player never knows which goat is the one revealed.。Don't let the jews fool you.. If life is an endless choice, then if we face it with this way of dealing with things, then what aspect of human nature is eternal? Some people always want to control the world and act as the so-called "God"?

so many math nerds in the comments with nothing better to do on saturday night

Oooooooooooooooooooh

Because you are likely wrong, switch.

i didnt get it..... i hate probability.

@SelfReflective

5 жыл бұрын

Then, try it. Find someone who puts the prize in one of three boxes. Pick a box. The helpful person reveals one box (not the one you chose) to be empty. Try sticking with your original box 20 times. Then try switching 20 times. See how well you do with either strategy.

Fuck me I finally understood it now. there's only 1/3 chance of you picking the correct door the first time, meaning there's only 1/3 chance that both doors left over (one of witch he opens) are empty. That means in 2/3 of the choosing of the first door (before the opening) the prize WILL BE in the remaining two doors, hence making it a 2/3 chance the unopened door you DIDN't chose contains the prize. Ofc, if he DIDN'T open one of the doors, you'd have equal chance of winning if you switch or stick. Ahhh feels like I've been reborn.

He's usually pretty good at explaining concepts, but in this case he did a terrible job. I doubt anyone in the class and probably most of you, are convinced that it is better to switch doors. Unfortunately they chose the winning door, so sticking with it was the right choice which makes it that much more difficult t convince them otherwise. He should have taken the time to show a few scenarios to convince them. The prize is behind door 2. Let's start with the contestant choosing door 1. He has to reveal door 3. Switching wins. If the contestant choose door 2 as happened, He can reveal either door . In this case switching loses. Now if the contestant chooses door 3, then he must reveal door 1, Switching wins again. Notice switching wins 2 out of 3 times. Put another way. You only have a 1/3 chance of picking the correct door. The probability of the prize being behind one of the other doors is 2/3. If the host said nothing, then there would be no advantage to switching doors. But, the host just showed you that one of the other doors is also a loser. You know that the probability that the prize is behind one of the doors you did not chooses is 2/3, since there is only one door that could be a winning door, the probability that it is a winner is still 2/3. The thing about probabilities is that they work with large numbers of trials. So if a 300 contestants played, then we would expect 2/3 of them (200) to win by switching, but that means 1/3 (100) would lose, so there is no guarantee that switching doors will win, as you saw in the demonstration with his class. all you are doing is giving yourself a better chance to win. I taught Physics which uses a lot of math, but there would be no reason to teach this problem. That being said, if I were teaching this lesson, I would have tried the experiment with the students several times. I would have had them stick to their choice the first time through, then have them change ever time on the next set of trials. Wayne Y. Adams B.S. Chemistry M.S. Physics

@philip5940

Жыл бұрын

However the removal of an option now gives a new game for a pick between two options for the one concealed prize . If the non winning option had been removed backstage say and the contestant presented the two options on stage without mentioning that a third option had once existed , then there is no argument that the probability is 50/50 and everyone would agree .The host removing an option under witness doesn't change anything. After a long run of say heads in a coin toss, the probability of tails for the next toss hasn't increased for a known randomised situation.

Edit: I understand so have removed this comment, see replies for someone who cannot read “I understand now”

@klaus7443

5 жыл бұрын

But you don't have two boxes to begin with, so what's your point?

@evaahh9584

5 жыл бұрын

Klaus 74 my point was (after having people describe it more clearly) that since you know that one door does no have a prize, you can forget about that door. After you reply imma delete this comment

@evaahh9584

5 жыл бұрын

Klaus 74 I know why it works now, hence the “was” and “after having people describe it more clearly”. Specifically my point was that if you start out with 2 boxes and can switch, then there is a 50/50 chance. If you are 3 boxes, and one is eliminated now you have two boxes and there is a 50/50 chance, and chance does not change based on what you know.

@evaahh9584

5 жыл бұрын

Klaus 74 do you? “I know how it works now” Stop calling me an idiot for not understanding when I have CLEARLY stated that I understand NOW and didn’t EARLIER key work: EARLIER, I repeat EARLIER do I need to spell it out for you?

@klaus7443

5 жыл бұрын

You repeated what you said in the beginning. So how the fuck is anyone to know what you say you understand without you describing what it is that you understand.....understand??

Hey Mr. Eddie Woo.. first off thanks for your videos. They are very informative. BUT this particular segment of Monty hall problem is not rightly explained. You should consider redoing it. No offence.

what ever i switch i not pick the one i no my luck

I, too, enjoy giving classroom lectures in front of an imaginary audience.

Hey I think probability does not tell us anything exactly. At starting we don't know anything about boxes so they have equal one thirds probability. But after we remove 1 box not having prize then according to you probability of box 2 is two thirds but I think that Box 2 still have chances that prize could be behind it because it's probability is still not '0'. So we may get the prize behind it. This I think I can say to justify my first statement that 'probability does not tell us anything'. I think it gives a rough estimation that one thing might have higher chances for something but still there are chances of getting the desired thing from the box having less probability. Plz reply

@edukid1984

6 жыл бұрын

Siddharth Singh Of course there's a chance the prize is still behind the door you stick with. BUT the chance is 1/3, whereas the chance of winning if you switch is 2/3. That's what probability tells us - you have a HIGHER chance of winning by switching, but a win is NOT GUARANTEED (because still 1/3 chance of losing).

@sophiebobb4010

6 жыл бұрын

Siddharth Singh yeah, you're right. When the enactment is done in video 1, the class gets the box right, which was and is always a possibility; the difference is that the chance is less, not that it is nonexistent.

sorry but i dont agree with this theory.

@RonaldABG

5 жыл бұрын

It's actually simple, the point is the contestant did it by random but the host did it purposely with knowledge, and that's what makes the difference. To make an analogy, you could see the two doors that remain closed as if they were picked by two different contestants, and the revealed one in Monty Hall is which neither of them selected here. So, suppose those two players are you and me. We both want to win the car; one of we must pick a door and then the other must pick another different one, but I will put this condition: I know where the car is. If I was the first to select, it would be a 100% chance to win for me, right? As I know where the car is, there is nothing that prevents me from selecting it. I would select the car for sure and you would be forced to pick a wrong door, because you cannot repeat my selection. So, let's change the rules: you select first and I select second. Do you think the game is completely fair now, and we both have 50% chance to win the car? Note that you can only win if you pick the correct at first, having 3 possibilities (1/3 chance) but I can win if you selected a wrong door and so the car was left in any of the other two (2/3 chance), because if that is the case, I know in which of those two the car is hiding. It would be the same as if you could only take one door but I could take the other two and look for the car in both. Wouldn't you prefer to be that second player, who knows the positions? This last case is exactly what occurs in Monty Hall problem: if you didn't select the car (2 out of 3 games), the other option the host leaves available (closed) is which has the prize, and that occurs because he knows the positions and is forced by the rules to always reveal a goat from the two doors you did not select. The host is basically "cleaning" the rest, and we know the car was more likely to be in one of the rest than in yours. So, to switch is like to bet on the second player who knew the positions, and to stay is like to bet on the first one.

@Somerandomdude-ev2uh

5 жыл бұрын

Not a theory, it is 100% true. -----If you choose door 1 If prize is behind door 1 you should stick If it is behind door 2 you should switch If it is behind door 3 you should switch -----If you choose door 2 If prize is behind door 2 you should stick If it is behind door 3 you should switch If it is behind door 1 you should switch -----If you choose door 3 If prize is behind door 3 you should stick If it is behind door 1 you should switch If it is behind door 2 you should switch Regardless of what door you chose, switching is better

The question is flawed from the very beginning, because it does not state the behavior of the Host. You are ignorant if you think this is how the Host behaved in the real show. Stop being lazy and do your research. This has already been debunked by real mathematicians and math doesn't care about the gullible fools.

The host needs to also randomly select their door ... if that reveals nothing. Then it is 50 to 50 chance.

@klaus7443

Жыл бұрын

Needs to randomly select their door?

@philip5940

Жыл бұрын

Does it matter? If the contestant doesn't have any idea where the prize is, then that is as good as random. There is no cause and effect that results from host knowingly or unknowingly eliminating the booby prize . It's always 50/50 probability when selecting between two choices on basis of guesswork . Occam's Razor kinda says "why make a mountain out of a moleHall .

@willoughbykrenzteinburg

Жыл бұрын

@@philip5940 I have two bags of marbles. There are 500 marbles in one bag and 1 marble in the other bag. There are 500 white marbles and 1 black marble. You win if you choose the bag with the black marble. According to you, it's always 50/50 probability when selecting between two choices......so is this 50/50? "The contestant doesn't have any idea where the prize is" is false. At least a competent contestant has an idea....

The answer to this riddle is stupidly simple "if you switch, you get two doors for one". WTF? Your answer in this video is idiotic. The empty door "transfers" to the other door? WTF? Your simple mind is fooled when you are "shown" what you already KNOW; that one or both of the other doors is empty or a zonk.

@klaus7443

5 жыл бұрын

So by getting two doors for your one you have gained a door with a goat behind it.

Very long winded. This could have been explained in 1 minute.

Discounts don't add up, probabilities don't jump over. You're making two distinct choices, not a pre-choice and post-choice. The problem here is that once one the two empty doors were unveiled, we introduced a bias, we made the one who needs to choose more prone to take action due to stress and emotion. Therefore, of course, if he switches and wins, then he's the vigilant, and if he sticks and wins, then he's the patient despite the odds. Notice in the enactment of the problem how the class split exactly 50-50 once you introduced the empty box.

@Copperhell144

6 жыл бұрын

It's actually not two distinct choices, though. The choice you made before the reveal changes the probability of being right after the reveal. If you want a practical proof, you can try the game yourself, but with ten or so doors instead. Choose a door, eliminate eight of the empty doors, then see how often switching gets you the win. For a more theoretical proof... I'm just gonna shamelessly copy-paste Michele Serra's comment here: "The point here is comparing the two game-strategies: stick vs switch. With the stick-strategy, you only win if your original choice was right, and that happens 1/3 of the times. With the switch-strategy, you win every time your original choice was wrong, which happens two times out of three. So switching flips things over: after you made your first choice out of the three doors, you switch so that if at first you were aiming at picking the one door with the prize now you are aiming at HAVING PICKED either of the two doors without."

6 жыл бұрын

Toss a coin, and tell me if you manage to get 1/3 change of heads in 100 tosses. Two doors - two choice options.

@infortality

6 жыл бұрын

You CAN get 33 heads and 67 tails. The probability of that happening for every 100 coins flipped is 1/4302 ≈ 0.0232%

6 жыл бұрын

as compared to 1/2 chance for each individual toss, or 0.07959 for 50 heads out of 100 tosses. So are we going to say now that there's a bias to the doors?

@infortality

6 жыл бұрын

Wait, I'm hearing two different arguments from you. When you say "Discounts don't add up, probabilities don't jump over. You're making two distinct choices, not a pre-choice and post-choice" and "Two doors - two choice options", you seem to be arguing that switching and staying make no difference. This is wrong. You can run a simulation where you always switch or always stay and see the 2/3 1/3 probabilities emerge. The other argument I'm hearing from you is that there is a psychological reason people are more likely to switch or stay. I guess you're saying that if someone sees how many votes there are for each door, and they are in the minority, then there is a larger pressure to switch, and if they voted with the majority then there is less pressure to switch. I think there could be some truth to that, but I'm doubtful that there would be enough pressure to equalize the votes. Each person who has to switch or stay should know that everyone else voted randomly to begin with. However, even if your second argument is correct, your first argument would still be false in the case of no feedback from other participants who are going through it with you. So switching DOES make you twice as likely to win -if you're doing it alone or you can't tell what other people are voting on.- (Correction: Switching would still make you twice as likely to win in all cases. It's just that your argument about bias might be true only in the case of a group of people who (1) get feedback on the number of votes for each door and (2) don't know the strategy of always switching.)

Sorry Eddie but I don't agree with your interpretation here. When you eliminated the third choice by opening third door you automatically made all other options equally more probable and not just some or those you haven't chosen. Probability is not bias to your choices or lack thereof. The probability does not depend on what you have chosen but on how many items are to choose from. What if I told you that I've chosen one. Will 2/3 go to option two instead? What if I didn't tell you at all what option have I chosen or what if I never made any choice. For probability it doesn't make any difference because option one isn't more special then option two so it cannot have different probability

@hydrogen538

6 жыл бұрын

You need Jesus.

@progzmasters

6 жыл бұрын

No, I am just trying to learn math here and you are spamming me with irrelevant content

@Drof1107

6 жыл бұрын

I’m sure you can agree when there are three boxes that the probability is 1 in 3 that you chose the right box. Another way of looking at it is if you started with 100 boxes. You choose and lock into one box so a 1 in 100 chance. Then he opens 98 boxes that are losing boxes. Between the two boxes would it make sense that you have a 50% chance of having the winning box? Considering it was 1/100 to start? Or do you think it’s more likely that you didn’t choose the right box to begin with?

@progzmasters

6 жыл бұрын

Jonatan Barreto This is exactly how I also see it

@dyseskytten1

6 жыл бұрын

The fact, I think you're missing is that the host KNOWS where the price is. If the price is in box 100 then in order to win with sticking you have to pick box 100 to start with. BUT if you switch you can choose either of the boxes from 1-99 and get the price. So with 100 boxes your chances of winning with sticking is 1 % but your chances with switching is 99 %. This is different from "lets make a deal" where nobody knows where the grand price is. In that case, if you are down to two briefcases and one of them happen to be the grand price, it is truly 50/50. Also there is no reason to dispute it because it has been proven. Try doing it 100 times with you and a friend. You can simply use playing cards. The data will prove it.

Your explanation is not logical, Here is the deal, there are three doors, one of the three doors has the prize, You have to choose only 1 door to win the prize, what will be your probability to win? it is 1/3 or 33.33%, Now one door was opened and show a goat... So there are two doors left... So which door has the car? you do not know but you know it will be in door 1 or door 2, because door 3 was proven that there is no car... So the present situation now is 2 doors, the same as the game started with only 2 doors, so your probability to win with two doors to choose whether you switch your choice or not is the same which is 1/2 or 50%.... I will give you another example: Suppose you are in a combat with two enemies, the three of you have the same skill in fighting, what is your chance to win? The same with the door problem 1/3, However you are lucky that one of your 2 enemies was shot by a sniper and died... Now you have one enemy left to fight and you have the same skill in fighting... What do you think your chance of winning? Definitely 1/2 or 50% because you have only one opponent left... It is the same situation with the door....

@klaus7443

2 жыл бұрын

Don't be stupid. If the host has two doors with goats he reveals one and leaves one. The probability that he has two doors with goats is 1/3, not 1/2.

@reynaldodavid2913Jo

2 жыл бұрын

@@klaus7443, It is you who are unbelievably stupid, you are babbling nonsense...

@Araqius

Жыл бұрын

@@reynaldodavid2913Jo Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick. Basic math/logic kids understand, idiot among idiot does not.

@reynaldodavid2913Jo

Жыл бұрын

@@Araqius, Are you a mormn? or a monkey?

@Araqius

Жыл бұрын

@@reynaldodavid2913Jo Are you fucking stupid? Oh, yes you are.

this is false. amazing how many people get sucked into switch

@Araqius

Жыл бұрын

Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick. Basic math/logic kids understand, idiot among idiot does not.

@philip5940

Жыл бұрын

It gives an insight into the power of spin and propaganda. Folk need to feel part of a community and stuff like that.

The host knowingly or unknowingly eliminating a bad choice doesn't add any great new information except to simply let you know that your probability of sitting on a winning pick is now 50/50 compared to the earlier ⅓ probability. It doesn't matter if host knows something. The contestant is not more wise to anything.

@Stubbari

Жыл бұрын

Wrong. The host, knowingly, eliminating a bad door doesn't give any more information about the contestant's door. The contestant's door still has 1/3 chance of having the prize. The host, unknowingly, eliminating a bad door does increase the chance of the contestant's door from 1/3 to 1/2.

@philip5940

Жыл бұрын

@@Stubbari what if the contestant is unaware of if the host knows or doesn't know location of prize before he eliminated a choice ? Are we suggesting that quantum phenomena extends into our macro world ? What if a choice is eliminated before the contestant chooses from remaining two options ? What if contestant is removed after the reveal from the stage for some kinda good reason and another contestant from the street walks in ; what then are the probabilities for a fresh contestant who now faces two options?..⅓ and ⅔ ? Or ½ and ½ ? . There's a couple other what ifs still .

@Stubbari

Жыл бұрын

@@philip5940 In the description of the puzzle is stated that the host knows where the prize is. If the wrong door is revealed before a player picks door then the odds are 50-50. Same goes for a "new player" who chooses between 2 doors. Neither of those cases is the MHP though.

You have a 50% chance - no matter what you do. You are discounting Monty having an option when you pick the prize.

@klaus7443

2 жыл бұрын

You are ignoring the fact that they have probabilities of 1/6 and 1/6.

@michaelgarrow3239

2 жыл бұрын

Klaus 74 - no…

@klaus7443

2 жыл бұрын

@@michaelgarrow3239 1/6 Pick Car, host shows Goat A 1/6 Pick Car, host shows Goat B 1/3 Pick Goat A, host shows Goat B 1/3 Pick Goat B, host shows Goat A Probability of winning by staying 1/6+1/6=1/3 Probability of winning by switching 1/3+1/3=2/3

@michaelgarrow3239

2 жыл бұрын

Klaus 74 - 100% host shows goat. You have 2 choices 1 goat 1 car. Otherwise you run into causality problems. I.E.- if you choose A it is most likely C. And if you choose C it is most likely A. If Monty opens door B with a goat. And what if you don’t choose any door at first? Which door gets 66%? Lots of problems with the theory.

@klaus7443

2 жыл бұрын

@@michaelgarrow3239 My list is correct.

Sorry but this is just objectively not how math works. Probability is "single-snapshot" by definition. The minute you REVEAL anything you are asking a completely different question. And the problem reduces from 1/3 probability of getting the correct answer to 1/5 probability of getting the right answer. Any mention of the possibility of being right 2/3 of the time is merely an artifact of a misunderstanding of the definition of probability.

@moocowman0

6 жыл бұрын

In other words, your interpretation LITERALLY only works if the prize is DEFINITELY not behind your original pick. That is the only way the other two doors maintain a 2/3 chance of having the prize.

@PENTA1907

6 жыл бұрын

Yeah you definitely know more than this math-enthusiast/teacher and THE ENTIRE MATHEMATICS COMMUNITY that have came to a conclusion (the one in the video) on this problem. This isn't something you can discover more on and eventually get a different result. There is a correct answer, which is proven quite well in the video, and your opinion on the matter won't change it.

@xzockerx5197

6 жыл бұрын

I think u missed something important. Yes, he asked a completely different question, but in the case of u, sticking to ur answer, u answered the question, before he asked it (at the start of the game, where the probability for everey chest was 1/3). Lets look at this scenario from a different prospective. The gamehost knows where the price is (this is important). U increase the number of chests from 3 up to 100. U chose one chest. The gamehost opens 98 boxes/ chests/whatever (keep in mind, that he knows where the price is!). Is it more likely that the number u said at the beginning of the gameshow will include the price or the box with the number, the gamehost didnt open. If this isnt clear enough, continue and lets play the game again. Again, chose a number. The gamehost now opens 98 boxes (he knows where the price is). Seems pretty unlikely, that u have chosen the right number 2 times in a row from 100 boxes, isnt it?

@sophiebobb4010

6 жыл бұрын

It's okay to not understand; it's called a paradox for a reason (please note that this is not the true definition of a paradox, simply one that means despite the fact that it doesn't look like it's true, it is). When this problem was first solved, there were heaps of very well-regarded mathematicians who rejected the solution - however, all of them eventually realised their error. Because Eddie's solution does make sense; the game is just constructed in a way that is inherently confusing. Say a robot with no biases plays the show, and say, for simplicity's sake, the correct door is door 1. (Of course, this can be repeated with the other two doors each being the correct; I'm sure we can all agree that they will have the same result, it is merely that result that we are arguing!!) Here is a probability tree. Line 1: choice A. Line 2: opened (no prize) door. Line 3: choice B. 1. 2. 3. / \ | | 2. 3. 3. 2. / \ / \ / \ / \ 1. 3. 1. 2. 1. 2. 1. 3. In all of these, your original chance of picking the correct door is 1/3. If you originally chose door 1 and you switch, you lose. If you originally chose either of the other two and you switch, you win. It is the fact that the same action gives the same result in two cases but gives a different result in one that skews the probability. Hope this helps!

@AnshulRaman

5 жыл бұрын

Instead of thinking of it from the contestant's side, let's look at it from Eddie's perspective. At the start, you choose a box. Once you choose a box, Eddie must reveal another one. The box he reveals must fulfill 2 requirements. 1, it has to be empty, and 2, you can't have picked it. If the prize is behind box 1 and you choose box 1, Eddie gets to pick if he wants to reveal box 2 or box 3. On the other hand, if the prize is in box 1 and you choose box 2, Eddie must reveal box 3. Similarly, if the prize is behind box 1 and you pick box 3, Eddie must reveal box 2. Therefore, there is a 1/3 chance that Eddie gets to pick which box to reveal, and there is a 2/3 chance that Eddie was forced to reveal the box he did. Once he reveals the box, you must think, what are the chances that Eddie was forced to show me the box that he did? Well, as I said above, there is a 2/3 chance that he had to pick that certain box that he did. The only way he would be forced to pick a box is if you choose the wrong box in the first place. Therefore, there is a 2/3 chance that you picked the wrong box at the start which means that there is a 2/3 chance that Eddie was forced to reveal a certain box, and if he was forced to reveal a certain box, that means there is a 2/3 chance that the other box is the one with the prize. I initially thought the same thing as you. "Once a box is revealed to be empty, why can't you just completely ignore it? It should be irrelevant, right?" The issue with that train of thought is that which box he reveals is also a vital piece of information. Out of 2 boxes that he could've revealed, why did he reveal the one he did? Well, there is a 2/3 chance that I chose the wrong box, which means there is a 2/3 chance that he didn't have a choice but to reveal the box that he did, which ultimately means that there is a 2/3 chance that the prize is in the box I didn't choose.

Let's say you were kidnapped by the SAW guy and he placed 100 boxes in front of you, which 99 of them had bombs and 1 had a delicious cake. After you picked an initial box, he would blow out 98 of the bombs and toss those 98 bombs with the boxes away. Would you want to let go of your initial box and grab the other box remained? Focus on how much chance you would have picked the cake from the start. Now play this scenario in your head multiple times. Then now this: Suppose there are 100 tickets with one having a prize. You and 98 other people each picked a ticket. The other 98 people opened theirs and all of them didn't get the prize. At this moment, you still haven't opened yours, and you have the chance to swap the ticket with the one no one picked. You should always switch in this situation.

@klaus7443

Жыл бұрын

"Then now this: Suppose there are 100 tickets with one having a prize. You and 98 other people each picked a ticket. The other 98 people opened theirs and all of them didn't get the prize. At this moment, you still haven't opened yours, and you have the chance to swap the ticket with the one no one picked. You should always switch in this situation." Nope, there is no advantage in switching in that case.

I never accepted this. This is another example of fascinating "imaginary math". It follows the rules, but it is not correct. You did not gain more information. The only information you gained is that one of the 3 is NOT the correct one. Your first choice is 1/3. Your second choice is 1/2. Even if you chose to keep your original choice, you have made a choice. And each time you make a choice the odds are based off of what information is available in that instance. Instance 1: 1/3 Instance 2: 1/2 No additional information was given other than reducing the pool to 2. The host had 2 possible wrong doors to reveal. If you chose the wrong door that will force him to open the other wrong door... if not his choice is also 50:50. You do not know if he skipped your door because it was wrong or correct. So this information does not apply. In math all information is usable. In real life information must be relevant to be used. So even though the math is correct the result is not.

@toolworks

4 жыл бұрын

3 Doors [ ] [ ] [ ] Each door has 1/3 chance of having a car behind it. There is therefore a 1/3 chance of the car being to the left of the x I have added, and 2/3 chance that it is to the right of the x I have added: [ ] x [ ] [ ] I choose door number 1, and the host then reveals that one of the other doors doesn't have the car: [ original choice ] x [ ] [ no car ] Nothing has changed about the original fact that there is a 2/3 chance of the car being to the right of the x which I added, yet there is only 1 door it could be behind. I could have placed the x anywhere, so the game is symmetrical, therefore switching wins 2/3 of the time.

@toolworks

4 жыл бұрын

To be clear, I am illustrating that there is definitely new information gained by the revelation. The new information is what takes you from the original knowledge (there is a 2/3 chance that the car is behind one of these 2 doors to the right of the x) to the new state (there is a 2/3 chance that the car is behind this specific door to the right of the x)

@RonaldABG

4 жыл бұрын

If you actually play it a lot of times, you win by switching about 2/3 of them, not 1/2, and that is real life.

@Grimlock1979

4 жыл бұрын

I think it's easier to understand if you use 100 doors instead of 3. You choose a door, then 98 doors without a prize are opened. Do you still think the chance is 50%?

@philip5940

Жыл бұрын

​@@RonaldABG alotta people somehow fall into that trap.

This is a conditional probability problem, so the ONLY proper explanation is to explain how the condition affects the outcome: Since Monty is restricted from revealing the car and gets two doors of the three, there is a 2/3 probability it will be behind the door he doesn't open. That's all. FINISHED.

It's now 8 years ago. So I guess there's no likelyhood of informing Eddie that too many conundrums and paradoxes result from a ⅓ to ⅔ probability spread . They don't arise for ½ to ½ probability spread which happens to be the correct answer. Let's not forget Occam's Razor .

@Stubbari

Жыл бұрын

There's no need to inform Eddie about anything. His answer is correct.

@klaus7443

Жыл бұрын

If it's taking you eight years to understand the problem then you have some serious issues. You could have taken three cards, two 'Goat', one 'Car', and simply written down what you first picked. That is what you would win by staying.

uh-hmmm ( vos savant and others were NOT correct.) after the initial door selection, the game is, effectively, RESET. that reset resulted in improved odds of winning; to one in Two. imo, (an improvement to) fifty-fifty does not support making a change in door selection. googletranslate

@RonaldABG

Жыл бұрын

No, the host's closed door is more likely to be the correct than the player's choice, because the host did not act randomly; he did it already knowing the locations and deliberately avoiding to reveal the prize. That means that the switching door is correct as long as the player starts failing, which happens most of the time, not 1/2. Consider you repeated the game 900 times. In about 1/3 of them you would start selecting the door that hides each content, so about 300 times which hides goat1, 300 times which hides goat2, and 300 times which hides the car. In total, 300 times the car and 600 times a goat: Your door The other two ============================ 1) 300 games -> Car two goats 2) 600 games -> goat Car and goat Now, remember that the host always reveals a goat from the two options you did not pick regardless of what you caught. That does not change the positions of the contents, so adding the revelation in the same 900 cases: Your door Revealed door Switching door ========================================== 1) 300 games -> Car goat goat 2) 600 games -> goat goat Car So, if you always stay, you can only win the car 300 times, that are 1/3 of the total 900, but if you always switch, you win it 600 times, that are 2/3 of 900.

@klaus7443

Жыл бұрын

"after the initial door selection, the game is, effectively, RESET." Right....the probabilities of the two doors the host has are either 0 and 2/3, or 2/3 and 0. So he leaves a door with a probability of 2/3 after opening the one with 0.

@philip5940

Жыл бұрын

And psychologicaly it hurts more to switch and lose than to not switch and lose. I wouldn't switch even if it really was the illusional ⅔ to ⅓ probability spread . There is here one of two random choices to score the prize , so it's 50/50 . It's always 50/50 . Take coin tossing , it's 50/50 for heads versus tails however it's hard to have some folk see that after a long run of say heads that it doesn't increase the probability of a tail occuring on the next toss .